# Finding the average energy from the superposition of state?

If I have two energy eigenstates $\psi_1(x)$ and $\psi_2(x)$ (corresponding to energy $E_1$ and $E_2$ respectively) and we prepare a particle in the superposition of both such that it is described by the state: $$\psi(x) =\frac{1}{\sqrt{2}}(\psi_1(x)+\psi_2(x))$$ at $t=0$ how would I find the average energy (or momentum say) as a function of time?

• Each eigenstate will evolve in time differently, complex exponential with frequency corresponding to the energy of each state – danimal May 13 '15 at 13:00

A slight expansion on danimal's comment: you can generally get the state $\psi(x,t)$ from the $\psi(x,0)$ you provided by operating on it with the unitary time evolution operator $\exp(-i \hat{H} t/\hbar)$. Since you know the eigenstates, you can write the Hamiltonian in a diagonal basis and this operator will appear to multiply $\psi_n$ simply by $e^{-i E_n t/\hbar}$.
To find the expectation value of any Hermitian observable $\hat{A}$ corresponding to your state at time $t$, you can simply compute $A(x,t) = \langle \psi(x,t) \vert \hat{A} \vert \psi(x,t)\rangle$. To find the time average, you could simply treat this expectation value as a classical function and average it over an interval, say by integration.
• Am I correct in saying that $\langle\psi_1 |\hat A|\psi_2 \rangle$ is 0? If this is the case I am totally confused of where time dependence comes from! If not please can you explain what it is. assuming that $\hat A$ is the energy operator – Quantum spaghettification May 14 '15 at 14:14
• If $\hat{A}$ is the energy operator, then what you wrote down is indeed zero, since it's diagonal in the basis of energy eigenstates. When you apply $\exp(-iHt/\hbar)$ to the initial eigenstate, the $\psi_1$ component picks up a complex phase $\exp(-iE_1 t/\hbar)$ and $\psi_2$ picks up $\exp(-iE_2 t/\hbar)$. So the relative phase between them is different, so the wavefunction is actually different. You can convince yourself of where the time dependence comes from by solving the time-depenendent Schrodinger equation, $i\hbar \dot{\psi} = \hat{H}\psi$. – danielsmw May 14 '15 at 14:33