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Infinite square potential well

  1. Jan 28, 2016 #1
    1. The problem statement, all variables and given/known data
    I think this is a square well potential problem. The question asks me to sketch the ground-state probability density, for the following situation:

    A quasielectron moves in a 'quantum dot' device. The potential V(x) = 0 for 0 ≤ x < L, and is infinite otherwise.

    2. Relevant equations


    3. The attempt at a solution
    I'm going to need to solve the Schrodinger equation within the device. I have no idea what a quantum dot is, but presumably it doesn't actually matter! I think you can just treat it like an infinite square potential. The Schrodinger equation is

    ##-\frac{\hbar^2}{2m} \frac{\partial \psi}{\partial x^2} = E \psi##
    Within the device. And I also need boundary conditions. One of these boundary conditions is ##\psi(L)=0##, since the potential is infinite at L. Annoyingly, the potential is 0 at x=0, so I can't use the same reasoning there. So my first question is, how do I get my second boundary condition? I don't see how the potential being zero at x=0 in any way restricts what ##\psi## could be...
     
  2. jcsd
  3. Jan 28, 2016 #2

    Vanadium 50

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    A. I wouldn't solve anything. It says "sketch".

    B. If I were going to solve something, I'd make the problem more symmetric.
     
  4. Jan 28, 2016 #3
    Don't I need to calculate the probability density in order to sketch it?
     
  5. Jan 28, 2016 #4

    PeroK

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    Specifying ##V(x) = 0 \ (0 \le x < 1)## is equivalent to ##V(x) = 0 \ (0 < x < 1)##.
     
  6. Jan 28, 2016 #5
    Why? In the first inequality V(x) is zero at x = 0, but in the second inequality the potential is infinite at x = 0. So I can't say that at x = 0, the wavefunction is zero, and I don't know where else to get a second boundary condition from.
     
  7. Jan 28, 2016 #6

    PeroK

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    Whether the boundary points are in or out doesn't affect the functions or the integrals. If the wavefunction is not ##0## at ##x=0## then it is discontinuous when viewed as a function for all ##x##.
     
  8. Jan 28, 2016 #7
    So if I said the wavefunction is zero at x = 0, then I'm excluding the electron from being at x = 0, even though at x = 0 the potential is zero... that's ok? I thought it would have something to do with continuity! I was trying to find a way to use the derivative of ##\psi##.

    Vanadium's post makes it sound like I don't actually need to solve for ##\psi## though to sketch the probability distribution. I know it's zero at the edges, but don't I need to integrate the magnitude between infinity and negative infinity to work out the probability density? How could I sketch it without an equation?
     
  9. Jan 28, 2016 #8

    PeroK

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    Yes. I suspect whover set the question didn't give it a second thought!
     
  10. Jan 28, 2016 #9
    I know I'd only actually have to integrate between L and 0, and it should equal 1. And it's zero at both edges. Where does the whole 'ground state' thing come into it, anyway?

    If this is actually a really easy question, I must be missing some fairly basic information.
     
  11. Jan 28, 2016 #10

    PeroK

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    Are you supposed to know the ground state of the infinite square well? If not, then to sketch it, you'd have to solve the Schrodinger equation, I imagine.
     
  12. Jan 28, 2016 #11
    OK. We haven't done anything specifically about ground states of infinite square wells. I'd better solve the Schrodinger equation then!
     
  13. Jan 28, 2016 #12

    PeroK

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    It can't do any harm!
     
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