Consider two electrons in a square 2D infinite potential well i.e $V=0\ for \ 0<x<a, 0<y<a, \ \ V=\infty$ everywhere else. Determine the energy and wavefunction(s) for the first excited state.
I did a similar problem for a 1D potential box but i am uncertian surrounding the degeneration of the energy when there are multiple possible combinations. The wave functions for a single particle would be: $\psi _{n_{x},n_{y}}(x,y)=\frac{2}{a}sin\left (\frac{n_{x}\pi}{a}x \right )sin\left (\frac{n_{y}\pi}{a}y \right )$ and the first excited state for the entire system would be when $n_{x}=n_{y}=1$ for one of them and $n_{x}=2 \ or \ 1 \ and \ n_{y} = 1 \ or \ 2 \ $ for the other.
My main question is whether it matters which one of $n_{x}$ or $n_{y}$ has the value 2 or if they are considered the same state regardless. As a concrete example, is $\ \psi_{1,1}(x_{1},y_{1})\psi_{1,2}(x_{2},y_{2})= \psi_{1,1}(x_{1},y_{1})\psi_{2,1}(x_{2},y_{2})$ ?
My intuitive guess would be no but this would give a lot of different possible total wavefunctions and even more when you take spin into account.
Thanks in advance
Best Answer
Yes it does matter whether it's $n_x$ or $n_y$ with the excited state, which means you've got 2 symmetric and 2 anti-symmetric linearly-independent orbital wavefunctions for the first excited state: $$\psi_{2,1}(r_1)\psi_{1,1}(r_2) \pm \psi_{1,1}(r_1)\psi_{2,1}(r_2)$$ $$\psi_{1,2}(r_1)\psi_{1,1}(r_2) \pm \psi_{1,1}(r_1)\psi_{1,2}(r_2)$$ Taking into account the spins you've got: $$|0,0\rangle[\psi_{2,1}(r_1)\psi_{1,1}(r_2) + \psi_{1,1}(r_1)\psi_{2,1}(r_2)]$$ $$|1,s_z\rangle[\psi_{2,1}(r_1)\psi_{1,1}(r_2) - \psi_{1,1}(r_1)\psi_{2,1}(r_2)]$$ $$|0,0\rangle[\psi_{1,2}(r_1)\psi_{1,1}(r_2) + \psi_{1,1}(r_1)\psi_{1,2}(r_2)]$$ $$|1,s_z\rangle[\psi_{1,2}(r_1)\psi_{1,1}(r_2) - \psi_{1,1}(r_1)\psi_{1,2}(r_2)]$$ Which is a total of 8 states, all of which are degenerate.