# Momentum operator on positon/momentum representation

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Hi. I have come across the following step in a derivation of the harmonic oscillator groundstate wavefunction using ladder operators

∫ <x | p | p><p | o > dp = ∫ p<x | p><p | o > dp = -iħ d/dx ∫ <x | p><p | 0>dp

I am confused about how the -iħ d/dx arises. I thought the p produced when the p operator acts on |p> is an eigenvalue not an operator ? And it also seems as though the equation is in the momentum representation not the position one and -iħ d/dx is the momentum operator in the position representation ?
Thanks

blue_leaf77
Homework Helper
how the -iħ d/dx arises.
##p \langle x|p \rangle = \hbar k e^{ikx} = -i\hbar \partial_x e^{ikx} = -i\hbar \partial_x \langle x|p \rangle##.
it also seems as though the equation is in the momentum representation not the position one and -iħ d/dx is the momentum operator in the position representation ?
It's in position representation because you project the state onto a position basis ##\langle x|##.

Thanks for your reply but i'm still confused. I understand how to get to the p <x |p > but what is the p ? An eigenvalue or an operator ?

blue_leaf77
Homework Helper
An eigenvalue or an operator ?
Yes, an eigenvalue and hence not an operator.

Thanks again. So how does the eigenvalue p become the momentum operator on position space ?

blue_leaf77
Homework Helper
Thanks again. So how does the eigenvalue p become the momentum operator on position space ?
That's what I wrote in post #2.

Thanks. I thought ħk is the momentum eigenvalue for a free particle and eikx is the momentum eigenfunction for a free particle ? Why do these still apply to a harmonic oscillator ?

blue_leaf77
Homework Helper
ħk is the momentum eigenvalue for a free particle and eikx is the momentum eigenfunction for a free particle ?
Yes.
Why do these still apply to a harmonic oscillator ?
As the poster, you should be the one who knows in which occasion the author of the resource from which you found that equation did this derivation. I can only guess that probably he only wants to show that acting a momentum operator to a ground state of harmonic oscillator (basically can also be to any state) and then projecting the resulting state onto position basis translates into a calculus operation, namely the first derivative of the ground state wavefunction in position basis.

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I found the equation in some notes deriving the ground state wavefunction of the harmonic oscillator using ladder operators. I can see how the maths works but I don't understand why free particle momentum eigenvalues and eigenfunctions can be applied to a harmonic oscillator.

blue_leaf77
$$\langle x |\hat{p}| 0\rangle = -i\hbar \frac{\partial}{\partial x} \langle x | 0\rangle$$
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