# Two distinguishable particles space-spin wavefunctions

1. Dec 30, 2004

### yxgao

Hi!

Two (distinguishable) non-interacting electrons are in an infinite square well with hard walls at x=0 and x=a, so that the one particle states are
$$\phi_n(x)=\sqrt{\frac{2}{a}} sin(\frac{n\pi}{a}x), E_n=n^2K$$ where $$K=\pi^2\hbar^2/(2ma^2)$$

My question is what are the spins and space-spin wavefunctions for the state with the lowest two energy eigenvalues?

The answer for the ground system is:

$$S=0$$

$$\phi_1(x_1,x_2)=\frac{2}{L} sin(\frac{\pi x_1}{L}) sin(\frac{\pi x_2}{L}) * \frac{1}{\sqrt{2}} (\chi_{up}(1)\chi_{down}(2) - \chi_{down}(1) \chi_{up}(2))$$

(the first part denotes the space wave function, the second part denotes the spin wavefunction)

$$E = 2 K$$ where
$$K = \frac{\pi^2\hbar^2}{2mL^2}$$

Is this becuase since the wavefunction is symmetric, the spin wavefunction must be antinsymmetric (singlet). Therefore, the electrons are in opposite spins and S=0. However, I'm confused about how to find the spin-state of the non-ground system. I know that S=1 and it must be a symmetric configuration but there are three of them. How do I know which one it is in?
Thanks.

2. Jan 14, 2005

### clive

If I well understood, you have a state with a given energy but three different wave functions (a degenerate state). Then, the system reaches all these states with a certain probability. The total wave function will be a linear combination of those three functions:

psi=c1*psi_1+c2*psi_2+c3*psi_3.

The system can be found in one of its degenerate state with a PROBABILITY given by coefficients c1, c2 and c3.

http://electron6.phys.utk.edu/qm1/modules/m1/assumptions.htm

Last edited: Jan 14, 2005