Determine the energies of the three lowest energy states.

[qweazy]

Homework Statement

A particle is confined to a two-dimensional box defined by the following boundary conditions: U(x, y) = 0 for \frac{-L}{2} ≤ x ≤ \frac{L}{2} and
\frac{-3L}{2} ≤ y ≤ \frac{3L}{2}, and U(x, y) = ∞ outside these ranges. Determine the energies of the three lowest energy states
Just want the setup right

Homework Equations

ψ(x,y,z)=Asin(k_{1}x)sin(k_{2}y)sin(k_{3}z)

The Attempt at a Solution

So I first started off with
ψ(x,y)=0 at x=\frac{L}{2} and y=\frac{3L}{2}

\Rightarrow k_{1}=\frac{2n_{1}π}{L} and k_{2}=\frac{2n_{2}π}{3L}

So then the energy =
\frac{h^{2}}{8π^{2}m}(k^{2}_{1}+k^{2}_{2})
=\frac{h^{2}}{8π^{2}m}(\frac{4n^{2}_{1}π^{2}}{L^{2}}+\frac{4n^{2}_{2}π^{2}}{9L^{2}})

π^{2} cancels out, factor out 4 and L^{2}

\frac{h^{2}}{2mL^{2}}(n^{2}_{1}+\frac{n^{2}_{2}}{9})

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

\frac{h^{2}}{8mL^{2}}(n^{2}_{1}+\frac{n^{2}_{2}}{9})

 

[clamtrox]

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.

 

[qweazy]

Oh ok I see. I just thought that Asin(kx) applied for everything. It makes sense now, thanks for your help!