# Momentum operator on positon/momentum representation

• I
• dyn
In summary: So when you shift back to momentum space, you're still dealing with a function that's dominated by the free particle wavefunction.

#### dyn

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

dyn said:
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##.
dyn said:
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 ?

dyn said:
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 ?

dyn said:
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 ?

dyn said:
ħk is the momentum eigenvalue for a free particle and eikx is the momentum eigenfunction for a free particle ?
Yes.
dyn said:
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.

dyn said:
I don't understand why free particle momentum eigenvalues and eigenfunctions can be applied to a harmonic oscillator.
The equation you found is just a mathematical manipulation to show that
$$\langle x |\hat{p}| 0\rangle = -i\hbar \frac{\partial}{\partial x} \langle x | 0\rangle$$
and in doing so it turns out that you end up using the wavefunction for a free particle. Anyway, a wavefunction written in position or momentum basis can never get away completely from the free particle wavefunction. If you remember, in shifting from momentum to position space or vice versa, you have to Fourier transform the wavefunction and the complex exponential function of the form ##\exp(ikx)## is inherent in the Fourier transform formulas.

dyn

## 1. What is the momentum operator in position/momentum representation?

The momentum operator in position/momentum representation is a mathematical operator that describes the momentum of a particle in quantum mechanics. It is represented by the symbol p and is defined as the derivative of the position operator with respect to time.

## 2. How is the momentum operator represented in position/momentum representation?

In position/momentum representation, the momentum operator is represented by the symbol p, which acts on the wave function of a particle to calculate its momentum.

## 3. What is the relationship between the momentum and position operators in position/momentum representation?

In position/momentum representation, the momentum and position operators are related by the Heisenberg uncertainty principle, which states that the more precisely the position of a particle is known, the less precisely its momentum can be known, and vice versa.

## 4. How does the momentum operator behave in position/momentum representation?

The momentum operator in position/momentum representation behaves as a differential operator, meaning it operates on the wave function of a particle to determine its momentum. It is also a Hermitian operator, meaning its eigenvalues are real and its eigenvectors are orthogonal.

## 5. How is the momentum operator used in position/momentum representation?

The momentum operator in position/momentum representation is used to calculate the average momentum of a particle in a given system, as well as to determine the uncertainty in the measurement of momentum. It is also a crucial component in the Schrödinger equation, which describes the time evolution of a quantum system.