Quantum mechanics is not weird (locality and non-locality weirdness)

[stevendaryl]

How do you do the calculation in the Schroedinger picture?

I think he's just saying, in the EPR case, that pure quantum mechanics, without collapse makes the prediction in an EPR-type experiment:

"The probability that Alice will measures spin-up along axis \vec{\alpha} and Bob will measure spin-up along axis \vec{\beta} is: \frac{1}{2} (1 - \vec{\alpha} \cdot \vec{\beta}) (or whatever it is). There is no need to talk about the state after Alice's measurement but before Bob's measurement, so there is no need to invoke collapse. You view it as a single, two-part measurement, rather than a sequence of measurements. Similarly, if you want to include more measurements after Bob, you formulate it as a 3-part measurement or 4-part measurement, or whatever. There is no need to ever invoke collapse in order to compute probabilities.

But this approach is silent on the question of "What is the state after Alice's measurement, but before Bob's measurement?"

 

[zonde]

There is no need to talk about the state after Alice's measurement but before Bob's measurement, so there is no need to invoke collapse.

There is simple counterexample. We measure Alice's particle and depending on outcome we select Bob's measurement angle. Maybe measurement is not very interesting but anyways: how would one calculate probabilities?

 

[stevendaryl]

There is simple counterexample. We measure Alice's particle and depending on outcome we select Bob's measurement angle. Maybe measurement is not very interesting but anyways: how would one calculate probabilities?

Well, the noncollapse way of doing it would involve treating Alice herself as quantum-mechanical system, which is of course impractical. But one need not consider collapse to be a physical thing; it could be just an approximation to avoid infeasible mathematics.

 

[atyy]

I was talking about Alice making two spin measurements in succession, with a timelike, not spacelike, separation between them. I wasn't talking about Alice's measurement followed by Bob's.

Anyway, this alternative way of looking at it basically amounts to (as I understand it) the "consistent histories" interpretation of QM, which is like Many-Worlds in avoiding a notion of wave function collapse. The fact that Alice measured spin-up doesn't imply anything about the "wave function of the universe", it just says something about which history she is on.

If you do two measurements in succession, then you do need collapse. A single two part measurement is not the same as two measurements in succession. The probabilities are the same, but the observed events in the invariant sense of classical relativity (which is preserved in quantum field theory) are not the same.

One can certainly try to avoid collapse by using interpretations such as MWI or consistent histories. They are certainly not standard, since there is no agreement as to whether they work or not.

 

[vanhees71]

How do you do the calculation in the Schroedinger picture?

The picture doesn't matter. If you have the state (I use the usual simplification to only show the polarization part)
$$|\Psi \rangle=\frac{1}{\sqrt{2}} (|HV \rangle - |VH \rangle),$$
then the reduced probabilities for A and B to find an arbitrary polarization state is
$$\hat{\rho}_A=\hat{\rho}_B=\frac{1}{2} \mathbb{1}_2,$$
nevertheless you have 100% correlations, e.g., the probability for both A and B finding horizontally or vertically polarized photons is
$$|\langle HH |\Psi \rangle|^2=|\langle VV|\Psi \rangle|^2=0$$
but that A finds H and B finds V is or vice versa is
$$|\langle VH|\Psi \rangle|^2=|\langle HV|\Psi \rangle|^2=1/2.$$
So you have always the correlation, but both A and B for themselves find just unpolarized photons when measuring on an ensemble. To realize the correlations they have to exchange their measurement protocols. Nowhere do you need any collapse hypothesis to predict these probabilities, and that's what can be measured.

The same works for any other measurement of the polarization, particularly also where Bell's inequality is violated, for which you need to measure the polarization in other than the same direction.

 

[atyy]

The picture doesn't matter. If you have the state (I use the usual simplification to only show the polarization part)
$$|\Psi \rangle=\frac{1}{\sqrt{2}} (|HV \rangle - |VH \rangle),$$
then the reduced probabilities for A and B to find an arbitrary polarization state is
$$\hat{\rho}_A=\hat{\rho}_B=\frac{1}{2} \mathbb{1}_2,$$
nevertheless you have 100% correlations, e.g., the probability for both A and B finding horizontally or vertically polarized photons is
$$|\langle HH |\Psi \rangle|^2=|\langle VV|\Psi \rangle|^2=0$$
but that A finds H and B finds V is or vice versa is
$$|\langle VH|\Psi \rangle|^2=|\langle HV|\Psi \rangle|^2=1/2.$$
So you have always the correlation, but both A and B for themselves find just unpolarized photons when measuring on an ensemble. To realize the correlations they have to exchange their measurement protocols. Nowhere do you need any collapse hypothesis to predict these probabilities, and that's what can be measured.

The same works for any other measurement of the polarization, particularly also where Bell's inequality is violated, for which you need to measure the polarization in other than the same direction.

That only works if Alice and Bob measure simultaneously. Can you do the calculation in a frame in which Alice measures first?

 

[A. Neumaier]

We measure Alice's particle and depending on outcome we select Bob's measurement angle.

In the usual scenarios this would require faster than light communication to B of the outcome of A.

 

[zonde]

Well, the noncollapse way of doing it would involve treating Alice herself as quantum-mechanical system, which is of course impractical.

How would one treat measurement angle as quantum-mechanical variable? It is classical variable and yet it shows up in calculations of amplitudes.

 

[A. Neumaier]

How would one treat measurement angle as quantum-mechanical variable?

The correct way to do it is described in the following paper:
Kastrup, Hans A. "Quantization of the canonically conjugate pair angle and orbital angular momentum." Physical Review A 73.5 (2006): 052104.

 

[zonde]

The correct way to do it is described in the following paper:
Kastrup, Hans A. "Quantization of the canonically conjugate pair angle and orbital angular momentum." Physical Review A 73.5 (2006): 052104.

I don't think this addresses my doubts:
"The problem is that the angle is a multivalued or discontinuous variable on the corresponding phase space. The remedy is to replace ϕ by the smooth periodic functions cos ϕ and sin ϕ."

Entanglement predictions are expressed using sine and cosine of relative angle not relative sine or cosine of absolute angles.

 

[A. Neumaier]

Entanglement predictions are expressed using sine and cosine of relative angle not relative sine or cosine of absolute angles.

what is different?

 

[zonde]

I thought it over and the uneasiness that I get about angle being classical parameter is that if we have superposition of measurement angles we get superposition of superpositions as we have different superpositions at different real angles.

 

[A. Neumaier]

we get superposition of superpositions as we have different superpositions at different real angles.

?

You need an observable $X=f(\phi)$ with proper commutation rules and spectrum so that solving $f(\phi)=X$ gives you the natural multivalued function representing the noncanonical variable $\phi$. Then the eigenvectors of a maximal commuting family of Hermitian operators containing $X$ will form an orthonormal basis representing (unnormalized) states of definite angle. Their superpositions produce every vector in the Hilbert space of the representation, hence superpositions of superpositions are just again simple superpositions.

If $f(\phi)$ for a given angle $\phi$ doesn't determine the system completely, there are additional commuting observables that provide the distinguishing labels.