What happed to the other, if one entangled particle is destroyed

[ouacc]

what happed to the other, if one entangled particle is "destroyed"

if one entangled particle is "destroyed"
what is the state of the OTHER particle?


Situation 1
two entangled particles, A and B(assume have mass).
if A is "destroyed", say it hit a wall :).
What happen to the B's momentum and location?



Situation 2
two entangled photons, A and B.
if A is "destroyed", assume photoelectric effect.
What happen to the other photon's spin? Would B return to NO SPIN? Keep its spin? or other results?


In a word,
Is "destruction" considered as some-sort-of "measurement"?
Does the wave function, in a destruction, collapse the same way as in a measurement?
What is the mathematical operator for "destruction"(if it is considered as a measurement)?

 

[Demystifier]

The destruction of the particle A always occurs due to an interaction with another particle, call it C. Namely, the destruction will be accompanied with a creation or increase of energy of another particle C. Thus, C will be entangled with B. It can be viewed as a sort of measurement.

 

[peter0302]

if one entangled particle is "destroyed"
what is the state of the OTHER particle?


Situation 1
two entangled particles, A and B(assume have mass).
if A is "destroyed", say it hit a wall :).
What happen to the B's momentum and location?

If A hits a wall, this would be a position measurement essentially, so you would know B would hit a similar wall in the same place, but you won't be able to predict its momentum.

Situation 2
two entangled photons, A and B.
if A is "destroyed", assume photoelectric effect.
What happen to the other photon's spin? Would B return to NO SPIN? Keep its spin? or other results?

Its spin would be unpredictable.

In a word,
Is "destruction" considered as some-sort-of "measurement"?
Does the wave function, in a destruction, collapse the same way as in a measurement?

Yes. There is no such thing as a purely passive measurement. Even a polarization filter, which you would think would result in the photons that were let through being undisturbed, actually does "destroy" the incoming photons and produces new photons, at least in the mathematical sense.

 

[reilly]

Situations 1 and 2 are standard experimental situations. Further, application of QM will tell you what happens -- that is, work out the scattering problem.

For 3:Check out creation and destruction operators in QM; best to start with the non-rel oscillator

Regards,
Reilly Atkinson

if one entangled particle is "destroyed"
what is the state of the OTHER particle?


Situation 1
two entangled particles, A and B(assume have mass).
if A is "destroyed", say it hit a wall :).
What happen to the B's momentum and location?



Situation 2
two entangled photons, A and B.
if A is "destroyed", assume photoelectric effect.
What happen to the other photon's spin? Would B return to NO SPIN? Keep its spin? or other results?


In a word,
Is "destruction" considered as some-sort-of "measurement"?
Does the wave function, in a destruction, collapse the same way as in a measurement?
What is the mathematical operator for "destruction"(if it is considered as a measurement)?

 

[ouacc]

Its spin would be unpredictable.

A related argument.

let's assume a photon pair from positronium annihilation(Wu&Shaknov 1949). They SHOULD have opposite polarization. Then, photon A involves in a photoelectric effect. Polarization of A cease to exist. SHOULDN'T the other photon B do the same? To achieve NO polarization for the system(considering the photoelectric effect as part of the system)?

What is wrong with the above argument? Thanks.

 

[vanesch]

A related argument.

let's assume a photon pair from positronium annihilation(Wu&Shaknov 1949). They SHOULD have opposite polarization. Then, photon A involves in a photoelectric effect. Polarization of A cease to exist. SHOULDN'T the other photon B do the same? To achieve NO polarization for the system(considering the photoelectric effect as part of the system)?
What is wrong with the above argument? Thanks.

Well, first of all, photon A will not just "cease to exist" without something else, because of course there will be conservation of angular momentum in the interaction photon A undergoes. So its spin-1 will go *somewhere* (for instance, by flipping the spin of the emitted electron, or I don't know what).

In those interpretations with strict unitary evolution (MWI, Bohmian mechanics...), we now simply have that photon B is entangled with whatever system photon A interacted with (and will soon end up being entangled with the whole environment).

In those interpretations that give a special status to measurement (Copenhagen...), you can consider the interaction as a measurement on photon A (and now the difficulty comes: WHAT measurement ? )

But PRACTICALLY, this doesn't matter. What happens now is that you can consider photon B not to be entangled anymore, but in a statistical mixture corresponding to its formerly entangled states.

What I mean is this:
If the initial state is:
|A up> |B down> + |A down> |B up>, then afterwards, we can consider the "state" of B to be:

50% |B down> and 50% |B up>

for instance, described by a density matrix.

If, from the interaction, we can DERIVE whether it was A-up, or A-down, or rather, A-left, or A-right, then it was a measurement which gave us some information, and B will then be in a pure state, as if we applied a measurement to A with result.