Position operator in momentum space (and vice-versa)

[smiler2505]

Hi all,

I understand how to transform between position space and momentum space; it's a Fourier transform:
\varphi|p>=\frac{1}sqrt{2\hbar\pi}\int_{\infty}^{\infty} <x|\varphi> exp(-ipx/\hbar)dx

But I can't figure out how to transform the operators. I know what they transform into (e.g., the p operator in position space goes to 'p' in momentum space), but not how.

Any help? Thanks

 

[A. Neumaier]

Hi all,

I understand how to transform between position space and momentum space; it's a Fourier transform:
\varphi|p>=\frac{1}sqrt{2\hbar\pi}\int_{\infty}^{\infty} <x|\varphi> exp(-ipx/\hbar)dx

But I can't figure out how to transform the operators. I know what they transform into (e.g., the p operator in position space goes to 'p' in momentum space), but not how.

Any help? Thanks

Apply the original definition of the momentum operator to the |p> just defined, and simplify, and you'll see that the effect is just multiplication with p. Similarly, you can verify the transformed formula for position.

 

[matonski]

Here is a detailed explanation of why p = -i\hbar \partial / \partial x in position space, starting from the commutators of x and p.