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Homework Statement
Given the following wave function valid over [itex]-a \le x \le a[/itex] and which is 0 elsewhere,
[tex]\psi(x) = 1/\sqrt{2a}[/tex]
Find the uncertainty in [itex]\left<\left(\Delta p\right)^2\right>[/itex] momentum, and the uncertainty product [itex]\left<\left(\Delta x\right)^2\right>\left<\left(\Delta p\right)^2\right>[/itex]
Homework Equations
Expectation value of an observable [itex]\underline{\alpha}[/itex]
[tex]\left<\alpha\right> = \int_{-\infty}^{\infty} \psi^*\alpha\psi\; dx[/tex]
Heaviside / unit step function
[tex]H(x) = \begin{cases}1, x\ge 0\\0, \mathrm{otherwise}\end{cases}[/tex]
Relationship between Heaviside and Dirac delta function
[tex]\delta(x) = \frac{dH(x)}{dx}[/tex]
Uncertainty of an observable [itex]\underline{\alpha}[/itex]
[tex]\begin{align*}
\left<\left(\Delta\alpha\right)^2\right> &= \left<\left(\alpha -
\left<\alpha\right>\right)^2\right>\\
&= \left<\alpha^2\right> - \left<\alpha\right>^2
\end{align*}[/tex]
The Attempt at a Solution
To find the uncertainty in momentum, which is just the difference between the expectation value of the square of momentum and the expectation value of the momentum, I need to find [itex]\left<p\right>[/itex] and [itex]\left<p^2\right>[/itex]
First, the wave function and its conjugate can be written in terms of Heaviside functions.
[tex]\begin{align*}\psi(x) &= \frac{1}{\sqrt{2a}}\left[H(x+a)-H(x-a)\right]\\\psi^*(x) &= \frac{1}{\sqrt{2a}}\left[H(x+a)-H(x-a)\right]\end{align*}[/tex]
Next, [itex]\left<p\right>[/itex]
[tex]\left<p\right> = \int_{-\infty}^{\infty} \psi^*(x)\left[\frac{\hbar}{i}\frac{d}{dx}\right]\psi(x)\; dx[/tex]
The derivative of [itex]\psi(x)[/itex] is
[tex]\begin{align*}\frac{d\psi(x)}{dx} &= \frac{d}{dx}\left(\frac{1}{\sqrt{2a}}\left[H(x+a)-H(x-a)\right]\right)\\
&= \frac{1}{\sqrt{2a}}\left[\delta(x+a)-\delta(x-a)\right]\end{align*}[/tex]
So, [itex]\left<p\right>[/itex] is
[tex]\begin{align*}\left<p\right> &= \int_{-\infty}^{\infty} \psi^*(x)\left[\frac{\hbar}{i}\frac{d}{dx}\right]\psi(x)\; dx\\
&= \int_{-\infty}^{\infty} \frac{1}{\sqrt{2a}}\left[H(x+a)-H(x-a)\right]\frac{\hbar}{i} \frac{1}{\sqrt{2a}}\left[\delta(x+a)-\delta(x-a)\right]\; dx\\
&= \frac{\hbar}{i2a}\int_{-\infty}^{\infty} \left[H(x+a)-H(x-a)\right]\left[\delta(x+a)-\delta(x-a)\right]\; dx\end{align*}[/tex]
Distributing out results (FOILing), in all but the [itex]H(x-a)\delta(x+a)[/itex] integrand component producing 1. The [itex]H(x-a)\delta(x+a)[/itex] part produces 0 in the course of integration. In other words,
[tex]\begin{align*}\left<p\right> &= \frac{\hbar}{i2a}(1 - 1 + 0 + 1)\end{align*} = \frac{\hbar}{i2a}[/tex]
First of all, is this reasonable for the momentum to have an expectation value that is imaginary?
Second, how would I compute [itex]\left<p^2\right>[/itex] given that the second derivative of the wave function, which is summation of Heavisides, would result in a derivative of a Dirac delta function?
Thanks very much everyone! :)