Showing that Hamiltonian expectation value is time independent I want to check that I am getting the concept right here, and my question is: if the expectation value of a Hamiltonian is the same whether you use the time dependent version or not. I thought I had it right initially -- maybe I did -- but I wanted to make sure I didn't go off the rails somewhere. 
We have a wave function: $\psi = \alpha \phi_1 + \beta \phi_2$ and normalized it's $\frac{1}{(\alpha^2 + \beta^2)}(\alpha \phi_1 + \beta \phi_2)$. 
The time evolution of that state will be $(\frac{1}{\alpha^2 + \beta^2})(\alpha \phi_1 e^{-i\omega_1 t} + \beta \phi_2 e^{-i\omega_2 t})$
And $\hat H \psi = i\hbar \frac{\partial}{\partial t}\psi(x,t)$ and $\langle \hat H \rangle = \int \psi \hat H \psi dx$
$\frac{\partial \psi}{\partial t} = \frac{1}{\alpha^2 + \beta^2} (-i\omega_1 \alpha \phi_1 e^{-i\omega_1 t}-i\omega_2 \beta e^{-i\omega_2 t}\phi_2)$
So the integral is $$\int \psi \hat H \psi dx = \int (\frac{1}{\alpha^2 + \beta^2})(\alpha \phi_1 e^{-i\omega_1 t} + \beta \phi_2 e^{-i\omega_2 t})i \hbar (-i\omega \alpha \phi_1 e^{-i\omega_1 t}-i\omega_2 \beta e^{-i\omega_2 t}\phi_2)dx$$
$$=-\hbar\frac{1}{\alpha^2 + \beta^2}\int(\alpha \phi_1 e^{-i\omega_1 t} + \beta \phi_2 e^{-i\omega_2 t}) (\alpha \omega_1 \phi_1 e^{-i\omega_1 t}-\beta \omega_2e^{-i\omega_2 t}\phi_2)dx$$
$$=-\hbar\frac{1}{\alpha^2 + \beta^2}\int (\alpha^2 \omega_1 \phi_1^2 e^{-2i\omega_1 t}-\beta^2 \omega_2 \phi_2^2e^{-2i\omega_2 t}\phi_2)dx$$
Using a 0 to L limit (we're doing a particle in a box here), and taking the sinusoidal form of $\phi_n$: 
$$=\frac{-2\hbar}{L (\alpha^2 + \beta^2)}\int^L_0 \alpha^2 \omega_1 e^{-2i\omega_1 t}\sin^2({\frac{\pi x}{L})}-\beta^2 \omega_2 e^{-2i\omega_2 t}\sin^2({\frac{2\pi x}{L})}dx$$
applying a trig identity 
$$=\frac{-2\hbar}{L (\alpha^2 + \beta^2)}\int^L_0 \alpha^2 \omega_1 e^{-2i\omega_1 t}\left( \frac{1}{2}-\frac{1}{2}\cos({\frac{2\pi x}{L})} \right)-\beta^2 \omega_2 e^{-2i\omega_2 t}\left( \frac{1}{2}-\frac{1}{2}\cos({\frac{4\pi x}{L})} \right)dx$$
and doing the integral 
$$\frac{-2\hbar}{L (\alpha^2 + \beta^2)}\left[ \frac{\alpha^2 \omega_1 e^{-2i\omega_1 t}x}{2}-\frac{\alpha^2 \omega_1 e^{-2i\omega_1 t}L}{2\pi}\sin{\frac{2\pi x}{L}}- \frac{\beta^2 \omega_2 e^{-2i\omega_2 t}}{2}-\frac{\beta^2 \omega_2 e^{-2i\omega_2 t}L}{4 \pi} \sin{\frac{4\pi x}{L}} \right]_0^L$$
$$=\frac{-\hbar \alpha^2 \omega_1 e^{-2i\omega_1 t}}{(\alpha^2 + \beta^2)}+ \frac{\hbar \beta^2 \omega_2 e^{-2i\omega_2 t}}{L(\alpha^2 + \beta^2)} $$
THis is all very well, but it still looks like it depends on time, unless it's because of one of a couple of things: 1. The difference between the exponentials is a phase difference, so they might be equivalent, or since these are stationary states we're talking about we can treat them as constants. But I wanted to make sure there wasn't some mathematical point I wasn't missing. I feel like I am almost there but not quite. 
I also suspect I didn't need to do the full integration but I am not so expert with Dirac notation. I also kind of wanted to see what was "under the hood" so to speak. 
and sorry for the long post. 
 A: The expectation value is actually
$$
\langle\hat H\rangle = \int \psi^* \hat H\psi{\text d}x
$$
where $\psi^*$ denotes the complex conjugate. This will only affect the exponential here, and $\exp(ix)^*\exp(ix) = \exp(-ix)\exp(ix) = \exp(ix-ix) = 1$. A tiny bit more algebra and you're done.
I think you also dropped a factor of $L$ in the denominator of one of your terms. Also, I don't understand where the factor of $2$ comes in in the top. Shouldn't the time derivative be
$$
\frac{\partial\psi}{\partial t}
= -\frac i{\alpha^2 + \beta^2}
\left(\alpha\omega_1\phi_1 e^{-i\omega_1 t} + \alpha\omega_2\phi_2 e^{-i\omega_2 t}\right)
$$
You also dropped factors of $\alpha$ and $\beta$, and introduced a negative sign when simplifying the integrand.
The following bits of intuition ought to help:


*

*$\alpha$ and $\beta$ are amplitudes. They only make sense in the "quantum" world. When you take the expectation value, you're looking at a "classical" value, and they won't appear except as probabilities, when they're squared. (I'm think they never appear except when squared, but I'm not certain.)

*The hamiltonian is energy. It's positive, unless you have a good excuse for it not being positive (like if you're examining tunnelling.)

*The hamiltonian is energy. Check your units. The second term doesn't have the right ones.

*Physics is simple. Complexity generally doesn't appear out of nowhere. If an answer you get is complicated, you should see one or more clear reasons why it should be. In this case, the answer you got was asymmetric (one term has a factor of $\frac 2 L$, the other didn't). Looking at the original problem, it's not clear why this should be the case, so the $\frac 2 L$ naturally falls under suspicion.

A: In addition to bytbox's answer, here is an explanation using matrix notation. The state $\psi$ in the $\{\phi_1,\phi_2\}$ basis is given by the vector
$$\boldsymbol{\psi}=\begin{pmatrix}\alpha\exp\left(\frac{E_1t}{i\hbar}\right)\\\beta\exp\left(\frac{E_2t}{i\hbar}\right)\end{pmatrix}$$
where WLOG $|\alpha|^2+|\beta|^2=1$. In the $\{\phi_1,\phi_2\}$ basis $H$ admits the matrix representation
$$\mathbf{H}=\begin{pmatrix}
E_1&0\\
0&E_2
\end{pmatrix}
$$
and thus the problem becomes a simple matrix multiplication,
$$\frac{d}{dt}\langle E\rangle=\frac{d}{dt}\left(\boldsymbol{\psi}^\dagger\mathbf{H}\boldsymbol{\psi}\right)=\frac{d}{dt}\left(|\alpha|^2E_1+|\beta|^2E_2\right)=0.$$
