Problem 2:
The expectation value of p in eigenstate.
⟨ Ψ n | p | Ψ n ⟩ = − i m ℏ ω 2 ⟨ Ψ n | a ^ − a ^ † | Ψ n ⟩ {\displaystyle \langle \Psi _{n}|p|\Psi _{n}\rangle =-i{\sqrt {\frac {m\hbar \omega }{2}}}\langle \Psi _{n}|{\hat {a}}-{\hat {a}}^{\dagger }|\Psi _{n}\rangle }
== − i m ℏ ω 2 ( ⟨ Ψ n | a ^ Ψ n ⟩ − ⟨ Ψ n | a ^ † Ψ n ⟩ ) {\displaystyle ==-i{\sqrt {\frac {m\hbar \omega }{2}}}(\langle \Psi _{n}|{\hat {a}}\Psi _{n}\rangle -\langle \Psi _{n}|{\hat {a}}^{\dagger }\Psi _{n}\rangle )}
== − i m ℏ ω 2 ( n ⟨ Ψ n | Ψ n − 1 ⟩ − n + 1 ⟨ Ψ n | Ψ n + 1 ⟩ ) {\displaystyle ==-i{\sqrt {\frac {m\hbar \omega }{2}}}({\sqrt {n}}\langle \Psi _{n}|\Psi _{n}-1\rangle -{\sqrt {n+1}}\langle \Psi _{n}|\Psi _{n}+1\rangle )}
= 0 {\displaystyle =0}
