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Inverse of linear momentum?

  • Thread starter ghazal-sh
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  • #1
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hi every body!
I'm looking for inverse of momentum operator in one dimensional problem.I have no idea to solve it!!please help me!
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so sorry about my bad speaking!

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Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

hi every body!
I'm looking for inverse of momentum operator in one dimensional problem.I have no idea to solve it!!please help me!
hi ghazal-sh! Welcome to PF! :smile:

Thankyou for the PM.

Can you show us the whole problem? :smile:
 
  • #3
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Can you show us the whole problem?
thanks for your attention!::smile:
the problem is just finding inverse of P!
 
  • #4
gabbagabbahey
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Start by considering the action of the momentum operator [itex]P[/itex] on an arbitrary wavefunction [itex]\psi(x)[/itex]...What is that?
 
  • #5
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Start by considering the action of the momentum operator LaTeX Code: P on an arbitrary wavefunction LaTeX Code: \\psi(x) ...What is that?
thanks!how can you reach to 1/p with this approach?
of course I found the answer.start by calculating expectation value of 1/p in momentum space (so easy)and then use Fourier transform of the wave functions.after so simple calculation you can see that:1/p =integral of dx
 
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  • #6
gabbagabbahey
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thanks!how can you reach to 1/p with this approach?
of course I found the answer.start by calculating expectation value of 1/p in momentum space (so easy)and then use Fourier transform of the wave functions.after so simple calculation you can see that:1/p =integral of dx
Well, the action of [itex]P[/itex] on [itex]\psi(x)[/itex] is of course just [itex]P\psi(x)=-i\hbar \frac{d}{dx} \psi(x)[/itex]...what do you get when you multiply both sides of this equation by the inverse of P (from the left), [itex]P^{-1}[/itex]?... compare that to the fundamental theorem of calculus and it should be apparent what [itex]P^{-1}[/itex] is.

P.S. using 1/p to represent the inverse is usually bad notation when dealing with operators.
 

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