QM/EPR/relativity

[daniel@elit.net]

Consider an EPR experimental set up, with a singlet state of two spin
1/2 particles. Measure the spin of particle 1 along the z-axis, and
let's suppose it is spin up. Proceed as follows:

1) Measure the spin of particle 1 along the z-axis (again). You'll get
spin up for certain.
2) Then measure the spin of particle 2 along the x-axis. You'll get
spin up or spin down with equal probability.

Set up a new experiment. Do everything as above, except reverse the
order of 1) and 2). For particle 2, you'll get the same probabilities.
For particle 1, you'll get spin up or spin down with equal probability.

We can change the order of things by a Lorentz boost. For example, we
could set it up so that observer A is in the rest frame of particle 1
and observer B is in the rest frame of particle 2, and BOTH A and B
think THEY are the ones making the first measurement. Leading to a
contradiction. How is it all resolved?

[Ilja Schmelzer]

<daniel@elit.net> schrieb
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No. You'll get spin up for certain. The EPR state between the
two particles is already destroyed by the first measurement of particle 1.

Ilja

[Ilja Schmelzer]

<daniel@elit.net> schrieb
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No. You'll get spin up for certain. The EPR state between the
two particles is already destroyed by the first measurement of particle 1.

Ilja

[Ilja Schmelzer]

<daniel@elit.net> schrieb
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No. You'll get spin up for certain. The EPR state between the
two particles is already destroyed by the first measurement of particle 1.

Ilja

[Ilja Schmelzer]

<daniel@elit.net> schrieb
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No. You'll get spin up for certain. The EPR state between the
two particles is already destroyed by the first measurement of particle 1.

Ilja

[Ilja Schmelzer]

<daniel@elit.net> schrieb
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No. You'll get spin up for certain. The EPR state between the
two particles is already destroyed by the first measurement of particle 1.

Ilja

[Ilja Schmelzer]

<daniel@elit.net> schrieb
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No. You'll get spin up for certain. The EPR state between the
two particles is already destroyed by the first measurement of particle 1.

Ilja

[Ilja Schmelzer]

<daniel@elit.net> schrieb
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No. You'll get spin up for certain. The EPR state between the
two particles is already destroyed by the first measurement of particle 1.

Ilja

[Ilja Schmelzer]

<daniel@elit.net> schrieb
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No. You'll get spin up for certain. The EPR state between the
two particles is already destroyed by the first measurement of particle 1.

Ilja

[Ilja Schmelzer]

<daniel@elit.net> schrieb
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No. You'll get spin up for certain. The EPR state between the
two particles is already destroyed by the first measurement of particle 1.

Ilja

[Ilja Schmelzer]

<daniel@elit.net> schrieb
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No. You'll get spin up for certain. The EPR state between the
two particles is already destroyed by the first measurement of particle 1.

Ilja

[Ilja Schmelzer]

<daniel@elit.net> schrieb
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No. You'll get spin up for certain. The EPR state between the
two particles is already destroyed by the first measurement of particle 1.

Ilja

[Ilja Schmelzer]

<daniel@elit.net> schrieb
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No. You'll get spin up for certain. The EPR state between the
two particles is already destroyed by the first measurement of particle 1.

Ilja

[Ilja Schmelzer]

<daniel@elit.net> schrieb
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No. You'll get spin up for certain. The EPR state between the
two particles is already destroyed by the first measurement of particle 1.

Ilja

[Ilja Schmelzer]

<daniel@elit.net> schrieb
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No. You'll get spin up for certain. The EPR state between the
two particles is already destroyed by the first measurement of particle 1.

Ilja

[Ilja Schmelzer]

<daniel@elit.net> schrieb
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No. You'll get spin up for certain. The EPR state between the
two particles is already destroyed by the first measurement of particle 1.

Ilja

[Ilja Schmelzer]

<daniel@elit.net> schrieb
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No. You'll get spin up for certain. The EPR state between the
two particles is already destroyed by the first measurement of particle 1.

Ilja

[Ilja Schmelzer]

<daniel@elit.net> schrieb
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No. You'll get spin up for certain. The EPR state between the
two particles is already destroyed by the first measurement of particle 1.

Ilja

[Ilja Schmelzer]

<daniel@elit.net> schrieb
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No. You'll get spin up for certain. The EPR state between the
two particles is already destroyed by the first measurement of particle 1.

Ilja

[Ilja Schmelzer]

<daniel@elit.net> schrieb
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No. You'll get spin up for certain. The EPR state between the
two particles is already destroyed by the first measurement of particle 1.

Ilja

[Ilja Schmelzer]

<daniel@elit.net> schrieb
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No. You'll get spin up for certain. The EPR state between the
two particles is already destroyed by the first measurement of particle 1.

Ilja

[Ilja Schmelzer]

<daniel@elit.net> schrieb
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No. You'll get spin up for certain. The EPR state between the
two particles is already destroyed by the first measurement of particle 1.

Ilja

[Ilja Schmelzer]

<daniel@elit.net> schrieb
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No. You'll get spin up for certain. The EPR state between the
two particles is already destroyed by the first measurement of particle 1.

Ilja

[Ilja Schmelzer]

<daniel@elit.net> schrieb
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No. You'll get spin up for certain. The EPR state between the
two particles is already destroyed by the first measurement of particle 1.

Ilja

[Ilja Schmelzer]

<daniel@elit.net> schrieb
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No. You'll get spin up for certain. The EPR state between the
two particles is already destroyed by the first measurement of particle 1.

Ilja

[Ilja Schmelzer]

<daniel@elit.net> schrieb
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No. You'll get spin up for certain. The EPR state between the
two particles is already destroyed by the first measurement of particle 1.

Ilja

[Ilja Schmelzer]

<daniel@elit.net> schrieb
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No. You'll get spin up for certain. The EPR state between the
two particles is already destroyed by the first measurement of particle 1.

Ilja

[Andreas Most]

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you'll get spin up for particle 1 along the z-axis regardless of when
or how you perform the measurement on particle 2.
QM does not state that one measurement changes the measurement on the
other particle. Instead, QM tells you that the measurements on the two
particles must be consistent.

If you think of the QM wave function as the set of information you
possess about a system, it is clear that with the knowledge of
the state of one particle in EPR you can deduce the state of the
second particle. But there is no exchange of "information" between the
two particles after you performed a measurement on one particle.

>
> We can change the order of things by a Lorentz boost. For example, we
> could set it up so that observer A is in the rest frame of particle 1
> and observer B is in the rest frame of particle 2, and BOTH A and B
> think THEY are the ones making the first measurement. Leading to a
> contradiction. How is it all resolved?
>

There is no contradiction. Einstein said that already when the EPR paper
was published.

Regards, Andreas.

[Andreas Most]

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you'll get spin up for particle 1 along the z-axis regardless of when
or how you perform the measurement on particle 2.
QM does not state that one measurement changes the measurement on the
other particle. Instead, QM tells you that the measurements on the two
particles must be consistent.

If you think of the QM wave function as the set of information you
possess about a system, it is clear that with the knowledge of
the state of one particle in EPR you can deduce the state of the
second particle. But there is no exchange of "information" between the
two particles after you performed a measurement on one particle.

>
> We can change the order of things by a Lorentz boost. For example, we
> could set it up so that observer A is in the rest frame of particle 1
> and observer B is in the rest frame of particle 2, and BOTH A and B
> think THEY are the ones making the first measurement. Leading to a
> contradiction. How is it all resolved?
>

There is no contradiction. Einstein said that already when the EPR paper
was published.

Regards, Andreas.

[Andreas Most]

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you'll get spin up for particle 1 along the z-axis regardless of when
or how you perform the measurement on particle 2.
QM does not state that one measurement changes the measurement on the
other particle. Instead, QM tells you that the measurements on the two
particles must be consistent.

If you think of the QM wave function as the set of information you
possess about a system, it is clear that with the knowledge of
the state of one particle in EPR you can deduce the state of the
second particle. But there is no exchange of "information" between the
two particles after you performed a measurement on one particle.

>
> We can change the order of things by a Lorentz boost. For example, we
> could set it up so that observer A is in the rest frame of particle 1
> and observer B is in the rest frame of particle 2, and BOTH A and B
> think THEY are the ones making the first measurement. Leading to a
> contradiction. How is it all resolved?
>

There is no contradiction. Einstein said that already when the EPR paper
was published.

Regards, Andreas.

[Andreas Most]

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you'll get spin up for particle 1 along the z-axis regardless of when
or how you perform the measurement on particle 2.
QM does not state that one measurement changes the measurement on the
other particle. Instead, QM tells you that the measurements on the two
particles must be consistent.

If you think of the QM wave function as the set of information you
possess about a system, it is clear that with the knowledge of
the state of one particle in EPR you can deduce the state of the
second particle. But there is no exchange of "information" between the
two particles after you performed a measurement on one particle.

>
> We can change the order of things by a Lorentz boost. For example, we
> could set it up so that observer A is in the rest frame of particle 1
> and observer B is in the rest frame of particle 2, and BOTH A and B
> think THEY are the ones making the first measurement. Leading to a
> contradiction. How is it all resolved?
>

There is no contradiction. Einstein said that already when the EPR paper
was published.

Regards, Andreas.

[Andreas Most]

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you'll get spin up for particle 1 along the z-axis regardless of when
or how you perform the measurement on particle 2.
QM does not state that one measurement changes the measurement on the
other particle. Instead, QM tells you that the measurements on the two
particles must be consistent.

If you think of the QM wave function as the set of information you
possess about a system, it is clear that with the knowledge of
the state of one particle in EPR you can deduce the state of the
second particle. But there is no exchange of "information" between the
two particles after you performed a measurement on one particle.

>
> We can change the order of things by a Lorentz boost. For example, we
> could set it up so that observer A is in the rest frame of particle 1
> and observer B is in the rest frame of particle 2, and BOTH A and B
> think THEY are the ones making the first measurement. Leading to a
> contradiction. How is it all resolved?
>

There is no contradiction. Einstein said that already when the EPR paper
was published.

Regards, Andreas.

[Andreas Most]

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you'll get spin up for particle 1 along the z-axis regardless of when
or how you perform the measurement on particle 2.
QM does not state that one measurement changes the measurement on the
other particle. Instead, QM tells you that the measurements on the two
particles must be consistent.

If you think of the QM wave function as the set of information you
possess about a system, it is clear that with the knowledge of
the state of one particle in EPR you can deduce the state of the
second particle. But there is no exchange of "information" between the
two particles after you performed a measurement on one particle.

>
> We can change the order of things by a Lorentz boost. For example, we
> could set it up so that observer A is in the rest frame of particle 1
> and observer B is in the rest frame of particle 2, and BOTH A and B
> think THEY are the ones making the first measurement. Leading to a
> contradiction. How is it all resolved?
>

There is no contradiction. Einstein said that already when the EPR paper
was published.

Regards, Andreas.

[Andreas Most]

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you'll get spin up for particle 1 along the z-axis regardless of when
or how you perform the measurement on particle 2.
QM does not state that one measurement changes the measurement on the
other particle. Instead, QM tells you that the measurements on the two
particles must be consistent.

If you think of the QM wave function as the set of information you
possess about a system, it is clear that with the knowledge of
the state of one particle in EPR you can deduce the state of the
second particle. But there is no exchange of "information" between the
two particles after you performed a measurement on one particle.

>
> We can change the order of things by a Lorentz boost. For example, we
> could set it up so that observer A is in the rest frame of particle 1
> and observer B is in the rest frame of particle 2, and BOTH A and B
> think THEY are the ones making the first measurement. Leading to a
> contradiction. How is it all resolved?
>

There is no contradiction. Einstein said that already when the EPR paper
was published.

Regards, Andreas.

[Andreas Most]

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you'll get spin up for particle 1 along the z-axis regardless of when
or how you perform the measurement on particle 2.
QM does not state that one measurement changes the measurement on the
other particle. Instead, QM tells you that the measurements on the two
particles must be consistent.

If you think of the QM wave function as the set of information you
possess about a system, it is clear that with the knowledge of
the state of one particle in EPR you can deduce the state of the
second particle. But there is no exchange of "information" between the
two particles after you performed a measurement on one particle.

>
> We can change the order of things by a Lorentz boost. For example, we
> could set it up so that observer A is in the rest frame of particle 1
> and observer B is in the rest frame of particle 2, and BOTH A and B
> think THEY are the ones making the first measurement. Leading to a
> contradiction. How is it all resolved?
>

There is no contradiction. Einstein said that already when the EPR paper
was published.

Regards, Andreas.

[Andreas Most]

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you'll get spin up for particle 1 along the z-axis regardless of when
or how you perform the measurement on particle 2.
QM does not state that one measurement changes the measurement on the
other particle. Instead, QM tells you that the measurements on the two
particles must be consistent.

If you think of the QM wave function as the set of information you
possess about a system, it is clear that with the knowledge of
the state of one particle in EPR you can deduce the state of the
second particle. But there is no exchange of "information" between the
two particles after you performed a measurement on one particle.

>
> We can change the order of things by a Lorentz boost. For example, we
> could set it up so that observer A is in the rest frame of particle 1
> and observer B is in the rest frame of particle 2, and BOTH A and B
> think THEY are the ones making the first measurement. Leading to a
> contradiction. How is it all resolved?
>

There is no contradiction. Einstein said that already when the EPR paper
was published.

Regards, Andreas.

[Andreas Most]

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you'll get spin up for particle 1 along the z-axis regardless of when
or how you perform the measurement on particle 2.
QM does not state that one measurement changes the measurement on the
other particle. Instead, QM tells you that the measurements on the two
particles must be consistent.

If you think of the QM wave function as the set of information you
possess about a system, it is clear that with the knowledge of
the state of one particle in EPR you can deduce the state of the
second particle. But there is no exchange of "information" between the
two particles after you performed a measurement on one particle.

>
> We can change the order of things by a Lorentz boost. For example, we
> could set it up so that observer A is in the rest frame of particle 1
> and observer B is in the rest frame of particle 2, and BOTH A and B
> think THEY are the ones making the first measurement. Leading to a
> contradiction. How is it all resolved?
>

There is no contradiction. Einstein said that already when the EPR paper
was published.

Regards, Andreas.

[Andreas Most]

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you'll get spin up for particle 1 along the z-axis regardless of when
or how you perform the measurement on particle 2.
QM does not state that one measurement changes the measurement on the
other particle. Instead, QM tells you that the measurements on the two
particles must be consistent.

If you think of the QM wave function as the set of information you
possess about a system, it is clear that with the knowledge of
the state of one particle in EPR you can deduce the state of the
second particle. But there is no exchange of "information" between the
two particles after you performed a measurement on one particle.

>
> We can change the order of things by a Lorentz boost. For example, we
> could set it up so that observer A is in the rest frame of particle 1
> and observer B is in the rest frame of particle 2, and BOTH A and B
> think THEY are the ones making the first measurement. Leading to a
> contradiction. How is it all resolved?
>

There is no contradiction. Einstein said that already when the EPR paper
was published.

Regards, Andreas.

[Andreas Most]

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you'll get spin up for particle 1 along the z-axis regardless of when
or how you perform the measurement on particle 2.
QM does not state that one measurement changes the measurement on the
other particle. Instead, QM tells you that the measurements on the two
particles must be consistent.

If you think of the QM wave function as the set of information you
possess about a system, it is clear that with the knowledge of
the state of one particle in EPR you can deduce the state of the
second particle. But there is no exchange of "information" between the
two particles after you performed a measurement on one particle.

>
> We can change the order of things by a Lorentz boost. For example, we
> could set it up so that observer A is in the rest frame of particle 1
> and observer B is in the rest frame of particle 2, and BOTH A and B
> think THEY are the ones making the first measurement. Leading to a
> contradiction. How is it all resolved?
>

There is no contradiction. Einstein said that already when the EPR paper
was published.

Regards, Andreas.

[Andreas Most]

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you'll get spin up for particle 1 along the z-axis regardless of when
or how you perform the measurement on particle 2.
QM does not state that one measurement changes the measurement on the
other particle. Instead, QM tells you that the measurements on the two
particles must be consistent.

If you think of the QM wave function as the set of information you
possess about a system, it is clear that with the knowledge of
the state of one particle in EPR you can deduce the state of the
second particle. But there is no exchange of "information" between the
two particles after you performed a measurement on one particle.

>
> We can change the order of things by a Lorentz boost. For example, we
> could set it up so that observer A is in the rest frame of particle 1
> and observer B is in the rest frame of particle 2, and BOTH A and B
> think THEY are the ones making the first measurement. Leading to a
> contradiction. How is it all resolved?
>

There is no contradiction. Einstein said that already when the EPR paper
was published.

Regards, Andreas.

[Andreas Most]

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you'll get spin up for particle 1 along the z-axis regardless of when
or how you perform the measurement on particle 2.
QM does not state that one measurement changes the measurement on the
other particle. Instead, QM tells you that the measurements on the two
particles must be consistent.

If you think of the QM wave function as the set of information you
possess about a system, it is clear that with the knowledge of
the state of one particle in EPR you can deduce the state of the
second particle. But there is no exchange of "information" between the
two particles after you performed a measurement on one particle.

>
> We can change the order of things by a Lorentz boost. For example, we
> could set it up so that observer A is in the rest frame of particle 1
> and observer B is in the rest frame of particle 2, and BOTH A and B
> think THEY are the ones making the first measurement. Leading to a
> contradiction. How is it all resolved?
>

There is no contradiction. Einstein said that already when the EPR paper
was published.

Regards, Andreas.

[Andreas Most]

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you'll get spin up for particle 1 along the z-axis regardless of when
or how you perform the measurement on particle 2.
QM does not state that one measurement changes the measurement on the
other particle. Instead, QM tells you that the measurements on the two
particles must be consistent.

If you think of the QM wave function as the set of information you
possess about a system, it is clear that with the knowledge of
the state of one particle in EPR you can deduce the state of the
second particle. But there is no exchange of "information" between the
two particles after you performed a measurement on one particle.

>
> We can change the order of things by a Lorentz boost. For example, we
> could set it up so that observer A is in the rest frame of particle 1
> and observer B is in the rest frame of particle 2, and BOTH A and B
> think THEY are the ones making the first measurement. Leading to a
> contradiction. How is it all resolved?
>

There is no contradiction. Einstein said that already when the EPR paper
was published.

Regards, Andreas.

[Andreas Most]

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you'll get spin up for particle 1 along the z-axis regardless of when
or how you perform the measurement on particle 2.
QM does not state that one measurement changes the measurement on the
other particle. Instead, QM tells you that the measurements on the two
particles must be consistent.

If you think of the QM wave function as the set of information you
possess about a system, it is clear that with the knowledge of
the state of one particle in EPR you can deduce the state of the
second particle. But there is no exchange of "information" between the
two particles after you performed a measurement on one particle.

>
> We can change the order of things by a Lorentz boost. For example, we
> could set it up so that observer A is in the rest frame of particle 1
> and observer B is in the rest frame of particle 2, and BOTH A and B
> think THEY are the ones making the first measurement. Leading to a
> contradiction. How is it all resolved?
>

There is no contradiction. Einstein said that already when the EPR paper
was published.

Regards, Andreas.

[Andreas Most]

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you'll get spin up for particle 1 along the z-axis regardless of when
or how you perform the measurement on particle 2.
QM does not state that one measurement changes the measurement on the
other particle. Instead, QM tells you that the measurements on the two
particles must be consistent.

If you think of the QM wave function as the set of information you
possess about a system, it is clear that with the knowledge of
the state of one particle in EPR you can deduce the state of the
second particle. But there is no exchange of "information" between the
two particles after you performed a measurement on one particle.

>
> We can change the order of things by a Lorentz boost. For example, we
> could set it up so that observer A is in the rest frame of particle 1
> and observer B is in the rest frame of particle 2, and BOTH A and B
> think THEY are the ones making the first measurement. Leading to a
> contradiction. How is it all resolved?
>

There is no contradiction. Einstein said that already when the EPR paper
was published.

Regards, Andreas.

[Andreas Most]

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you'll get spin up for particle 1 along the z-axis regardless of when
or how you perform the measurement on particle 2.
QM does not state that one measurement changes the measurement on the
other particle. Instead, QM tells you that the measurements on the two
particles must be consistent.

If you think of the QM wave function as the set of information you
possess about a system, it is clear that with the knowledge of
the state of one particle in EPR you can deduce the state of the
second particle. But there is no exchange of "information" between the
two particles after you performed a measurement on one particle.

>
> We can change the order of things by a Lorentz boost. For example, we
> could set it up so that observer A is in the rest frame of particle 1
> and observer B is in the rest frame of particle 2, and BOTH A and B
> think THEY are the ones making the first measurement. Leading to a
> contradiction. How is it all resolved?
>

There is no contradiction. Einstein said that already when the EPR paper
was published.

Regards, Andreas.

[Andreas Most]

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you'll get spin up for particle 1 along the z-axis regardless of when
or how you perform the measurement on particle 2.
QM does not state that one measurement changes the measurement on the
other particle. Instead, QM tells you that the measurements on the two
particles must be consistent.

If you think of the QM wave function as the set of information you
possess about a system, it is clear that with the knowledge of
the state of one particle in EPR you can deduce the state of the
second particle. But there is no exchange of "information" between the
two particles after you performed a measurement on one particle.

>
> We can change the order of things by a Lorentz boost. For example, we
> could set it up so that observer A is in the rest frame of particle 1
> and observer B is in the rest frame of particle 2, and BOTH A and B
> think THEY are the ones making the first measurement. Leading to a
> contradiction. How is it all resolved?
>

There is no contradiction. Einstein said that already when the EPR paper
was published.

Regards, Andreas.

[Andreas Most]

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you'll get spin up for particle 1 along the z-axis regardless of when
or how you perform the measurement on particle 2.
QM does not state that one measurement changes the measurement on the
other particle. Instead, QM tells you that the measurements on the two
particles must be consistent.

If you think of the QM wave function as the set of information you
possess about a system, it is clear that with the knowledge of
the state of one particle in EPR you can deduce the state of the
second particle. But there is no exchange of "information" between the
two particles after you performed a measurement on one particle.

>
> We can change the order of things by a Lorentz boost. For example, we
> could set it up so that observer A is in the rest frame of particle 1
> and observer B is in the rest frame of particle 2, and BOTH A and B
> think THEY are the ones making the first measurement. Leading to a
> contradiction. How is it all resolved?
>

There is no contradiction. Einstein said that already when the EPR paper
was published.

Regards, Andreas.

[Andreas Most]

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you'll get spin up for particle 1 along the z-axis regardless of when
or how you perform the measurement on particle 2.
QM does not state that one measurement changes the measurement on the
other particle. Instead, QM tells you that the measurements on the two
particles must be consistent.

If you think of the QM wave function as the set of information you
possess about a system, it is clear that with the knowledge of
the state of one particle in EPR you can deduce the state of the
second particle. But there is no exchange of "information" between the
two particles after you performed a measurement on one particle.

>
> We can change the order of things by a Lorentz boost. For example, we
> could set it up so that observer A is in the rest frame of particle 1
> and observer B is in the rest frame of particle 2, and BOTH A and B
> think THEY are the ones making the first measurement. Leading to a
> contradiction. How is it all resolved?
>

There is no contradiction. Einstein said that already when the EPR paper
was published.

Regards, Andreas.

[Andreas Most]

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you'll get spin up for particle 1 along the z-axis regardless of when
or how you perform the measurement on particle 2.
QM does not state that one measurement changes the measurement on the
other particle. Instead, QM tells you that the measurements on the two
particles must be consistent.

If you think of the QM wave function as the set of information you
possess about a system, it is clear that with the knowledge of
the state of one particle in EPR you can deduce the state of the
second particle. But there is no exchange of "information" between the
two particles after you performed a measurement on one particle.

>
> We can change the order of things by a Lorentz boost. For example, we
> could set it up so that observer A is in the rest frame of particle 1
> and observer B is in the rest frame of particle 2, and BOTH A and B
> think THEY are the ones making the first measurement. Leading to a
> contradiction. How is it all resolved?
>

There is no contradiction. Einstein said that already when the EPR paper
was published.

Regards, Andreas.

[Andreas Most]

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you'll get spin up for particle 1 along the z-axis regardless of when
or how you perform the measurement on particle 2.
QM does not state that one measurement changes the measurement on the
other particle. Instead, QM tells you that the measurements on the two
particles must be consistent.

If you think of the QM wave function as the set of information you
possess about a system, it is clear that with the knowledge of
the state of one particle in EPR you can deduce the state of the
second particle. But there is no exchange of "information" between the
two particles after you performed a measurement on one particle.

>
> We can change the order of things by a Lorentz boost. For example, we
> could set it up so that observer A is in the rest frame of particle 1
> and observer B is in the rest frame of particle 2, and BOTH A and B
> think THEY are the ones making the first measurement. Leading to a
> contradiction. How is it all resolved?
>

There is no contradiction. Einstein said that already when the EPR paper
was published.

Regards, Andreas.

[Andreas Most]

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you'll get spin up for particle 1 along the z-axis regardless of when
or how you perform the measurement on particle 2.
QM does not state that one measurement changes the measurement on the
other particle. Instead, QM tells you that the measurements on the two
particles must be consistent.

If you think of the QM wave function as the set of information you
possess about a system, it is clear that with the knowledge of
the state of one particle in EPR you can deduce the state of the
second particle. But there is no exchange of "information" between the
two particles after you performed a measurement on one particle.

>
> We can change the order of things by a Lorentz boost. For example, we
> could set it up so that observer A is in the rest frame of particle 1
> and observer B is in the rest frame of particle 2, and BOTH A and B
> think THEY are the ones making the first measurement. Leading to a
> contradiction. How is it all resolved?
>

There is no contradiction. Einstein said that already when the EPR paper
was published.

Regards, Andreas.

[Andreas Most]

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you'll get spin up for particle 1 along the z-axis regardless of when
or how you perform the measurement on particle 2.
QM does not state that one measurement changes the measurement on the
other particle. Instead, QM tells you that the measurements on the two
particles must be consistent.

If you think of the QM wave function as the set of information you
possess about a system, it is clear that with the knowledge of
the state of one particle in EPR you can deduce the state of the
second particle. But there is no exchange of "information" between the
two particles after you performed a measurement on one particle.

>
> We can change the order of things by a Lorentz boost. For example, we
> could set it up so that observer A is in the rest frame of particle 1
> and observer B is in the rest frame of particle 2, and BOTH A and B
> think THEY are the ones making the first measurement. Leading to a
> contradiction. How is it all resolved?
>

There is no contradiction. Einstein said that already when the EPR paper
was published.

Regards, Andreas.

[Andreas Most]

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you'll get spin up for particle 1 along the z-axis regardless of when
or how you perform the measurement on particle 2.
QM does not state that one measurement changes the measurement on the
other particle. Instead, QM tells you that the measurements on the two
particles must be consistent.

If you think of the QM wave function as the set of information you
possess about a system, it is clear that with the knowledge of
the state of one particle in EPR you can deduce the state of the
second particle. But there is no exchange of "information" between the
two particles after you performed a measurement on one particle.

>
> We can change the order of things by a Lorentz boost. For example, we
> could set it up so that observer A is in the rest frame of particle 1
> and observer B is in the rest frame of particle 2, and BOTH A and B
> think THEY are the ones making the first measurement. Leading to a
> contradiction. How is it all resolved?
>

There is no contradiction. Einstein said that already when the EPR paper
was published.

Regards, Andreas.

[Andreas Most]

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you'll get spin up for particle 1 along the z-axis regardless of when
or how you perform the measurement on particle 2.
QM does not state that one measurement changes the measurement on the
other particle. Instead, QM tells you that the measurements on the two
particles must be consistent.

If you think of the QM wave function as the set of information you
possess about a system, it is clear that with the knowledge of
the state of one particle in EPR you can deduce the state of the
second particle. But there is no exchange of "information" between the
two particles after you performed a measurement on one particle.

>
> We can change the order of things by a Lorentz boost. For example, we
> could set it up so that observer A is in the rest frame of particle 1
> and observer B is in the rest frame of particle 2, and BOTH A and B
> think THEY are the ones making the first measurement. Leading to a
> contradiction. How is it all resolved?
>

There is no contradiction. Einstein said that already when the EPR paper
was published.

Regards, Andreas.

[Ralph Hartley]

Is there a FAQ entry for this? We see it all the time.

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you get spin up with probability 1.

Lets do the calculation. Initial state (ignoring factors of sqrt(2)):

|Up>|Down> - |Down>|Up>

after the first measurement:

|Up>|Down>

We want to measure the second particle in a different direction, so we
will use

|Up> = |Left> + |Right> (Sill ignoring factors of sqrt(2))
|Down> = |Left> - |Right> (They can take care of themselves)
and
|Left> = |Up> + |Down>
|Right> = |Up> - |Down>

to re-write the state as:

|Up>|Down> = |Up>(|Left> - |Right>) = |Up>|Left> - |Up>|Right>

After the second measurement it is either |Up>|Left> or |Up>|Right>. In
both cases measuring the first particle again gives up.

Lets try again without the first measurement:

|Up>|Down> - |Down>|Up>
= |Up>(|Left> - |Right>) - |Down>(|Left> + |Right>)
= |Up>|Left> - |Up>|Right> - |Down>|left> - |Down>|Right>
= (|Up> - |Down>)|Left> - (|Up> + |Down>)|Right>
= |Right>|Left> - |Left>|Right>

Measuring the second particle makes the state either
|Right>|Left> or |Left>|Right>. Now measuring the first particle (in the
z axis) *does* give 50-50. Of course, if we had made that measurement
first it still would be 50-50. So it was the extra measurement at the
start that made the difference.

Entanglement is a one time thing. Once you do a measurement on either
particle, you don't have an entangled pair any more, you have a pair of
particles, both in known states.

You might think that the loss of entanglement itself could be used to
send a signal. It can't, because entanglement cannot be detected
locally. To tell if the particles are still entangled you have to
compare measurements. You can't do that instantaneously unless you can
*already* send an instantaneous signal, by some other means.

Ralph Hartley

[Ralph Hartley]

Is there a FAQ entry for this? We see it all the time.

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you get spin up with probability 1.

Lets do the calculation. Initial state (ignoring factors of sqrt(2)):

|Up>|Down> - |Down>|Up>

after the first measurement:

|Up>|Down>

We want to measure the second particle in a different direction, so we
will use

|Up> = |Left> + |Right> (Sill ignoring factors of sqrt(2))
|Down> = |Left> - |Right> (They can take care of themselves)
and
|Left> = |Up> + |Down>
|Right> = |Up> - |Down>

to re-write the state as:

|Up>|Down> = |Up>(|Left> - |Right>) = |Up>|Left> - |Up>|Right>

After the second measurement it is either |Up>|Left> or |Up>|Right>. In
both cases measuring the first particle again gives up.

Lets try again without the first measurement:

|Up>|Down> - |Down>|Up>
= |Up>(|Left> - |Right>) - |Down>(|Left> + |Right>)
= |Up>|Left> - |Up>|Right> - |Down>|left> - |Down>|Right>
= (|Up> - |Down>)|Left> - (|Up> + |Down>)|Right>
= |Right>|Left> - |Left>|Right>

Measuring the second particle makes the state either
|Right>|Left> or |Left>|Right>. Now measuring the first particle (in the
z axis) *does* give 50-50. Of course, if we had made that measurement
first it still would be 50-50. So it was the extra measurement at the
start that made the difference.

Entanglement is a one time thing. Once you do a measurement on either
particle, you don't have an entangled pair any more, you have a pair of
particles, both in known states.

You might think that the loss of entanglement itself could be used to
send a signal. It can't, because entanglement cannot be detected
locally. To tell if the particles are still entangled you have to
compare measurements. You can't do that instantaneously unless you can
*already* send an instantaneous signal, by some other means.

Ralph Hartley

[Ralph Hartley]

Is there a FAQ entry for this? We see it all the time.

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you get spin up with probability 1.

Lets do the calculation. Initial state (ignoring factors of sqrt(2)):

|Up>|Down> - |Down>|Up>

after the first measurement:

|Up>|Down>

We want to measure the second particle in a different direction, so we
will use

|Up> = |Left> + |Right> (Sill ignoring factors of sqrt(2))
|Down> = |Left> - |Right> (They can take care of themselves)
and
|Left> = |Up> + |Down>
|Right> = |Up> - |Down>

to re-write the state as:

|Up>|Down> = |Up>(|Left> - |Right>) = |Up>|Left> - |Up>|Right>

After the second measurement it is either |Up>|Left> or |Up>|Right>. In
both cases measuring the first particle again gives up.

Lets try again without the first measurement:

|Up>|Down> - |Down>|Up>
= |Up>(|Left> - |Right>) - |Down>(|Left> + |Right>)
= |Up>|Left> - |Up>|Right> - |Down>|left> - |Down>|Right>
= (|Up> - |Down>)|Left> - (|Up> + |Down>)|Right>
= |Right>|Left> - |Left>|Right>

Measuring the second particle makes the state either
|Right>|Left> or |Left>|Right>. Now measuring the first particle (in the
z axis) *does* give 50-50. Of course, if we had made that measurement
first it still would be 50-50. So it was the extra measurement at the
start that made the difference.

Entanglement is a one time thing. Once you do a measurement on either
particle, you don't have an entangled pair any more, you have a pair of
particles, both in known states.

You might think that the loss of entanglement itself could be used to
send a signal. It can't, because entanglement cannot be detected
locally. To tell if the particles are still entangled you have to
compare measurements. You can't do that instantaneously unless you can
*already* send an instantaneous signal, by some other means.

Ralph Hartley

[Ralph Hartley]

Is there a FAQ entry for this? We see it all the time.

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you get spin up with probability 1.

Lets do the calculation. Initial state (ignoring factors of sqrt(2)):

|Up>|Down> - |Down>|Up>

after the first measurement:

|Up>|Down>

We want to measure the second particle in a different direction, so we
will use

|Up> = |Left> + |Right> (Sill ignoring factors of sqrt(2))
|Down> = |Left> - |Right> (They can take care of themselves)
and
|Left> = |Up> + |Down>
|Right> = |Up> - |Down>

to re-write the state as:

|Up>|Down> = |Up>(|Left> - |Right>) = |Up>|Left> - |Up>|Right>

After the second measurement it is either |Up>|Left> or |Up>|Right>. In
both cases measuring the first particle again gives up.

Lets try again without the first measurement:

|Up>|Down> - |Down>|Up>
= |Up>(|Left> - |Right>) - |Down>(|Left> + |Right>)
= |Up>|Left> - |Up>|Right> - |Down>|left> - |Down>|Right>
= (|Up> - |Down>)|Left> - (|Up> + |Down>)|Right>
= |Right>|Left> - |Left>|Right>

Measuring the second particle makes the state either
|Right>|Left> or |Left>|Right>. Now measuring the first particle (in the
z axis) *does* give 50-50. Of course, if we had made that measurement
first it still would be 50-50. So it was the extra measurement at the
start that made the difference.

Entanglement is a one time thing. Once you do a measurement on either
particle, you don't have an entangled pair any more, you have a pair of
particles, both in known states.

You might think that the loss of entanglement itself could be used to
send a signal. It can't, because entanglement cannot be detected
locally. To tell if the particles are still entangled you have to
compare measurements. You can't do that instantaneously unless you can
*already* send an instantaneous signal, by some other means.

Ralph Hartley

[Ralph Hartley]

Is there a FAQ entry for this? We see it all the time.

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you get spin up with probability 1.

Lets do the calculation. Initial state (ignoring factors of sqrt(2)):

|Up>|Down> - |Down>|Up>

after the first measurement:

|Up>|Down>

We want to measure the second particle in a different direction, so we
will use

|Up> = |Left> + |Right> (Sill ignoring factors of sqrt(2))
|Down> = |Left> - |Right> (They can take care of themselves)
and
|Left> = |Up> + |Down>
|Right> = |Up> - |Down>

to re-write the state as:

|Up>|Down> = |Up>(|Left> - |Right>) = |Up>|Left> - |Up>|Right>

After the second measurement it is either |Up>|Left> or |Up>|Right>. In
both cases measuring the first particle again gives up.

Lets try again without the first measurement:

|Up>|Down> - |Down>|Up>
= |Up>(|Left> - |Right>) - |Down>(|Left> + |Right>)
= |Up>|Left> - |Up>|Right> - |Down>|left> - |Down>|Right>
= (|Up> - |Down>)|Left> - (|Up> + |Down>)|Right>
= |Right>|Left> - |Left>|Right>

Measuring the second particle makes the state either
|Right>|Left> or |Left>|Right>. Now measuring the first particle (in the
z axis) *does* give 50-50. Of course, if we had made that measurement
first it still would be 50-50. So it was the extra measurement at the
start that made the difference.

Entanglement is a one time thing. Once you do a measurement on either
particle, you don't have an entangled pair any more, you have a pair of
particles, both in known states.

You might think that the loss of entanglement itself could be used to
send a signal. It can't, because entanglement cannot be detected
locally. To tell if the particles are still entangled you have to
compare measurements. You can't do that instantaneously unless you can
*already* send an instantaneous signal, by some other means.

Ralph Hartley

[Ralph Hartley]

Is there a FAQ entry for this? We see it all the time.

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you get spin up with probability 1.

Lets do the calculation. Initial state (ignoring factors of sqrt(2)):

|Up>|Down> - |Down>|Up>

after the first measurement:

|Up>|Down>

We want to measure the second particle in a different direction, so we
will use

|Up> = |Left> + |Right> (Sill ignoring factors of sqrt(2))
|Down> = |Left> - |Right> (They can take care of themselves)
and
|Left> = |Up> + |Down>
|Right> = |Up> - |Down>

to re-write the state as:

|Up>|Down> = |Up>(|Left> - |Right>) = |Up>|Left> - |Up>|Right>

After the second measurement it is either |Up>|Left> or |Up>|Right>. In
both cases measuring the first particle again gives up.

Lets try again without the first measurement:

|Up>|Down> - |Down>|Up>
= |Up>(|Left> - |Right>) - |Down>(|Left> + |Right>)
= |Up>|Left> - |Up>|Right> - |Down>|left> - |Down>|Right>
= (|Up> - |Down>)|Left> - (|Up> + |Down>)|Right>
= |Right>|Left> - |Left>|Right>

Measuring the second particle makes the state either
|Right>|Left> or |Left>|Right>. Now measuring the first particle (in the
z axis) *does* give 50-50. Of course, if we had made that measurement
first it still would be 50-50. So it was the extra measurement at the
start that made the difference.

Entanglement is a one time thing. Once you do a measurement on either
particle, you don't have an entangled pair any more, you have a pair of
particles, both in known states.

You might think that the loss of entanglement itself could be used to
send a signal. It can't, because entanglement cannot be detected
locally. To tell if the particles are still entangled you have to
compare measurements. You can't do that instantaneously unless you can
*already* send an instantaneous signal, by some other means.

Ralph Hartley

[Ralph Hartley]

Is there a FAQ entry for this? We see it all the time.

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you get spin up with probability 1.

Lets do the calculation. Initial state (ignoring factors of sqrt(2)):

|Up>|Down> - |Down>|Up>

after the first measurement:

|Up>|Down>

We want to measure the second particle in a different direction, so we
will use

|Up> = |Left> + |Right> (Sill ignoring factors of sqrt(2))
|Down> = |Left> - |Right> (They can take care of themselves)
and
|Left> = |Up> + |Down>
|Right> = |Up> - |Down>

to re-write the state as:

|Up>|Down> = |Up>(|Left> - |Right>) = |Up>|Left> - |Up>|Right>

After the second measurement it is either |Up>|Left> or |Up>|Right>. In
both cases measuring the first particle again gives up.

Lets try again without the first measurement:

|Up>|Down> - |Down>|Up>
= |Up>(|Left> - |Right>) - |Down>(|Left> + |Right>)
= |Up>|Left> - |Up>|Right> - |Down>|left> - |Down>|Right>
= (|Up> - |Down>)|Left> - (|Up> + |Down>)|Right>
= |Right>|Left> - |Left>|Right>

Measuring the second particle makes the state either
|Right>|Left> or |Left>|Right>. Now measuring the first particle (in the
z axis) *does* give 50-50. Of course, if we had made that measurement
first it still would be 50-50. So it was the extra measurement at the
start that made the difference.

Entanglement is a one time thing. Once you do a measurement on either
particle, you don't have an entangled pair any more, you have a pair of
particles, both in known states.

You might think that the loss of entanglement itself could be used to
send a signal. It can't, because entanglement cannot be detected
locally. To tell if the particles are still entangled you have to
compare measurements. You can't do that instantaneously unless you can
*already* send an instantaneous signal, by some other means.

Ralph Hartley

[Ralph Hartley]

Is there a FAQ entry for this? We see it all the time.

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you get spin up with probability 1.

Lets do the calculation. Initial state (ignoring factors of sqrt(2)):

|Up>|Down> - |Down>|Up>

after the first measurement:

|Up>|Down>

We want to measure the second particle in a different direction, so we
will use

|Up> = |Left> + |Right> (Sill ignoring factors of sqrt(2))
|Down> = |Left> - |Right> (They can take care of themselves)
and
|Left> = |Up> + |Down>
|Right> = |Up> - |Down>

to re-write the state as:

|Up>|Down> = |Up>(|Left> - |Right>) = |Up>|Left> - |Up>|Right>

After the second measurement it is either |Up>|Left> or |Up>|Right>. In
both cases measuring the first particle again gives up.

Lets try again without the first measurement:

|Up>|Down> - |Down>|Up>
= |Up>(|Left> - |Right>) - |Down>(|Left> + |Right>)
= |Up>|Left> - |Up>|Right> - |Down>|left> - |Down>|Right>
= (|Up> - |Down>)|Left> - (|Up> + |Down>)|Right>
= |Right>|Left> - |Left>|Right>

Measuring the second particle makes the state either
|Right>|Left> or |Left>|Right>. Now measuring the first particle (in the
z axis) *does* give 50-50. Of course, if we had made that measurement
first it still would be 50-50. So it was the extra measurement at the
start that made the difference.

Entanglement is a one time thing. Once you do a measurement on either
particle, you don't have an entangled pair any more, you have a pair of
particles, both in known states.

You might think that the loss of entanglement itself could be used to
send a signal. It can't, because entanglement cannot be detected
locally. To tell if the particles are still entangled you have to
compare measurements. You can't do that instantaneously unless you can
*already* send an instantaneous signal, by some other means.

Ralph Hartley

[Ralph Hartley]

Is there a FAQ entry for this? We see it all the time.

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you get spin up with probability 1.

Lets do the calculation. Initial state (ignoring factors of sqrt(2)):

|Up>|Down> - |Down>|Up>

after the first measurement:

|Up>|Down>

We want to measure the second particle in a different direction, so we
will use

|Up> = |Left> + |Right> (Sill ignoring factors of sqrt(2))
|Down> = |Left> - |Right> (They can take care of themselves)
and
|Left> = |Up> + |Down>
|Right> = |Up> - |Down>

to re-write the state as:

|Up>|Down> = |Up>(|Left> - |Right>) = |Up>|Left> - |Up>|Right>

After the second measurement it is either |Up>|Left> or |Up>|Right>. In
both cases measuring the first particle again gives up.

Lets try again without the first measurement:

|Up>|Down> - |Down>|Up>
= |Up>(|Left> - |Right>) - |Down>(|Left> + |Right>)
= |Up>|Left> - |Up>|Right> - |Down>|left> - |Down>|Right>
= (|Up> - |Down>)|Left> - (|Up> + |Down>)|Right>
= |Right>|Left> - |Left>|Right>

Measuring the second particle makes the state either
|Right>|Left> or |Left>|Right>. Now measuring the first particle (in the
z axis) *does* give 50-50. Of course, if we had made that measurement
first it still would be 50-50. So it was the extra measurement at the
start that made the difference.

Entanglement is a one time thing. Once you do a measurement on either
particle, you don't have an entangled pair any more, you have a pair of
particles, both in known states.

You might think that the loss of entanglement itself could be used to
send a signal. It can't, because entanglement cannot be detected
locally. To tell if the particles are still entangled you have to
compare measurements. You can't do that instantaneously unless you can
*already* send an instantaneous signal, by some other means.

Ralph Hartley

[Ralph Hartley]

Is there a FAQ entry for this? We see it all the time.

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you get spin up with probability 1.

Lets do the calculation. Initial state (ignoring factors of sqrt(2)):

|Up>|Down> - |Down>|Up>

after the first measurement:

|Up>|Down>

We want to measure the second particle in a different direction, so we
will use

|Up> = |Left> + |Right> (Sill ignoring factors of sqrt(2))
|Down> = |Left> - |Right> (They can take care of themselves)
and
|Left> = |Up> + |Down>
|Right> = |Up> - |Down>

to re-write the state as:

|Up>|Down> = |Up>(|Left> - |Right>) = |Up>|Left> - |Up>|Right>

After the second measurement it is either |Up>|Left> or |Up>|Right>. In
both cases measuring the first particle again gives up.

Lets try again without the first measurement:

|Up>|Down> - |Down>|Up>
= |Up>(|Left> - |Right>) - |Down>(|Left> + |Right>)
= |Up>|Left> - |Up>|Right> - |Down>|left> - |Down>|Right>
= (|Up> - |Down>)|Left> - (|Up> + |Down>)|Right>
= |Right>|Left> - |Left>|Right>

Measuring the second particle makes the state either
|Right>|Left> or |Left>|Right>. Now measuring the first particle (in the
z axis) *does* give 50-50. Of course, if we had made that measurement
first it still would be 50-50. So it was the extra measurement at the
start that made the difference.

Entanglement is a one time thing. Once you do a measurement on either
particle, you don't have an entangled pair any more, you have a pair of
particles, both in known states.

You might think that the loss of entanglement itself could be used to
send a signal. It can't, because entanglement cannot be detected
locally. To tell if the particles are still entangled you have to
compare measurements. You can't do that instantaneously unless you can
*already* send an instantaneous signal, by some other means.

Ralph Hartley

[Ralph Hartley]

Is there a FAQ entry for this? We see it all the time.

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you get spin up with probability 1.

Lets do the calculation. Initial state (ignoring factors of sqrt(2)):

|Up>|Down> - |Down>|Up>

after the first measurement:

|Up>|Down>

We want to measure the second particle in a different direction, so we
will use

|Up> = |Left> + |Right> (Sill ignoring factors of sqrt(2))
|Down> = |Left> - |Right> (They can take care of themselves)
and
|Left> = |Up> + |Down>
|Right> = |Up> - |Down>

to re-write the state as:

|Up>|Down> = |Up>(|Left> - |Right>) = |Up>|Left> - |Up>|Right>

After the second measurement it is either |Up>|Left> or |Up>|Right>. In
both cases measuring the first particle again gives up.

Lets try again without the first measurement:

|Up>|Down> - |Down>|Up>
= |Up>(|Left> - |Right>) - |Down>(|Left> + |Right>)
= |Up>|Left> - |Up>|Right> - |Down>|left> - |Down>|Right>
= (|Up> - |Down>)|Left> - (|Up> + |Down>)|Right>
= |Right>|Left> - |Left>|Right>

Measuring the second particle makes the state either
|Right>|Left> or |Left>|Right>. Now measuring the first particle (in the
z axis) *does* give 50-50. Of course, if we had made that measurement
first it still would be 50-50. So it was the extra measurement at the
start that made the difference.

Entanglement is a one time thing. Once you do a measurement on either
particle, you don't have an entangled pair any more, you have a pair of
particles, both in known states.

You might think that the loss of entanglement itself could be used to
send a signal. It can't, because entanglement cannot be detected
locally. To tell if the particles are still entangled you have to
compare measurements. You can't do that instantaneously unless you can
*already* send an instantaneous signal, by some other means.

Ralph Hartley

[Ralph Hartley]

Is there a FAQ entry for this? We see it all the time.

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you get spin up with probability 1.

Lets do the calculation. Initial state (ignoring factors of sqrt(2)):

|Up>|Down> - |Down>|Up>

after the first measurement:

|Up>|Down>

We want to measure the second particle in a different direction, so we
will use

|Up> = |Left> + |Right> (Sill ignoring factors of sqrt(2))
|Down> = |Left> - |Right> (They can take care of themselves)
and
|Left> = |Up> + |Down>
|Right> = |Up> - |Down>

to re-write the state as:

|Up>|Down> = |Up>(|Left> - |Right>) = |Up>|Left> - |Up>|Right>

After the second measurement it is either |Up>|Left> or |Up>|Right>. In
both cases measuring the first particle again gives up.

Lets try again without the first measurement:

|Up>|Down> - |Down>|Up>
= |Up>(|Left> - |Right>) - |Down>(|Left> + |Right>)
= |Up>|Left> - |Up>|Right> - |Down>|left> - |Down>|Right>
= (|Up> - |Down>)|Left> - (|Up> + |Down>)|Right>
= |Right>|Left> - |Left>|Right>

Measuring the second particle makes the state either
|Right>|Left> or |Left>|Right>. Now measuring the first particle (in the
z axis) *does* give 50-50. Of course, if we had made that measurement
first it still would be 50-50. So it was the extra measurement at the
start that made the difference.

Entanglement is a one time thing. Once you do a measurement on either
particle, you don't have an entangled pair any more, you have a pair of
particles, both in known states.

You might think that the loss of entanglement itself could be used to
send a signal. It can't, because entanglement cannot be detected
locally. To tell if the particles are still entangled you have to
compare measurements. You can't do that instantaneously unless you can
*already* send an instantaneous signal, by some other means.

Ralph Hartley

[Ralph Hartley]

Is there a FAQ entry for this? We see it all the time.

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you get spin up with probability 1.

Lets do the calculation. Initial state (ignoring factors of sqrt(2)):

|Up>|Down> - |Down>|Up>

after the first measurement:

|Up>|Down>

We want to measure the second particle in a different direction, so we
will use

|Up> = |Left> + |Right> (Sill ignoring factors of sqrt(2))
|Down> = |Left> - |Right> (They can take care of themselves)
and
|Left> = |Up> + |Down>
|Right> = |Up> - |Down>

to re-write the state as:

|Up>|Down> = |Up>(|Left> - |Right>) = |Up>|Left> - |Up>|Right>

After the second measurement it is either |Up>|Left> or |Up>|Right>. In
both cases measuring the first particle again gives up.

Lets try again without the first measurement:

|Up>|Down> - |Down>|Up>
= |Up>(|Left> - |Right>) - |Down>(|Left> + |Right>)
= |Up>|Left> - |Up>|Right> - |Down>|left> - |Down>|Right>
= (|Up> - |Down>)|Left> - (|Up> + |Down>)|Right>
= |Right>|Left> - |Left>|Right>

Measuring the second particle makes the state either
|Right>|Left> or |Left>|Right>. Now measuring the first particle (in the
z axis) *does* give 50-50. Of course, if we had made that measurement
first it still would be 50-50. So it was the extra measurement at the
start that made the difference.

Entanglement is a one time thing. Once you do a measurement on either
particle, you don't have an entangled pair any more, you have a pair of
particles, both in known states.

You might think that the loss of entanglement itself could be used to
send a signal. It can't, because entanglement cannot be detected
locally. To tell if the particles are still entangled you have to
compare measurements. You can't do that instantaneously unless you can
*already* send an instantaneous signal, by some other means.

Ralph Hartley

[Ralph Hartley]

Is there a FAQ entry for this? We see it all the time.

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you get spin up with probability 1.

Lets do the calculation. Initial state (ignoring factors of sqrt(2)):

|Up>|Down> - |Down>|Up>

after the first measurement:

|Up>|Down>

We want to measure the second particle in a different direction, so we
will use

|Up> = |Left> + |Right> (Sill ignoring factors of sqrt(2))
|Down> = |Left> - |Right> (They can take care of themselves)
and
|Left> = |Up> + |Down>
|Right> = |Up> - |Down>

to re-write the state as:

|Up>|Down> = |Up>(|Left> - |Right>) = |Up>|Left> - |Up>|Right>

After the second measurement it is either |Up>|Left> or |Up>|Right>. In
both cases measuring the first particle again gives up.

Lets try again without the first measurement:

|Up>|Down> - |Down>|Up>
= |Up>(|Left> - |Right>) - |Down>(|Left> + |Right>)
= |Up>|Left> - |Up>|Right> - |Down>|left> - |Down>|Right>
= (|Up> - |Down>)|Left> - (|Up> + |Down>)|Right>
= |Right>|Left> - |Left>|Right>

Measuring the second particle makes the state either
|Right>|Left> or |Left>|Right>. Now measuring the first particle (in the
z axis) *does* give 50-50. Of course, if we had made that measurement
first it still would be 50-50. So it was the extra measurement at the
start that made the difference.

Entanglement is a one time thing. Once you do a measurement on either
particle, you don't have an entangled pair any more, you have a pair of
particles, both in known states.

You might think that the loss of entanglement itself could be used to
send a signal. It can't, because entanglement cannot be detected
locally. To tell if the particles are still entangled you have to
compare measurements. You can't do that instantaneously unless you can
*already* send an instantaneous signal, by some other means.

Ralph Hartley

[Ralph Hartley]

Is there a FAQ entry for this? We see it all the time.

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you get spin up with probability 1.

Lets do the calculation. Initial state (ignoring factors of sqrt(2)):

|Up>|Down> - |Down>|Up>

after the first measurement:

|Up>|Down>

We want to measure the second particle in a different direction, so we
will use

|Up> = |Left> + |Right> (Sill ignoring factors of sqrt(2))
|Down> = |Left> - |Right> (They can take care of themselves)
and
|Left> = |Up> + |Down>
|Right> = |Up> - |Down>

to re-write the state as:

|Up>|Down> = |Up>(|Left> - |Right>) = |Up>|Left> - |Up>|Right>

After the second measurement it is either |Up>|Left> or |Up>|Right>. In
both cases measuring the first particle again gives up.

Lets try again without the first measurement:

|Up>|Down> - |Down>|Up>
= |Up>(|Left> - |Right>) - |Down>(|Left> + |Right>)
= |Up>|Left> - |Up>|Right> - |Down>|left> - |Down>|Right>
= (|Up> - |Down>)|Left> - (|Up> + |Down>)|Right>
= |Right>|Left> - |Left>|Right>

Measuring the second particle makes the state either
|Right>|Left> or |Left>|Right>. Now measuring the first particle (in the
z axis) *does* give 50-50. Of course, if we had made that measurement
first it still would be 50-50. So it was the extra measurement at the
start that made the difference.

Entanglement is a one time thing. Once you do a measurement on either
particle, you don't have an entangled pair any more, you have a pair of
particles, both in known states.

You might think that the loss of entanglement itself could be used to
send a signal. It can't, because entanglement cannot be detected
locally. To tell if the particles are still entangled you have to
compare measurements. You can't do that instantaneously unless you can
*already* send an instantaneous signal, by some other means.

Ralph Hartley

[Ralph Hartley]

Is there a FAQ entry for this? We see it all the time.

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you get spin up with probability 1.

Lets do the calculation. Initial state (ignoring factors of sqrt(2)):

|Up>|Down> - |Down>|Up>

after the first measurement:

|Up>|Down>

We want to measure the second particle in a different direction, so we
will use

|Up> = |Left> + |Right> (Sill ignoring factors of sqrt(2))
|Down> = |Left> - |Right> (They can take care of themselves)
and
|Left> = |Up> + |Down>
|Right> = |Up> - |Down>

to re-write the state as:

|Up>|Down> = |Up>(|Left> - |Right>) = |Up>|Left> - |Up>|Right>

After the second measurement it is either |Up>|Left> or |Up>|Right>. In
both cases measuring the first particle again gives up.

Lets try again without the first measurement:

|Up>|Down> - |Down>|Up>
= |Up>(|Left> - |Right>) - |Down>(|Left> + |Right>)
= |Up>|Left> - |Up>|Right> - |Down>|left> - |Down>|Right>
= (|Up> - |Down>)|Left> - (|Up> + |Down>)|Right>
= |Right>|Left> - |Left>|Right>

Measuring the second particle makes the state either
|Right>|Left> or |Left>|Right>. Now measuring the first particle (in the
z axis) *does* give 50-50. Of course, if we had made that measurement
first it still would be 50-50. So it was the extra measurement at the
start that made the difference.

Entanglement is a one time thing. Once you do a measurement on either
particle, you don't have an entangled pair any more, you have a pair of
particles, both in known states.

You might think that the loss of entanglement itself could be used to
send a signal. It can't, because entanglement cannot be detected
locally. To tell if the particles are still entangled you have to
compare measurements. You can't do that instantaneously unless you can
*already* send an instantaneous signal, by some other means.

Ralph Hartley

[Ralph Hartley]

Is there a FAQ entry for this? We see it all the time.

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you get spin up with probability 1.

Lets do the calculation. Initial state (ignoring factors of sqrt(2)):

|Up>|Down> - |Down>|Up>

after the first measurement:

|Up>|Down>

We want to measure the second particle in a different direction, so we
will use

|Up> = |Left> + |Right> (Sill ignoring factors of sqrt(2))
|Down> = |Left> - |Right> (They can take care of themselves)
and
|Left> = |Up> + |Down>
|Right> = |Up> - |Down>

to re-write the state as:

|Up>|Down> = |Up>(|Left> - |Right>) = |Up>|Left> - |Up>|Right>

After the second measurement it is either |Up>|Left> or |Up>|Right>. In
both cases measuring the first particle again gives up.

Lets try again without the first measurement:

|Up>|Down> - |Down>|Up>
= |Up>(|Left> - |Right>) - |Down>(|Left> + |Right>)
= |Up>|Left> - |Up>|Right> - |Down>|left> - |Down>|Right>
= (|Up> - |Down>)|Left> - (|Up> + |Down>)|Right>
= |Right>|Left> - |Left>|Right>

Measuring the second particle makes the state either
|Right>|Left> or |Left>|Right>. Now measuring the first particle (in the
z axis) *does* give 50-50. Of course, if we had made that measurement
first it still would be 50-50. So it was the extra measurement at the
start that made the difference.

Entanglement is a one time thing. Once you do a measurement on either
particle, you don't have an entangled pair any more, you have a pair of
particles, both in known states.

You might think that the loss of entanglement itself could be used to
send a signal. It can't, because entanglement cannot be detected
locally. To tell if the particles are still entangled you have to
compare measurements. You can't do that instantaneously unless you can
*already* send an instantaneous signal, by some other means.

Ralph Hartley

[Ralph Hartley]

Is there a FAQ entry for this? We see it all the time.

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you get spin up with probability 1.

Lets do the calculation. Initial state (ignoring factors of sqrt(2)):

|Up>|Down> - |Down>|Up>

after the first measurement:

|Up>|Down>

We want to measure the second particle in a different direction, so we
will use

|Up> = |Left> + |Right> (Sill ignoring factors of sqrt(2))
|Down> = |Left> - |Right> (They can take care of themselves)
and
|Left> = |Up> + |Down>
|Right> = |Up> - |Down>

to re-write the state as:

|Up>|Down> = |Up>(|Left> - |Right>) = |Up>|Left> - |Up>|Right>

After the second measurement it is either |Up>|Left> or |Up>|Right>. In
both cases measuring the first particle again gives up.

Lets try again without the first measurement:

|Up>|Down> - |Down>|Up>
= |Up>(|Left> - |Right>) - |Down>(|Left> + |Right>)
= |Up>|Left> - |Up>|Right> - |Down>|left> - |Down>|Right>
= (|Up> - |Down>)|Left> - (|Up> + |Down>)|Right>
= |Right>|Left> - |Left>|Right>

Measuring the second particle makes the state either
|Right>|Left> or |Left>|Right>. Now measuring the first particle (in the
z axis) *does* give 50-50. Of course, if we had made that measurement
first it still would be 50-50. So it was the extra measurement at the
start that made the difference.

Entanglement is a one time thing. Once you do a measurement on either
particle, you don't have an entangled pair any more, you have a pair of
particles, both in known states.

You might think that the loss of entanglement itself could be used to
send a signal. It can't, because entanglement cannot be detected
locally. To tell if the particles are still entangled you have to
compare measurements. You can't do that instantaneously unless you can
*already* send an instantaneous signal, by some other means.

Ralph Hartley

[Ralph Hartley]

Is there a FAQ entry for this? We see it all the time.

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you get spin up with probability 1.

Lets do the calculation. Initial state (ignoring factors of sqrt(2)):

|Up>|Down> - |Down>|Up>

after the first measurement:

|Up>|Down>

We want to measure the second particle in a different direction, so we
will use

|Up> = |Left> + |Right> (Sill ignoring factors of sqrt(2))
|Down> = |Left> - |Right> (They can take care of themselves)
and
|Left> = |Up> + |Down>
|Right> = |Up> - |Down>

to re-write the state as:

|Up>|Down> = |Up>(|Left> - |Right>) = |Up>|Left> - |Up>|Right>

After the second measurement it is either |Up>|Left> or |Up>|Right>. In
both cases measuring the first particle again gives up.

Lets try again without the first measurement:

|Up>|Down> - |Down>|Up>
= |Up>(|Left> - |Right>) - |Down>(|Left> + |Right>)
= |Up>|Left> - |Up>|Right> - |Down>|left> - |Down>|Right>
= (|Up> - |Down>)|Left> - (|Up> + |Down>)|Right>
= |Right>|Left> - |Left>|Right>

Measuring the second particle makes the state either
|Right>|Left> or |Left>|Right>. Now measuring the first particle (in the
z axis) *does* give 50-50. Of course, if we had made that measurement
first it still would be 50-50. So it was the extra measurement at the
start that made the difference.

Entanglement is a one time thing. Once you do a measurement on either
particle, you don't have an entangled pair any more, you have a pair of
particles, both in known states.

You might think that the loss of entanglement itself could be used to
send a signal. It can't, because entanglement cannot be detected
locally. To tell if the particles are still entangled you have to
compare measurements. You can't do that instantaneously unless you can
*already* send an instantaneous signal, by some other means.

Ralph Hartley

[Ralph Hartley]

Is there a FAQ entry for this? We see it all the time.

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you get spin up with probability 1.

Lets do the calculation. Initial state (ignoring factors of sqrt(2)):

|Up>|Down> - |Down>|Up>

after the first measurement:

|Up>|Down>

We want to measure the second particle in a different direction, so we
will use

|Up> = |Left> + |Right> (Sill ignoring factors of sqrt(2))
|Down> = |Left> - |Right> (They can take care of themselves)
and
|Left> = |Up> + |Down>
|Right> = |Up> - |Down>

to re-write the state as:

|Up>|Down> = |Up>(|Left> - |Right>) = |Up>|Left> - |Up>|Right>

After the second measurement it is either |Up>|Left> or |Up>|Right>. In
both cases measuring the first particle again gives up.

Lets try again without the first measurement:

|Up>|Down> - |Down>|Up>
= |Up>(|Left> - |Right>) - |Down>(|Left> + |Right>)
= |Up>|Left> - |Up>|Right> - |Down>|left> - |Down>|Right>
= (|Up> - |Down>)|Left> - (|Up> + |Down>)|Right>
= |Right>|Left> - |Left>|Right>

Measuring the second particle makes the state either
|Right>|Left> or |Left>|Right>. Now measuring the first particle (in the
z axis) *does* give 50-50. Of course, if we had made that measurement
first it still would be 50-50. So it was the extra measurement at the
start that made the difference.

Entanglement is a one time thing. Once you do a measurement on either
particle, you don't have an entangled pair any more, you have a pair of
particles, both in known states.

You might think that the loss of entanglement itself could be used to
send a signal. It can't, because entanglement cannot be detected
locally. To tell if the particles are still entangled you have to
compare measurements. You can't do that instantaneously unless you can
*already* send an instantaneous signal, by some other means.

Ralph Hartley

[Ralph Hartley]

Is there a FAQ entry for this? We see it all the time.

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you get spin up with probability 1.

Lets do the calculation. Initial state (ignoring factors of sqrt(2)):

|Up>|Down> - |Down>|Up>

after the first measurement:

|Up>|Down>

We want to measure the second particle in a different direction, so we
will use

|Up> = |Left> + |Right> (Sill ignoring factors of sqrt(2))
|Down> = |Left> - |Right> (They can take care of themselves)
and
|Left> = |Up> + |Down>
|Right> = |Up> - |Down>

to re-write the state as:

|Up>|Down> = |Up>(|Left> - |Right>) = |Up>|Left> - |Up>|Right>

After the second measurement it is either |Up>|Left> or |Up>|Right>. In
both cases measuring the first particle again gives up.

Lets try again without the first measurement:

|Up>|Down> - |Down>|Up>
= |Up>(|Left> - |Right>) - |Down>(|Left> + |Right>)
= |Up>|Left> - |Up>|Right> - |Down>|left> - |Down>|Right>
= (|Up> - |Down>)|Left> - (|Up> + |Down>)|Right>
= |Right>|Left> - |Left>|Right>

Measuring the second particle makes the state either
|Right>|Left> or |Left>|Right>. Now measuring the first particle (in the
z axis) *does* give 50-50. Of course, if we had made that measurement
first it still would be 50-50. So it was the extra measurement at the
start that made the difference.

Entanglement is a one time thing. Once you do a measurement on either
particle, you don't have an entangled pair any more, you have a pair of
particles, both in known states.

You might think that the loss of entanglement itself could be used to
send a signal. It can't, because entanglement cannot be detected
locally. To tell if the particles are still entangled you have to
compare measurements. You can't do that instantaneously unless you can
*already* send an instantaneous signal, by some other means.

Ralph Hartley

[Ralph Hartley]

Is there a FAQ entry for this? We see it all the time.

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you get spin up with probability 1.

Lets do the calculation. Initial state (ignoring factors of sqrt(2)):

|Up>|Down> - |Down>|Up>

after the first measurement:

|Up>|Down>

We want to measure the second particle in a different direction, so we
will use

|Up> = |Left> + |Right> (Sill ignoring factors of sqrt(2))
|Down> = |Left> - |Right> (They can take care of themselves)
and
|Left> = |Up> + |Down>
|Right> = |Up> - |Down>

to re-write the state as:

|Up>|Down> = |Up>(|Left> - |Right>) = |Up>|Left> - |Up>|Right>

After the second measurement it is either |Up>|Left> or |Up>|Right>. In
both cases measuring the first particle again gives up.

Lets try again without the first measurement:

|Up>|Down> - |Down>|Up>
= |Up>(|Left> - |Right>) - |Down>(|Left> + |Right>)
= |Up>|Left> - |Up>|Right> - |Down>|left> - |Down>|Right>
= (|Up> - |Down>)|Left> - (|Up> + |Down>)|Right>
= |Right>|Left> - |Left>|Right>

Measuring the second particle makes the state either
|Right>|Left> or |Left>|Right>. Now measuring the first particle (in the
z axis) *does* give 50-50. Of course, if we had made that measurement
first it still would be 50-50. So it was the extra measurement at the
start that made the difference.

Entanglement is a one time thing. Once you do a measurement on either
particle, you don't have an entangled pair any more, you have a pair of
particles, both in known states.

You might think that the loss of entanglement itself could be used to
send a signal. It can't, because entanglement cannot be detected
locally. To tell if the particles are still entangled you have to
compare measurements. You can't do that instantaneously unless you can
*already* send an instantaneous signal, by some other means.

Ralph Hartley

[Ralph Hartley]

Is there a FAQ entry for this? We see it all the time.

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you get spin up with probability 1.

Lets do the calculation. Initial state (ignoring factors of sqrt(2)):

|Up>|Down> - |Down>|Up>

after the first measurement:

|Up>|Down>

We want to measure the second particle in a different direction, so we
will use

|Up> = |Left> + |Right> (Sill ignoring factors of sqrt(2))
|Down> = |Left> - |Right> (They can take care of themselves)
and
|Left> = |Up> + |Down>
|Right> = |Up> - |Down>

to re-write the state as:

|Up>|Down> = |Up>(|Left> - |Right>) = |Up>|Left> - |Up>|Right>

After the second measurement it is either |Up>|Left> or |Up>|Right>. In
both cases measuring the first particle again gives up.

Lets try again without the first measurement:

|Up>|Down> - |Down>|Up>
= |Up>(|Left> - |Right>) - |Down>(|Left> + |Right>)
= |Up>|Left> - |Up>|Right> - |Down>|left> - |Down>|Right>
= (|Up> - |Down>)|Left> - (|Up> + |Down>)|Right>
= |Right>|Left> - |Left>|Right>

Measuring the second particle makes the state either
|Right>|Left> or |Left>|Right>. Now measuring the first particle (in the
z axis) *does* give 50-50. Of course, if we had made that measurement
first it still would be 50-50. So it was the extra measurement at the
start that made the difference.

Entanglement is a one time thing. Once you do a measurement on either
particle, you don't have an entangled pair any more, you have a pair of
particles, both in known states.

You might think that the loss of entanglement itself could be used to
send a signal. It can't, because entanglement cannot be detected
locally. To tell if the particles are still entangled you have to
compare measurements. You can't do that instantaneously unless you can
*already* send an instantaneous signal, by some other means.

Ralph Hartley

[Ralph Hartley]

Is there a FAQ entry for this? We see it all the time.

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you get spin up with probability 1.

Lets do the calculation. Initial state (ignoring factors of sqrt(2)):

|Up>|Down> - |Down>|Up>

after the first measurement:

|Up>|Down>

We want to measure the second particle in a different direction, so we
will use

|Up> = |Left> + |Right> (Sill ignoring factors of sqrt(2))
|Down> = |Left> - |Right> (They can take care of themselves)
and
|Left> = |Up> + |Down>
|Right> = |Up> - |Down>

to re-write the state as:

|Up>|Down> = |Up>(|Left> - |Right>) = |Up>|Left> - |Up>|Right>

After the second measurement it is either |Up>|Left> or |Up>|Right>. In
both cases measuring the first particle again gives up.

Lets try again without the first measurement:

|Up>|Down> - |Down>|Up>
= |Up>(|Left> - |Right>) - |Down>(|Left> + |Right>)
= |Up>|Left> - |Up>|Right> - |Down>|left> - |Down>|Right>
= (|Up> - |Down>)|Left> - (|Up> + |Down>)|Right>
= |Right>|Left> - |Left>|Right>

Measuring the second particle makes the state either
|Right>|Left> or |Left>|Right>. Now measuring the first particle (in the
z axis) *does* give 50-50. Of course, if we had made that measurement
first it still would be 50-50. So it was the extra measurement at the
start that made the difference.

Entanglement is a one time thing. Once you do a measurement on either
particle, you don't have an entangled pair any more, you have a pair of
particles, both in known states.

You might think that the loss of entanglement itself could be used to
send a signal. It can't, because entanglement cannot be detected
locally. To tell if the particles are still entangled you have to
compare measurements. You can't do that instantaneously unless you can
*already* send an instantaneous signal, by some other means.

Ralph Hartley

[Ralph Hartley]

Is there a FAQ entry for this? We see it all the time.

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you get spin up with probability 1.

Lets do the calculation. Initial state (ignoring factors of sqrt(2)):

|Up>|Down> - |Down>|Up>

after the first measurement:

|Up>|Down>

We want to measure the second particle in a different direction, so we
will use

|Up> = |Left> + |Right> (Sill ignoring factors of sqrt(2))
|Down> = |Left> - |Right> (They can take care of themselves)
and
|Left> = |Up> + |Down>
|Right> = |Up> - |Down>

to re-write the state as:

|Up>|Down> = |Up>(|Left> - |Right>) = |Up>|Left> - |Up>|Right>

After the second measurement it is either |Up>|Left> or |Up>|Right>. In
both cases measuring the first particle again gives up.

Lets try again without the first measurement:

|Up>|Down> - |Down>|Up>
= |Up>(|Left> - |Right>) - |Down>(|Left> + |Right>)
= |Up>|Left> - |Up>|Right> - |Down>|left> - |Down>|Right>
= (|Up> - |Down>)|Left> - (|Up> + |Down>)|Right>
= |Right>|Left> - |Left>|Right>

Measuring the second particle makes the state either
|Right>|Left> or |Left>|Right>. Now measuring the first particle (in the
z axis) *does* give 50-50. Of course, if we had made that measurement
first it still would be 50-50. So it was the extra measurement at the
start that made the difference.

Entanglement is a one time thing. Once you do a measurement on either
particle, you don't have an entangled pair any more, you have a pair of
particles, both in known states.

You might think that the loss of entanglement itself could be used to
send a signal. It can't, because entanglement cannot be detected
locally. To tell if the particles are still entangled you have to
compare measurements. You can't do that instantaneously unless you can
*already* send an instantaneous signal, by some other means.

Ralph Hartley

[Ralph Hartley]

Is there a FAQ entry for this? We see it all the time.

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you get spin up with probability 1.

Lets do the calculation. Initial state (ignoring factors of sqrt(2)):

|Up>|Down> - |Down>|Up>

after the first measurement:

|Up>|Down>

We want to measure the second particle in a different direction, so we
will use

|Up> = |Left> + |Right> (Sill ignoring factors of sqrt(2))
|Down> = |Left> - |Right> (They can take care of themselves)
and
|Left> = |Up> + |Down>
|Right> = |Up> - |Down>

to re-write the state as:

|Up>|Down> = |Up>(|Left> - |Right>) = |Up>|Left> - |Up>|Right>

After the second measurement it is either |Up>|Left> or |Up>|Right>. In
both cases measuring the first particle again gives up.

Lets try again without the first measurement:

|Up>|Down> - |Down>|Up>
= |Up>(|Left> - |Right>) - |Down>(|Left> + |Right>)
= |Up>|Left> - |Up>|Right> - |Down>|left> - |Down>|Right>
= (|Up> - |Down>)|Left> - (|Up> + |Down>)|Right>
= |Right>|Left> - |Left>|Right>

Measuring the second particle makes the state either
|Right>|Left> or |Left>|Right>. Now measuring the first particle (in the
z axis) *does* give 50-50. Of course, if we had made that measurement
first it still would be 50-50. So it was the extra measurement at the
start that made the difference.

Entanglement is a one time thing. Once you do a measurement on either
particle, you don't have an entangled pair any more, you have a pair of
particles, both in known states.

You might think that the loss of entanglement itself could be used to
send a signal. It can't, because entanglement cannot be detected
locally. To tell if the particles are still entangled you have to
compare measurements. You can't do that instantaneously unless you can
*already* send an instantaneous signal, by some other means.

Ralph Hartley

[Ralph Hartley]

Is there a FAQ entry for this? We see it all the time.

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you get spin up with probability 1.

Lets do the calculation. Initial state (ignoring factors of sqrt(2)):

|Up>|Down> - |Down>|Up>

after the first measurement:

|Up>|Down>

We want to measure the second particle in a different direction, so we
will use

|Up> = |Left> + |Right> (Sill ignoring factors of sqrt(2))
|Down> = |Left> - |Right> (They can take care of themselves)
and
|Left> = |Up> + |Down>
|Right> = |Up> - |Down>

to re-write the state as:

|Up>|Down> = |Up>(|Left> - |Right>) = |Up>|Left> - |Up>|Right>

After the second measurement it is either |Up>|Left> or |Up>|Right>. In
both cases measuring the first particle again gives up.

Lets try again without the first measurement:

|Up>|Down> - |Down>|Up>
= |Up>(|Left> - |Right>) - |Down>(|Left> + |Right>)
= |Up>|Left> - |Up>|Right> - |Down>|left> - |Down>|Right>
= (|Up> - |Down>)|Left> - (|Up> + |Down>)|Right>
= |Right>|Left> - |Left>|Right>

Measuring the second particle makes the state either
|Right>|Left> or |Left>|Right>. Now measuring the first particle (in the
z axis) *does* give 50-50. Of course, if we had made that measurement
first it still would be 50-50. So it was the extra measurement at the
start that made the difference.

Entanglement is a one time thing. Once you do a measurement on either
particle, you don't have an entangled pair any more, you have a pair of
particles, both in known states.

You might think that the loss of entanglement itself could be used to
send a signal. It can't, because entanglement cannot be detected
locally. To tell if the particles are still entangled you have to
compare measurements. You can't do that instantaneously unless you can
*already* send an instantaneous signal, by some other means.

Ralph Hartley

[Ralph Hartley]

Is there a FAQ entry for this? We see it all the time.

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you get spin up with probability 1.

Lets do the calculation. Initial state (ignoring factors of sqrt(2)):

|Up>|Down> - |Down>|Up>

after the first measurement:

|Up>|Down>

We want to measure the second particle in a different direction, so we
will use

|Up> = |Left> + |Right> (Sill ignoring factors of sqrt(2))
|Down> = |Left> - |Right> (They can take care of themselves)
and
|Left> = |Up> + |Down>
|Right> = |Up> - |Down>

to re-write the state as:

|Up>|Down> = |Up>(|Left> - |Right>) = |Up>|Left> - |Up>|Right>

After the second measurement it is either |Up>|Left> or |Up>|Right>. In
both cases measuring the first particle again gives up.

Lets try again without the first measurement:

|Up>|Down> - |Down>|Up>
= |Up>(|Left> - |Right>) - |Down>(|Left> + |Right>)
= |Up>|Left> - |Up>|Right> - |Down>|left> - |Down>|Right>
= (|Up> - |Down>)|Left> - (|Up> + |Down>)|Right>
= |Right>|Left> - |Left>|Right>

Measuring the second particle makes the state either
|Right>|Left> or |Left>|Right>. Now measuring the first particle (in the
z axis) *does* give 50-50. Of course, if we had made that measurement
first it still would be 50-50. So it was the extra measurement at the
start that made the difference.

Entanglement is a one time thing. Once you do a measurement on either
particle, you don't have an entangled pair any more, you have a pair of
particles, both in known states.

You might think that the loss of entanglement itself could be used to
send a signal. It can't, because entanglement cannot be detected
locally. To tell if the particles are still entangled you have to
compare measurements. You can't do that instantaneously unless you can
*already* send an instantaneous signal, by some other means.

Ralph Hartley

[Ralph Hartley]

Is there a FAQ entry for this? We see it all the time.

daniel@elit.net wrote:
> Consider an EPR experimental set up, with a singlet state of two spin
> 1/2 particles. Measure the spin of particle 1 along the z-axis, and
> let's suppose it is spin up. Proceed as follows:
>
> 1) Measure the spin of particle 1 along the z-axis (again). You'll get
> spin up for certain.
> 2) Then measure the spin of particle 2 along the x-axis. You'll get
> spin up or spin down with equal probability.
>
> Set up a new experiment. Do everything as above, except reverse the
> order of 1) and 2). For particle 2, you'll get the same probabilities.
> For particle 1, you'll get spin up or spin down with equal probability.

No, you get spin up with probability 1.

Lets do the calculation. Initial state (ignoring factors of sqrt(2)):

|Up>|Down> - |Down>|Up>

after the first measurement:

|Up>|Down>

We want to measure the second particle in a different direction, so we
will use

|Up> = |Left> + |Right> (Sill ignoring factors of sqrt(2))
|Down> = |Left> - |Right> (They can take care of themselves)
and
|Left> = |Up> + |Down>
|Right> = |Up> - |Down>

to re-write the state as:

|Up>|Down> = |Up>(|Left> - |Right>) = |Up>|Left> - |Up>|Right>

After the second measurement it is either |Up>|Left> or |Up>|Right>. In
both cases measuring the first particle again gives up.

Lets try again without the first measurement:

|Up>|Down> - |Down>|Up>
= |Up>(|Left> - |Right>) - |Down>(|Left> + |Right>)
= |Up>|Left> - |Up>|Right> - |Down>|left> - |Down>|Right>
= (|Up> - |Down>)|Left> - (|Up> + |Down>)|Right>
= |Right>|Left> - |Left>|Right>

Measuring the second particle makes the state either
|Right>|Left> or |Left>|Right>. Now measuring the first particle (in the
z axis) *does* give 50-50. Of course, if we had made that measurement
first it still would be 50-50. So it was the extra measurement at the
start that made the difference.

Entanglement is a one time thing. Once you do a measurement on either
particle, you don't have an entangled pair any more, you have a pair of
particles, both in known states.

You might think that the loss of entanglement itself could be used to
send a signal. It can't, because entanglement cannot be detected
locally. To tell if the particles are still entangled you have to
compare measurements. You can't do that instantaneously unless you can
*already* send an instantaneous signal, by some other means.

Ralph Hartley