How Does Changing Coordinates Affect the Momentum Space Wavefunction?

[ehrenfest]

For a wavefunction given by

$$\psi(x,t) = \sum a_n u_n(x) exp(-i E_n T/\hbar)$$ how would you show that a change of coordinates x > x + d does not affect the momentum space wavefunction phi(x) by more than a phase change?
You get phi(x) by Fourier transforming psi.
So, I do not see why it would affect psi at all because you are moving the origin d to the left but you are integrating over all pace in the Fourier transform.

 

[malawi_glenn]

For a wavefunction given by

$$\psi(x,t) = \sum a_n u_n(x) exp(-i E_n T/\hbar)$$ how would you show that a change of coordinates x > x + d does not affect the momentum space wavefunction phi(x) by more than a phase change?
You get phi(x) by Fourier transforming psi.
So, I do not see why it would affect psi at all because you are moving the origin d to the left but you are integrating over all pace in the Fourier transform.

You mean we get phi(p) when doing Fourier transformation with respect to p. How is E related to p? Have you tried doing the mathematics or are you just trying to solve it by looking at it? =P

 

[olgranpappy]

So, I do not see why it would affect psi at all because you are moving the origin d to the left but you are integrating over all pace in the Fourier transform.

What you are saying exactly is not clear. Do you want to show that for two functions (of 'x') f and g and their Fourier transforms (functions of 'p') F and G, if f and g obey

$$g(x)=f(x+a)$$

then F and G obey

$$G(p)=e^{ipa}F(p)\;.$$

Is this what you want to show?

 

[ehrenfest]

What you are saying exactly is not clear. Do you want to show that for two functions (of 'x') f and g and their Fourier transforms (functions of 'p') F and G, if f and g obey

$$g(x)=f(x+a)$$

then F and G obey

$$G(p)=e^{ipa}F(p)\;.$$

Is this what you want to show?

Exactly! That F and G only differ by a phase factor (so their moduli squared are the same).
$$\phi(p,t) = \int\psi(x+d,t) e^{-i p x/ \hbar}dx = \int\psi(u,t) e^{-i p (u - d)/ \hbar}du =e^{i p d/ \hbar} \int\psi(u,t) e^{-i p u/ \hbar}du$$

I think that rephrasing of the question helped me finish it!

 

[olgranpappy]

I think that rephrasing of the question helped me finish it!

I'm glad. Good luck w/ the rest of your work.