Are <x> and <p> the same in position and momentum space?

[jacobrhcp]

Homework Statement

show that <x> in position space is the same as <p> in momentum space.

Homework Equations

\Phi(p,t)=\frac{1}{\sqrt{2\pi\hbar}}{{\int_{-\infty}}^\infty} e^{-ipx} \Psi(x,t) dx (1
\Psi(x,t)=\frac{1}{\sqrt{2\pi\hbar}}{{\int_{-\infty}}^\infty} e^{ipx} \Phi(p,t) dp (2

The Attempt at a Solution

&lt;x&gt; = {{\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 (2

= \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

...

= ... by (1 = {{\int_{-\infty}}^\infty} \Phi^{*}(p,t) i \hbar (\frac{\partial}{\partial p} \Phi(p,t)) dp = &lt;p&gt; 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.

 

[kdv]

Homework Statement

show that <x> in position space is the same as <p> in momentum space.

Homework Equations

\Phi(p,t)=\frac{1}{\sqrt{2\pi\hbar}}{{\int_{-\infty}}^\infty} e^{-ipx} \Psi(x,t) dx (1
\Psi(x,t)=\frac{1}{\sqrt{2\pi\hbar}}{{\int_{-\infty}}^\infty} e^{ipx} \Phi(p,t) dp (2

The Attempt at a Solution

&lt;x&gt; = {{\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 (2

= \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

...

= ... by (1 = {{\int_{-\infty}}^\infty} \Phi^{*}(p,t) i \hbar (\frac{\partial}{\partial p} \Phi(p,t)) dp = &lt;p&gt; 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.

Your mistake is that when you write the spatial wavefunctions in terms of their momentum space expressions you use the same variable "p" in each case. You can't do that. Use two different variables for each Fourier transform (say p and p'). Then it will work out. (indeed you will have to do an integration by parts and then carry out the integral over x and then the integral over one of your momentum variables).

 

[jacobrhcp]

Thanks =) after you told me this, I tried it and it worked.

But why are they two different variables? If you tell me this, it will be completely solved.