Query of a problem once emerged in Physics GRE practice

[diliu]

I have a doubt at the answer of the following problem:

Two spin-1/2 particles, 1 and 2, have spins in a singlet state with spin wave function
Ψ(1,2)=1/sqrt(2)[α(1) β(2) −α(2) β(1)]
where α and β refer to up and down spins, respectively, along any chosen axis. The spin of
particle 1 is measured along the z-axis and found to be up.
55. A simultaneous measurement of the spin of particle 2 along the z-axis would yield which of the
following results?
(A) Up with 100% probability
(B) Down with 100% probability
(C) Up with 25% probability, down with 75%
probability
(D) Up with 50% probability, down with 50%
probability
(E) Up with 75% probability, down with 25%
probability

The answer is "D", but my answer is "B". Because I think this is a quantum entanglement state, when the spin of particle "1" is measured, the state of particle "2" is completely assured. It must be "Down".

[bfollinprm]

If they're measured simultaneously, you will break the coupled state.

From Wiki:

Now suppose Alice is an observer for system A, and Bob is an observer for system B. If in the entangled state given above Alice makes a measurement in the eigenbasis of A, there are two possible outcomes, occurring with equal probability:[31]
Alice measures 0, and the state of the system collapses to .
Alice measures 1, and the state of the system collapses to .
If the former occurs, then any subsequent measurement performed by Bob, in the same basis, will always return 1. If the latter occurs, (Alice measures 1) then Bob's measurement will return 0 with certainty. Thus, system B has been altered by Alice performing a local measurement on system A. This remains true even if the systems A and B are spatially separated. This is the foundation of the EPR paradox.

[diliu]

Well, thanks. It can't be considered classically, but I also suspect the definition of "simultaneous". What if we describe this in different reference systems. I think It is a big issue I need to read more and consider more.

[CarlBrannen]

I agree with the answer: (B) Down with 100% probability

Are you sure that both measurements were taken to be "z"? Maybe one of them was "x".

[diliu]

I'm quite sure of that, I have thought of this question for a long time.

I think there is a mechanism that can't be understood classically acting on this process, so that when you measure this couple at the same time, you will break their entanglement state, and the spin probability of particle 2 is Up with 50% and Down with 50%. But how does this break happen? I'm not sure, I would like to spend some time to search this and reply some time later.

Another question you mentioned: "A simultaneous measurement of the spin of particle 2 along the x-axis". The answer is: (D) Up with 50% probability, down with 50% probability. Because the average value of spin particle 2 along this direction is zero.

[CarlBrannen]

Simultaneous measurements of the spin (in any single chosen direction) of a spin-0 state (such as the singlet given in the problem) have to give total spin = 0 because of conservation of angular momentum. That is, the system begins in an eigenstate of angular momentum (in the z-direction or in any other, in this case) and so the result from measuring angular momentum is not a matter of probabilities. When you measure both particle's spin in the z-direction you're doing a measurement of total spin in the z-direction. They have to give opposite results. 100% of the time. This is the basis of the EPR experiment.

By the way, I'm not guessing on this. In addition to a 990 on the physics GRE, I published a physics paper on spin:
Spin Path Integrals and Generations
http://arxiv.org/abs/1006.3114

[physicsworks]

I agree that the answer is (B). You can actually prove it:
if a system is in a state $$\lvert \psi \rangle$$, then the probability for getting an eigenvalue $$\lambda$$ of an operator $$F$$ is
$$\displaystyle P(\lambda) = \sum_{i=1}^N \lvert \langle f \rvert \psi \rangle \rvert^2$$,
where $$\lvert f \rangle$$ are a basis for the N-dimensional subspace of vectors with eigenvalue $$\lambda$$:
$$F\lvert f \rangle = \lambda\lvert f \rangle$$.
We have a singlet state $$\lvert \psi \rangle = \frac{1}{\sqrt{2}} (\lvert +- \rangle - \lvert -+ \rangle)$$. The probability that we measure the spin of particle 1 to be up is:
$$P\left(S_{z}^{(1)} = \hbar/2 \right) = \lvert \langle ++ \rvert \psi \rangle \rvert^2 + \lvert \langle +- \rvert \psi \rangle \rvert^2 = \frac{1}{2}$$
At the same time, the probability for getting particle 1 with "spin up" and particle 2 with "spin down" is
$$P\left(S_{z}^{(1)}= \hbar/2; S_{z}^{(2)} = -\hbar/2 \right) = \lvert \langle +- \rvert \psi \rangle \rvert^2 = \frac{1}{2}$$.
From these two results, it's easy to see that when particle 1 is measured to have spin up along z, then particle 2 will have spin down along z with certainty.

P.S. Where did you get this problem?

[diliu]

Thanks CarlBrannen, the measurement of the two spins in the same direction (z-axle) is compatible, so the result is certain.

[diliu]

Thanks physicsworks, your neat proof helps me understand it more precisely.
This problem comes from a Chinese web-set: http://wenku.baidu.com/view/aed54368011 ... 390ed.html.

[SSM]

Two spin-1/2 particles, 1 and 2, have spins in a singlet state with spin wave function
Ψ(1,2)=1/sqrt(2)[α(1) β(2) −α(2) β(1)]
where α and β refer to up and down spins, respectively, along any chosen axis.

In physicsworks proof, and I think Carl's argument, you assumed that the eigenstates were chosen along the z axis, right? The wording of the problem is horrendous, but maybe they meant that in general you don't measure the down state with 100% probability, only if you choose the up/down eigenstates to be pointed along the z axis to begin with. I think it's just a wording issue with the problem. If you choose these eigenstates to be pointed along z, then of course you have to measure the down state with 100% probability.

[baconface]

Wait wait wait. After typing a defense for the question, I just reread it... and it seems they DO specify the z direction.

Errr... does anyone have any clue if this is some kind of word game or just plain wrong? As stated, it just seems wrong to me.

[CarlBrannen]

In physicsworks proof, and I think Carl's argument, you assumed that the eigenstates were chosen along the z axis, right?

Suppose you make a spin-0 state by combining spin-1/2 up and spin-1/2 down in the z direction. But the resulting state does have spin 0 and consequently the fact that you used the "z-direction" is information that is no longer present in the wave function.

In short, if you made a spin-0 state by combining spin up and spin down in some other direction, perhaps the x direction, you'll also end up with a spin-0 state. And it will be the same state as what results from the choice of z-direction.

Let's do the x-axis problem.

Spin up is $$(1,1)/\sqrt{2}$$ and spin down is $$(1,-1)/\sqrt{2}$$. The spin-0 state is:
$$\sqrt{1/2}[ (1,1)\times (1,-1)/2 - (1,-1)\times (1,1)/2]$$
where the "/2" comes from having two factors of $$1/\sqrt{2}$$. Multiplying this out we get:
$$\sqrt{1/2}[(1,-1,1,-1)-(1,1,-1,-1)]/2 = (0,-1,1,0)/\sqrt{2}$$.

We wish to compare this with
$$\sqrt{1/2}[(1,0)\times (0,1) - (0,1)\times(1,0)] = \sqrt{1/2}[(0,1,0,0) - (0,0,1,0)]$$
$$= (0,1,-1,0)/\sqrt{2}$$
We see that they are the same, except for an overall phase factor of -1.

[baconface]

I agree with the answer: (B) Down with 100% probability

Are you sure that both measurements were taken to be "z"? Maybe one of them was "x".

I'm with CarlBrannen. It's either that, a typo, or some kind of trick. I can't really find any ambiguities or tricks in the problem statement though. But I mean, I'm pretty sure problems like this (with solution B) are used to introduce the EPR paradox in some QM textbooks. I also don't think Quantum Cryptography would work either...

[Chiron]

I realize that this is an old post, but I will post my thoughts on this in the hopes it may be helpful.

I think that the correct answer to this problem very well could be D.

If the problem had said that they measured particle 1 to be up, and what is the spin and probability of that spin for particle 2, then the answer would be down with 100% probability. However, the question specifically mentioned that they made a new measurement.

While it is true that the operators for both operators commute, the trick may be in assuming that the two particles are indistinguishable. If the 2 particles are indistinguishable then perhaps what the question is really getting at is that the second measurement only has a 50% chance of actually being a measurement of the other particle.

If this is not what the problem is getting at then I believe a mistake may have been made.