# Indistinguishable particles inside infinite potential well, QM

1. Nov 26, 2011

### fluidistic

1. The problem statement, all variables and given/known data
Assume that inside an infinite potential well there are 2 identical particles that doesn't interactuate between themselves and that have spin 1/2 (for instance electrons).
1)Write down the Schrödinger's equation associated with such a system. Write the eigenfunctions in terms of the wavefunction of a single particle.
2)Make a diagram of the seven first energy levels, indicating the degeneracy and the total spin for each state.
3)Calculate the expectation value of the energy for a particle in the ground state and the other in the first excited state. Compare with the case of distinguishable particles.

2. Relevant equations
Coming in attempt part.

3. The attempt at a solution
1)Non interacting particles implies that $V(x_1,x_2)=V(x_1)+V(x_2)=0+0=0$ in this exercise.
I consider electrons, they are fermions so the total wavefunction $\Psi$ must be antisymmetric. Its spatial part can be either antisymmetric ($\frac{1}{\sqrt 2 } \left [ \psi _1 (x_1)\psi _2 (x_2) - \psi _2 (x_1)\psi _1 (x_2) \right ]$ ) in wich case the spin part must be symmetric ($\chi ^S$). Or the spatial part can be symmetric ($\frac{1}{\sqrt 2 } \left [ \psi _1 (x_1)\psi _2 (x_2) + \psi _2 (x_1)\psi _1 (x_2) \right ]$ ) and then the spin part must be antisymmetric ($\chi ^A$).
Where $\chi ^A (1,2)=\frac{1}{\sqrt 2} (\uparrow _1 \downarrow _2- \downarrow _1\uparrow _2 )$.
$\chi _1 ^S (1,2)=\uparrow _1\uparrow _2$
$\chi _0 ^S (1,2)=\frac{1}{\sqrt 2 } (\uparrow _1 \downarrow _2+ \downarrow _1\uparrow _2 )$
$\chi _{-1}^S (1,2)=\downarrow _1\downarrow _2$.
Also the energy of a single particle is $E_n=\alpha n^2$ where $\alpha = \frac{h^2}{8L^2m}$ where L is the length of the potential well. The energy of the system in this exercise is the sum of the energy of each particle.
So my answer to part 1 is:
$-\frac{h^2}{2m} \frac{d^2}{dx^2} \Psi (x_1,x_2)=[E(1)+E(2)] \Psi (x_1, x_2)$.
Where $E(1)+E(2)=\alpha (n_1 ^2+n_2^2)$.
And $\Psi (x_1,x_2)= \begin{cases} \frac{1}{\sqrt 2 } \left [ \psi _1 (x_1)\psi _2 (x_2) - \psi _2 (x_1)\psi _1 (x_2) \right ] \chi ^S \\ \frac{1}{\sqrt 2 } \left [ \psi _1 (x_1)\psi _2 (x_2) + \psi _2 (x_1)\psi _1 (x_2) \right ] \chi ^A \end{cases}$.

2)Ground state: $n_1+n_2=2 \Rightarrow E=2\alpha$.
First excited state: $n_1+n_2=3 \Rightarrow E= 3 \alpha$. This can be done with 2 different ways (without considering the spin): $n_1=1$ and $n_2=2$ or $n_1=2$ and $n_2=1$.
I've noticed that without considering the spin, for the nth excited state, $E_n= n \alpha$ with $n-1$ degeneracies.
I am not sure what to do with the spin... By intuition I know that if both particles are in the ground state, they can only have opposite spin. I am not sure I can distinguish between "the spin of particle 1 is up and the spin of particle 2 is down" and "the spin of particle 1 is down and the spin of particle 2 is up". So I am not sure what is the degeneracy of the ground state. Maybe I should use the wavefunction of part 1), but I don't know how.

3)Rather than calculating $\int _{0}^L \Psi (x_1,x_2) ^ * \hat E \Psi (x_1, x_2)dx$, I guess I can just say that $\langle E \rangle = |c_1|^2E(1)+|c2|^2E(2)=\frac{E(1)+E(2)}{2}$.
If those particles were distinguishable like in classical mechanics, $\langle E \rangle = (E(1)+E(2))$.

I'd appreciate if someone could correct me for any error(s) and for part 2) especially. Thank you!