Exploring Quantum Physics - Week 1 Question 4

Problem:

The ground state wavefunction for a particle in a shallow potential well in 1 dimension is of the form: $A e^{-\frac{|x|}{2d}}$. Given that the particle must be found somewhere in the range $x \in (-\infty, \infty)$, the born rule then places a constraint on the modulus of A. Assuming that A is real and positive, what is the value of A?

Solution:

The born rule requires the squared wavefunction is the probability density function of finding the particle, which means it is simply an integration problem.

$\begin{eqnarray*} \int\limits_{-\infty}^{\infty}{\psi^2(x)dx} &=& 1 \\ \int\limits_{-\infty}^{\infty}{(Ae^{-\frac{|x|}{2d}})^2dx} &=& 1 \\ A^2\int\limits_{-\infty}^{\infty}{e^{-\frac{|x|}{d}}dx} &=& 1 \\ 2A^2\int\limits_{0}^{\infty}{e^{-\frac{x}{d}}dx} &=& 1 \\ 2A^2(-de^{-\frac{x}{d}})|_{0}^{\infty} &=& 1 \\ 2A^2(0 - (-d)) &=& 1 \\ A^2 &=& \frac{1}{2d} \\ A &=& \frac{1}{\sqrt{2d}} \end{eqnarray*}$

That is the answer we wanted.