# Determine the energies of the three lowest energy states.

1. Nov 28, 2013

### qweazy

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 $\frac{-L}{2}$ ≤ 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

2. Relevant equations
ψ(x,y,z)=Asin(k$_{1}$x)sin(k$_{2}$y)sin(k$_{3}$z)

3. 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$^{2}_{1}$+$\frac{n^{2}_{2}}{9}$)

2. Nov 28, 2013

### 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.

3. Nov 28, 2013

### qweazy

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