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Quantum Harmonic Oscillator

  1. Oct 17, 2013 #1
    1. The problem statement, all variables and given/known data
    Compute ##\left \langle x^2 \right\rangle## for the states ##\psi _0## and ##\psi _1## by explicit integration.

    2. Relevant equations
    ##\xi\equiv \sqrt{\frac{m \omega}{\hbar}}x##
    ##α \equiv (\frac{m \omega}{\pi \hbar})^{1/4}##
    ##\psi _0 = α e^{\frac{\xi ^2}{2}}##


    3. The attempt at a solution
    I'm not getting the right answer here:

    $$\left \langle x^2 \right\rangle= α^2\int ^{\infty} _{-\infty} \psi ^* x^2 \psi dx = α^2\frac{\hbar}{2m\omega}\int ^{\infty} _{-\infty} \psi ^* [(a_+)^2 + (a_+ a_- )+(a_- a_ )+(a_- )^2]\psi dx$$

    This follows from the fact that ##x^2## can be defined by ##x=\sqrt{\frac{\hbar}{2m\omega}}(a_+ +a_- )##

    Because of the orthogonality of ##\psi _n## with ##\psi _m## we can cancel the first and last terms out, leaving ##(a_+ a_-)+(a_- a_ )##

    If I'm not mistaken, this should mean the integral goes to zero because ##a_{\pm}\psi _0 = 0##?

    I know the correct answer is ##\frac{\hbar}{2m\omega}##

    If somebody could point me in the right direction, I would appreciate it.
     
  2. jcsd
  3. Oct 17, 2013 #2

    DrClaude

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    Staff: Mentor

    You are mistaken. There is an error in that last equation.

    My interpretation of "explicit integration" is that you have to integrate functions of x, not work with operators.
     
  4. Oct 17, 2013 #3
    If I'm not suppose to use operators, that would mean I would just directly integrate through with ##x^2##?

    I'm fairly certain I'm expected to use operators. From a few pages back (direct quote):

    There's a beautiful device for evaluating integrals of this kind (involving powers of ##x## or ##p##): use the definite to express ##x## and ##p## in terms of the raising and lowering operators.

    I think I noticed my mistake, which was that the final product should be ##\psi ^* [(a_+ a_- )\psi + (a_- a_+ )\psi] = \psi ^2##

    $$α^2\frac{\hbar}{2m\omega}\int ^{\infty} _{-\infty}e^{-\xi ^2}dx$$

    Something doesn't feel right here.
     
  5. Oct 17, 2013 #4

    vanhees71

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    The notation is a bit inconvenient. It's much easier to use the bra-ket notation. So let [itex]|n \rangle[/itex] be the normalized eigenvectors of the number operator [itex]\hat{N}=\hat{a}^{\dagger} \hat{a}[/itex].

    Further you need
    [tex]\hat{a} |n \rangle=\sqrt{n} |n-1 \rangle, \quad \hat{a}^{\dagger} |n \rangle=\sqrt{n+1} |n+1 \rangle.[/tex]
    Then it's easy to evaluate
    [tex]\langle n|(\hat{a}+\hat{a}^{\dagger})^2 |n \rangle =\langle n|(\hat{a}^2+\hat{a}^{\dagger 2} + \hat{a} \hat{a}^{\dagger} + \hat{a}^{\dagger} \hat{a} |n \rangle ,[/tex]
    using the above equations and the orthonormality of the eigenstates,
    [tex]\langle n'|n \rangle=\delta_{nn'}.[/tex]
     
  6. Oct 17, 2013 #5

    DrClaude

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    It is just my interpretation of the question. It could be that the question is asking to use explicit in order to contrast with the same calculation using operators.

    That is not correct. You have to get a different result for ##\psi_0## and ##\psi_1##. (And note that ##\psi^2 \neq \psi^* \psi = |\psi|^2##).
     
  7. Oct 17, 2013 #6

    vanhees71

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    Argh. Sorry, I overread that you are forced to solve the question by "direct integration".

    The trick is to use a generating function for all the moments of [itex]x[/itex] needed. In this case it's simly
    [tex]f(z)=\int_{-\infty}^{\infty} \exp(-z x^2)=\sqrt{\frac{\pi}{z}}.[/tex]
    Then you get all moments by differentiation (for odd powers of [itex]x[/itex] you get of course 0 anyway),
    [tex]\int_{-\infty}^{\infty} x^{2n} \exp(-z x^2)=(-1)^n \frac{\mathrm{d}^n}{\mathrm{d} z^n} f(z).[/tex]
     
  8. Oct 17, 2013 #7
    I worked out the integral in post #3 and it came out correct for ##\psi _0##, so I'm not sure what's going on now, I haven't tried it yet for ##\psi _1 ##, but I think it'll work out.
     
  9. Oct 18, 2013 #8

    DrClaude

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    Yes
    $$
    \psi_0^* (a_+ a_- + a_- a_+) \psi_0 = \psi_0^* \psi_0
    $$
    but
    $$
    \psi_1^* (a_+ a_- + a_- a_+) \psi_1 \neq \psi_1^* \psi_1
    $$
     
  10. Oct 18, 2013 #9
    You are missing an x^2 in #3.
     
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