Finding energy of a superposition of wavefunctions

1. May 19, 2013

phosgene

1. The problem statement, all variables and given/known data

Consider a particle in an infinite square well described initially by a wave that is a superposition of the ground and first excited states of the well

$ψ(x,0) = C[ψ_{1}(x) + ψ_{2}(x)]$

Show that the superposition is not a stationary state, but that the average energy in this state is the arithmetic mean $(E_{1} + E_{2})/2$ of the ground and first excited state energies $E_{1}$ and $E_{2}$.

2. Relevant equations

Time-independant Schroedinger equation (potential set to zero):

$Eψ(x) = -\frac{\hbar^{2}∂^{2}}{2m∂x^{2}}$

3. The attempt at a solution

Assuming that the well is from x=0 to x=L, my wavefunction is $ψ(x) = \sqrt{\frac{1}{L}}(sin(\frac{\pi x}{L}) + sin(\frac{2\pi x}{L}))$

Then using the Schroedinger equation

$-\frac{\hbar^{2}∂^{2}}{2m∂x^{2}}\sqrt{\frac{1}{L}}(sin(\frac{\pi x}{L}) + sin(\frac{2\pi x}{L}))=Eψ(x)$

$=\frac{\hbar^{2}}{2m}\sqrt{\frac{1}{L}}\frac{\pi ^{2}}{L^{2}}(sin(\frac{\pi x}{L}) + 4sin(\frac{2\pi x}{L}))$

But now I'm stuck because of the 4 in front of $4sin(\frac{2\pi x}{L})$

2. May 19, 2013

TSny

Your wavefunction is a superposition of two states of different energies. So, your wavefunction does not have a definite energy and therefore will not itself be an eigenstate of the Hamiltonian (i.e., it's not a "stationary state"). So, your wavefunction will not satisfy the time-independent Schrodinger equation.

The average value of the energy for your wavefunction is the "expectation value" of the Hamiltonian for your wavefunction.

3. May 19, 2013

phosgene

Ah, I see. I'll give that a try. Thanks :)