- #1
Dazed&Confused
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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.
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.
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