# Quantum oscillator from position to momentum space

1. Jul 7, 2015

### Dazed&Confused

So I've read you can get the corresponding wave function of a quantum harmonic oscillator in momentum space from position space by making the substitution $x \to k$ and $m \omega \to 1/m \omega$.

However in deriving the TISE for momentum space, I seem to be making a mistake. In momentum space $P$ acts like $\hbar k$ and $X$ acts like $i d/dk.$

The TISE is $$\left ( \frac{P^2}{2m} + \frac12 m \omega^2 X^2 \right) \psi = \left ( \frac{\hbar^2 k^2}{2m} - \frac12 m \omega^2 \frac{d^2}{dk^2} \right ) \psi = E\psi.$$

Therefore $$\psi'' + \frac{2}{m \omega^2} \left ( E - \frac{\hbar^2 k^2}{2m} \right) = 0.$$

The equivalent equation in position space is $$\psi'' + \frac{2m}{\hbar^2} \left ( E -\frac12 m\omega^2x^2 \right ) \psi = 0.$$

I'm not sure what I'm doing wrong. Edit: incorrect operators, can be deleted.

Last edited: Jul 7, 2015
2. Jul 7, 2015

3. Jul 7, 2015

### Avodyne

It's not $x\to k$, it's $x\to p$. So replace $P$ with $p$ and $X$ with $i\hbar d/dp$, and see what you get.