- #1
Bill Foster
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Homework Statement
Evaluate the expectation value of p and p² using the momentum-space wave function
Homework Equations
Momentum-space wave function:
[tex]\sqrt{\frac{d}{\hbar\sqrt{\pi}}}e^{\frac{-\left(p'-\hbar k\right)^2d^2}{2\hbar^2}}[/tex]
The Attempt at a Solution
I can get [tex]\langle p \rangle [/tex], so that's not a problem.
[tex]\langle p^2 \rangle = \int_{-\infty}^{\infty}\sqrt{\frac{d}{\hbar\sqrt{\pi}}}e^{\frac{-\left(p'-\hbar k\right)^2d^2}{2\hbar^2}}p'^2\sqrt{\frac{d}{\hbar\sqrt{\pi}}}e^{\frac{-\left(p'-\hbar k\right)^2d^2}{2\hbar^2}}dp'[/tex]
[tex]= \frac{d}{\hbar\sqrt{\pi}}\int_{-\infty}^{\infty}e^{\frac{-\left(p'-\hbar k\right)^2d^2}{2\hbar^2}}p'^2e^{\frac{-\left(p'-\hbar k\right)^2d^2}{2\hbar^2}}dp'[/tex]
[tex]= \frac{d}{\hbar\sqrt{\pi}}\int_{-\infty}^{\infty}e^{\frac{-\left(p'-\hbar k\right)^2d^2}{\hbar^2}}p'^2dp'[/tex]
This has the form:
[tex]\int_{-\infty}^{\infty}e^{-a\left(x-b\right)^2}xdx = b\sqrt{\frac{\pi}{a}}[/tex]
So I'll integrate it by parts:
[tex]u=p'[/tex]
[tex]du=dp'[/tex]
[tex]dv=e^{\frac{-\left(p'-\hbar k\right)^2d^2}{\hbar^2}}p' dp'[/tex]
[tex]v=\hbar k\frac{\hbar}{d}\sqrt{\pi}[/tex]
[tex]p'\hbar k\frac{\hbar}{d}\sqrt{\pi}|_{-\infty}^{\infty}-\int_{-\infty}^{\infty}\hbar k\frac{\hbar}{d}\sqrt{\pi}dp'[/tex]
[tex]=p'\hbar k\frac{\hbar}{d}\sqrt{\pi}|_{-\infty}^{\infty}-p'\hbar k\frac{\hbar}{d}\sqrt{\pi}|_{-\infty}^{\infty}=0[/tex]
But it should be:
[tex]\langle p^2 \rangle = \frac{\hbar^2}{2d^2}+\left(\hbar k\right)^2[/tex]
What am I doing wrong?