Let's say the particle is in the state $| \psi(0) \rangle = \exp(-i\alpha p/\hbar) |0 \rangle$, where $p$ is the momentum operator.
I have to show that $| \psi(0) \rangle$ is a coherent state and to compute the expectation value of $H = \hbar \omega (n+1/2)$.
The first question is simple and I obtain
$$| \psi(0) \rangle = \exp(-|\alpha|^2/2) \sum_{n=0}^{+\infty} \frac{\alpha^n}{\sqrt{n!}}| n \rangle$$
and, of course, it's easy to verify that:
$$a| \psi (0) \rangle = \alpha |\psi(0) \rangle$$
and this verify that $| \psi(0) \rangle$ is an eigenket of the annihilation operator, by definition of coherent state.
I don't understand entirely, instead, the physical mean of the second question.
Firstly, we have that $$| \psi(t) \rangle = \exp(-|\alpha|^2/2) \sum_{n=0}^{+\infty} \frac{\alpha^n}{\sqrt{n!}}\exp(-i\omega(n+1/2)t)| n \rangle$$
Noting that $a| \psi(t) \rangle = \alpha\cdot \exp(-i\omega t)|\psi(t) \rangle$ for compute the expectation value of hamiltonian we have $$\langle \psi(t)|H| \psi(t) \rangle = \langle \psi(t) | \hbar\omega(a^{\dagger}a+1/2)|\psi(t) \rangle = \hbar \omega(|\alpha|^2+1/2).$$
Is this correct? By the way, if it's is correct, it means that the expectation value of $H$ in a coherent state is independent of time. But why?
The coherent states does not lose their shape on time and I think: is this the answer? Because it remain the same over time?
Best Answer
As the Hamilton operator $H$ and the time-evolution operator $U(t)=e^{-iHt/\hbar}$ commute, the expectation value of $H$ in any state $|\psi(t)\rangle = U(t) | \psi(0)\rangle$ is independent of $t$. This has nothing to with $|\psi(0)\rangle$ being a coherent state or not.