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Infinite Square Well - Particle in linear combination of states

  1. Oct 6, 2013 #1
    A particle of mass m is trapped in a one-dimensional infinite square well running from x= -L/2 to L/2. The particle is in a linear combination of its ground state and first excited state such that its expectation value of momentum takes on its largest possible value at t=0.

    I know the process of solving PDE's clear as day, thats not the issue. The problem is that I'm tripping myself out on how to write [itex]\Psi(x,t)[/itex] as a linear combination of its ground state + first excited state.

    My hunch is to approach the problem like this :


    where 0 and 1 represent the ground state and first state, respectively

    I'm confusing myself on what needs to fill in the equations! Any help would be appreciated.
    Last edited: Oct 6, 2013
  2. jcsd
  3. Oct 6, 2013 #2


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    The clever answer is that this problem is ill-posed, because strictly speaking a particle in an infinite-square well has no momentum observable, but I'm pretty sure that this is not what the professor wants to know ;-).

    So guess you are supposed to find [itex]c_0[/itex] and [itex]c_1[/itex] such that the naive "momentum"-expectation value
    [tex]\int_{-L/2}^{L/2} \mathrm{d} x \; \psi^*(x,t) \left (-\hbar \frac{\mathrm{d}}{\mathrm{d} x} \psi(x,t) \right )=\text{max},[/tex]
    where [itex]\psi[/itex] must be properly normalized, i.e.,
    [tex]\int_{-L/2}^{L/2} \mathrm{d} x \; |\psi(x,t)|^2=0.[/tex]
  4. Oct 6, 2013 #3
    thanks! that makes sense. By the way, the problem wants me to find the nomalized ψ(x,t) and the usual dynamical quantities (<x>, <p>, check m(d/dx)<x>=p, Δx, Δp, check ΔxΔp satisfies uncertainty principle)

    how do we write ψ(x,t) as a linear combination of its ground and first excited state like the problem asks??
    Last edited: Oct 6, 2013
  5. Oct 6, 2013 #4


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    Just like you did in the original post.
  6. Oct 7, 2013 #5
    I'm stuck on the fact that were given the phrase "expectation value of the momentum is a maximum when t=0."

    the energy of the second state is higher than that of the first, which implies the momentum of the second state which is more than that in the first (ps- I'm just talking about expectation values)

    Does anyone know what this means ?
  7. Oct 8, 2013 #6
    Still confused about that phrase, if anyone has ideas please let me know ! Thanks
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