understanding 1. single photon and 2. entanglement states


by San K
Tags: entanglement, photon, single, states
San K
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#1
Jun23-11, 05:11 AM
P: 915
understanding 1. single photon and 2. entanglement ........states

a photon spin

is not know till we measure it.

once we measure it we get some value (L or R, V or H etc)?

does this value change if we were to measure it again after a few seconds? (assuming no interaction in-between)?

now we move to entangled pair as a single system

is not know till we measure it.

can we measure the spin of the single (two-photon) system? how?

literature says entanglement can be measured, purified, teleported etc. what properties of entanglement can be measured?

does this value change if we were to measure it again after a few seconds? (assuming no interaction in-between)?
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SpectraCat
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Jun23-11, 07:53 AM
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Quote Quote by San K View Post
understanding 1. single photon and 2. entanglement ........states

a photon spin

is not know till we measure it.

once we measure it we get some value (L or R, V or H etc)?

does this value change if we were to measure it again after a few seconds? (assuming no interaction in-between)?

now we move to entangled pair as a single system

is not know till we measure it.

can we measure the spin of the single (two-photon) system? how?

literature says entanglement can be measured, purified, teleported etc. what properties of entanglement can be measured?

does this value change if we were to measure it again after a few seconds? (assuming no interaction in-between)?
San K .. you keep asking the same questions over and over again. This strongly indicates that you have not understood our previous answers, because you don't appear to realize that they are the same questions.

So let's try something different ... please answer the questions below, which will help you to understand the significance of a measurement basis for polarization measurements on photons . Let's start with the simpler case of single photon measurements (let's run before we walk .. we'll get to entanglement later).

You seem to know that in any polarization basis, there are two possible basis states, typically labeled |H> and |V>. Let's assume all measurements are done with polarizing beamsplitters using two detectors, one for each basis state, so that we record 100% of the photons. Once we choose the measurement basis (i.e. the angle), the result of any measurement on a single photon must be either |H> or |V>.

Ok .. assuming you understand all of that, here are the questions.

Suppose that you make a polarization measurement on a photon with some basis at t=0, and then make another polarization measurement of the same photon with the same basis at some later time t=[itex]\tau[/itex]. What is the relationship between the possible results (|H> and |V>) of the two measurements?

Suppose now that you make a polarization measurement on a photon with some basis at t=0, and then rotate the basis by and angle [itex]\theta[/itex] and measure the polarization of the same photon at some later time t=[itex]\tau[/itex]. What is the relationship between the possible results of the two measurements? (Call the first basis |H> and |V>, and the rotated basis |H'> and |V'>).
San K
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Jun23-11, 09:30 AM
P: 915
Quote Quote by SpectraCat View Post
San K .. you keep asking the same questions over and over again. This strongly indicates that you have not understood our previous answers, because you don't appear to realize that they are the same questions.

So let's try something different ... please answer the questions below, which will help you to understand the significance of a measurement basis for polarization measurements on photons . Let's start with the simpler case of single photon measurements (let's run before we walk .. we'll get to entanglement later).

You seem to know that in any polarization basis, there are two possible basis states, typically labeled |H> and |V>. Let's assume all measurements are done with polarizing beamsplitters using two detectors, one for each basis state, so that we record 100% of the photons. Once we choose the measurement basis (i.e. the angle), the result of any measurement on a single photon must be either |H> or |V>.

Ok .. assuming you understand all of that, here are the questions.

Suppose that you make a polarization measurement on a photon with some basis at t=0, and then make another polarization measurement of the same photon with the same basis at some later time t=[itex]\tau[/itex]. What is the relationship between the possible results (|H> and |V>) of the two measurements?

Suppose now that you make a polarization measurement on a photon with some basis at t=0, and then rotate the basis by and angle [itex]\theta[/itex] and measure the polarization of the same photon at some later time t=[itex]\tau[/itex]. What is the relationship between the possible results of the two measurements? (Call the first basis |H> and |V>, and the rotated basis |H'> and |V'>).
Sorry if I have asked too many times. I think the confusion is about the indeterminacy/indeterminacy of the spin after/before measurement.

my question was triggered by the idea that photon state is indeterminate and I was wondering does it revert back to indeterminate after measurement? and the second one was that how/if can the spin of an entangled state be measured?

To answer your questions (and I don't fully understand them):

The photon, I guess, won't change it's polarization. If it's detected V it will remain V.

Now if we change the basis I guess polarization would still remain the same because I thought the polarizer is a filter and does not change the polarization of the photon but simply allows those photons to pass that are "aligned" with the axis of the polarizer.

here's another view and I am reading your response to it....

Quote Quote by v4theory View Post
When you measure the spin of a particle its spin changes to a random spin again. You can't know which way it will be spinning the next time you measure it. The magnetic force you apply changes the spin. The change of spin releases the energy of the force you applied and that's how we know what its spin was before we measured it. But then we don't know what it is after we measure it. When we measure the spin of the other particle of the pair we also know what its spin was before we measured it. But then its spin changes after.

.
Maybe I am getting this all mixed up.

SpectraCat
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Jun23-11, 10:04 AM
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understanding 1. single photon and 2. entanglement states


Quote Quote by San K View Post
I confess that I don't have much knowledge.

my question was triggered by the idea that photon state is indeterminate and I was wondering does it revert back to indeterminate after measurement?

To answer your questions (and I don't fully understand them):

The photon, I guess, won't change it's polarization. If it's detected V it will remain V.
Correct ... in quantum mechanical terms, the first measurement projected the photon polarization into one of the eigenstates of the measurement basis. Since eigenstates are time-independent, if you then measure the photon in the same basis at a later time, you will get the same result. Note that you can also understand this in a completely classical context from looking at the equations for linearly or circularly polarized light. The description above is precisely equivalent to saying, "If I polarize a classical light beam so that it is vertical, and then pass that light beam through a second vertically-oriented polaroid, then I will not see any change in the intensity."

Now if we change the basis I guess polarization would still remain the same because I thought the polarizer is a filter and does not change the polarization of the photon but simply allows those photons to pass that are "aligned" with the axis of the polarizer.

Maybe I am getting this all mixed up.
You are getting it mixed up, but hopefully we can fix that.

When you do the second measurement in a rotated polarization basis, you can only observe one of the eigenstates of the new polarization basis, which we are calling |H'> and |V'>. The probability of obtaining one or the other of those results is related by a well-known relation called Malus' law (you may have seen this mentioned previously on these boards). We define the rotation angle [itex]\theta[/theta] such that [itex]\theta=0[/itex] corresponds to the case where |H> and |H'> are identical. Then, if the first polarization measurement found the photon to be in state |H>, then the second polarization measurement would observe |H'> or |V'> with the following probabilities:

[itex]P_{|H'>}=cos^2\theta[/itex]
[itex]P_{|V'>}=1-cos^2\theta=sin^2\theta[/itex]

If the first measurement found the photon to be in state |V>, then you would just switch the expressions on the right-hand sides of the equations above to get the probabilities for the second measurement.

Classically, Malus' law says that, "If I pass a linear polarized light beam through a polarizer oriented at angle [itex]\theta[/itex] (where [itex]\theta=0[/itex] equals the polarization angle of the original beam), then the relationship between the intensity of the beam before (I0) and after (I) the polarizer will be given by: [itex]I=I_0 cos^2\theta[/itex]".

Is all of that clear?
San K
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Jun23-11, 10:20 AM
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Quote Quote by SpectraCat View Post

When you do the second measurement in a rotated polarization basis, you can only observe one of the eigenstates of the new polarization basis, which we are calling |H'> and |V'>. The probability of obtaining one or the other of those results is related by a well-known relation called Malus' law (you may have seen this mentioned previously on these boards). We define the rotation angle [itex]\theta[/theta] such that [itex]\theta=0[/itex] corresponds to the case where |H> and |H'> are identical. Then, if the first polarization measurement found the photon to be in state |H>, then the second polarization measurement would observe |H'> or |V'> with the following probabilities:

[itex]P_{|H'>}=cos^2\theta[/itex]
[itex]P_{|V'>}=1-cos^2\theta=sin^2\theta[/itex]

If the first measurement found the photon to be in state |V>, then you would just switch the expressions on the right-hand sides of the equations above to get the probabilities for the second measurement.

Classically, Malus' law says that, "If I pass a linear polarized light beam through a polarizer oriented at angle [itex]\theta[/itex] (where [itex]\theta=0[/itex] equals the polarization angle of the original beam), then the relationship between the intensity of the beam before (I0) and after (I) the polarizer will be given by: [itex]I=I_0 cos^2\theta[/itex]".

Is all of that clear?
Not all clear yet. To understand this will require a bit of back and forth, so bear with me.

let's assume on first measurement we got H..... as the spin-direction of the photon.

i thought -- If the polarization angle is other than horizontal, then the photon (that was measured first measured say |H> ) should not even pass through.

But I now realize it NOT all or nothing?

an H polarized photon can still pass through (a "rotated relative to base" polarizer), however

1. with certain probability of passing through? i.e. certain probability that it will have state H and certain probability it will have state V?
2. it will loose it's intensity per Malus' law?
DrChinese
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Jun23-11, 10:41 AM
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Quote Quote by San K View Post
an H polarized photon can still pass through (a "rotated relative to base" polarizer), however

1. with certain probability of passing through? i.e. certain probability that it will have state H and certain probability it will have state V?
2. it will loose it's intensity per Malus' law?
No intensity loss for a single photon passing the polarizer as 1 photon is 1 photon until absorbed. The intensity is more a measure of a group of photons.


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