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I What is entanglement?

  1. Mar 10, 2017 #1

    edguy99

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    In an effort clarify confusion in my own mind on the definition of entanglement, I looked at wiki and found this:

    As an example of entanglement: a subatomic particle decays into an entangled pair of other particles. The decay events obey the various conservation laws, and as a result, the measurement outcomes of one daughter particle must be highly correlated with the measurement outcomes of the other daughter particle (so that the total momenta, angular momenta, energy, and so forth remains roughly the same before and after this process).

    Which I completely agree with, followed by this:

    For instance, a spin-zero particle could decay into a pair of spin-½ particles. Since the total spin before and after this decay must be zero (conservation of angular momentum), whenever the first particle is measured to be spin up on some axis, the other, when measured on the same axis, is always found to be spin down. (This is called the spin anti-correlated case; and if the prior probabilities for measuring each spin are equal, the pair is said to be in the singlet state.)

    Which I feel is not correct. As I understand it, if the spin axis of the original particle was the same as the measurement axis, this would be true (ie, prepare vertical and measure vertical), but if you measure on a different axis, you only have probabilities of seeing this based on cos^2 of the difference between the measurement axis and the original particle axis. Many of the particles prepared in this manner, will not be anti-correlated if measured off of the original particles spin axis.
     
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  3. Mar 10, 2017 #2
    I think the detections of both sides would correlate following cos2(φ), but this would be a consequence of the spins being opposite. Or at least that view would produce the correlation.
     
  4. Mar 10, 2017 #3
    But they always are. It's strange but true! However....the measurement of the spin does affect the spin direction so you are not necessarily measuring what you started with. The final direction is function of both the initial spin and the orientation of the detection equipment.​
     
  5. Mar 10, 2017 #4

    vanhees71

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    In the example you have an original spin state ##|s=0,\sigma_z=0 \rangle##. If the particle decays in two spin-1/2 particles, the angular-momentum state must be still such that the total angular momentum is ##|J=0,J_z=0 \rangle##, but this is uniquely given in terms of the product states of the two spins (check a QT textbook about addition of angular momenta or Clebsch-Gordan coefficients etc.):
    $$|J=0,J_z=0 \rangle=\frac{1}{\sqrt{2}} (|s=1/2,\sigma_z=1/2 \rangle \otimes |s=1/2,\sigma_z=-1/2 \rangle-|s=1/2,\sigma_z=-1/2 \rangle \otimes |s=1/2,\sigma_z=1/2 \rangle.$$
    This implies that, if you find when measuring ##\sigma_z## of particle 1 to be ##+1/2## then necessarily particle 2 must have ##\sigma_z=-1/2## etc.

    The spin-z component of each single particle is maximally indetermined. The reduced statistical operator for particle 1 is
    $$\hat{\rho}_1=\mathrm{Tr}_1 |J=0,J_z=0 \rangle \langle J=0,J_z=0| = \frac{1}{2} \mathbb{1},$$
    i.e., the particle is unpolarized. The same is true for particle ##2##.
     
  6. Mar 10, 2017 #5

    DrChinese

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    The conservation rule applies here. If the probabilities were as you mention, you would have a Product state and the conservation rule would not hold.
     
  7. Mar 10, 2017 #6
    QM can tell you about what happens at detection but not about what happens in between. I would be suspicious that where external fields are involved conservation laws could be relied on.
     
  8. Mar 10, 2017 #7

    Nugatory

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    Conservation laws still work; you just have to be careful to include the change in the external field as well. However, there is no external field involved in the decay interaction that @vanhees71 was describing, so this is a bit of a red herring. We started with an isolated quantum system prepared in a state of zero total angular momentum; as long as it remains isolated its angular momentum will remain zero.
     
  9. Mar 10, 2017 #8

    edguy99

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    Would this hold true if the basis vectors of the original spin angle or measuring angle were rotated?
     
  10. Mar 10, 2017 #9
    @Nugatory , I agree, but the measurement in the SG setup does include a field.
     
  11. Mar 10, 2017 #10

    Strilanc

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    The original particle is spin zero, it doesn't have a spin axis.

    One of the first things taught about the singlet state is that it doesn't have a preferred axis or orientation. We write it down in some basis, as if it had a preferred orientation, but when you switch to another basis it looks exactly the same.

    For example, suppose Alice and Bob share two qubits in the singlet state ##|01\rangle - |10\rangle##. If they measured their qubits right now, in the computational basis, which distinguishes between ##|0\rangle## and ##|1\rangle##, they will get opposite answers. But instead they are going to rotate their qubits before measuring, so they effectively measure along a different axis.

    The rotation operation ##U = \begin{bmatrix} a & b \\ c & d \end{bmatrix}## is a 2x2 unitary matrix. Because both Alice and Bob are applying ##U## to their qubit, we end up applying ##U \otimes U## to the two-qubit system. Now turn the crank:

    $$\begin{align}
    \psi_2
    &= (U \otimes U) \cdot \psi_1
    \\
    &= (U \otimes U) \cdot \big( |01\rangle - |10\rangle \big)
    \\
    &= (U \otimes U) \cdot |01\rangle - (U \otimes U) \cdot |10\rangle
    \\
    &= (U \otimes U) \cdot \big( |0\rangle \otimes |1\rangle \big) - (U \otimes U) \cdot \big( |1\rangle \otimes |0\rangle \big)
    \\
    &= \big( U |0\rangle \big) \otimes \big( U |1\rangle \big) - \big( U |1\rangle \big) \otimes \big( U |0\rangle \big)
    \\
    &= \big( a |0\rangle + c |1\rangle \big) \otimes \big( b |0\rangle + d |1\rangle \big) - \big( b |0\rangle + d |1\rangle \big) \otimes \big( a |0\rangle + c |1\rangle \big)
    \\
    &= \big( ab |00\rangle + ad |01\rangle + cb |10\rangle + cd |11\rangle \big) - \big( ba |00\rangle + bc |01\rangle + da |10\rangle + dc |11\rangle \big)
    \\
    &= (ab-ba) |00\rangle + (ad-bc) |01\rangle + (ca-da) |10\rangle + (cd-dc) |11\rangle
    \\
    &= (ad-bc) |01\rangle + (cb-da) |10\rangle
    \\
    &= \det(U) \big( |01\rangle - |10\rangle \big)
    \\
    &= e^{i \theta} \psi_1
    \end{align}$$

    As you can see, no matter what rotation Alice and Bob both apply, they end up back in the singlet state. If they limit themselves to operations from SU(2), they don't even pick up a global phase factor (not that it matters, since those have no observable effects).

    This proves that the singlet state has no preferred orientation.

    Some entangled states do change when the same operation is applied to both sides. For example, if Alice and Bob start in the singlet state and then only Alice flips her qubit over around the Z axis, then X and Y axis measurements would agree whereas the Z axis measurement still disagrees. The Z axis parity "stands out" from the other two. But this is still not the kind of pre-existing orientation that you're talking about. You can still use such states to violate Bell inequalities.
     
  12. Mar 10, 2017 #11

    edguy99

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    Absolutely, but the moment it decays we have a spin axis for each spin 1/2 particle on a set of basis vectors.
     
  13. Mar 10, 2017 #12

    Strilanc

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    Read the rest of my post. The singlet state doesn't have a preferred spin axis.
     
  14. Mar 10, 2017 #13
    @strianc, are you saying that there cannot be a state defined for each of the particles individually!?
     
  15. Mar 10, 2017 #14

    DrChinese

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    If they are entangled on that basis, no.
     
  16. Mar 10, 2017 #15

    Strilanc

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    Yes. This is a basic well known fact about entanglement. In fact, people often define "entangled" to mean states that can't be factored into parts.
     
  17. Mar 10, 2017 #16

    Nugatory

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    indeed it does.... But that's part of the measuring apparatus, and the original claim was that we couldn't trust conservation of momentum in an unmeasured system. As long as the quantum system that started as the parent particle remains isolated, it's momentum is constant and momentum is trivially conserved. Eventually that system interacts with the inhomogeneous magnetic field of the SG machines when we measure it; this may change the momentum of the particles, but also changes the momentum of the SG machines in such a way that momentum is still conserved.
     
  18. Mar 10, 2017 #17
    So if one particle for whatever reason ceased to exist, the other particle's state would immediately change?
     
  19. Mar 10, 2017 #18
    I didn't see anything in the earlier posts that talked about anything other than measurement.
     
  20. Mar 10, 2017 #19

    Strilanc

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    No, that's not how it works at all. Do you expect the color of your left sock to be affected by throwing your right sock into a fire?

    The particles are strongly correlated, more correlated than possible with socks, but doing something to one of them still has no observable effect on the other.
     
  21. Mar 10, 2017 #20
    I agree, although it doesn't explain the mismatches in my sock drawer!
     
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