 Solution

Writing $a=cx+idp$, where $c=\sqrt{m\omega /(2\hslash )}$ and $d=1/\sqrt{2m\omega \hslash }$, the canonical commutation relation givesConsequently $[a^{†}a,a]=-a$ and $\boxed{[H,a]=-\hslash \omega a}$. If $a|{\varphi }_0⟩$ were nonzero, this commutator would make it an energy eigenstate with energy $E_0-\hslash \omega$, contradicting the lowest-energy property. Therefore $a|{\varphi }_0⟩=0$.

The commutator product rule and induction giveActing on the vacuum and shifting the sum then shows that the unnormalized bosonic coherent state $|\tilde{\psi }⟩=e^{\alpha a^{†}}|0⟩$ satisfies $a|\tilde{\psi }⟩=\alpha |\tilde{\psi }⟩$.

The displayed series lacks its wavefunction normalization factor. Since $(a^{†}{)}^n|0⟩=\sqrt{n!}|n⟩$, its squared norm is ${\sum }_n|\alpha {|}^{2n}/n!=e^{|\alpha {|}^2}$. The normalized coherent state needed for the stated expectation values isUse $⟨a⟩=\alpha$, $⟨a^{†}⟩={\alpha }^{\ast }$, $⟨a^{†}a⟩=|\alpha {|}^2$ and $⟨a{a}^{†}⟩=|\alpha {|}^2+1$. For $x=x_0(a+a^{†})$, $x_0=\sqrt{\hslash /(2m\omega )}$,The displacement changes the mean position but not its ground-state uncertainty. Unnormalized bras and kets give the same physical expectation values only after division by their squared norm.