Answer to Question #163897 in Quantum Mechanics for Nidhin S

Question

 Show that < 𝑥𝑝 > − < 𝑝𝑥 > = 𝑖ħ for the ground state wave function of the Quantum 

Harmonic Oscillator (QHO)

Answer 1

The QHO ground state wave function is given by $\psi(x) = (\frac{m\omega}{\pi \hbar})^{1/4} e^{-\frac{m \omega x^2}{2\hbar}}$. In coordinate representation, $\hat{x} = x, \hat{p} =- i\hbar \frac{\partial}{\partial x}$ and thus we find :

$<\hat x \hat p >_\psi = \int_{-\infty}^{+\infty} x\cdot (i\hbar\frac{m\omega x}{\hbar})\cdot \psi^2(x) dx$, applying the integration by parts ($u = x, dv = im\omega x \psi^2 dx$) we find $<\hat x \hat p >_\psi = [\frac{1}{2}x\psi^2]^{+\infty}_{-\infty} + \frac{1}{2}i\hbar \int \psi^2 dx =\frac{ i\hbar}{2}$ as $\psi$ is a normalised wave function.

Now we calculate $<\hat p \hat x>_\psi = -i\hbar \int (\psi^2+x\psi \psi')dx$. The integral $-i\hbar \int \psi^2dx =-i\hbar$ is given by the normalisation of $\psi$. The second integral was calculated previously. Thus we find $<\hat p \hat x>_\psi = -i\hbar /2$ .

Thus $<xp-px>_\psi = i\hbar$ which is coherent with the fact that $[x,p]=i\hbar \cdot Id$ and thus $<[x,p]>_\psi = i\hbar$ for any normalised wave function $\psi$.