Phy5645/HO problem1

First, we rewrite the position operator in terms of the raising and lowering operators:

${\hat {x}}={\sqrt {\frac {\hbar }{2m\omega }}}({\hat {a}}+{\hat {a}}^{\dagger })$

Now we determine the expectation value of the position for an arbitrary harmonic oscillator eigenstate $|n\rangle :$

$\langle n|{\hat {x}}|n\rangle ={\sqrt {\frac {\hbar }{2m\omega }}}\langle n|({\hat {a}}+{\hat {a}}^{\dagger })|n\rangle$

$={\sqrt {\frac {\hbar }{2m\omega }}}(\langle n|{\hat {a}}|n\rangle +\langle n|{\hat {a}}^{\dagger }|n\rangle )$

$={\sqrt {\frac {\hbar }{2m\omega }}}({\sqrt {n}}\langle n|n-1\rangle +{\sqrt {n+1}}\langle n|n+1\rangle )$

$=0\!$

We may also see this intuitively; we know that, because the potential is an even function of $x,\!$ the eigenfunctions must be either even or odd functions of $x.\!$ This means that the probability density is always an even function, meaning that the particle is just as likely to be found at $-x\!$ as it is to be found at $x.\!$

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