Exploring Quantum Physics - Week 3 Question 5

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)