= Solution
\b[Hamiltonian and <commutators>.] Use the <canonical commutation relation> $[\widehat x,\widehat p]=i\hbar I$ and define $c_0=\sqrt{m\omega/(2\hbar)}$. Expanding the product of the given <creation and annihilation operators>,
$$
\widehat a^\dagger\widehat a
=c_0^2\left(\widehat x^2+\frac{\widehat p^2}{m^2\omega^2}
+\frac{i}{m\omega}[\widehat x,\widehat p]\right)
=\frac{m\omega}{2\hbar}\widehat x^2
+\frac{\widehat p^2}{2m\hbar\omega}-\frac12I.
$$
Therefore the <quantum harmonic oscillator> <Hamiltonian operator> is
$$
\boxed{\widehat H=\frac{\widehat p^2}{2m}+\frac12m\omega^2\widehat x^2
=\hbar\omega\left(\widehat a^\dagger\widehat a+\frac12I\right).}
$$
Similarly,
$$
[\widehat a,\widehat a^\dagger]
=c_0^2\left(-\frac{i}{m\omega}[\widehat x,\widehat p]
+\frac{i}{m\omega}[\widehat p,\widehat x]\right)=I.
$$
Using $[AB,C]=A[B,C]+[A,C]B$ gives
$$
\boxed{[\widehat a,\widehat a^\dagger]=I,\qquad
[\widehat H,\widehat a]=\hbar\omega[\widehat a^\dagger\widehat a,\widehat a]
=-\hbar\omega\widehat a.}
$$
\b[Lowering a normalized <energy> <eigenstate>.] The <commutator> implies
$$
\widehat H\widehat a|\phi_n\rangle
=(E_n-\hbar\omega)\widehat a|\phi_n\rangle
=E_{n-1}\widehat a|\phi_n\rangle.
$$
For $n>0$, nondegeneracy of this one-dimensional <energy> level means $\widehat a|\phi_n\rangle=c_n|\phi_{n-1}\rangle$. Its norm is fixed by the <number operator>:
$$
|c_n|^2=\langle\phi_n|\widehat a^\dagger\widehat a|\phi_n\rangle
=\frac{E_n}{\hbar\omega}-\frac12=n.
$$
Choosing successive <eigenstate> phases in the conventional way makes $c_n=\sqrt n$, so
$$
\boxed{\widehat a|\phi_n\rangle=\sqrt n\,|\phi_{n-1}\rangle.}
$$
For $n=0$, the same norm is zero, hence $\widehat a|\phi_0\rangle=0$. A nonzero lowered <ground state> would have <energy> below the lowest allowed level, and positivity of $\widehat a^\dagger\widehat a$ forbids that possibility.
\b[Coherent-state <eigenvalue>.] Applying the <annihilation operator> term by term, with the $n=0$ contribution zero, gives
$$
\begin{aligned}
\widehat a|\chi_\alpha\rangle
&=e^{-|\alpha|^2/2}\sum_{n=1}^\infty
\frac{\alpha^n\sqrt n}{\sqrt{n!}}|\phi_{n-1}\rangle\\
&=\alpha e^{-|\alpha|^2/2}\sum_{r=0}^\infty
\frac{\alpha^r}{\sqrt{r!}}|\phi_r\rangle
=\alpha|\chi_\alpha\rangle.
\end{aligned}
$$
Thus
$$
\boxed{\widehat a|\chi_\alpha\rangle=\alpha|\chi_\alpha\rangle.}
$$
The <coherent state> is normalized because $e^{-|\alpha|^2}\sum_n|\alpha|^{2n}/n!=1$. Its <annihilation operator> is not Hermitian, so the <eigenvalue> $\alpha$ may be complex.
\b[Time evolution.] Under the time-independent <Hamiltonian operator>, each <energy> component acquires its own phase:
$$
\begin{aligned}
|\chi_\alpha;t\rangle
&=e^{-|\alpha|^2/2}\sum_{n=0}^\infty
\frac{\alpha^n}{\sqrt{n!}}e^{-i(n+1/2)\omega t}|\phi_n\rangle\\
&=e^{-i\omega t/2}e^{-|\alpha|^2/2}\sum_{n=0}^\infty
\frac{(\alpha e^{-i\omega t})^n}{\sqrt{n!}}|\phi_n\rangle.
\end{aligned}
$$
Since $|\alpha e^{-i\omega t}|=|\alpha|$, the normalization prefactor is unchanged. Hence
$$
\boxed{|\chi_\alpha;t\rangle=e^{-i\omega t/2}|\chi_\beta\rangle,
\qquad\beta=\alpha e^{-i\omega t}.}
$$
The state remains a <coherent state>, with its complex parameter rotating at the oscillator's <angular frequency>; the displayed extra factor is a global phase from the zero-point <energy>.
Back to article page
