What is the Inverse of Momentum Operator in One Dimensional Problems?

[ghazal-sh]

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!

 

[tiny-tim]

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!

Thankyou for the PM.

Can you show us the whole problem?

 

[ghazal-sh]

Can you show us the whole problem?

thanks for your attention!:
the problem is just finding inverse of P!

 

[gabbagabbahey]

Start by considering the action of the momentum operator $P$ on an arbitrary wavefunction $\psi(x)$...What is that?

 

[ghazal-sh]

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

 

[gabbagabbahey]

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 $P$ on $\psi(x)$ is of course just $P\psi(x)=-i\hbar \frac{d}{dx} \psi(x)$...what do you get when you multiply both sides of this equation by the inverse of P (from the left), $P^{-1}$?... compare that to the fundamental theorem of calculus and it should be apparent what $P^{-1}$ is.

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