# Infinite square well with triangular initial state using delta function

Required math: calculus

Required physics: Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.48.

Earlier, we looked at the case of a particle in the infinite square well with an initial wave function that was triangular:

$\displaystyle \Psi(x,0)=\begin{cases} Ax & 0\le x\le\frac{a}{2}\\ A(a-x) & \frac{a}{2}\le x\le a \end{cases} \ \ \ \ \ (1)$

We find ${A}$ from normalization:

$\displaystyle \int_{0}^{a}|\Psi|^{2}dx=A^{2}a^{3}/12=1 \ \ \ \ \ (2)$

so

$\displaystyle A=\frac{\sqrt{12}}{a^{3/2}} \ \ \ \ \ (3)$

Because the first derivative of ${\Psi(x,0)}$ is discontinuous at ${x=0}$, we might encounter problems in calculating the second derivative, which we need to find the mean value of the energy if we use integration, since this is

$\displaystyle \left\langle E\right\rangle =\frac{\hbar^{2}}{2m}\int_{0}^{a}\Psi^*(x,0)\frac{d^{2}}{dx^{2}}\Psi(x,0)dx \ \ \ \ \ (4)$

We can express the first derivative of wave function as a step function ${\theta(x)}$:

$\displaystyle \frac{d\Psi(x,0)}{dx}=\begin{cases} A & 0

where

$\displaystyle \theta(x)=\begin{cases} 1 & x<0\\ 0 & x>0 \end{cases} \ \ \ \ \ (6)$

We’ve seen that the derivative of the step function can be taken as the delta function, so

$\displaystyle \frac{d^{2}\Psi(x,0)}{dx^{2}}=-2A\delta(x-a/2) \ \ \ \ \ (7)$

Using the delta function directly, we get

$\displaystyle \left\langle H\right\rangle =-\frac{\hbar^{2}}{2m}\int_{0}^{a}\Psi^*\frac{d^{2}\Psi}{dx^{2}}dx=\frac{\hbar^{2}A^{2}}{m}\frac{a}{2}=\frac{6\hbar^{2}}{ma^{2}} \ \ \ \ \ (8)$

using ${A^{2}=12/a^{3}}$ from above. The final result is the same as that from summing the series as we did earlier.