We know that $[\hat{H}, \hat{P}]=i\hbar\frac{\mathrm{d}V}{\mathrm{d}x}$, therefore $\hat{H}$ and $\hat{P}$ commute if $\frac{\mathrm{d}V}{\mathrm{d}x}=0$, which is true for $V=0$. In that case the operators share the same eigenstates. As a matter of fact, both of the operators should return an eigenvalue if they operate on an eigenstate.
Now if we take a plane wave state $|\psi\rangle = Ae^{ikx}$, we see the above argument holds: \begin{align} \hat{P}\text{ operator gives:}& & -i\hbar \frac{\mathrm{d}}{\mathrm{d}x}|\psi\rangle &=\hbar k|\psi\rangle \\ \hat{H}\text{ operator gives:}& & -\frac{\hbar^2}{2m}\frac{\mathrm{d}^2}{\mathrm{d}x^2}|\psi\rangle &=\frac{\hbar^2k^2}{2m}|\psi\rangle \end{align}
But if we go to the particle in a box problem, then $|\psi\rangle =\sqrt{\frac{2}{a}}\sin(\frac{n\pi}{a}x)$, where $k=\frac{n\pi}{a}$, so \begin{align} \hat{P}\text{ operator gives:}& & -i\hbar \frac{\mathrm{d}}{\mathrm{d}x}|\psi\rangle &= -i\hbar\frac{n\pi}{a}\sqrt{\frac{2}{a}}\cos\left(\frac{n\pi}{a}x\right) \neq\hbar\frac{n\pi}{a}|\psi\rangle \\ \hat{H}\text{ operator gives:}& & -\frac{\hbar^2}{2m}\frac{\mathrm{d}^2}{\mathrm{d}x^2}|\psi\rangle &= -\frac{\hbar^2}{2m}\left(\frac{n\pi}{a}\right)\left(-\frac{n\pi}{a}\right) \sqrt{\frac{2}{a}} \sin\left(\frac{n\pi}{a}x\right) =\frac{\hbar^2k^2}{2m}|\psi\rangle \end{align}
So in a bound state, $\hat{H}$ returns an eigenvalue but $\hat{P}$ doesn't, although the commutation condition is satisfied. (For the harmonic oscillator potential, I am not surprised by a similar result as the commutation criteria are not satisfied in the first place). Why is this?
No comments:
Post a Comment