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Ground State Energy for infinte potential well

  1. Jun 7, 2012 #1
    1. The problem statement, all variables and given/known data

    A particle of mass (m) moves in the one-dimensional potential

    V(x) = V0 0 ≤ x ≤ a
    = ∞ otherwise

    Wave function of the particle is ψ(x,t) = C sin ([itex]\frac{x\pi}{a}[/itex]exp[-iωt]
    Determine V0

    2. Relevant equations

    Schrödinger's Equations

    Time-independent

    Hψ(x) + V(x)ψ=i[itex]\hbar[/itex][itex]\frac{\partialψ}{\partial t}[/itex]

    3. The attempt at a solution

    I got extremely confused because usually when I run into these questions, the wave function is not a function of time.

    So I don't actually know "how" to solve this equation. Do I separate the variables in the wave function and try to solve for E?

    Do I use the time-dependent equation? If so how do I solve for E.

    No need to show the math, but please take me through every step.

    Thanks
     
  2. jcsd
  3. Jun 7, 2012 #2
    Oh and if someone can point me towards a video/website showing how to separate the variables in the wave function if I need to do so.
     
  4. Jun 7, 2012 #3
    This is an important concept to get your head around: the difference between the time-dependent and the time-independent Schrodinger Equation. The key insight here is that since you're dealing with eigenstates of the Hamiltonian, their time behavior is very simple.

    Specifically, since [itex]i\frac{d}{dt}\psi(x,t) = \hat{H}\psi(x,t)[/itex], and [itex]\hat{H}\psi(x,t) = E\psi(x,t)[/itex], we have [itex]i\frac{d}{dt}\psi(x, t) = E\psi(x, t)[/itex]. That means that the time derivative of the wave function is a constant--it depends neither on time nor position. Therefore, the phase of the wavefunction just oscillates at a constant rate, which is proportional to the energy. So if the wavefunction at [itex]t=0[/itex] is [itex]\psi_0(x)[/itex], the wavefunction at other times is [itex]\psi(x, t) = e^{-iEt}\psi_0(x)[/itex].

    Since we know the time behavior of the wavefunction can be easily deduced simply by finding the energy, there's really no point in solving the full differential equation with time and space. It's enough to simply solve the spatial equation [itex]\psi_0(x)[/itex] only, and then we can reconstruct the full equation off of that. In this case, you should be able to use the boundary conditions plus the time-independent SE to solve for [itex]\psi_0(x)[/itex], and then you should be able to find the energy [itex]E[/itex] off of that.
     
  5. Jun 8, 2012 #4
    Thanks for the quick reply. I'll try it out and see if I run into problems. I'm still trying to get my head around what you said.

    Thanks again!
     
  6. Jun 9, 2012 #5
    It worked out fine, and I "think" (very sure I do) I understand the concept.

    Thank you so much for your help!!!
     
  7. Jun 10, 2012 #6
    If Iv got this right, your saying that ψ0(x) = C sin (xπ/a)exp[it(E-ω)]
     
    Last edited: Jun 10, 2012
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