It seems that expectation value of $H$ on coherent states is independent of time? But why?

Question

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?

Answer 1

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.

Answer 2

The Hamiltonian (assumed to be time independent) has trivial Heisenberg evolution: it is a constant of motion, exactly as in classical Hamiltonian mechanics. Therefore a fortiori the expectation value with respect to every pure (or also mixed) state is constant in time. However the state can change, and it generally happens, its shape in time. In fact the considered property is a property of the Hamiltonian, not of the state.

Answer 3

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?

Yes, it is correct.

By the way, if it's is correct...

The correctness of your expression does not have anything to do with why the expectation value of the Hamiltonian is independent of time.

But why?

Because the energy is a constant of the motion. Just like in classical mechanics. This is true when the Hamiltonian does not explicitly depend on time. Again, like in the classical case.

The expectation for any state is constant, e.g., a state characterized by an arbitrary state vector $|\chi\rangle$.

For any state vector $|\chi\rangle$ and any operator $A$ we have: $$ |\chi(t)\rangle = e^{-iHt}|\chi(0)\rangle $$ so that: $$ A(t) \equiv \langle \chi(t)|\hat A|\chi(t)\rangle $$ $$ =\langle \chi(0)|e^{iHt}\hat Ae^{-iHt}|\chi(0)\rangle\;. $$

In your case, the operator $\hat A$ is actually the Hamiltonian $\hat H$ and therefore it should be obvious that: $$ [\hat H, e^{-i\hat Ht}] = 0 $$ so we have: $$ H(t)=\langle \chi(0)|e^{iHt}\hat He^{-iHt}|\chi(0)\rangle\;. $$ $$ =\langle \chi(0)|e^{iHt}e^{-iHt}\hat H|\chi(0)\rangle $$ $$ =\langle \chi(0)|\hat H|\chi(0)\rangle = H(0) $$

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Update to generalize a bit:

The way that $H(t) = \langle \hat H(t)\rangle$ can fail to be constant is if $\hat H(t)$ explicitly depends on time.

For example, in the general case of a (potentially) mixed system and a (potentially) explicitly time-dependent Hamiltonian, we have: $$ \hat\rho(t) = \hat U(t,0)\hat\rho(0)\hat U^\dagger(t,0)\;, $$ where $\hat \rho$ is the density matrix and where $$ \frac{d\hat U(t,0)}{dt} = -i\hat H\hat U(t,0) $$

In this notation, we have: $$ H(t) = \langle \hat H(t)\rangle = Tr\left( \hat\rho(t)\hat H(t) \right) $$ and $$ \frac{d\langle\hat H(t)\rangle}{dt} = Tr\left( -i[\hat H(t),\hat \rho(t)]\hat H(t)+\hat\rho(t)\frac{\partial \hat H}{\partial t} \right)\;. $$ The commutator trace term is zero because of the cyclic property of the trace. Thus, in general: $$ \frac{d\langle\hat H(t)\rangle}{dt} =\langle \frac{\partial \hat H(t)}{\partial t}\rangle\;, $$ which is zero unless the Hamiltonian has explicit time dependence.