Question

P7D.6 Consider a particle of mass m confined to a one-dimensional box of length L and in a state with normalized wavefunction y,. (a) Without evaluating any integrals, explain why(- L/2. (b) Without evaluating any integrals, explain why (p)-0. (c) Derive an expression for ) (the necessary integrals will be found in the Resource section). (d) For a particle in a box the energy is given by En =n2h2 /8rnf and, because the potential energy is zero, all of this energy is kinetic. Use this observation and, without evaluating any integrals, explain why 〈2)=n2h2 /4L2

Answer 1

Part A

The expected value for position x is:

$\left \langle x \right \rangle\equiv \int_{-\infty}^{\infty }\Psi ^{\ast }x\Psi dx=\frac{L}{2}$

This result shows that the expected value of x is correct in the center of the box, which is to be expected from the symmetry of the square of the wave function around the center. The probability density of finding the particle in L / 2 is maximum.

Part B

$\left \langle p_{x} \right \rangle=0$

Since the particle can go from left to right or from right to left with equal probability, the average is necessarily zero.

Part C

To solve the problem, we will use the following equation:

$\left \langle x^{2} \right \rangle=\int_{-\infty}^{\infty}x^{2}\Psi ^{2}(x)dx$

$\left \langle x^{2} \right \rangle=\int_{0}^{L}x^{2}sin^{2}(\frac{\pi x}{L})dx$

to solve the integral we made the following change of variable

$\theta =\frac{\pi x}{L}$

$\left \langle x^{2} \right \rangle=\frac{2}{L}(\frac{L}{\pi})^{3}\int_{0}^{\pi}\theta ^{2}sin^{2}\theta d\theta$

$\left \langle x^{2} \right \rangle=\frac{2L^{2}}{\pi^{3}}\int_{0}^{\pi}\theta ^{2}sin^{2}\theta d\theta$

Calculating the integral with the help of tables of integrals, you get:

$\int_{0}^{\pi}\theta ^{2}sin^{2}\theta d\theta =\left [ \frac{\theta ^{3}}{6}-(\frac{\theta ^{2}}{4}-\frac{1}{8})sin2\theta -\frac{\theta cos2\theta }{4} \right ]_{0}^{\pi}=\frac{\pi^{3}}{6}-\frac{\pi}{4}$

$\left \langle x^{2} \right \rangle=\frac{2L^{2}}{\pi^{3}}(\frac{\pi^{3}}{6}-\frac{\pi}{4})=(\frac{1}{3}-\frac{1}{2\pi^{2}})L^{2}$