- #1
jacobrhcp
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Homework Statement
show that <x> in position space is the same as <p> in momentum space.
Homework Equations
[tex] \Phi(p,t)=\frac{1}{\sqrt{2\pi\hbar}}{{\int_{-\infty}}^\infty} e^{-ipx} \Psi(x,t) dx [/tex] (1
[tex] \Psi(x,t)=\frac{1}{\sqrt{2\pi\hbar}}{{\int_{-\infty}}^\infty} e^{ipx} \Phi(p,t) dp [/tex] (2
The Attempt at a Solution
[tex] <x> = {{\int_{-\infty}}^\infty} \Psi^{*}(x,t) x \Psi(x,t) dx = \frac{1}{2\pi\hbar}{{\int_{-\infty}}^\infty} [{{\int_{-\infty}}^\infty} e^{ipx} \Phi(p,t) dp]^{*} [{{\int_{-\infty}}^\infty} x e^{ipx} \Phi(p,t) dp] dx [/tex] (2
[tex] = \frac{1}{2\pi\hbar}{{\int_{-\infty}}^\infty} [{{\int_{-\infty}}^\infty} e^{ipx} \Phi(p,t) dp]^{*} [{{\int_{-\infty}}^\infty} i \hbar (\frac{\partial}{\partial p}e^{ipx}) \Phi(p,t) dp] dx = \frac{1}{2\pi\hbar}{{\int_{-\infty}}^\infty} [{{\int_{-\infty}}^\infty} e^{-ipx} \Phi^{*}(p,t) dp] [{{\int_{-\infty}}^\infty} i \hbar (\frac{\partial}{\partial p}e^{ipx}) \Phi(p,t) dp] dx [/tex]
...
= ... by (1 = [tex] {{\int_{-\infty}}^\infty} \Phi^{*}(p,t) i \hbar (\frac{\partial}{\partial p} \Phi(p,t)) dp = <p> [/tex] in momentum space.
I could use some help filling in the blanks. I was thinking integration by parts, which I tried on my paper sheet, but it didn't quite work out. I once was told that if you're stuck on an integral problem, try integration by parts or substitution of variables, but I couldn't make either one work.