# 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

## 1 Answer

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.

• Given my set up are we expecting a result that is in fact not a function of time? – Quantum spaghettification May 14 '15 at 9:51
• I assume that the wavefunction you provided is the initial wavefunction at time zero. As I mentioned, the wavefunction at time t is then given by operating on the initial wave function with exp(-iHt/h). So in general, it seems that the result should be time dependent. Say that the two eigenstates you provided are the only ones. Even if H isn't time dependent, the time evolution could change the coefficients in front of each eigenstate contribution, which would change the weighted average of the expected energy. – danielsmw May 14 '15 at 14:05
• 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
• Of course, there will be no change in the total energy with time (conservation of energy); sorry, I realized I might have confused you there. I just mean that the wavefunction will change with time, but the expectation value of the energy will stay constant since cross terms will cancel, as you pointed out in your comment. – danielsmw May 14 '15 at 14:36