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Two-dimensional infinite potential box

  1. Dec 4, 2014 #1
    1. The problem statement, all variables and given/known data
    A particle in two-dimensional infinite potential well $$
    H=\frac{p^2}{2m}+\left\{\begin{matrix}
    0, & |x|<\frac{a}{2}\text{ and }|y|<\frac{a}{2}\\
    \infty , & \text{otherwise}
    \end{matrix}\right.$$
    a) Find eigenfunctions and their energies. Also describe the degeneration of ground and first excited state of the system.
    b) How does the first excited state "split" (is that even the right expression?) if a weak potential is added ##V(r)=\lambda sin(\frac{\pi }{a}x)sin(\frac{\pi }{a}y)##?

    2. Relevant equations


    3. The attempt at a solution
    First part is rather easy, I will only write the few steps:
    a) Let ##\psi =X(x)Y(y)## than Schrödinger equation looks something like $$-\frac{\hbar ^2}{2m}\frac{{X}''}{X}-\frac{\hbar ^2}{2m}\frac{{Y}''}{Y}+V_x+V_y=E_x+E_y$$ We can notice that ##x## and ##y## can be completely separated, therefore our solution is $$\psi(x,y)=Asin(\frac{n_x\pi }{a}x)sin(\frac{n_y\pi }{a}y)$$ any energies should be $$E=\frac{\hbar^2}{2m}(k_x^2+k_y^2)=\frac{\hbar^2\pi^2}{2ma^2}(n_x^2+n_y^2)$$ There is no degeneration of the ground state, while we have two possible states in the first excited state.

    b) Ammmm? Some help, please?
     
  2. jcsd
  3. Dec 4, 2014 #2

    ShayanJ

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  4. Dec 4, 2014 #3

    TSny

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    Hello, skrat. Did you check to see if your wavefunctions satisfy the appropriate boundary conditions?
     
  5. Dec 5, 2014 #4
    Am... Not really. But they should.
    Why wouldn't they? We get a completely separated equations for ##x## and ##y## variables as if we only had a one-dimensional potential. We know that in one dimension the solution is ##\psi =Asin(\frac{n\pi}{a}x)##.
     
  6. Dec 5, 2014 #5

    TSny

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    For 1D, the solution ##\psi =Asin(\frac{n\pi}{a}x)## is for a well that extends from x = 0 to x = a.
     
  7. Dec 5, 2014 #6
    ##\psi (0,y)=Asin(0)sin(\frac{n_y\pi }{a}y)=0##

    ##\psi (a,y)=Asin(\frac{n_x\pi }{a}a)sin(\frac{n_y\pi }{a}y)=Asin(n_x\pi)sin(\frac{n_y\pi }{a}y)=0##

    or am I missing something?
     
  8. Dec 5, 2014 #7

    TSny

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    Your 2D well extends from x = - a/2 to x = a/2 rather than x = 0 to x = a. Similarly for y.
     
  9. Dec 5, 2014 #8
    Am, ok that's true but the well is the same. Is it that important not to move the origin to one corner?
     
  10. Dec 5, 2014 #9

    TSny

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    You can move the well so that the origin is at one corner. Then your wavefunctions will be valid. But you will then have to "move" your potential V(r) in a corresponding manner when working out part (b).
     
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