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Driven Quantum harmonic oscillator by way of the S-Matrix

  1. Aug 23, 2015 #1
    1. The problem statement, all variables and given/known data
    We have the lagragian [tex] L = \frac{m}{2} \dot{x}^2 - \frac{m \omega x^2}{2} + f(t) x(t)[/tex]
    where [tex] f(t) = f_0 [/tex] for [tex] 0 \le t \le T [/tex] 0 otherwise. The only diagram that survives in the s -matrix expansion when calculating <0|S|0> is [tex] D = \int dt dt' f(t)f(t') <0|T x(t)x(t')|0> [/tex]. (This is easily seen by looking at the magnus expansion with interacting hamiltonian in the interaction picture being f(t)x(t)). Show that in frequency space [tex] D = \int d\nu \frac{-1}{4m \pi} f(-\nu) \frac{i}{\nu^2 - \omega^2 + i \epsilon} f(\nu) [/tex]
    2. Relevant equations


    3. The attempt at a solution
    [tex] <0| T x(t)x(t')|0> = \frac{1}{2m\omega}e^{i(t-t')\omega} [/tex]
    where [tex] x(t) = \frac{1}{\sqrt{2m\omega}}(a e^{-i\omega t}+ a^{\dagger}e^{i\omega t} )[/tex]. I got the result by applying wick's theorem. I assumed that t > t' and merely calculated [tex]<0| [a e^{-i \omega t}, a^{\dagger}e^{i \omega t'}] |0> \frac{1}{2m \omega}[/tex]. Now putting in all the fourier expressions [tex]D = \frac{1}{(2 \pi)^2} \int dt dt' d\nu d\nu' f(\nu) e^{-i\nu t}\frac{1}{2m\omega}e^{i(t-t')\omega} e^{-i \nu' t' }f(\nu') [/tex] . There is no need of going on because integrating out t and t' merely gives delta functions which will not give the result. The only place where I can see I have gone wrong is in calculating the VEV but then I do not know what I am doing wrong.
     
    Last edited: Aug 23, 2015
  2. jcsd
  3. Aug 24, 2015 #2

    fzero

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    Since we have a time ordered product, we also have to consider ##t<t'##, so you should find that
    $$\langle 0| T( x(t)x(t') )|0\rangle \propto e^{i\omega|t-t'|}.$$
    The absolute value has consequences below.

    When you include the absolute value, we have different expressions for ##t <t'## and ##t> t'##. One way to account for this is to split the integrals
    $$ \int dt dt' e^{i\omega|t-t'| - i \nu t -i \nu' t' } = \int_{-\infty}^\infty dt \left[ e^{i(\omega-\nu)t } \int_{-\infty}^t e^{-i(\omega +\nu' + i\epsilon) t'} + e^{-i(\omega+\nu)t }\int^{\infty}_t e^{i(\omega -\nu' + i\epsilon) t'}\right],$$
    where the ##i\epsilon##s have been added to make the ##t'## integrals converge. If you follow along the calculation, you should be able to reproduce the denominator from the suggested result.
     
  4. Aug 24, 2015 #3
    Thanks for the reply, the absolute value does accomplish the trick with the slight modification that I ignored the epsilons in the exponential when I did the integrals over t, otherwise I could not get the delta functions that I know I need.
     
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