# Confusion for Part of Bogolyubov Transform

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.

• 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? Commented Apr 23, 2021 at 20:37

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

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

• Thank you! Am I correct that there is a typo on (1), and it should be omega prime? Commented Apr 23, 2021 at 22:07
• @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. Commented Apr 23, 2021 at 22:08
• 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. Commented Apr 23, 2021 at 22:10
• @epjmm15 Yes, you're correct. Thanks for pointing that out! Great, so actually the result agrees with the correct one? Commented Apr 23, 2021 at 22:11
• Yes. I just saw your comment above, and fixed the question. Commented Apr 23, 2021 at 22:11