Question:
Recall that in the lecture 5 we found the Fourier transform of the wavefunction for a bound state of the delta function potential, V(x)=−\alpha \delta(x) , in 1 dimension. Our result was \tilde{\phi}_k=\frac{\alpha\psi(0)}{\frac{\hbar^2 k^2}{2m}−E} , where in the bound state, E = −\frac{m\alpha^2}{2\hbar^2} . What is the wavefunction of the bound state in real space, \psi(x) ? (Don't forget to normalize the wavefunction. You may take \psi(0) to be real and positive.)
Solution:
It is pretty obvious that the question is asking for the inverse Fourier transform of the function. Working backwards from the solution, I need to prove two lemmas about Fourier transform for myself, first, let look at this Fourier transform pair with a < 0 .
\begin{eqnarray*}
\mathcal{F}(e^{a|t|}) &=& F(\omega) \\
&=& \int\limits_{-\infty}^{\infty}{e^{a|t|}e^{-i\omega t}dt} \\
&=& \int\limits_{-\infty}^{0}{e^{-at}e^{-i\omega t}dt} + \int\limits_{0}^{\infty}{e^{at}e^{-i\omega t}dt} \\
&=& \int\limits_{-\infty}^{0}{e^{(-a-i\omega) t}dt} + \int\limits_{0}^{\infty}{e^{(a-i\omega) t}dt} \\
&=& \frac{e^{(-a-i\omega)t}}{-a-i\omega}|_{-\infty}^{0} + \frac{e^{(a-i\omega) t}}{a-i\omega}|_{0}^{+\infty} \\
&=& \frac{1}{-a-i\omega} + \frac{-1}{a-i\omega} \\
&=& \frac{-1}{a+i\omega} + \frac{-1}{a-i\omega} \\
&=& \frac{-1(a-i\omega)}{(a+i\omega)(a-i\omega)} + \frac{-1(a+i\omega)}{(a+i\omega)(a-i\omega)} \\
&=& \frac{-2a}{a^2+\omega^2}
\end{eqnarray*}
It does look like our form, but well, our k^2 comes with a coefficient, so how do we fix this? Let consider frequency scaling as follow:
\begin{eqnarray*}
F(\omega) &=& \mathcal{F}(f(t)) \\
&=& \int\limits_{-\infty}^{\infty}{f(t)e^{-i\omega t}dt} \\
F(a\omega) &=& \int\limits_{-\infty}^{\infty}{f(t)e^{-i a \omega t} dt} \\
\end{eqnarray*}
If a > 0 , we can do a substitution T = at , we have dT = a dt , and so we have
\begin{eqnarray*}
F(a\omega) &=& \int\limits_{-\infty}^{\infty}{f(t)e^{-i a \omega t} dt} \\
&=& \int\limits_{-\infty}^{\infty}{f(\frac{T}{a})e^{-i \omega T} \frac{1}{a}dT} \\
&=& \mathcal{F}(\frac{1}{a}f(\frac{T}{a}))
\end{eqnarray*}
Otherwise, we can still do the substitution, but the limit changes sign
\begin{eqnarray*}
F(a\omega) &=& \int\limits_{-\infty}^{\infty}{f(t)e^{-i a \omega t} dt} \\
&=& \int\limits_{\infty}^{-\infty}{f(\frac{T}{a})e^{-i \omega T} \frac{1}{a}dT} \\
&=& -\mathcal{F}(\frac{1}{a}f(\frac{T}{a}))
\end{eqnarray*}
So we can compactly represent this as
\begin{eqnarray*}
F(a\omega) &=& \mathcal{F}(\frac{1}{|a|}f(\frac{T}{a}))
\end{eqnarray*}
With the two lemmas, we can now proceed to the problem as follow:
\begin{eqnarray*}
F(e^{a|t|}) &=& \frac{-2a}{a^2+\omega^2} \\
F(\frac{1}{|b|}e^{a|\frac{t}{b}|}) &=& \frac{-2a}{a^2+(b\omega)^2} \\
\end{eqnarray*}
Since we will need to normalize anyway, we will simply ignore any constant factor.
By just pattern matching, we have got b = \frac{\hbar}{\sqrt{2m}} > 0 , a = -\sqrt{-E} . Plugging back in we can get:
\begin{eqnarray*}
F(e^{a|\frac{t}{b}|}) & \propto & \frac{-2a}{a^2+(b\omega)^2} \\
F(e^{-\sqrt{-E}\frac{|t|}{\frac{\hbar}{\sqrt{2m}}}}) & \propto & \frac{a\psi(0)}{E+\frac{\hbar^2\omega^2}{2m}}
\end{eqnarray*}
Let's see what is -\sqrt{-E}\frac{1}{\frac{\hbar}{\sqrt{2m}}} , putting back the solution -E = \frac{m\alpha^2}{2\hbar^2} , we get
\begin{eqnarray*}
-\sqrt{-E}\frac{1}{\frac{\hbar}{\sqrt{2m}}} &=& \sqrt{\frac{m\alpha^2}{2\hbar^2}}\frac{1}{\frac{\hbar}{\sqrt{2m}}} \\
&=& -\frac{m\alpha}{\hbar^2} \\
\end{eqnarray*}
We are almost there, now we know \psi(x) \propto e^{\frac{-m\alpha |x|}{\hbar^2}} , it remains to normalize it
\begin{eqnarray*}
\int\limits_{-\infty}^{\infty}{(e^{\frac{-m\alpha |x|}{\hbar^2}})^2 dt} &=& 2 \int\limits_{0}^{\infty}{(e^{\frac{-m\alpha x}{\hbar^2}})^2 dt} \\
&=& 2 \int\limits_{0}^{\infty}{e^{\frac{-2m\alpha x}{\hbar^2}} dt} \\
&=& 2 \frac{e^{\frac{-2m\alpha x}{\hbar^2}}}{\frac{-2m\alpha}{\hbar^2}}|_0^{\infty} \\
&=& \frac{\hbar^2}{m\alpha}
\end{eqnarray*}
Now it is obvious the solution is \frac{\sqrt{m\alpha}}{\hbar}e^{\frac{-m\alpha |x|}{\hbar^2}} .
Phew - now that's a lot of calculation - I have to confess, that I engineered this presentation towards the answer (because there are only 4 options) I knew it is right. I'd have made tons of mistakes without that (well you may want to take into account it is 1:48 am my time)
