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

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?

• Is the $\delta$ in your first line supposed to be an $\alpha$? Or maybe the alphas in your subsequent lines are supposed to be deltas?
– hft
Commented Nov 3, 2023 at 19:56
• I guess it must be, I made that edit
– hft
Commented Nov 3, 2023 at 20:17

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.

• Mathematically it's okay. But what is the physical meaning of this? if $|\psi \rangle$ is an eigenket of $H$ I see clear the physical meaning (is an eigenstate, of course, when you measure the state of particle you obtain those eigenstate). But coherent state is not an eigenstates of $H$. It is a superposition. And it evolve over time. Why it's expectation value of energy is not time dependent? Commented Nov 3, 2023 at 16:45
• @Damark The expectation value of the Hamiltonian is always time-independent in any state, because it represents the total energy and is hence conserved. This is the physical content of this answer. (If the Hamiltonian is time-dependent, this isn't true, though.) Commented Nov 3, 2023 at 16:49
• Oh, I'm very stupid. Sorry, and thank you Marh and Hyperon. (Should I cancel the question?) Commented Nov 3, 2023 at 16:59
• @Damark Never ever cancel a question! Someone else might find the question/answer useful. Commented Nov 3, 2023 at 17:02
• @Damark No, do not cancel such questions (for the reason others have stated already). However, consider to accept this answer. Commented Nov 3, 2023 at 17:23

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.

• I find it curious that the Italian indipendente swaps out the first e for an i when compared to the English independent. Commented Nov 3, 2023 at 20:00
• It is Latin essentially. Indipendes, indipendentis. The ablative case is indipendente and often, the ablative case passed to the modern Italian. Thanks for pointing out the typo generated by my mother tongue. Commented Nov 3, 2023 at 20:05
• I took a mistake, it is an adjective, therefore the ablative case is indipendenti differently from the nouns of the third declination. It does not matter I can use standard Italian in case I have to speak to the Pope! Commented Nov 3, 2023 at 20:29
• Hello Valter, thank you for your answer, very useful. Commented Nov 3, 2023 at 22:47

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: $$H(t)=\langle \chi(0)|e^{iHt}\hat He^{-iHt}|\chi(0)\rangle\;.$$ $$=\langle \chi(0)|e^{iHt}e^{-iHt}\hat H|\chi(0)\rangle$$ $$=\langle \chi(0)|\hat H|\chi(0)\rangle = H(0)$$

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.

• Thank you hft, very clarifying answer. Commented Nov 3, 2023 at 22:48