Help with Fourier integral using contour

[ognik]

Find $ F=\frac{\hbar}{2\pi i} \int_{-\infty}^{\infty} \frac{e^{-i \omega t}}{E_0-\frac{i\Gamma}{2} -\hbar \omega} \,d\omega $ using contour integration. I have a couple questions I'd like some help with please...

Taking out the $\hbar$ in the denominator, which cancels with the $\hbar$ outside the integral, the denominator becomes $(a-\omega)$ where $a=\frac{1}{\hbar}(E_0-\frac{i\Gamma}{2}) $

Then $F = \frac{1}{2\pi i} \int_{-\infty}^{\infty} \frac{e^{-i \omega t}}{a-\omega} \,d\omega $ with $f(\omega) = \frac{e^{-i \omega t}}{a-\omega}=\frac{cos \omega t}{a-\omega}-i\frac{Sin \omega t}{a-\omega} $

Q1: so if $f(\omega)$ is analytic, a is a simple pole and I can use residues, but I'm not sure how to prove $f(\omega)$ is analytic, ie. how to write $\omega$ in terms of x and y, so I can use the Cauchy Riemann conditions?

Assuming analytic, I can use Cauchy's residue theorem (with a simple pole): $ \oint f(\omega) d\omega=2\pi i \: res(f, a)$, where $res(f, a)=\lim_{{\omega }\to{a}} (\omega - a)f(\omega) = e^{-iat}$

$ \therefore F= \frac{\hbar}{2\pi i} \int_{-\infty}^{\infty} \frac{e^{-i \omega t}}{E_0-\frac{i\Gamma}{2} -\hbar \omega} \,d\omega = e^{-it \left[ \frac{1}{\hbar}(E_0-\frac{i\Gamma}{2}) \right] }$ $= {e}^\left[ {\frac{-iE_{0 }t}{\hbar}} \right] {e}^\left[ {\frac{-t \Gamma}{2\hbar}}\right] $

but Q2: I am not sure about the contour. I assumed a semi-circle, but am not sure where the pole is? $f(\omega)$ looks like 4th quadrant? I also need to consider t>0 and t<0 ...

Q3: I think the contour would be $ \oint_C = \lim_{{R}\to{\infty}}\int_{-R}^{R} +\int_{2\pi}^{\pi} $, but for this function I have no idea how to show $\int_{2\pi}^{\pi} = 0$

 

[jvicens]

Hi there! It's great to see that you are using contour integration to solve this problem. To answer your questions:

Q1: To prove that $f(\omega)$ is analytic, you can show that it satisfies the Cauchy-Riemann conditions. The Cauchy-Riemann conditions state that a function $f(x+iy)$ is analytic if and only if it satisfies the following conditions:

$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$

where $u$ and $v$ are the real and imaginary parts of $f(x+iy)$, respectively. In this case, $f(\omega)$ can be written as $f(\omega) = u(\omega) + iv(\omega)$, where $u(\omega) = \frac{cos \omega t}{a-\omega}$ and $v(\omega) = -\frac{sin \omega t}{a-\omega}$. You can then calculate the partial derivatives of $u$ and $v$ with respect to $x$ and $y$ and show that they satisfy the Cauchy-Riemann conditions.

Q2: The pole of $f(\omega)$ is at $\omega = a$, which is given by $a=\frac{1}{\hbar}(E_0-\frac{i\Gamma}{2})$. To determine the contour, you can use a semi-circle in the lower half plane, since $f(\omega)$ has a pole in the lower half plane. This is because the integral along the semi-circle will tend to zero as the radius goes to infinity, as long as $t>0$.

Q3: Yes, the contour would be $\oint_C = \lim_{{R}\to{\infty}}\int_{-R}^{R} +\int_{C_R}$, where $C_R$ is the semi-circle in the lower half plane. To show that $\int_{C_R} = 0$, you can use the Cauchy integral formula, which states that for a function $f(z)$ that is analytic in a region containing a closed contour $C$, the integral around $C$ is equal to $2\pi i$ times the sum of the residues of $f(z)$ at its poles inside