Finding energy of a superposition of wavefunctions

[phosgene]

Homework Statement

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}.

Homework Equations

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

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

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})

 

[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.

 

[phosgene]

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