Position operator in momentum space (and vice-versa)

  • Thread starter smiler2505
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  • #1
Hi all,

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

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
 

Answers and Replies

  • #2
A. Neumaier
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Hi all,

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

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.
 
  • #3
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Here is a detailed explanation of why [itex]p = -i\hbar \partial / \partial x[/itex] in position space, starting from the commutators of x and p.
 

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