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Finding energy of a superposition of wavefunctions

  1. May 19, 2013 #1
    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

    [itex]ψ(x,0) = C[ψ_{1}(x) + ψ_{2}(x)][/itex]

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

    2. Relevant equations

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

    [itex]Eψ(x) = -\frac{\hbar^{2}∂^{2}}{2m∂x^{2}}[/itex]

    3. The attempt at a solution

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

    Then using the Schroedinger equation

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

    [itex]=\frac{\hbar^{2}}{2m}\sqrt{\frac{1}{L}}\frac{\pi ^{2}}{L^{2}}(sin(\frac{\pi x}{L}) + 4sin(\frac{2\pi x}{L}))[/itex]

    But now I'm stuck because of the 4 in front of [itex]4sin(\frac{2\pi x}{L})[/itex]
     
  2. jcsd
  3. May 19, 2013 #2

    TSny

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    Homework Helper
    Gold Member

    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.
     
  4. May 19, 2013 #3
    Ah, I see. I'll give that a try. Thanks :)
     
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