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## Homework Statement

The (normalized) wavefunction of the particle is defined as:

[tex]\psi_n = \sqrt{\frac{2}{L}} \sin \left ({\frac{\pi n}{L} x} \right )[/tex] for [tex]0 \leqslant x \leqslant L[/tex] and [tex]\psi_n = 0[/tex] otherwise, where L is the length of the one-dimensional box and n is the particle's energy level (so n is a nonzero positive integer).

1. Find the expectation values for the particle's position.

2. Find the expectation values for the particle's momentum.

## Homework Equations

Quantum operators on the wavefunction.

Calculus.

## The Attempt at a Solution

1. We integrate over the wavefunction squared using the position operator (which is just x) to obtain the expectation values for the particle's position:

\begin{align*}\int_0^L | \psi_n |^2 x ~~ dx &= \int_0^L x \frac{2}{L} \sin^2 \left ({\frac{\pi n}{L} x} \right ) ~~ dx \\

&= \frac{2}{L} \int_0^L \frac{x}{2} \left (1 - \cos \left ({\frac{2 \pi n}{L} x} \right ) \right ) ~~ dx = \frac{1}{L} \int_0^L x - x\cos \left ({\frac{2 \pi n}{L} x} \right ) ~~ dx \\

&= \frac{1}{L} \left [ \frac{x^2}{2} \right ]_{0}^{L} - \frac{1}{L} \left [ x\sin \left ({\frac{2 \pi n}{L} x} \right ) + \cos \left ({\frac{2 \pi n}{L} x} \right ) \right ]_0^L ~~ dx = \frac{L}{2}\end{align*} So the expectation value for position is always the middle of the box for any energy level.

2. We use the momentum operator which is -ih d/dx, so:

[tex]\frac{d}{dx} = \sqrt{2} \pi \left ( \frac{1}{L} \right )^\frac{3}{2} n \cos{ \left ( \frac{\pi n}{L} x \right ) }[/tex]

\begin{align*}-ih \int_0^L | \psi_n |^2 \frac{d}{dx} ~~ dx &= -ih \int_0^L \frac{2}{L} \sin^2 \left ({\frac{\pi n}{L} x} \right ) \sqrt{2} \pi \left ( \frac{1}{L} \right )^\frac{3}{2} n \cos{ \left ( \frac{\pi n}{L} x \right ) } ~~ dx \\

&= -ih \frac{2}{L} \sqrt{2} \pi \left ( \frac{1}{L} \right )^\frac{3}{2} n \int_0^L \sin^2 \left ({\frac{\pi n}{L} x} \right ) \cos{ \left ( \frac{\pi n}{L} x \right ) } ~~ dx \\

&= c \int_0^L \sin^2 \left ({\frac{\pi n}{L} x} \right ) \cos{ \left ( \frac{\pi n}{L} x \right ) } ~~ dx

\end{align*} where [itex]c = -ih \frac{2}{L} \sqrt{2} \pi \left ( \frac{1}{L} \right )^\frac{3}{2} n[/itex]

Using substitution, we obtain:

[tex]c \left [ \frac{\sin^3{ \frac{\pi n}{L} x }}{3 \frac{\pi n}{L}} \right ]_0^L = 0[/tex]So the momentum is zero for any energy level.

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Is the reasoning and maths correct? And does anyone have any tips to simplify the integrals, they are incredibly messy I find - would it help to just always bunch up constants into a single letter like I did for 2) and are there dangers in doing this?

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