Measurement and expectation value

[hokhani]

TL;DR

Does a measurement affect the expectation value?

If measurement of an observable $\hat{A}$ is done over an ensembles of large number of copies of the system, then the average value of these measurements gives $\langle A \rangle$. If we do this measurement over this ensemble again, we expect to get to the last result $\langle A \rangle$. So, it seems that consequent measurements disturb the time dependency of $\langle A \rangle$. Doesn't it?

 

[anuttarasammyak]

TL;DR: Does a measurement affect the expectation value?

If we do this measurement over this ensemble again, we expect to get to the last result

I do not think so in general. Some counter examples.
Photons disappear on the screen leaving dots in double slit experiment. We cannot repeat the experiment on the same photons.
Measurement of electron momentum by scattering ends to scattered states which does not hold the property of the original state.
In another thread of you, it was mentioned that in loose
position observation center of dispersing Gaussian packet does inertial motion in general so <x> changes with time.(the immediate second measurement would give the same value with the first one)

Measurement is performed by interaction between the object system and measuring apparatus. Some measurement process destroy or disturb the object system.

[edit]

TL;DR: Does a measurement affect the expectation value?

So, it seems that consequent measurements disturb the time dependency of ⟨A⟩. Doesn't it

Quantum Zeno effect might be of your interest.

 

[Roberto Pavani]

Do you mean repeating the measurement on a freshly prepared ensemble, or on the same ensemble after the first measurement? The answer is different in the two cases.

 

[hokhani]

Do you mean repeating the measurement on a freshly prepared ensemble, or on the same ensemble after the first measurement? The answer is different in the two cases.

On the same ensemble after the first measurement.

 

[Roberto Pavani]

The answer may also depend on how invasive the measurement is.
Some measurements strongly disturb the system (or even destroy it), while others can have a much smaller back-action.
I'm not sure that "measuring the same ensemble again" has a unique answer without specifying the measurement model.

I'm sure that some expert here can provide a better answer.

 

[Matterwave]

On the same ensemble after the first measurement.

Others have discussed some practical issues with doing this experiment but if we ignore those, we can analyze this experiment based on the pure mathematical formalism. At least it's relatively easy to do in standard QM, ignoring QFT effects of particle annihilation, creation etc.

From that perspective, we have to ask a critical question, which is, how long after the first measurement on any given system is the second measurement made?

If the second measurement is made immediately after the first, then you will get the same answer as your first measurement. This is just the projection/collapse postulate (process 1 according to Von Neumann). If the second measurement takes some time (and remember, in QM typical time scales can be quite short) then the wave function will have already evolved via the Schroedinger equation (assuming $A$ doesn't commute with the Hamiltonian) and you will have to solve that for the new distribution.

 

[Matterwave]

So to answer the top level question, yes measurements will affect $\langle A\rangle$ because they affect $| \psi\rangle$ and $\langle A\rangle=\langle\psi|A| \psi\rangle$.

 

[hokhani]

So to answer the top level question, yes measurements will affect $\langle A\rangle$ because they affect $| \psi\rangle$ and $\langle A\rangle=\langle\psi|A| \psi\rangle$.

So, we cannot consider the classical measurements like the expectation values in QM because we may have classical measurements which don't affect the system. But, from Ehrenfest theorem, we expect the wave pocket to behave like a classical particle!

 

[Matterwave]

But, from Ehrenfest theorem, we expect the wave pocket to behave like a classical particle!

2 points.

1. You have to consider my answer in light of my longer explanation above.
2. That's not what the Ehrenfest theorem says.

Ehrenfests theorem is a statement about the time evolution of expectation values assuming the underlying states evolve according to the Schrodinger equation. Your question expressly brought up measurement, which means the underlying states in your question do not simply undergo Schroedinger evolution.

 

[hokhani]

If the second measurement takes some time (and remember, in QM typical time scales can be quite short) then the wave function will have already evolved via the Schroedinger equation (assuming $A$ doesn't commute with the Hamiltonian) and you will have to solve that for the new distribution.

You mean that if the second measurement is done long after the first one, the second result may be different. But I haven't seen anywhere that the time between the two consequent measurements of commuting operators matters. Could you please explain more?

 

[Matterwave]

You mean that if the second measurement is done long after the first one, the second result may be different. But I haven't seen anywhere that the time between the two consequent measurements of commuting operators matters. Could you please explain more?

A more complete exposition of the situation might require a really long post. Perhaps you could elucidate further your core question or confusion?

 

[PeterDonis]

I haven't seen anywhere that the time between the two consequent measurements of commuting operators matters.

It does if the state in between is not an eigenstate of the Hamiltonian. Then the state will change by time evolution in between the measurements. That's what @Matterwave was talking about in post #6, which you quoted from.

 

[hokhani]

It does if the state in between is not an eigenstate of the Hamiltonian. Then the state will change by time evolution in between the measurements. That's what @Matterwave was talking about in post #6, which you quoted from.

Thank you and Matterwave. We are at a good point to review my problem again. We would like to measure the quantity $\hat{A}$ and $[\hat{H}, \hat{A}]\ne0$. By the first measurement which is done at time $t_1$ the system collapses to one of the eigenfunctions of $\hat{A}$, say $|a_1\rangle$. Then at time $t_2$ we repeat the measurement of $\hat{A}$ again. With a comparison by Stern-Gerlach experiment we expect the system to be still in the state $|a_1\rangle$ no matter how long is $t_2-t_1$. But you believe that if $t_2-t_1$ is long, our second measurement may result in $|a_2\rangle$. I think the problem is clearly explained now and appreciate if help me with that.

 

[anuttarasammyak]

S-G Hamiltonian is proportional to Sz. [H,Sz]=0. Sz is conserved,I.e.,we get the same result in measurement performed in any future. What observable A is actually?

 

[hokhani]

S-G Hamiltonian is proportional to Sz. [H,Sz]=0. Sz is conserved,I.e.,we get the same value in measurement performed in any future. What observable A is actually?

Right, so for example in the S-G experiment, before the first measurement we have the Hamiltonian $\frac{\hat{P}^2}{2m}$, at first measurement at $t_1$ we have Hamiltonian in the form $\frac{(\hat{P}-eA/c)^2}{2m}+S_zB$ and between $t_1$ and $t_2$ we have again $\frac{\hat{P}^2}{2m}$ and so in all of the routes $[H,S_z]=0$.

 

[anuttarasammyak]

Right, but how about the interaction of magmatic field with translational motion of the electron?

I do not get what is A from your suggestion. More clear and direct explanation will be appreciated.

 

[hokhani]

I do not get what is A from your suggestion. More clear and direct explanation will be appreciated.

You are right, I edited the post #15.

 

[anuttarasammyak]

I still do not find what is A. A=P^2/2m ? If so $[H,A]\neq 0$ ?

 

[hokhani]

You mean A=P^2/2m ? If so where $[H,A]\neq 0$ ?

No, I approved your comment about S-G experiment by taking $\hat{A}=\hat{S_z}$ and $A$ in the Hamiltonian is not operator but the vector potential for magnetic field.

 

[anuttarasammyak]

No, I approved your comment about S-G experiment by taking A^=Sz^

Then [H^,A^]=0 and post #13 does not stand.

 

[hokhani]

Then [H^,A^]=0 and post #15 does not stand.

Ok, and I told since $[H,\hat{A}=\hat{S_z}]=0$, in the Stern-Gerlach experiment after the time $t_1$ the system is always in the specific spin, say up, and this way my problem in post #13 is resolved. I hadn't considered there the commutation of $[\hat{A} , H]=0$.