Solving Fourier Transform of T(t) to Find b(w) and Plot |b(w)|^2

[v_pino]

Homework Statement

An atom raised at t=0 to an excited state with energy [itex] E_0= \hbar \omega_0 [itex] has the time dependence [itex] T(t)=\frac{1}{\sqrt \tau}e^{-t/ 2 \tau} [itex] for t>0 and T(t)=0 for t<0. Thus the probability of being in an excited state decays exponentially with time.

[itex] T(t)^2= \ frac {1} {\tau} e^ {-t/ \tau} [itex]

a.) Find the transform b(w) of T(t).

('w' is omega, frequency)

b.) Plot |b(w)|^2 as a function of w.

c.) show that b(w) traces a circle on the complex plane as w runs from well below w_0 to well above it.

Homework Equations

$$b(\omega)=\frac{1}{\sqrt{2 \pi}}\int{T(t)e^{i \omega t}}$$

The integral runs from -infinity to infinity.

The Attempt at a Solution

$$b(\omega)=-\frac{1}{\sqrt{2 \pi \tau}}\frac{1}{i(\omega - \omega _0)-\frac{1}{2 \tau}}$$

When I do plot for part b, I get exponential incerase. Does that sound right?

But when I do part (c), I don't get a circle nor elipse.

 

[anathema]

First, let's clarify the problem statement. We are given a time dependence T(t) and we are asked to find its transform b(w). The transform is defined as:

b(w)=\frac{1}{\sqrt{2 \pi}}\int{T(t)e^{i \omega t}} dt

So, in order to find b(w), we need to plug in the given T(t) into this integral and solve it. Let's do that:

b(w)=\frac{1}{\sqrt{2 \pi}}\int_{0}^{\infty}\frac{1}{\sqrt{\tau}}e^{-t/2\tau}e^{i \omega t} dt

Next, we can simplify this integral by using the property of exponential functions that e^a * e^b = e^(a+b). So, we can write the integral as:

b(w)=\frac{1}{\sqrt{2 \pi}}\int_{0}^{\infty}\frac{1}{\sqrt{\tau}}e^{-t/2\tau + i \omega t} dt

Now, we can use the exponential rule that e^(a+b) = e^a * e^b, to rewrite the integral as:

b(w)=\frac{1}{\sqrt{2 \pi}}\int_{0}^{\infty}\frac{1}{\sqrt{\tau}}e^{-(t/2\tau - i \omega t)} dt

Next, we can pull out the constant term (1/sqrt(tau)) from the integral and we are left with:

b(w)=\frac{1}{\sqrt{2 \pi \tau}}\int_{0}^{\infty}e^{-(t/2\tau - i \omega t)} dt

Now, we can use the rule that the integral of e^x is just e^x, to solve this integral:

b(w)=\frac{1}{\sqrt{2 \pi \tau}} \Big[e^{-(t/2\tau - i \omega t)}\Big]_{0}^{\infty}

Plugging in the upper and lower limits of the integral, we get:

b(w)=\frac{1}{\sqrt{2 \pi \tau}} \Big[0 - e^{-(0/2\tau - i \omega 0)}