Collapse of Wave: Effects on Energy & Position Measurements

[jimgraber]

I think you are right that this is not new, but a restatement of an existing established position.

 

[Mentz114]

It's fine to agree to disagree, but I would like to know what, specifically, you're disagreeing with.

Thanks to @jimgraber for the reference to the Griffith paper (still wet from the printer !).

Steven, my objections to Bell are stated there

However the derivation of the CHSH version of a Bell inequality, which is Eq. (1) in [1], has as one of its assumptions that S_z and S_y, and other components of spin can be replaced by classical, which is to say commuting, quantities, ...

I don't think I ever understood the maths Bell wrote, but the distinction between quantum and classical reckoning is brought into focus by his multicolored socks.

 

[zonde]

I don't think I ever understood the maths Bell wrote, but the distinction between quantum and classical reckoning is brought into focus by his multicolored socks.

Yes, quantum particle states are described using Hilbert space, but arguments based on this is just attacks on strawmen as detection events are classical facts. There is no need to make assumptions about microscopic reality when we derive Bell type inequality just from measurement results.

 

[DuckAmuck]

What is a "collapsed" or "non-collapsed" wave function? There's no such thing in standard quantum mechanics. The wave function describes a state in a certain basis, namely the (generalized) eigenbasis of the position operator (and perhaps other observables compatible with position like spin to have a complete basis). That's it. There's no way to say a wave-function is "collapsed" or "non-collapsed". It simply doesn't make any sense!

You can only say that the position of the particle is more or less determined, depending on the width of the probability distribution of position (which is given by the modulus of the wave function squared, due to Born's rule).

In the case of position (or any observable), you may start out with a state distributed in a sinusoidal fashion over x. If you measure the position, the wavefunction "collapses" to a delta function centered on the position x₀ you measured. Over time it will evolve back to a sinusoidal state.

 

[Mentz114]

Yes, quantum particle states are described using Hilbert space, but arguments based on this is just attacks on strawmen as detection events are classical facts. There is no need to make assumptions about microscopic reality when we derive Bell type inequality just from measurement results.

I have to say that I cannot discern what you mean. Mea culpa.

 

[zonde]

I have to say that I cannot discern what you mean.

Say you don't know anything about QM. You just have measurement results at two distant locations (produced by black-box). You can still derive Bell type inequalities for these measurements assuming they are produced by process that respects relativistic locality.

 

[Mentz114]

Say you don't know anything about QM. You just have measurement results at two distant locations (produced by black-box). You can still derive Bell type inequalities for these measurements assuming they are produced by process that respects relativistic locality.

The inequalities serve to define statistics (functions of the data) which are sensitive to the things we are looking for. But any number of different ones can be used if they are suitable for the experimental setup.

 

[vanhees71]

Ok but then, if we ignore everything else but spin, isn't the state of an atom before Stern-Gerlach given by a|spin up>+b|spin down> and after passing through, it is either |spin up> or |spin down>? So there is something collapsing here.

In the Stern-Gerlach experiment you cannot ignore position. It's the crucial point of this experiment that after running through the inhomogeneous magnetic field, you have a spin-position entangled state, namely
$$\psi^j(\vec{x})=\psi_{\text{left}}(\vec{x}) \chi_{\text{up}}^j + \psi_{\text{right}}(\vec{x}) \chi_{\text{down}}^j.$$
If you tailor your magnetic field right, the $\psi_{\text{left}}$ and $\psi_{\text{right}}$ functions peak practically in separated regions of space, and you can filter out one part by just blocking the corresponding partial beam of particles. Then you have prepared particles with a determined spin component in direction of the magnetic field, i.e., up or down.

The blocking is also due to the interactions of the particles with the medium used to block them. There's no other dynamics than quantum dynamics and no collapse hypothesis is necessary.

 

[vanhees71]

It does not follow the standard rules of probability unless one postulates collapse. That is the important point about Bayesian updating - the standard rules of probability are not enough without collapse.

I still don't understand this terminology. You have a well defined initial state, describing the statistical properties of the system, including the correlation between spins, and then it is clear to A what B must measure after A has measured her photon. This is not due to some mystical collapse but just due to the usual rules of quantum kinematics and dynamics (without any necessity for an additional collapse hypothesis).

 

[vanhees71]

Again, I wish to stress that your interpretation is very non-minimal. How can you be sure that "nothing has happened to the object"?

You replied saying QED obeys signal causality. Sure, but as I have stressed repeatedly, signal causality being respected does not mean that "nothing has happened to the object". A simple way to see that you lack an argument for your assertion that the updating is purely informational with nothing happening to the object, is that if collapse is physical, then something has happened to the object and yet faster than light signalling is prevented.

The minimal interpretation is agnostic, not confidently assertive of things it cannot show, unlike your claim that "nothing has happened to the object". If you read Cohen-Tannoudji, Diu and Laloe's famous text, you will see that they are not so cavalier as you are at this point.

Within QED if A's and B's registration events are spacelike separated, the registration event at A cannot have affected B's photon before he registered it and vice versa. Since QED describes all these experiments with very high significance right, I tend to say that's the correct description. If one day one finds a reproducible violation of any prediction within QED, one has to think about a new theory, assuming non-local actions at a distance and maybe even the very foundation of the space-time model, but before this is not the case, I don't see any necessity to give up the very successful standard paradigm of local microcausal relativistic QFT.

 

[martinbn]

In the Stern-Gerlach experiment you cannot ignore position. It's the crucial point of this experiment that after running through the inhomogeneous magnetic field, you have a spin-position entangled state, namely
$$\psi^j(\vec{x})=\psi_{\text{left}}(\vec{x}) \chi_{\text{up}}^j + \psi_{\text{right}}(\vec{x}) \chi_{\text{down}}^j.$$
If you tailor your magnetic field right, the $\psi_{\text{left}}$ and $\psi_{\text{right}}$ functions peak practically in separated regions of space, and you can filter out one part by just blocking the corresponding partial beam of particles. Then you have prepared particles with a determined spin component in direction of the magnetic field, i.e., up or down.

The blocking is also due to the interactions of the particles with the medium used to block them. There's no other dynamics than quantum dynamics and no collapse hypothesis is necessary.

That's the same just more unnecessary details. The question remains. If the state, you wrote above, is the state of the atom before going through the SG apparatus, then the state after is only one of the summands. So there is some kind of collapse?