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<x<a/2\\ -A & a/2<x<a \end{cases}=-A(2\theta(x-a/2)-1) \ \ \ \ \ (5)

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.

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