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Question on expectation value.

  1. Feb 23, 2012 #1
    1. The problem statement, all variables and given/known data

    Given an observable quantity A, when will it happen that the same value for A will be measured every time?

    What is the relationship between the operator [itex]\hat{A}[/itex] and [itex]\Psi[/itex] for this case?


    What is the relationship between [itex]\widehat{A}[/itex] and [itex]\widehat{H}[/itex], the hamiltonian, for this case?

    2. Relevant equations

    Expectation value equation:


    Also I feel like it may have something to do with the Schrodinger equation.

    3. The attempt at a solution

    I'm not sure if the first part is asking for a mathematical answer or not. The measured value would be the same every time only if there was a single allowed value for that physical quantity.

    I'm not sure about the last two parts either. Energy is still quantized, except that only one solution exists... what does this mean for the relationship between eigenfunction and operator? Does it mean that the operator gives the same function back, like with an eigenvalue of 1? That is just a shot in the dark, I'm kinda clueless over here.

    I'm not even sure how to relate A to H with an equation, let alone comment on this specific case.

  2. jcsd
  3. Feb 23, 2012 #2
  4. Feb 23, 2012 #3


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    First, the problem statement has nothing to do with the expectation value of ##\hat{A}##. It has to do with making a measurement of the observable. Those are two different things.

    Second, the problem has to do with what happens when you make a measurement. Read up on that in your QM textbook, and this problem will make more sense to you.
  5. Feb 23, 2012 #4
    Edit: deleted my post while I work on the problem a bit more.
  6. Feb 24, 2012 #5
    Sorry! Correct - the expectation value isn't relevant here :-)
  7. Mar 5, 2012 #6
    So... I can't find anything on the Ehrenfest theorem in my notes or textbook. I realize it may be a way of solving this, but my professor would not expect us to have to find it on our own. Any other hints on other methods I could use?
  8. Mar 5, 2012 #7


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    The Ehrenfest theorem doesn't apply here because the problem isn't about expectation values. It's more about understanding how states evolve with time. It's a conceptual problem. Don't assume ##\hat{A}## has only one allowed measurement result.
  9. Mar 5, 2012 #8
    Ok then I'm back to square one. I'm glad it is a conceptual problem though. This is literally the only problem left on my practice test that I haven't been able to figure out, but it is worrying me.

    How else could the same value be measured every time unless the wavefunction gave a probability of 1 at that point? If other results were possible, wouldn't they mean that the measurement would not be the same every time? Or am I approaching this from the wrong direction?

    From your last post... yes I know that the wavefunction "collapses" when you make a measurement (not from my textbook which doesn't really mention such a thing, but rather from reading these forums). What does this have to do with the specific case in the above problem?
  10. Mar 6, 2012 #9
    Ok so I went to talk to my professor today and he wouldn't help me with this question because he basically said it was going to be on the test, but that it should be a simple answer. So now I'm really desperate because I can't get help except from here, and it is on the test.

    A) If you get the same value for A every time, that means that ΔA=0. Is this the answer to the first part?

    B) [itex]\hat{A}\Psi = a\Psi[/itex] This is how you relate [itex]\hat{A}[/itex] and [itex]\Psi[/itex] What can I say about this relationship?

    C) If ΔA=0, does this tell you something about the commutator relationship [[itex]\widehat{A},\widehat{H}[/itex]]? I know for, say, position and momentum, if the uncertainty in x is 0, then the uncertainty in p is infinite because they don't commute. But I don't know if A and H commute, so I seem to be stuck.

    Also, lets say they didn't commute. When position and momentum don't commute I say "there is a heisenberg uncertainty relationship between the observables position and momentum." If A and H don't commute, would I say "there is a heisenberg uncertainty relationship between the observable A and the energy of the system"? I'm just checking I understand this correctly.

    Help me physics forums, you're my only hope.
    Last edited: Mar 6, 2012
  11. Mar 6, 2012 #10
    What is the result of observation A for a general state, and what is the new state after that?
  12. Mar 6, 2012 #11
    The wavefunction collapses into a single possible eigenstate.

    Same as above? A single eigenstate?
  13. Mar 6, 2012 #12
    And how does this eigenstate evolve until the next measure?
  14. Mar 6, 2012 #13
    Does it go back to being a set of possible states again? This is really stretching my grasp of this concept. Like I said, I've never actually seen "wavefunction collapse" in my textbook, only wikipedia....

    Obviously you are trying to get at something here. Could you show me how this relates to my question? I need a concrete hint to go on. So far I have been thinking about this for days and I can't figure it out.

    EDIT: If I get the same measurement for A each time, does it mean that [itex]\hat{A}\Psi = a\Psi[/itex] has only one possible eigenstate?

    EDIT 2: It would be helpful also if someone could give me a large enough hint so that I could actually try to solve the problem on my own rather than short posts with long stretches in between where I am no closer to an answer.... not trying to be rude, I'm just very anxious and frustrated with this "simple" question.
    Last edited: Mar 6, 2012
  15. Mar 6, 2012 #14
    I m not an expert... No it means that the state is an eigenstate each time. The evolution is given by Schrödinger eq. You must find an hamiltonian such that this state remains an eigenstate.
  16. Mar 6, 2012 #15
    I thought no matter what, it is an eigenstate when you measure something?

    I'm dying over here folks.

    Does (B) mean that Psi is an eigenfunction with the operator A?
    Last edited: Mar 6, 2012
  17. Mar 6, 2012 #16
    Before the measure generally not, just after surely yes.

    As long as you do not measure you can follow your state with the Schr. eq. and predict the probability to get one value or another by projecting the state one the respective eigenvectors of A. But by measuring A you pick up a value and the state becomes the corresponding eigenvector. Then you can again follow the state which may or may not remain an eigenvector.
  18. Mar 6, 2012 #17
    So.... how does this answer my question? Again, not trying to be rude, I just feel a little obtuse because I can't figure this out.
  19. Mar 6, 2012 #18


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    Say the observable ##\hat{A}## has eigenstates ##|a_1\rangle##, ##|a_2\rangle##, ##|a_3\rangle##, …. such that ##\hat{A}|a_i\rangle = a_i |a_i\rangle##. For simplicity, let's assume these eigenstates are non-degenerate. We can express the state of the system ##|\psi(t)\rangle## at time t as a linear combination of these eigenstates.
    $$|\psi(t)\rangle = c_1(t) |a_1\rangle + c_2(t) |a_2\rangle + \cdots $$ You're given that every time you measure ##\hat{A}##, you get the same result. What does that imply about the form ##|\psi(t)\rangle## must take?
  20. Mar 6, 2012 #19
    When you say you measure A and get the same result each time, does this mean the eigenstate is the same each time?

    Ok so ##\hat{A}## has various eigenstates. ψ at a certain time is a linear combination of these eigenstates. The fact that the eigenstate is the same each time (?), does this mean that a1, a2 etc. all specify the same state, i.e. they are equal?

    Sorry if my responses don't make much sense. But I really appreciate you trying to help me.
  21. Mar 6, 2012 #20


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    Let's backtrack a bit. Suppose the state is as given above. What's the probability of measuring ai at time t?

    The eigenstates are distinct. In fact, I suggested we look at the non-degenerate case, so we can go even better and say all the eigenvalues are distinct.
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