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Particle in one-dimensional box

  1. Nov 12, 2008 #1
    By separation of variables, I have found that


    [tex]\frac{-\hbar}{2mg(x)}\frac{d^{2}g(x)}{dx^{2}} = \frac{i}{h(t)}\frac{d h(t)}{dt}[/tex]

    Both sides there have to equal the same constant. But why is this constant the total energy of the particle?
     
  2. jcsd
  3. Nov 12, 2008 #2

    nicksauce

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    The idea is that you have that Hamiltonian [tex]H = \frac{p^2}{2m} = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial^2x}[/tex]. The energy of the particle are eigenvalues of the Hamiltonian, so [tex]H\psi=E\psi[/tex], so [tex]\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial^2x}\psi=E\psi[/tex].

    (And in the way you wrote it, the constant won't be the energy of the particle, but rather the energy divided by hbar)
     
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