Phy5645/HO problem1

We can compute the expectation value of x simply.

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

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

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

$=0$

We should have seen that coming. Since each term in the operator changes the eigenstate, the dot product with the original (orthogonal) state must give zero.

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