# I Quantum entanglement and parallel displacement

1. May 6, 2016

Suppose we fire two entangled particles in a tour round-flight around the galaxy and measure their spins using two Stern-Gerlach devices after returning back to the earth. Will the correlation between their spin measurement still obey quantum correlation?
According to General Relativity, parallel displacement of the spin vectors should bring the two vectors into different states with some displacement. But QM prediction does not consider such effect as long as the state of one particle is only revealed when it is measured. So will parallel displacement issue account here to change the state of particles in a way that may not follow what QM predicts?

2. May 6, 2016

### stevendaryl

Staff Emeritus
I'm not qualified to talk about QM in curved spacetime, but my guess about how GR affects EPR is the following:
• Alice measures the spin state of one particle: Say, spin-up along some axis $\vec{a}$
• This implies a different spin state, say spin-up along $\vec{a'}$, at the time the twin particles were created.
• This in turn implies that Bob's particle had spin-state: spin-down along $\vec{a'}$, at the time the particles were created.
• Finally, this implies that Bob's particle has some spin state, spin-down along a third axis, $\vec{a''}$ at the time he measures the spin.
So in curved spacetime, it will not necessarily be that Alice's particle will have a spin state that is opposite Bob's (you can't even define "opposite spin states" in a path-independent way), but it will be that Alice's spin state is still strongly correlated with Bob's spin state. The correlation is just more complicated.

The other complication is that it is possible that different paths to get to Bob result in different spin states $\vec{a''}$. In this case, I would think that you wouldn't get perfect correlations any more, because there would be interference between the spin states associated with different paths.

3. May 6, 2016

I dont understand this. For if Alice has to imply that, it would mean the particle has a definite spin at the time of creation which contradicts the collapse theory which says the particle has no definite state of spin until it is measured.
What if we choose a to be the third direction in Bells inequality. According to what I understood from you comment, there would be a perfect anti-correlation between Alices particle at a and Bobs particle at a which contradicts what QM says that the correlation depends on the angle between a and a.

4. May 6, 2016

### Strilanc

You're confusing QM's predictions for specifically the singlet state with its predictions for entangled states in general.

If we create an EPR pair, but I rotate your particle 10 degrees giving it to you, the spins are still entangled. All the correlations will be off by 10 degrees, but the entanglement is still there and still detectable. Fixing the offset is just a matter of performing the opposite rotation.

5. May 6, 2016

How can you rotate my particle and it is still entangled without collapsing its state to a state at the angle of rotation? Rotating the particle spin means you already measure it which is no longer entangled.

6. May 6, 2016

### Strilanc

You can rotate a particle's spin without measuring it.

7. May 6, 2016

How? what application would you use?

8. May 6, 2016

### stevendaryl

Staff Emeritus
Okay, let me try to get more rigorous. The way that QM predicts probabilities is this: The probability of getting a particular result is the absolute square of the amplitude for getting that result. So how does QM predict amplitudes? It's like this:

The amplitude for a result $R$ is given by: $\sum_\alpha C_\alpha \psi(R|\alpha)$

where $C_\alpha$ is the amplitude for being in initial state $|\chi_\alpha\rangle$, and $\psi(R|\alpha)$ is the amplitude for getting result $R$ given that the system starts off in initial state $|\chi_\alpha\rangle$. ($|\chi_\alpha\rangle$ is any complete set of states)

You have complete freedom for choosing your complete set of states $|\chi_\alpha\rangle$. I might as well choose the following basis:
1. $|\chi_1\rangle = |\vec{a'}\rangle |\vec{a'}\rangle$ (Alice's particle and Bob's particle are both initially spin-up in the a'-direction)
2. $|\chi_2\rangle = |\vec{a'}\rangle |-\vec{a'}\rangle$ (Alice's particle is initially spin-up in the a'-direction Bob's particle is initially spin-up in the negative a'-direction.)
3. $|\chi_3\rangle = |-\vec{a'}\rangle |\vec{a'}\rangle$ (Alice's particle is spin-up in the negative a'-direction and Bob's particle is spin-up in the a'-direction)
4. $|\chi_4\rangle = |-\vec{a'}\rangle |-\vec{a'}\rangle$ (Alice's particle and Bob's particle are both initially spin-up in the negative a'-direction)

Because for spin-1/2 EPR, the total spin is zero, we can immediately compute the amplitudes $C_\alpha$:
$C_1 = 0$
$C_2 = \frac{1}{\sqrt{2}}$
$C_3 = -\frac{1}{\sqrt{2}}$
$C_4 = 0$

(The overall phase is unobservable, so we're free to choose $C_2$ to be real and positive. This forces $C_3$ to be negative and real in order for the total spin to be zero.) So our probability amplitude simplifies to:

$\frac{1}{\sqrt{2}} \psi(R|\alpha=2) - \frac{1}{\sqrt{2}} \psi(R|\alpha=3)$

Now, if the result $R$ is that Alice measures spin-up along axis $\vec{a}$, while Bob measures spin-down along axis $\vec{b}$, then we can write:

$\psi(R|\alpha=2) = \psi_A(\vec{a}|\vec{a'})\psi_B(\vec{b}|-\vec{a'})$
$\psi(R|\alpha=3) = \psi_A(\vec{a}|-\vec{a'})\psi_B(\vec{b}|\vec{a'})$

where $\psi_A(\vec{a}|\vec{z})$ is the probability that Alice's particle will have spin-up in the $\vec{a}$ direction given that it initially had spin up in the $\vec{a'}$ direction, and $\psi_B(\vec{b}|-\vec{a'})$ is the probability that Bob's particle will have spin-up in the $\vec{b}$ direction given that it initially had spin up in the $-\vec{a'}$ direction. And similarly for the other terms.

What I was assuming in my first post was that there was a unique direction $\vec{a'}$ such that $\psi_A(\vec{a}|\vec{a'}) = 1$ and $\psi_A(\vec{a}|-\vec{a'}) = 0$

9. May 6, 2016

### Strilanc

Magnetic fields make the spin precess. Precession is a rotation.

This is a really basic fact about spins. Not something I should have to bring up in a thread marked "advanced" (i.e. grad-student-in-physics level).

10. May 6, 2016

I know Larmors equation but I dont know whether applying magnetic field on one of entangled particle would affect its state or no. If it rotates its state by a definite angle, then its state is changed relative to the initial state by the amount of that angle. If so, the state of its entangled particle must also be changed in order to keep the total spin=0. Again, arent we doing measurement here?

11. May 6, 2016

But:
,,, would imply that two particles have given directions at the moment they were created which is a local hidden variable!

12. May 6, 2016

### Mentz114

I don't agree. Every electron has spin but we don't know the direction so $\vec{a'}$ can have any value.

13. May 6, 2016

I dont agree too. According to QM, our knowledge of spin direction is only revealed at the time of measurement. That is how solving the eigen problem is interpreted. If we measure spin of a particle and it comes to be spin-up along some direction, does it mean the particle has spin up before the measurement? Particle spin is the quantum property which is only revealed after the measurement.

14. May 6, 2016

### Strilanc

How could any spin be precessed by a magnetic field if the process was measuring it? Measurements flatten spins into mixed states, so we wouldn't call the process "precession" if that's what was happening.

I actually don't know the answer to this apparent paradox of angular momentum conservation being violated, but it doesn't require entanglement. The solution for the unentangled case will apply to the entangled case.

15. May 6, 2016

I didn`t get you.
My questions was in that form: if we create a pair of entangled particles, one of them is allowed to pass under magnetic field of known strength and direction while the other one is set free. Would the change made on the state of the first particle after exiting the magnetic field affect the state of the second particle?

16. May 6, 2016

### stevendaryl

Staff Emeritus
That's just quantum mechanics. You have some initial state $|\psi_{initial}\rangle$. You're trying to compute the probability that you will end up in some final state $|\psi_{final}\rangle$. That probability is $|\langle \psi_{final}|U(t)|\psi_{initial}\rangle|^2$, where $U(t)$ is the time evolution operator, and $t$ is the time between the preparation of the initial state and the measurement of the final state. That's just basic QM.

Now, the next step is to compute $\langle \psi_{final}|U(t)|\psi_{initial}\rangle$. This step is pure mathematics--there is no additional assumptions or interpretations involved:

$\langle \psi_{final}|U(t)|\psi_{initial}\rangle = \sum_{\alpha} \langle \psi_{final} | U(t)| \chi_\alpha \rangle \langle \chi_\alpha | \psi_{initial} \rangle$

where $|\chi_\alpha\rangle$ is a complete sets of states.

Now, the various pieces of this expression can be given interpretations:

• $\langle \psi_{final}|U(t)|\chi_\alpha \rangle$ = the probability amplitude that a system, initially prepared in state $\chi_\alpha$, will be found in state $|\psi_{final}\rangle$ a time $t$ later.
• $\langle \chi_\alpha | \psi_{initial} \rangle$ = the probability amplitude that a system can be found in state $|\chi_\alpha\rangle$ given that it is prepared in state $|\psi_{initial}\rangle$ (or alternatively, the coefficients of $|\psi_{initial}\rangle$ when expressed as a superposition of the basis $|\chi_\alpha\rangle$)
The fact that you consider a complete set of initial states $|\chi_\alpha\rangle$ does not in any way imply that you think it is REALLY in one of those states, and you just don't know which. So it is not at all a hidden-variables assumption.

17. May 6, 2016

### stevendaryl

Staff Emeritus
No, it would not.

18. May 6, 2016

### Mentz114

.
You are confusing the property spin with its value ( a direction). Every electron has spin whether it has been measured or not. You actually say that !

19. May 6, 2016

### Strilanc

20. May 6, 2016

So, provided that process is not considered as measurement and provided that the second particle state is not affected, where is the law of conservation of total spin now? If we would write the state equation of the system, the first particle state is in the form of a new state while the second one is not, which violates the law of conservation of total spin!

21. May 6, 2016

### stevendaryl

Staff Emeritus
The deflection of an electron by the electromagnetic field conserves angular momentum. There is angular momentum in the electromagnetic field as well as the angular momentum associated with the particle.

22. May 6, 2016

But for a system of two entangled particles, in order to conserve the total spin, the same amount of deflection should be added to the second particle in opposite direction for the total system to be conserved, otherwise, they are no longer entangled any more. This should appear in the state component of the second particle as well. If not, that implies the conservation law was only between the first particle and the EM field. This means a sort of interaction between them happens which should be measured, against what is claimed here that there is no such measurement. The same reasoning applied to my original thought experiment considering two entangled particles in the gravitational field.
So we left here in one of two situations:
1) Either the two particles always remain entangled with total spin=0 which means the interaction between the first particle and the field (whether gravitational or EM) does not affect the physical state of the system if two particles. That means that physical interaction has no physical interaction!! like Barber paradox :)
2) There is a sort of interaction between the first particle and the field that can be measured and hence breaks the entanglement.

Last edited: May 6, 2016
23. May 6, 2016

### Strilanc

Okay, I figured out why precession doesn't measure the state in general. Basically, because there's already quite a lot of uncertainty in the exact angular momentum in practice, the transfer of angular momentum to the magnetic-field apparatus is washed out in the noise. The amount of decoherence is non-zero, but negligible.

So, to a good approximation, you can rotate a spin without measuring it. This applies to both unentangled and entangled spins.

24. May 7, 2016