Entanglement, magnetic field and conservation law

[Adel Makram]

I am not sure that I fully understand even the basic aspects of the Quantum measurement and entanglement but I just came across this thought experiment and I wish to resolve it.
In a setting of two entangled spin-1/2 particles, suppose that Alice applies a uniform magnetic field $B_0$ along $z$ direction at her location. She dose not measure the spin of her particle but because the particle passes through $B_0$ the spin should align along that direction as this is the lowest energy the particle may have. And if she is 100% sure that her unmeasured particle is spin-up, she must conclude that Bob`s particle is spin-down due to conservation of the angular momentum. Now if Bob, however, measures his particle spin and finds that it is spin-down, then he must conclude that Alice uniform magnetic field is directed up and vice versa. Then based on the outcome of his result, he knows the direction of Alice`s $B_0$ which means that Alice can encode any message in the form of 1 and 0 where 1 means $B_0$ is toward the +ve Z-direction and 0 means that $B_0$ is along the -ve Z-direction and this message can be delivered faster than light.
Now suppose a different scenario, 100 pairs of entangled particles are emitted into 2 directions. Alice measures her particles at the same time and it happens that she finds 55 out of 100 particles are spin-up and 45 are down. Bob, who does not measure anything, should have 45 up and 55 down. If Bob put a little coil near his particles, he can picks up a signal which is induced into the coil by that 10 difference in the magnetic angular momentum. So, despite that the Bob system is undisturbed, how the can the laws of angular momentum and energy conservation hold?

 

[Nugatory]

She dose not measure the spin of her particle but because the particle passes through $B_0$ the spin should align along that direction as this is the lowest energy the particle may have.

That interaction is sufficient to end the entanglement. Alice ends up with a particle that will always be spin-up if measured on the z-axis, Bob ends up with a particle that will be 50/50 up or down if measured on any randomly chosen axis, there is no correlation between their measurements, and no way of passing information.

 

[PeterDonis]

suppose that Alice applies a uniform magnetic field $B_0$ along $z$ direction at her location. She dose not measure the spin of her particle

Yes, she does: putting the particle through the magnetic field is measuring its spin (specifically, measuring its spin about the $z$ axis). You've just described a Stern-Gerlach spin measuring device.

because the particle passes through $B_0$ the spin should align along that direction

Yes, in one of two ways: either up or down. Note that this is not what classical physics predicts will happen: the Stern-Gerlach experiment was one of the key experiments where the quantum result was very different from the classical prediction.

Alice can encode any message in the form of 1 and 0

No, she can't, because she can't control what results she gets from her spin measurements.

 

[Adel Makram]

That interaction is sufficient to end the entanglement. Alice ends up with a particle that will always be spin-up if measured on the z-axis, Bob ends up with a particle that will be 50/50 up or down if measured on any randomly chosen axis, there is no correlation between their measurements, and no way of passing information.

But what does "end of entanglement" mean? Can passing the particle through the Stern-Gerlach device be considered as another sort of interaction? Why does not the Stern-Gerlach device end the entanglement and passing through B₀ does?

 

[Adel Makram]

Yes, she does: putting the particle through the magnetic field is measuring its spin (specifically, measuring its spin about the $z$ axis). You've just described a Stern-Gerlach spin measuring device.

But Stern-Gerlach device applies a non-uniform magnetic field.

 

[Nugatory]

But what does "end of entanglement" mean? Can passing the particle through the Stern-Gerlach device be considered as another sort of interaction? Why does not the Stern-Gerlach device end the entanglement and passing through B₀ does?

As PeterDonis said... They are both measurements. However, the mechanics of the measurements are somewhat different because you are using a strong homogeneous magnetic field to force the particle into a particular orientation ("Alice is 100% sure that her particle is spin-up"), whereas an SG device uses an inhomogeneous magnetic field to steer the particle upwards (measured spin-up) or downwards (measured spin-down) without forcing it into spin-up.

In your case, if Bob measures spin-up, Alice's particle will have emitted a photon (it was down and flipped up into a lower energy state); while if Bob measures spin-down Alice's particle will not emit a photon because it was already up (no flip needed to get it into the lower energy state). It is irrelevant whether Alice actually observes the emission or non-emission of the photon - the interaction happened, it measured the spin of Alice's particle to be consistent with Bob's observation, and it also left the post-measurement state of Alice's particle spin-up.

 

[PeterDonis]

Stern-Gerlach device applies a non-uniform magnetic field.

Oops, you're right. But as Nugatory says, they are both measurements.

 

[Nugatory]

As PeterDonis said... They are both measurements. However, the mechanics of the measurements are somewhat different because you are using a strong homogeneous magnetic field to force the particle into a particular orientation ("Alice is 100% sure that her particle is spin-up"), whereas an SG device uses an inhomogeneous magnetic field to steer the particle upwards (measured spin-up) or downwards (measured spin-down) without forcing it into spin-up.

In your case, if Bob measures spin-up, Alice's particle will have emitted a photon (it was down and flipped up into a lower energy state); while if Bob measures spin-down Alice's particle will not emit a photon because it was already up (no flip needed to get it into the lower energy state). It is irrelevant whether Alice actually observes the emission or non-emission of the photon - the interaction happened, it measured the spin of Alice's particle to be consistent with Bob's observation, and it also left the post-measurement state of Alice's particle spin-up.

Although the mechanics are different, the outcome is the same - if one of them measures down the other measures up, and after that the particles are no longer entangled.

(You should be aware that when I say that Alice's particle "was spin down and flipped up" I am perpetrating a terrible misdescription of the pre-measurement state. It doesn't matter here, and it's hard to do better using just English words... But at some point you will want to understand the mathematical formalism behind what I said).

 

[Adel Makram]

In your case, if Bob measures spin-up, Alice's particle will have emitted a photon (it was down and flipped up into a lower energy state); while if Bob measures spin-down Alice's particle will not emit a photon because it was already up (no flip needed to get it into the lower energy state). It is irrelevant whether Alice actually observes the emission or non-emission of the photon - the interaction happened, it measured the spin of Alice's particle to be consistent with Bob's observation, and it also left the post-measurement state of Alice's particle spin-up.

No, the position of the uniform magnetic field is before the location of Alice which means Bob can not measure his particle before Alice particle already passes though B₀. See the picture.

 

[Nugatory]

No, the position of the uniform magnetic field is before the location of Alice which means Bob can not measure his particle before Alice particle already passes though B₀. See the picture.

The relative timing of Bob's and Alice's measurements is irrelevant. Both Alice's magnetic field and Bob have a 50% chance of measuring spin-up, and the entanglement only becomes apparent when they get together afterwards, compare their measurements, and find that one of them measured spin-up and the other measured spin-down.

As PeterDonis and I have already pointed out, passing through Alice's magnetic field is a measurement; that measurement will be properly correlated with Bob's measurement no matter which came first. If Alice measures again after the particle has left the magnetic field (which would be something of a wasted effort - we already know that the particle left the magnetic field in the spin-up state) she will get spin-up, but that result will not be correlated with Bob's - the entanglement was already broken by the first measurement.

(It's worth noting that even if the particles were not entangled, there is a 50% chance of getting opposite results. Alice and Bob would have to run the experiment many times with many pairs of particles and see opposite results every time before they could conclude that the particle source between them is generating entangled pairs).

 

[Adel Makram]

The relative timing of Bob's and Alice's measurements is irrelevant.

The relative time reflects the order of measurement in the frame of reference of Alice-Bob. Even in the classical entanglement experiment, when Alice measures her particle first, the wave-function collapses and the state of Bob particle is reduced to the one with opposite spin along the same direction. If Bob, however, does it first, Alice particle state will be correlated with what Bob comes up with. Having said that the overall correlation is independent on which particle is measured first does not mean that the order of the measurement in each individual event is not important.

Both Alice's magnetic field and Bob have a 50% chance of measuring spin-up, and the entanglement only becomes apparent when they get together afterwards, compare their measurements, and find that one of them measured spin-up and the other measured spin-down.

But how Alice magnetic field has a 50% chance of measuring spin-up and you said in post #2 that Alice particles will be always ended up spin-up which means there is a 100% chance spin-up.
In addition, when you agreed that Alice magnetic field is a sort of measurement and if the outcome of that measurement is always spin-up, why the entanglement disappeared at the very moment of the interaction with $B_0$? Why does not Bob obtain spin-down all times? Because of the interaction ends the entanglement! then why the Stern-Gerlach device does not end the entanglement and we still have a correlated outcomes between both sides? Because SG device applies a weaker field! does the strength of the field has anything to do with the entanglement?

 

[Vanadium 50]

The details obscure what is going on.

1. We start with an entangled state.
2. Alice takes one electron and changes its state.
3. The system is no longer entangled.

That's it.

 

[Nugatory]

But how Alice magnetic field has a 50% chance of measuring spin-up and you said in post #2 that Alice particles will be always ended up spin-up which means there is a 100% chance spin-up.
In addition, when you agreed that Alice magnetic field is a sort of measurement and if the outcome of that measurement is always spin-up,

You have misunderstood what was said in #6. The interaction with the magnetic field is measuring the spin of Alice's particle as it enters the magnetic field, before it it has been forced to spin-up. If a photon is emitted, the particle was spin-down when it entered the field and was flipped up; if no photon was emitted the particle was spin-up when it entered the field. Either way, we have measured the spin of the particle as it was when it entered the magnetic field, and either way the particle leaves the magnetic field spin-up.

 

[Adel Makram]

You have misunderstood what was said in #6. The interaction with the magnetic field is measuring the spin of Alice's particle as it enters the magnetic field, before it it has been forced to spin-up. If a photon is emitted, the particle was spin-down when it entered the field and was flipped up; if no photon was emitted the particle was spin-up when it entered the field. Either way, we have measured the spin of the particle as it was when it entered the magnetic field, and either way the particle leaves the magnetic field spin-up.

Thank you, yes I misunderstood the act of measurement as you said. I thought the outcome of that interaction, which is always spin-up, is the result of the measurement.
Just another point related to the interpretation of the meaning of the entangled state. You have pointed that the emission of the photon is related to the state of the particle when it enters the field, but does it mean that the particle state is predetermined before the measurement? I always think about it as a superposition of both spin-up and spin-down not just at a probabilistic level but at a physical level as well. I think that the particle before the measurement has no definite state at all so we can not say that this particle was spin-up when it enters the field.

 

[Nugatory]

that the emission of the photon is related to the state of the particle when it enters the field, but does it mean that the particle state is predetermined before the measurement?

No. That's why there's that warning in parentheses in post #8.

 

[Adel Makram]

No. That's why there's that warning in parentheses in post #8.

So what is the true mathematical description of the quantum state interacting with the field? will this be under quantization of EM field?