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Electron in a well defined by three plates

  1. Jul 10, 2008 #1
    I would like to ask for your help to solve a problem
    concerning the well-known case of an electron inside a well of defined potential.
    I've already studied the case of the 1dimensional infinite potential well,
    but my case is a bit more complicate since it includes a 2D space and
    different finite potentials.

    My problem goes as follows.
    We have 3 metal plates (2D), one of which is infinite.
    Two of them are placed in the same planar y=0
    and the third one at y=h.
    Plate one has potential V1, plate two V2
    and the infinite one zero.
    If we place an electron at a random point P=(xo,yo)
    find the mean value of the possibility to find the electron at position (x,y)

    I start from the potential function which I suppose is the following

    [tex] V(x,y)=\begin{cases}
    V_{1} & 0 \le x < a , y=0 \\
    V_{2} & b \le x = <c, y=0 \\
    a \ linear \ function \ of \ y \ like \ \frac {V_{1}+ V_{2}}{2} -const*y & 0>y \ge h \\
    0 & else
    \end{cases}
    [/tex]

    where a<b<c,\ and\ Vo>V1>0
    and my space extends from

    [tex]
    -\infty < x<+ \infty\ and\ 0<y<h

    [/tex]


    From the classical approach electrons will be moved from
    y=0 to y=h.

    From the quantum approach we shall begin by solving the independent Schrodinger equation.

    First of all am I correct till here?
    Secondary I would appreciate some tips for how to solve the above equation.
    Is it solve analytically for such a problem?
     
  2. jcsd
  3. Jul 11, 2008 #2
    anyone please?
     
  4. Jul 11, 2008 #3
    I'll take a crack at it. This is a far cry from your 1D infinite well problem as far as difficulty.

    Overview:
    You need to first solve Poisson's equation: [tex]\nabla^{2} \Phi(x,y) = \frac{\rho}{\epsilon_0}= 0[/tex] because there are no charges in the middle other than the electron we are putting in there. We are solving this subject to the boundary conditions that you gave us. [tex]\Phi(x,y)[/tex] will be the potential that we plug into Schroedinger's equation and then solve to find the wave function of the electron. Square the wave function and you've got the probability that you're looking for.

    Okay, that sounds easy, right? Not really. First of all let's look at Poisson's equation. It will give us [tex]\newcommand{\pd}[3]{ \frac{ \partial^{#3}{#1} }{ \partial {#2}^{#3} } } \pd{\Phi(x,y)}{x}{2}=-\pd{\Phi(x,y)}{y}{2}[/tex]. Of course, this is also satisfied is both sides are 0. Here's my intuition as to what to use but I don't think that it quite fits the bill in my opinion. I would separate variables: [tex]\Phi(x,y)\equiv X(x)Y(y)[/tex]. Looking at the X part, I would say that it should be constant except where the plates meet. At that point it should be a constant plus a step (Heaviside) function. Maybe it could be something like [tex]X(x)=V_{1}-(V_{1}-V_{2})\Theta (x)[/tex] where I have set the x=0 as the point where the plates meet. This is because right where they meet there is a discontinuity in the potential right near the plates. Also, I have been treating the plates as being infinite in extent. I think it would be easier to solve this way.

    Then, I would make the y components be piecewise, one for the x < 0 (plate 1) and one for x>0. For x <0 it might be given by [tex]Y(y)= C_{1}(1-\frac{y}{h})[/tex] and likewise for [tex]V_{2}[/tex] in the x>0 region. Actually, now that I think about it, you might not need to make this piecewise since the C in front will actually be 1. The problem with my guessed solution is that there will be a big discontinuity at x=0, but this is my best guess at how to solve this.

    Now that you have [tex]\Phi(x)[/tex] you plug it into Schroedinger's equation. Maybe you can separation of variables but you're going to have to deal with the derivative of a dirac function in the x direction. If you can get it separated you should be able to solve the y component by doing the WKB method or something like that. I'm probably overthinking this whole problem, but that's my try at it. Good luck!
     
  5. Jul 14, 2008 #4
    Thank you badphysicist.
    I'm working on your advice.
    I'll come back with feedback when I find some sort of solution
     
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