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Check my working please? Expectation of momentum

  1. Jun 8, 2014 #1
    1. The problem statement, all variables and given/known data

    A state at time t is given by:

    [tex]|\psi\rangle = \frac{1}{\sqrt 2}\left[ e^{-\frac{i\omega t}{2}}|0\rangle + e^{-i\frac{3\omega t}{2}}|1\rangle \right] [/tex]

    Where eigenfunctions are ##\phi_0 = \left(\frac{1}{a^2 \pi}\right)^{\frac{1}{4}}e^{-\frac{x^2}{2a^2}}## and ##\phi_1 = \left(\frac{4}{a^6 \pi}\right)^{\frac{1}{4}} x \space e^{-\frac{x^2}{2a^2}}##.

    I'm supposed to find ##\langle \psi|\hat p|\psi \rangle##.

    2. Relevant equations

    [tex]\hat p = -i\hbar \frac{\partial}{\partial x}[/tex]

    3. The attempt at a solution

    By symmetry, we know that ##\langle 0|\hat p|0 \rangle = \langle 1 | \hat p | 1 \rangle = 0##.


    [tex]\frac{1}{2} \langle 0 | e^{-i\omega t} \hat p |1\rangle [/tex]
    [tex] = -\frac{1}{2}i\hbar \left(\frac{4}{a^8 \pi^2}\right)^{\frac{1}{4}}e^{-i\omega t} \int_{-\infty}^{\infty} e^{-\frac{x^2}{2a^2}} \frac{\partial}{\partial x} \left(x e^{-\frac{x}{2a^2}}\right) dx[/tex]
    [tex]= -\frac{1}{2}i\hbar \left(\frac{4}{a^8 \pi^2}\right)^{\frac{1}{4}}e^{-i\omega t} \int_{-\infty}^{\infty} e^{-\frac{x^2}{2a^2}}\left(e^{-\frac{x^2}{2a^2}} - \frac{x^2}{a^2} e^{-\frac{x^2}{2a^2}}\right) dx [/tex]
    [tex] = -\frac{1}{2}i\hbar \left(\frac{4}{a^8 \pi^2}\right)^{\frac{1}{4}}e^{-i\omega t} \int_{-\infty}^{\infty} e^{-\frac{x^2}{a^2}} \left(1 - \frac{x^2}{a^2}\right) dx [/tex]
    [tex] = -\frac{1}{2}i\hbar \left(\frac{4}{a^8 \pi^2}\right)^{\frac{1}{4}}e^{-i\omega t} \left[ \sqrt \pi a - \frac{1}{a^2} \frac{1}{2} \sqrt {\pi} a^3 \right] [/tex]
    [tex] = -\frac{i\hbar}{2} \left(\frac{4}{a^8 \pi^2}\right)^{\frac{1}{4}} \left(\frac{1}{2} \sqrt \pi a \right) e^{-i\omega t} [/tex]
    [tex] = -\frac{i\hbar}{2\sqrt 2 a} e^{-i\omega t} [/tex]

    Similarly, for ##\frac{1}{2}\langle 1 | e^{i\omega t} \hat p | 0 \rangle##, we take the complex conjugate of above.

    Adding them both together, we get:

    [tex]-\frac{\hbar}{\sqrt 2 a} sin (\omega t) [/tex]

    Given ## \langle \psi |x|\psi\rangle = \frac{a}{\sqrt 2} cos (\omega t) ##, then I'm supposed to find:
    [tex]\frac{\partial}{\partial t} \left( \langle \psi|p|\psi\rangle + m\omega^2 \langle \psi |x|\psi\rangle \right) [/tex]

    They don't add up to zero, which is strange since the first term is simply the force, and the second term is the restoring force..
    Last edited: Jun 8, 2014
  2. jcsd
  3. Jun 9, 2014 #2
    Surely this would be much easier if you just express [itex]X[/itex] and [itex]P[/itex] in terms of creation & annihilation operators?
  4. Jun 9, 2014 #3
    They aren't given in the question at all, so I don't think we need to use them.
  5. Jun 10, 2014 #4


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    From your other thread...
    Yeah, sorry. I did look at this one, but couldn't see anything obviously wrong with your working.

    But you make a valid point, so I'm trying again to understand what's going on/wrong here...

    Isn't there a dimensional problem here? Momentum has dimensions ##MLT^{-1}## and ##\omega## has dimensions ##T^{-1}##.
    So ##m\omega^2 x## has dimensions ##MLT^{-2}##.
    If I got that right, then you can't add ##p## and ##m\omega^2 x##.
    (In this problem, ##a## must have dimensions ##L## for consistency, so that seems consistent with the expectation values you found.)

    Are you perhaps mixing up the partial time derivative with the total time derivative? Cf. Ehrenfest's theorem .
  6. Jun 10, 2014 #5
    Perhaps you mean

    [tex]\frac{d}{dt} \left( \langle \psi|P|\psi\rangle\right) + m\omega^2 \langle \psi |X|\psi\rangle = 0[/tex]

    (which would be true)
  7. Jun 10, 2014 #6
    Yes, I meant that, would that make a difference to the answer? I know it's true, because it commutes with ##hat H##, but the problem is how to show it..
  8. Jun 10, 2014 #7


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    From this, I guess you didn't bother to read the link I suggested earlier at the end of post #4 ?
  9. Jun 11, 2014 #8
    Sorry, I do know Ehrenfest's theorem. What I was trying to say is since all the terms commute with the hamiltonian it doesn't make a difference whether it was a partial or total derivative.
  10. Jun 11, 2014 #9

    I don't know what you mean here, since both [itex]X[/itex] and [itex]P[/itex] fail to commute with [itex]H[/itex].

    But the main point is that in the version I posted, the time derivative only acts on the first term. That fixes (among other things) the dimensional inconsistency pointed out by strangerep.
  11. Jun 11, 2014 #10


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    As with Oxvillian, I don't know what you mean by this. Neither x nor p commute with H.

    So what is the question?? In your opening post you just wanted to calculate the ##p## expectation. Is there another part to the question that you haven't shown us?
  12. Jun 12, 2014 #11
    Here is the full question:


    I'm trying to find the link between finding ##\langle x \rangle##, ##\langle p\rangle## and the last part of the question.
  13. Jun 12, 2014 #12


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    OK, the actual question is simpler than I thought.

    So... have you evaluated $$\frac{d}{dt} \langle p \rangle ~=~ ~~?$$If so, what do you get? Also, what classical dynamical quantity does this correspond to?

    And what do you get when you add it to ##m\omega^2 \langle x \rangle## ? (And what classical quantity does that correspond to?)

    (I'm still having trouble understanding what your problem really is.)
    Last edited: Jun 12, 2014
  14. Jun 13, 2014 #13
    My problem is that they don't sum to zero, as forces should balance.

    [tex]\frac{d\langle p \rangle}{dt} = -\frac{\hbar \omega}{\sqrt 2 a} cos (\omega t)[/tex]

    [tex] m\omega^2 \frac{d \langle x \rangle }{dt} = -\frac{ma\omega^3}{\sqrt 2} sin (\omega t)[/tex]

    They are wildly different.
  15. Jun 13, 2014 #14
    unscientific - I think you have us pulling out our hair in frustration here :smile:

    Why do you still insist on differentiating [itex]\langle x \rangle[/itex]?
  16. Jun 13, 2014 #15


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    He's seeing parentheses where there aren't any.
  17. Jun 14, 2014 #16
    But even still, they don't sum to zero:

    [tex]\frac{d\langle p \rangle}{dt} = -\frac{\hbar \omega}{\sqrt 2 a} cos (\omega t)[/tex]

    [tex]m\omega^2 \langle \psi |x|\psi\rangle = \frac{ma\omega^2}{\sqrt 2} cos (\omega t)[/tex]
  18. Jun 14, 2014 #17
    Almost there! Now use the definition of [itex]a[/itex] given in the question.
  19. Jun 15, 2014 #18
    ##a## is a constant, it's not defined in the question.
  20. Jun 15, 2014 #19
    Unscientific, your second long equation is not correct. Try the differentiation again. : )

    Edit, oh I see the first equation is mis-typed? Belay that!
    Last edited: Jun 15, 2014
  21. Jun 15, 2014 #20

    Look about 2/3 of the way through the question, right after the wavefunctions. It's the characteristic length scale of the harmonic oscillator, for what it's worth.
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