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Homework Help: Determine the energies of the three lowest energy states.

  1. Nov 28, 2013 #1
    1. The problem statement, all variables and given/known data
    A particle is confined to a two-dimensional box defined by the following boundary conditions: U(x, y) = 0 for [itex]\frac{-L}{2}[/itex] ≤ x ≤ [itex]\frac{L}{2}[/itex] and
    [itex]\frac{-3L}{2}[/itex] ≤ y ≤ [itex]\frac{3L}{2}[/itex], and U(x, y) = ∞ outside these ranges. Determine the energies of the three lowest energy states
    Just want the setup right

    2. Relevant equations

    3. The attempt at a solution
    So I first started off with
    ψ(x,y)=0 at x=[itex]\frac{L}{2}[/itex] and y=[itex]\frac{3L}{2}[/itex]

    [itex]\Rightarrow[/itex] k[itex]_{1}[/itex]=[itex]\frac{2n_{1}π}{L}[/itex] and k[itex]_{2}[/itex]=[itex]\frac{2n_{2}π}{3L}[/itex]

    So then the energy =

    π[itex]^{2}[/itex] cancels out, factor out 4 and L[itex]^{2}[/itex]


    I have it wrong, but I don't know why
    the actual one is

  2. jcsd
  3. Nov 28, 2013 #2
    You are not calculating the ground state energy! Your wave function is for the first excited state!

    Note that there is a difference between boxes [0,L] and [-L/2, L/2]. In the former, the solution is A sin(kx), as B cos(kx) can't satisfy the boundary conditions (0 when x=0). In the latter case however, both solutions must be considered, and in fact, the ground state is the first cosine state.
  4. Nov 28, 2013 #3
    Oh ok I see. I just thought that Asin(kx) applied for everything. It makes sense now, thanks for your help!
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