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Electron in potential well equation

  1. Oct 22, 2008 #1
    Hφ(x) =((p2op)/2m + V(x))φ(x) = (-hbar2/2m.d2/dx2+ V(x)φ(x) = E φ(x)

    This equation is supposed to relate to an electron confined in a one-dimensional potential well. I'm really confused about where it comes from and I do not know what V(x) represents. If anyone could help explain this, I'd be very grateful :D
  2. jcsd
  3. Oct 22, 2008 #2
    umm no that's not the eqn for a particle in a 1d potential well

    -hbar^2/2m.d^2/dx^2 = E φ(x) with the apropo boundary conditions is the eqn that describes a particle in a 1d well.
  4. Oct 22, 2008 #3


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    *scratch my head*

    But of course it is. Ok, there is a closing bracket missing after the V(x), but I guess it is a typo.

    This is the standard Schrödinger equation for a particle. I do not know, on which level your knowledge of physics is, but I suppose: Either you know, what a Schrödinger equation is or you need a crash course in quantum mechanics.

    The V(x) is the potential you have, so in case of a potential well it will be 0 for x inside the potential well and will be infinite or have some fixed value (depending on whether your potential well has finite or infinte height) for x outside the potential well.
  5. Oct 22, 2008 #4
    :rolleyes: and what does v(x) = 0 imply for the shrodinger eqn?

    nowhere in the actual shrodinger differential eqn is the fact that the particle is in a square well expressed except that v(x) = 0. hence what i said stands.
  6. Oct 23, 2008 #5


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    V(x) is 0 for x>a and x<b if a and b denote the left and right border of the well. However V(x) has some constant value C or is infinite outside the potential well, which is the reason for the "short version" of just applying boundary conditions in the infinite case.

    I agree, that this is not very different from using a Schrödinger equation without potential and just applying some boundary conditions in the case of the infinite potential well. In the potential well with finite height however, for the bound state you will find wavefunctions, which are not 0 outside the potential well, but fall off exponentially and the wave function must be continuuous and differentiable at the borders of the well. In this case the exact height of the potential outside the well is needed to determine the normalization constants of the piecewise defined wave function.
  7. Oct 29, 2008 #6
    Thanks for the help :D. I see now that it is just the time-independent schrodinger equation, and that the potential is zero inside the well.
  8. Oct 29, 2008 #7
    That's the same equation that I had. The term -hbar^2/2m.d^2/dx^2 was written as the mometum operator squared, pop^2. I see where it comes from now, thanks :)
  9. Oct 30, 2008 #8
    now that is actually something i did not know till yesterday.
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