3
$\begingroup$

Consider the operator (written in terms of a mode expansion) given by $$\hat P(u) = \int_0^\infty \frac{d\Omega}{\sqrt{2\pi}}\frac{1}{\sqrt{2\Omega}}\left(\hat b_\Omega e^{-i\Omega u} + \hat b_\Omega^\dagger e^{i\Omega u}\right)$$ I would like to compute the Fourier transform $$\int_{-\infty}^\infty \frac{du}{\sqrt{2\pi}}e^{i\Omega u}\hat P(u) = \frac{1}{\sqrt{2|\Omega|}}\begin{cases}\hat b_\Omega & \Omega > 0\\\hat b_{|\Omega|}^\dagger & \Omega < 0\end{cases}$$ I would like some help getting to that result. I anticipate it is a simple answer, but I do not directly see how to get there. I would greatly appreciate any help for getting to that result. I think it may be as simple as the Fourier Transform of an inverse Fourier Transform, but again, I do not clearly see how to get there.

$\endgroup$
1
  • $\begingroup$ I have been thinking about this for many days, however after posting this I think I thought of the right way to do this? Should I be rewriting P as an integral from - infinity to infinity and use Step functions to split it so that it is a Fourier transform? $\endgroup$
    – epjmm15
    Apr 23, 2021 at 20:37

1 Answer 1

1
$\begingroup$

Let us define $$K(\Omega) \equiv \frac{1}{2\pi} \int\limits_0^\infty \int\limits_{-\infty}^\infty \mathrm{d}\Omega^\prime\, \mathrm{d}u \, \frac{1}{\sqrt{2\Omega^\prime}}\, \left( \hat{b}_{\Omega^\prime} \,e^{-i\Omega^\prime u}+\hat{b}_{\Omega^\prime}^\dagger \,e^{i\Omega^\prime u}\right)\, e^{i\Omega u} \tag{1}$$

and recall that $$ \delta(x-x^\prime) \equiv \frac{1}{2 \pi} \int\limits_{-\infty}^\infty \mathrm{d}k\, e^{ik(x-x^\prime)} \quad . \tag{2}$$

Using equation $(2)$ in $(1)$ and performing the integration with respect to $u$ then yields

$$K(\Omega) = \int\limits_0^\infty \mathrm{d}\Omega^\prime\, \frac{1}{\sqrt{2 \Omega^\prime}}\, \left(\hat{b}_{\Omega^\prime} \, \delta(\Omega - \Omega^\prime)+ \hat{b}_{\Omega^\prime}^\dagger \, \delta(\Omega + \Omega^\prime)\right) \quad . \tag{3}$$

Now it is easy to see that for $\Omega >0$, only the first term in the integral of equation $(3)$ will contribute. Similarly, for $\Omega <0$ only the second term will contribute.

We eventually find

$$K(\Omega) = \frac{1}{\sqrt{2|\Omega|}}\,\begin{cases}\hat b_\Omega & \Omega > 0\\\hat b_{|\Omega|}^\dagger & \Omega < 0\end{cases} \quad . $$

$\endgroup$
6
  • 1
    $\begingroup$ Thank you! Am I correct that there is a typo on (1), and it should be omega prime? $\endgroup$
    – epjmm15
    Apr 23, 2021 at 22:07
  • $\begingroup$ @epjmm15 Do you mean in the exponential at the very right? No, its an $\Omega$. $K$ is a function of $\Omega$. But please also note my recent edit. My result differs from the one you gave. I obtained an $\hat b_{|\Omega|}^\dagger$ in the $\Omega <0$ case. $\endgroup$ Apr 23, 2021 at 22:08
  • $\begingroup$ No, I mean the omega in the square root of the first equation. Also, I found a typo in my question, and the coefficient $\hat b_\Omega^\dagger$ should have an absolute value around the Omega. So I believe you are correct. $\endgroup$
    – epjmm15
    Apr 23, 2021 at 22:10
  • $\begingroup$ @epjmm15 Yes, you're correct. Thanks for pointing that out! Great, so actually the result agrees with the correct one? $\endgroup$ Apr 23, 2021 at 22:11
  • $\begingroup$ Yes. I just saw your comment above, and fixed the question. $\endgroup$
    – epjmm15
    Apr 23, 2021 at 22:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.