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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?

Um, 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.

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

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.

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?

Um, 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.

The expectation for any pure state (any state that can be characterized by a vector in Hilbert space) is constant (as explained in other answers and explained again below).

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

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?

Um, 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.

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

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?

Um, 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.

What did you do wrong? You forgot about the action of $a^\dagger$.

Expand out your $\Psi(0)$ or $\Psi(t)$ and use the fact that: $$ a^\dagger a|n\rangle = n|n\rangle $$

But why?

Because the energy is a constant of the motion. Just like in classical mechanics. The expectation for any pure state (state that can be characterized by a vector in hilbert space) is constant (as explained in other answers here).

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?

Um, 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. 

The expectation for any pure state (any state that can be characterized by a vector in Hilbert space) is constant (as explained in other answers and explained again below).

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