Driven Quantum harmonic oscillator by way of the S-Matrix

[Henriamaa]

Homework Statement

We have the lagragian L = \frac{m}{2} \dot{x}^2 - \frac{m \omega x^2}{2} + f(t) x(t)
where f(t) = f_0 for 0 \le t \le T 0 otherwise. The only diagram that survives in the s -matrix expansion when calculating <0|S|0> is D = \int dt dt&#039; f(t)f(t&#039;) &lt;0|T x(t)x(t&#039;)|0&gt;. (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 D = \int d\nu \frac{-1}{4m \pi} f(-\nu) \frac{i}{\nu^2 - \omega^2 + i \epsilon} f(\nu)

Homework Equations

The Attempt at a Solution

&lt;0| T x(t)x(t&#039;)|0&gt; = \frac{1}{2m\omega}e^{i(t-t&#039;)\omega}
where x(t) = \frac{1}{\sqrt{2m\omega}}(a e^{-i\omega t}+ a^{\dagger}e^{i\omega t} ). I got the result by applying wick's theorem. I assumed that t > t' and merely calculated &lt;0| [a e^{-i \omega t}, a^{\dagger}e^{i \omega t&#039;}] |0&gt; \frac{1}{2m \omega}. Now putting in all the Fourier expressions D = \frac{1}{(2 \pi)^2} \int dt dt&#039; d\nu d\nu&#039; f(\nu) e^{-i\nu t}\frac{1}{2m\omega}e^{i(t-t&#039;)\omega} e^{-i \nu&#039; t&#039; }f(\nu&#039;) . 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.

 

[fzero]

&lt;0| T x(t)x(t&#039;)|0&gt; = \frac{1}{2m\omega}e^{i(t-t&#039;)\omega}
where x(t) = \frac{1}{\sqrt{2m\omega}}(a e^{-i\omega t}+ a^{\dagger}e^{i\omega t} ). I got the result by applying wick's theorem. I assumed that t > t' and merely calculated &lt;0| [a e^{-i \omega t}, a^{\dagger}e^{i \omega t&#039;}] |0&gt; \frac{1}{2m \omega}.

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.

Now putting in all the Fourier expressions D = \frac{1}{(2 \pi)^2} \int dt dt&#039; d\nu d\nu&#039; f(\nu) e^{-i\nu t}\frac{1}{2m\omega}e^{i(t-t&#039;)\omega} e^{-i \nu&#039; t&#039; }f(\nu&#039;) . 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.

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.

 

[Henriamaa]

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.