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Expectation values linear harmonic oscillator

  1. Aug 18, 2016 #1
    hello :-)
    here is my problem...:

    1. The problem statement, all variables and given/known data


    For a linear harmonic oscillator, [tex] \hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2} m \omega^2x^2[/tex]
    a) show that the expectation values for position,[tex] \bar{x},[/tex] and momentum [tex] \bar{p}[/tex] oscillate around zero with angular frequency [tex] \omega.[/tex] Hint: Use Ehrenfest Theorem

    2. Relevant equations

    Ehrenfest Theorem: [tex] \frac {d}{dt} \bar{A}= \frac{1}{i\hbar} \bar{[A,H]} = \frac{1}{i\hbar} \langle [A,H] \rangle [/tex]

    3. The attempt at a solution

    So I have to make some calculations with the result [tex] \bar{x} = \bar{p} = 0.[/tex]
    And I want and have to use the Ehrenfest Theorem.
    With the Ehrenfest Theorem I calculated the temporal derivative of the expectation value of the position [tex] \frac {d}{dt} \bar{x}
    [/tex]
    and the result is zero. But that is just the derivative of the result, which I actually want to know! So.. maybe someone give me another hint? :-)
    Thank you
     
  2. jcsd
  3. Aug 18, 2016 #2

    DrClaude

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    So ##\hat{x}## commutes with ##\hat{H}##?
     
  4. Aug 18, 2016 #3

    BvU

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    Hi,
    No, you certainly don't.
    For what states, exactly, are you doing this ?
     
  5. Aug 18, 2016 #4
    oh. no. [tex] \frac{d}{dt} \bar{x} = \frac{1}{m} \bar{p} = \frac{\hbar}{im} \frac{\bar{d}}{dx} [/tex]

    So I write it explicit:

    [tex] \frac{d}{dt} \int_a^b \psi^*(x,t) x \psi(x,t) dx = \frac{\hbar}{im} \int_a^b dx \psi^*(x,t) \frac{d}{dx} \psi(x,t) [/tex]

    And now, go on calculating this way? Is that the right way? Doesn`t feel like that...


    I thought the "problem statement" says that? that it oscillates around zero... so zero has to be the expectation value.
    For which states? Do you want to know more about the "problem statement"? I have no more information. If you want to know what I think, then I would say... I shall do it for every state...?
     
    Last edited: Aug 18, 2016
  6. Aug 18, 2016 #5

    BvU

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    Check out ##{d\over dt} \bar p## too ... recognize anything ? Writing out in full is hard work, better think twice.
    Remember how an arbitrary state develops ?

    I ask 'for which states' because the steady states aren't oscillating in an interesting way...:rolleyes:
     
  7. Aug 18, 2016 #6
    I already checked it:
    [tex] \frac{d}{dt} \bar{p} = -\langle \frac{dU}{dx} \rangle\quad\quad\quad U = \frac{1}{2}m\omega^2x^2[/tex]
    I recognize nothing. you mean, when I compare it with: [tex] \frac{d}{dt} \bar{x} = \frac{1}{m} \bar{p} = \frac{\hbar}{im} \frac{\bar{d}}{dx} ?[/tex]

    I don`t know what you want to say me with your questions...
    so they are depending on time.. that means "not steady"
     
    Last edited: Aug 18, 2016
  8. Aug 18, 2016 #7

    BvU

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    So you have $$
    \frac{d}{dt} \bar{x} ={1\over m} \bar{p} \quad {\text {and }} \quad \frac{d}{dt} \bar{p} = - m\omega^2 \bar{x} $$ the equations for a classical harmonic oscillator.

    Steady states have ##\bar x = 0 \ ## (check this !), but a state with ##\bar x \ne 0 ## will exhibit the desired time dependence.
    The exercise doesn't ask for specific states, but: can you think of such a state ?
     
  9. Aug 22, 2016 #8
    Sorry for not answering,
    I had to make a break with that...

    ok. here is my steady state: [tex] \psi(x,r) = \phi(x)e^{-iEt/\hbar} [/tex]

    [tex] \bar{x} = \int_{-\infty}^{+\infty} dx \phi^*(x)e^{iEt/\hbar} x \phi(x)e^{-iEt/\hbar} = = \int_{-\infty}^{+\infty} |\phi(x)|^2 x [/tex]
    And this has to be zero. Because an integral is just the area between the grap of a function and the x-axis. and the "x" brings a "point reflection.

    okay. So the result can be derived temporal and that would be the same than [tex] 1/m * \bar{p} [/tex]

    no specific one... but a state which has not the form [tex] \psi(x,r) = \phi(x)e^{-iEt/\hbar} [/tex] or more general: a state, which propability density [tex] | \psi(x)|^2 [/tex] doesn`t depend from time.

    for example....: [tex] \psi(x,r) = \phi(x)(Et/\hbar) [/tex]

    sorry, but I am not joking :rolleyes::biggrin:
     
  10. Aug 22, 2016 #9

    BvU

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    From your answer it's not clear to me you now understand the exercise, I'm afraid. Wat is ##\phi(x)## and what is ##E## in your steady state ##\psi(x,t)## (not ##\psi(x,r)## ! ) ?

    And what do you mean with ##\psi(x,r) = \phi(x)(Et/\hbar)## ? It's not a solution to the time-dependent SE !
     
  11. Aug 23, 2016 #10
    In my first part, I just checked, whether [tex] \bar{x} = 0 [/tex] for steady states.

    And yes, [tex] \psi(x,t) = \phi(x)(Et/\hbar) [/tex] is not a solution of time-dependent SE.
    And also yes, I wanted to write [tex] \psi(x,t) [/tex] and not [tex] \psi(x,r) [/tex], but the second time I wrote it I just copied it, so the mistake happened twice.

    I try what to explain, what to do in this exercise and what we have reached so far:smile:

    We have to show, that the expectation value for position and for momentum oscillate around zero with angular frequency omega.

    We have the first half of our solution: The solution for steady states: The expectation values for position [tex] \bar{x} [/tex] is zero, (I showed that graphical) and because of [tex] \frac{d}{dt} \bar{x} = \frac{1}{m} \bar{p} ,[/tex] also the expectation value of the momentum is zero.

    Now we have to do the second half: Show the expectation values for not-steady states?

    My last guess was bad. I have no second guess...
     
  12. Aug 23, 2016 #11

    BvU

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    Steady states for the harmonic oscillator are eigenfunctions of the Hamiltonian: ##H\phi(x) = E\phi(x)##. The solutions of the time-independen Schroedinger equation are distinct wavefunctions ##\phi_n(r)## with distinct energies ##E_n##. Their expectation value for ##\bar x## is zero because ##\phi_n^* \phi_n ## is symmetric and x is antisymmetric.

    What about e.g. ##\phi_1^* \phi_0 ## ?

    Arbitrary general states ##\Phi(r)## can be written as ##\sum_n C_n\phi_n## with ##C_n## derived from ##\Phi(r)## at t= 0 and they develop in time as
    ##\Phi(r,t) = \sum_n C_n\phi_n e^{-{i\over \hbar} E_n t}## .

    A simple case is ##\Phi(r) = {1\over \sqrt 2} \left ( \phi_0 + \phi_1 \right )## where the time development of ##\bar x## no longer vanishes. Doing the integrals with the Hermite polynomials is hard work (but worth it). Easier to use ladder operators, but you may not have seen those yet.

    Check the first picture in wikipedia
     
  13. Aug 24, 2016 #12
    That is the same, I wanted to say with that, just graphical:

    I make an example. I calculate the expectation value for the position for ## \psi_1 ## which is:

    [tex] \psi_1(\zeta) = (\frac{m\omega}{\pi\hbar})^{1/4} \frac{1}{\sqrt{2^11!}} H_1(\zeta) e^{-0,5\zeta^2}
    = (\frac{m\omega}{\pi\hbar})^{1/4} \frac{1}{\sqrt{2}} 2\zeta e^{-0,5\zeta^2} = (\frac{m\omega}{\pi\hbar})^{1/4} \sqrt{2} \zeta e^{-0,5\zeta^2} [/tex]
    (Is that right? because my skript says sth. different and I don`t find a proper source elsewere)
    [tex] \rightarrow \bar{\zeta} = \int_{-\infty}^{+\infty} d\zeta (\frac{m\omega}{\pi\hbar})^{1/2} 2 \zeta^2 e^{-\zeta^2} \zeta = 0, [/tex] because of the two made explanations. (the one part of this integral is symmetric, the other one antisymmetric...) and because of [tex] \zeta = \sqrt{\frac{m\omega}{\hbar}} x [/tex] also the expectation value of x is zero.

    [tex] \rightarrow \bar{\zeta} = \int_{-\infty}^{+\infty} (\frac{m\omega}{\pi\hbar})^{1/4} \sqrt{2} \zeta e^{-0,5\zeta^2} \cdot \zeta \cdot (\frac{m\omega}{\pi\hbar})^{1/4} e^{-0,5\zeta^2} = \int_{-\infty}^{+\infty} (\frac{m\omega}{\pi\hbar})^{1/2} e^{-\zeta^2} \sqrt{2} \zeta^2 = (\frac{m\omega}{\pi\hbar})^{1/2} \frac{1}{\sqrt{2}} \sqrt{\pi} [/tex] The integral was solved with a gauss-integral... so not zero..
    is there a rule for that?

    just the integral ##\int \phi_1^* \phi_0 ## would be zero. The integral for Hermite polynomials says that.

    so the time development vanishes for ##\phi_n^* \phi_n ## and ##\phi_n^* \phi_m ## because they are steady states. ##\Phi(r) = {1\over \sqrt 2} \left ( \phi_0 + \phi_1 \right )## is also solution of schroedinger equation, but the time development doesn`t vanish.

    so you mean I shall calculate this:::


    [tex] \bar{\zeta} = \int_{-\infty}^{+\infty} \frac{1}{\sqrt{2} } (\frac{m\omega}{\pi\hbar})^{1/4} e^{-0,5\zeta^2} e^{iE_0t/\hbar} + (\frac{m\omega}{\pi\hbar})^{1/4} \sqrt{2} \zeta e^{-0,5\zeta^2} e^{iE_1t/\hbar} \cdot \zeta \cdot \frac{1}{sqrt{2}} (\frac{m\omega}{\pi\hbar})^{1/4} e^{-0,5\zeta^2} e^{-iE_0t/\hbar} + (\frac{m\omega}{\pi\hbar})^{1/4} \sqrt{2} \zeta e^{-0,5\zeta^2} e^{-iE_1t/\hbar} d\zeta [/tex]?

    So the answer to the question, how the expectation values for a linear harmonic oscillator are, is:

    ## \bar{x} ## is zero for ##\phi_n^* \phi_n ## and a constant for ##\phi_n^* \phi_m ##.
    ## \bar{p} ## is zero for ##\phi_n^* \phi_n ## and for ##\phi_n^* \phi_m ##
    and ## \bar{x} ## + ## \bar{p} ## aren`t zero, when we use a combination of different states?
     
  14. Aug 24, 2016 #13

    BvU

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    Oh boy, what a big post. You certainly did a lot of ##\TeX##ing there !

    Good work, but some things need to be revisited:

    Most importantly: we overlooked the time-dependent part when calculating ##\bar x##. No disaster for $$
    \left (\phi_0(x) e^{-{i\over \hbar}E_0t} \right )^* \;x\; \left (\phi_0(x) e^{-{i\over \hbar}E_0t} \right ) $$ but you can see a ##e^{-{i\over \hbar}(E_1-E_0) t} ## appearing for ##\phi_1^* \phi_0##

    And the only interesting terms in ##\bar x (t) ## for ##
    \Phi(r) = {1\over \sqrt 2} \left ( \phi_0 + \phi_1 \right )## are ##\phi_1^* \phi_0## and ##\phi_0^* \phi_1## which are equal, so you have to calculate only one.

    And Hermite polynomials have some useful properties that allow partial integration. But you can also be satisfied with an outcome like constant ##* \cos(\omega t)##
     
  15. Aug 24, 2016 #14
    Thank you for your recognition :redface:

    I see that. Just the "minus" is wrong I think.
    [tex] \rightarrow \bar{\zeta} = \int_{-\infty}^{+\infty} (\frac{m\omega}{\pi\hbar})^{1/4} \sqrt{2} \zeta e^{-0,5\zeta^2} e^{+iE_1t/\hbar}
    \cdot \zeta \cdot
    (\frac{m\omega}{\pi\hbar})^{1/4} e^{-0,5\zeta^2} e^{-iE_0t/\hbar}
    = \int_{-\infty}^{+\infty} (\frac{m\omega}{\pi\hbar})^{1/2} e^{-\zeta^2} \sqrt{2} \zeta^2 e^{i(E_1-E_0)t/\hbar} = (\frac{m\omega}{\pi\hbar})^{1/2} \frac{1}{\sqrt{2}} \sqrt{\pi} e^{i(E_1-E_0)t/\hbar} [/tex]

    [tex] \rightarrow \bar{x} = \frac{1}{\sqrt{2}} e^{i(E_1-E_0)t/\hbar} [/tex]
    [tex] \rightarrow \bar{p} = m \frac{i(E_1-E_0)}{\hbar} \frac{1}{\sqrt{2}} e^{i(E_1-E_0)t/\hbar} [/tex]

    So this are the both expectation values...

    How about more general: ## \int_{-\infty}^{+\infty} \phi_n^* \cdot \zeta \cdot \phi_m ##

    would it be: ## \bar{x} = \frac{1}{\sqrt{2}} e^{i(E_n-E_m)t/\hbar} ## ???

    My problem is now...
    I understand the calculations and what want to show for ## \phi_n^*\phi_n ##. And how their graphs look like.

    But what is with ## \phi_n^*\phi_m ##? The normalization "says", that it is zero. But what does that say? Is ##\int \phi_n^*\phi_m ## the probability that both states exist at the same time? Then zero would make sense. But what does the expectation value for ## \int \phi_n^*\phi_m ## say then say?

    fortunately that makes sense for me. So:

    [tex] \bar{\zeta} = 2 e^{i(E_= - E_1) t / \hbar} \int \phi_0 \phi_1 \zeta d\zeta = 2 e^{i(E_0 - E_1) t / \hbar} (\frac{m\omega}{\pi\hbar})^{1/2} \frac{1}{\sqrt{2}} \sqrt{\pi} [/tex]

    I already calculated it for a previous Thread. And I used a Gauss-integral. okay..

    [tex] \rightarrow \bar{x} = \sqrt{2} e^{i(E_0 - E_1) t / \hbar} [/tex]

    So. This is a next result. But there are a lot of more.. for example a combination of 3 different terms..
     
  16. Aug 24, 2016 #15

    BvU

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    Probably. But your ##
    \rightarrow \bar{x} = \frac{1}{\sqrt{2}} e^{i(E_1-E_0)t/\hbar}## doesn't ring too good either: there are two imaginary terms (one from ##\ \phi_1^*x\phi_0\ ## and one from ##\ \phi_0^* x \phi_1\ ##and together they should yield a real expectation value ##\bar{x} = \bar x_0 \cos \omega t ##.

    [edit] can't be all; there should also be some ## \bar p_0/m \sin \omega t ##.


    From the back of my mind (rusty and dusty there) I come up with the expectation :smile: that they all give 0 except when ##|n-m| = 1##. Will do some looking up later ...
     
    Last edited: Aug 24, 2016
  17. Aug 25, 2016 #16

    BvU

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    I googled around but can't find a treatment based on the explicit wave functions/Hermite polynomials: everybody (ucd, ucsd, usu, mit, Reed) uses ladder operators (for good reasons: much less tedious and very useful for later). Check out one or two of them to see if we didn't make too big mistakes in this thread :smile:

    In view of the fact that your exercise was effectively done already with post #6 I propose we call it a day :smile:

    Just one loose end to clear up:
    It says that eigenfunctions with different eigenvalues are orthogonal. The probablility to find eigenvalue n when the system is in the eigenstate m is zero when n and m are different. This is in line with the time-independency of eigenstates of the Hamiltonian. Important concept, so make sure you have no doubts or lingering questions left.

    [edit] looks like I skipped something: H needs to be unitary for orthogonality.

    But since the Schroedinger equation is linear, solutions (i.e. possible states) can be linear combinations of eigenstates. And such solutions are not eigenfunctions of the Hamiltonian, so they do develop in time (as we've seen in this exercise). And this time development is unique (fortunately) thanks to the orthogonalty of the eigenstates.

    I liked this exercise/thread a great deal (as a refresher also :rolleyes:).
     
    Last edited: Aug 25, 2016
  18. Aug 25, 2016 #17
    Yes, I have also googled a lot, and they almost use the ladder operators...

    Sure :smile:

    First half makes very much sense for me, and I have to read more about the second half :smile:



    Yes, for me too. You helped me a lot :smile:
    greetings from germany :smile:
     
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