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

[atyy]

But then you can just forget about this collapse assumption. It's empty, because non-observable!

Collapse is needed mathematically. You are welcome to consider it real or not. However, there is no problem with considering collapse to be real and also maintaining your definition of locality.

 

[vanhees71]

Why is it needed mathematically? I just do calculations in QT without ever using it...

 

[stevendaryl]

I do not see how a proof of microcausality (something along the lines that local field at spacelike separated events commute, or anti-commute) answers the question about whether QM (or QFT) is nonlocal. Whether we're talking about QM or QFT, the theory has two parts:

1. There is some notion of "state" that changes deterministically according to some mathematical evolution law.
2. Using the state, we compute probabilities for outcomes of measurements, and when we actually perform a measurement, we get definite values.

Microcausality is about the first aspect of quantum theory, the evolution equations, but the reason people suspect QM is nonlocal is because of the second aspect, measurement (and the Born interpretation of the quantum state). So a rigorous proof of microcausality cannot possibly resolve the issue.

Unless, of course, you adopt the Many Worlds Interpretation, and say that the state is everything, and that measurement and the Born interpretation can somehow be derived from the state evolution equations.

 

[atyy]

Why is it needed mathematically? I just do calculations in QT without ever using it...

How do you show that after Alice measures and receives the outcome up, the state jumps from |uu>+|dd> to |u>?

 

[vanhees71]

One cannot prove that (at least there's no proof known). See Weinberg, Lectures on Quantum Mechanics, Cambridge University Press, but why should such a proof solve the apparent problem. I don't know, where the problem is within the minimal interpretation, because I just take the minimal interpretation and Born's rule as it is. I can predict only probabilities for the outcome of measurements. Then I measure the quantity I've predicted the probabilities for on an ensemble of identically prepared setups and check whether my prediction is right. So far all predictions of QT agreed with experiment. That's it. That a measurement of an observable gives definite values is due to the construction of the measurement apparatus, because otherwise you'd not call it an apparatus that measures the quantity of interest. It's just a technical problem to construct the apparatus, and our experimental colleagues are real wizzards in doing astonishing constructions.

 

[vanhees71]

How do you show that after Alice measures and receives the outcome up, the state jumps from |uu>+|dd> to |u>?

Why do I need to show it? I can make all predictions about the outcome of measurements with the entangled state prepared in the very beginning. No need for the reduction of the state to $|u \rangle$. Why do you need that?

 

[stevendaryl]

How do you show that after Alice measures and receives the outcome up, the state jumps from |uu>+|dd> to |u>?

I think there is a sense in which it's not necessary to consider what happens after a measurement is performed. Alice performs a spin-measurement, and measures spin-up. A short while later, she performs a second measurement, and again measures spin-up. You want to say that the reason she measures spin-up the second time is because the state collapsed into a pure spin-up state following her first measurement. But you could avoid collapse by instead considering the two measurements to be a single, two-part measurement. Pure quantum mechanics without collapse would presumably predict that such a two-part measurement has essentially zero chance of giving the result "up, down" or "down, up".

 

[atyy]

I think there is a sense in which it's not necessary to consider what happens after a measurement is performed. Alice performs a spin-measurement, and measures spin-up. A short while later, she performs a second measurement, and again measures spin-up. You want to say that the reason she measures spin-up the second time is because the state collapsed into a pure spin-up state following her first measurement. But you could avoid collapse by instead considering the two measurements to be a single, two-part measurement. Pure quantum mechanics without collapse would presumably predict that such a two-part measurement has essentially zero chance of giving the result "up, down" or "down, up".

That is not possible. The two measurement outcomes are spacelike separated, so in anyone frame in which one considers them to be simultaneous, there will be another frame in which they are sequential.

 

[atyy]

Why do I need to show it? I can make all predictions about the outcome of measurements with the entangled state prepared in the very beginning. No need for the reduction of the state to $|u \rangle$. Why do you need that?

How do you do the calculation in the Schroedinger picture?

 

[stevendaryl]

That is not possible. The two measurement outcomes are spacelike separated, so in anyone frame in which one considers them to be simultaneous, there will be another frame in which they are sequential.

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.

 

[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.