Difference Between Collapse and Projection

[Demystifier]

I don't know. I do know that the detector used in the Stern Gerlach experiment did absorb the atoms. If you are proposing a different kind of detector in your thought experiment, it's up to you to specify what kind of detector it is.

Electron microscope can detect a single atom without absorbing it. But electron microscope probably can't detect an atom that moves, so it's still not adequate for measuring spin with a SG magnet. Any ideas? You are a nuclear engineer, right?

 

[PeterDonis]

Any ideas? You are a nuclear engineer, right?

Nuclear engineering doesn't really deal with manipulating single atoms (or single nuclei) at a time, unfortunately.

 

[vanhees71]

Sorry but you are wrong, in a way that has nothing to do with interpretations and philosophy. You must have misunderstood something purely physical about POVM measurements.

Let me explain how POVM works for photon detection. Let $\{|m\rangle\}$, $m\neq 0$, be a basis of 1-photon states and let $|m=0\rangle$ be the photon vacuum. The Kraus operators for ideal (perfectly efficient) photon detection can be taken to be
$$M_m=|0\rangle\langle m|$$
They satisfy
$$\sum_m M_m^{\dagger}M_m=\sum_m |m\rangle \langle m|=1$$
If the state before measurement is a superposition
$$|\psi\rangle=\sum_{m'} c_{m'}|m'\rangle$$
then after measurement (that is, when the result of measurement is known), the updated state is proportional to
$$M_m|\psi\rangle=c_m|0\rangle$$
So after the measurement we always have the vacuum, that is, the photon is destroyed (absorbed) by the detector.

@atyy I have significantly edited this post that you already liked. I hope you will still like it, perhaps now even more.

Ok, then it's fine for this case, and I stand corrected. So $\hat{M}_m$ describes effectively a transition matrix element between states of the measured object after interacting with the measurement device. Then it's fine, because you don't assume something outside the time-evolution formalism in the sense that the $\hat{M}_m$ can be derived in principle by the time evolution of an open quantum system.

This misunderstanding could have been avoided by clearly defining the meaning of the operators $\hat{M}_m$ in the beginning!

 

[Demystifier]

Ok, then it's fine for this case, and I stand corrected. So $\hat{M}_m$ describes effectively a transition matrix element between states of the measured object after interacting with the measurement device. Then it's fine, because you don't assume something outside the time-evolution formalism in the sense that the $\hat{M}_m$ can be derived in principle by the time evolution of an open quantum system.

This misunderstanding could have been avoided by clearly defining the meaning of the operators $\hat{M}_m$ in the beginning!

So we agree on physics. But there is one interpretation issue that, I think, you was still not completely clear about. If we consider a closed system, including the measuring apparatus, is everything deterministic?

 

[vanhees71]

The state evolution is deterministic but the state doesn't determine all possible observables but provide and only provide probabilities for the outcome of measurements of any observable, for whose test you need an ensemble of equally prepared individual systems.

 

[Demystifier]

only provide probabilities for the outcome of measurements

Suppose that there is no measurement at the time $t$. Consider two quantities
$$|\psi(t)\rangle=e^{-iHt}|\psi(0)\rangle \; \; \; \; (1)$$
$$p(t)=\langle\psi(t)|\pi|\psi(t)\rangle \; \; \; \; (2)$$
where $\pi$ is a projector (not associated with a measurement because, as I said, there is no measurement at time $t$). Since there is no measurement at $t$, does probability (2) have any physical meaning? If not, does the state (1) have any physical meaning?

A related question. How the fact that there is a measurement expressed mathematically? I would like an answer in the form: We say that an observable is measured when
$$some \;\; equation$$

 

[vanhees71]

If $\pi$ is a projector, it's of the form $|a \rangle \langle a|$ with a unit-vector $|a \rangle$. If you interpret $|a \rangle$ to be the eigenstate of some operator $\hat{A}$ that represents some observable $A$, $p(t)$ is of course the probability to find the value $a$ when you measure $A$ and the system is prepared in the state $\hat{\rho}=|\psi(t) \rangle \langle \psi(t)|$.

The physical meaning is the usual one $\hat{\rho}$ is the statistical operator in the Schrödinger picture of time evolution and $p(t)$ is the probability to get $a$ as the result of a measurement of $A$ when measured at time $t$.

The whole point is that you cannot give an answer to your related question in terms of a general postulate. It depends on the individual experiment, how the system evolves when interacting with the measurement device.

 

[vanhees71]

Yes, but assuming a measurement causes a collapse, i.e., a change of the state, implies a causal influence of the measurement on the state, and that's the problem particularly in this context. It's contradicting the very assumptions you make about the dynamics of the system (microcausality condition), which by construction cannot violate causality, i.e., space-like separated events cannot be causally connected.

 

[Demystifier]

Yes, but assuming a measurement causes a collapse, i.e., a change of the state, implies a causal influence of the measurement on the state, and that's the problem particularly in this context. It's contradicting the very assumptions you make about the dynamics of the system (microcausality condition), which by construction cannot violate causality, i.e., space-like separated events cannot be causally connected.

You reject that measurement causes collapse of the state, but you accept that measurement entails update of the state, am I right? But "causing collapse" and "entailing update" are described by the same mathematics and no experiment can distinguish one from the other. So when you insist that they are different, you are doing philosophy, not physics. And it would not be a problem if it was a consistent philosophy, but it's not. It's inconsistent because the update of the state also violates the Schrodinger equation (or its relativistic QFT analog), so you both accept and don't accept violation of the Schrodinger equation. So you do philosophy, and you do it inconsistently, but you are not disturbed because "that's just philosophy", so who cares.