Infinite square well – triangular initial state

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.7.

As another example of an explicit case of a particle in the infinite square well, we can consider a particle that starts off with a triangular wave function, of form

\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)

The general solution is a linear combination of the stationary states:

\displaystyle  \Psi(x,t)=\sum_{n=1}^{\infty}c_{n}\psi_{n}(x)e^{-iE_{n}t/\hbar} \ \ \ \ \ (4)

and at {t=0}, we get

\displaystyle  \Psi(x,0)=\sum_{n=1}^{\infty}c_{n}\psi_{n}(x) \ \ \ \ \ (5)

To find {\Psi(x,t)} we need the coefficients {c_{n}} which we get from the orthonormality of the stationary states:

\displaystyle   c_{n} \displaystyle  = \displaystyle  \sqrt{\frac{2}{a}}\int_{0}^{a}\sin\left(\frac{n\pi x}{a}\right)\Psi(x,0)dx \ \ \ \ \ (6)

Doing the integral (by hand or use Maple) gives:

\displaystyle  c_{n}=\frac{4\sqrt{6}}{n^{2}\pi^{2}}\sin\left(\frac{n\pi}{2}\right)=\begin{cases} 0 & n\;\mathrm{even}\\ \frac{4\sqrt{6}}{n^{2}\pi^{2}} & n=1,5,9,\ldots\\ -\frac{4\sqrt{6}}{n^{2}\pi^{2}} & n=3,7,11,\ldots \end{cases} \ \ \ \ \ (7)

The probability that the energy is {E_{1}} is {c_{1}^{2}=\frac{96}{\pi^{4}}=0.9855}. (Incidentally, since we know that {\sum_{n}\left|c_{n}\right|^{2}=1}, we get for free that {\sum_{n\; odd}\frac{1}{n^{4}}=\frac{\pi^{4}}{96}}.)

The average energy can be found by using the energy levels for the infinite square well: {E_{n}=(n\pi\hbar)^{2}/2ma^{2}} we have, using the formula {\sum_{n\; odd}\frac{1}{n^{2}}=\frac{\pi^{2}}{8}}

\displaystyle  \left\langle H\right\rangle =\sum_{n=0}^{\infty}|c_{n}|^{2}E_{n}=\frac{96}{\pi^{4}}\frac{\pi^{2}\hbar^{2}}{2ma^{2}}\sum_{n\text{ odd }}\frac{1}{n^{2}}=\frac{6\hbar^{2}}{ma^{2}}\approx1.216E_{1} \ \ \ \ \ (8)

3 thoughts on “Infinite square well – triangular initial state

  1. Pingback: Infinite square well with triangular initial state using delta function « Physics tutorials

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