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Expectation of the Momentum Operator

  1. Oct 30, 2008 #1
    1. The problem statement, all variables and given/known data
    Here is another True or False question from the same practice test.

    Since the expectation of the momentum operator <p>=<n|pn> is zero for an energy eigen state of the harmonic oscillator, a measurement of the momentum will give zero every time (True or False).


    2. Relevant equations
    N/A


    3. The attempt at a solution
    This question sounds very false to me, since an expectation value implies a probability and not an absolute. However, I'm thinking it might be a trick question...

    Thanks!
     
  2. jcsd
  3. Oct 30, 2008 #2

    olgranpappy

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    Maybe the best way to understand this problem is to rewrite the operator p as the commutator of x with the hamiltonian (times some factor). Can you do this?
     
  4. Oct 30, 2008 #3
    Do you mean like this?

    p = [x,H]g(x) = xHg(x) - Hxg(x) = 0
     
  5. Oct 30, 2008 #4

    olgranpappy

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    no. if I follow all those equal signs that equation says that the momentum operator p=0...

    Here. Just do this. Tell me, what is this
    [tex]
    \left[
    \hat x
    ,
    \hat H
    \right]
    [/tex]
    equal to? In the above [itex]\hat x[/itex] is the position operator and [itex]\hat H[/itex] is the hamiltonian
    [tex]
    \hat H=\frac{\hat p^2}{2m}+V(\hat x)\;.
    [/tex]
     
  6. Oct 30, 2008 #5
    [tex]\hat{x}\hat{H}f(x)-\hat{H}\hat{x}f(x)[/tex]
     
  7. Oct 30, 2008 #6
    Can you just substitute the Hamiltonian now?
     
  8. Oct 30, 2008 #7

    olgranpappy

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    presumably "f(x)" is some "test function" that you are letting the operators act on?

    Yes. go ahead and substitute in the hamiltonian now and see what you get. what do you get?
     
  9. Oct 30, 2008 #8
    Okay I'll omit the test function:

    [tex]\frac{\hat x\hat p^2}{2m}+\hat xV(\hat x) - \frac{\hat p^2\hat x}{2m}+V(\hat x) \hat x[/tex]

    This can't be simplified, can it?

    (Sorry - I'm very new to QM)
     
  10. Oct 30, 2008 #9

    olgranpappy

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    It certainly can.

    For one thing you have forgetting to distribute the minus sign into the last term. Your expression should read:
    [tex]
    \frac{\hat x\hat p^2}{2m}+\hat xV(\hat x) - \left(\frac{\hat p^2\hat x}{2m}+V(\hat x) \hat x\right)
    =
    \frac{\hat x\hat p^2}{2m}+\hat xV(\hat x) - \frac{\hat p^2\hat x}{2m}-V(\hat x) \hat x
    =
    \frac{\hat x\hat p^2}{2m} - \frac{\hat p^2\hat x}{2m}
    [/tex]

    Better bring back your test function if you want to continue simplifying the rest. Since p=-id/dx,
    you now want to determine
    [tex]
    -\frac{x}{2m}\frac{d^2f}{dx^2} + \frac{d^2}{dx^2}\left(\frac{x}{2m}f\right)
    [/tex]
     
  11. Oct 30, 2008 #10
    You don't need those manipulations at all: What would be the condition for the state |n> to *always* give the result p=0 for a measurement of momentum?
     
  12. Oct 30, 2008 #11

    olgranpappy

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    One does not *need* all "those manipulations". But "those manipualtions" are important from a pedagogical standpoint. Do you understand that?

    And besides... I believe that the way you are suggesting to think about this problem (at least in my estimation of what you are getting at) is incorrect.
     
  13. Oct 30, 2008 #12
    I don't agree: Poster says he's a beginner at qm. You're using an atomic bomb to kill a musquito.

    lol.
     
  14. Oct 30, 2008 #13

    olgranpappy

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    Perhaps. Perhaps not. But at least I am not misleading the OP.
     
  15. Oct 30, 2008 #14
    Misleading?? Soon I'll show my simple elegant pedagogical solution, maybe you'll understand it's not misleading, nor incorrect.
     
  16. Oct 30, 2008 #15

    olgranpappy

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    sure.
     
  17. Oct 30, 2008 #16
    Ok, daveyman, here we go:

    First, you write the operator p as a linear combination of the operators a^+ and a^-, that I think you will have heard of when discussing the harmonic oscillator.

    Second, the given statement that <n|pn>=0 states that p|n> is orthogonal to |n>. You can prove that is true [although the True/False question doesn't ask you to prove this] by using the properties of the afore-mentioned operators a^+ and a^-.

    Third, the statement that you'd *always* find zero momentum in the state |n> would be equivalent to stating that p|n>=0|n>. Using the preceding result, you'll find that p|n> is *not* equal to 0|n> for any value of n.

    Done.

    I think oldgran was letting you prove <n|p|n>=0 in a more complicated way. I also think that perhaps he was thinking that the equation p|\psi>=0 doesn't have a normalizable solution, but that is not relevant here: you only have to prove that |n> is *not* a solution to that equation.
     
  18. Oct 30, 2008 #17

    olgranpappy

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    sigh. I hate to say this... but you are actually correct; I was trying to help the OP prove that <n|p|n>=0 for eigenstates. But, now that I reread the question, that is not at all what he was asking.

    Now I look silly... and it serves me right for trying to answer the question before reading the whole question. I apologize.

    Anyways, OP. borgwal is correct to say that you should understand what it means that *every* momentum measurement yields zero. Namely, if this were true the state in question would be an eigenstate of p (with eigenvalue zero). And this is certainly not the case for |n> (which is an eigenstate of H not P).
     
  19. Oct 30, 2008 #18
    That's very gracious of you, thanks! Now let's hope daveyman is happy too :-)
     
  20. Oct 30, 2008 #19

    cepheid

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    I didn't read this whole freakin' thread, but basically eigenstates of the Hamiltonian are stationary states...states that are guaranteed to yield a certain value for the energy (the eigenvalue) when a measurement is taken. So, in order for the value of the observable p to always be the same whenever a measurement is taken, |n> would have to be an eigenstate of p-hat as well as of H-hat. If those operators don't commute, then simultaneous eigenstates of them both cannot be constructed. So maybe all of that fancy math was required to answer the problem, maybe it wasn't (if you already know whether those two operators commute...)
     
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