Calculate the Fourier transform of the susceptibility of an Oscillator

  • #1
Lambda96
217
74
Homework Statement
Calculate the Fourier transform of ##\frac{4 \pi d_0^2}{\hbar} \sin(\omega_0(t-t'))##
Relevant Equations
none
Hi,

I'm not sure if I have Fourier transformed the expression correctly

Bildschirmfoto 2024-12-03 um 14.56.35.png

For the Fourier transformation, I used the following formula ##\int_{-\infty}^{\infty} f(t) e^{i \omega t}dt##

$$\frac{4 \pi d_0^2}{\hbar}\int_{-\infty}^{\infty} \sin(\omega_0(t-t')) e^{i \omega t}dt$$
$$\frac{4 \pi d_0^2}{\hbar}\int_{-\infty}^{\infty} \frac{1}{2i} \Bigl( e^{i \omega_0(t-t')} -e^{-i \omega_0(t-t')} \Bigr) e^{i \omega t}dt$$
$$\frac{4 \pi d_0^2}{\hbar} \frac{1}{2i} \Bigl(e^{-i \omega_0t'} \int_{-\infty}^{\infty} e^{it(\omega_0+\omega)} -e^{i \omega_0t'} \int_{-\infty}^{\infty} e^{it(\omega-\omega_0)} \Bigr) dt$$

To solve the integrals, I then applied the identity ##\int_{-\infty}^{\infty} e^{ik(x-x')} dk=2\pi \delta(x-x')## and obtained the following solution:

$$\frac{4 \pi d_0^2}{\hbar} \frac{1}{2i} \Bigl(e^{-i \omega_0t'} \delta(\omega_0 + \omega) -e^{i \omega_0t'} \delta(\omega - \omega_0) \Bigr)$$

Is that correct?
 
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  • #2
Lambda96 said:
Homework Statement: Calculate the Fourier transform of ##\frac{4 \pi d_0^2}{\hbar} \sin(\omega_0(t-t'))##
Relevant Equations: none

Hi,

I'm not sure if I have Fourier transformed the expression correctly

View attachment 354068
For the Fourier transformation, I used the following formula ##\int_{-\infty}^{\infty} f(t) e^{i \omega t}dt##

$$\frac{4 \pi d_0^2}{\hbar}\int_{-\infty}^{\infty} \sin(\omega_0(t-t')) e^{i \omega t}dt$$
$$\frac{4 \pi d_0^2}{\hbar}\int_{-\infty}^{\infty} \frac{1}{2i} \Bigl( e^{i \omega_0(t-t')} -e^{-i \omega_0(t-t')} \Bigr) e^{i \omega t}dt$$
$$\frac{4 \pi d_0^2}{\hbar} \frac{1}{2i} \Bigl(e^{-i \omega_0t'} \int_{-\infty}^{\infty} e^{it(\omega_0+\omega)} -e^{i \omega_0t'} \int_{-\infty}^{\infty} e^{it(\omega-\omega_0)} \Bigr) dt$$

To solve the integrals, I then applied the identity ##\int_{-\infty}^{\infty} e^{ik(x-x')} dk=2\pi \delta(x-x')## and obtained the following solution:

$$\frac{4 \pi d_0^2}{\hbar} \frac{1}{2i} \Bigl(e^{-i \omega_0t'} \delta(\omega_0 + \omega) -e^{i \omega_0t'} \delta(\omega - \omega_0) \Bigr)$$

Is that correct?
Looks fine to me.
 
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