Wave equation for two Fermions

Question

I came across this basic exercise but I don't fully get the gist of it.

Consider two neutral particles in a 1D Box with the interval $0\leq x \leq L$. The interaction between the two particles is negligible.

We first compute the hamiltonian operator for the two non-interacting particles.

$$ \hat{H}=\frac{-\hbar}{2m_e}\big(\frac{d^2}{dx_1^2}+\frac{d^2}{dx_2^2}\big) + V(x_1, x_2) \tag{1} $$

with

$$V(x_1,x_2) = \begin{cases} 0, & x_1,x_2\in[0,L] \\ \infty, & \text{else}\end{cases} \tag{2}$$

We further calculate the eigenfunctions and the eigenvalues

$$E_{n_1,n_2}=\frac{h^2}{8mL^2}(n_1^2 + n_2^2) \tag{3.1}$$

$$\Psi_{n_1,n_2}(x_1,x_2) = \sqrt{\frac{2}{L}}\sin(\frac{n_1\pi x_1}{L})\sqrt{\frac{2}{L}}\sin(\frac{n_2\pi x_2}{L}) \tag{3.2}$$

We now consider two neutrons. Neutrons are Fermions with a spin quantum number of $s=1/2$. We know that the projection of the spin vector onto the z-Axis is given by:

$$\hat{S}_z | S, M_S\rangle = \hbar M_S|S,M_S\rangle \tag{4}$$

So for $S=1/2$ we get $M_S=\pm 1/2$ and thus we find the two possible projections

$$\pm\frac{\hbar}{2} \tag{5}$$

From the Pauli-Principle we know that the wave equation for Fermions must be anti-symmetric. Now the exercise wants me to modify the previously found wave equation such that

i) it is compatible with the Pauli-Principle

ii) it is a product of two functions whereas the first functions should be a function of the x-coordinate of both Neutrons and the second should be a function of the spin states of the two-neutron-system.

So they say that the wave equations can be written as a product of the "space-part" and the "spin-part" of the wave equation. For the "space-part" we can find an symmetric and an anti-symmetric function:

$$\Psi_{s/a}^{\text{Space}}(x_1, x_2) = \frac{1}{\sqrt{2}}\big( \Psi_{n_1, n_2}(x_1, x_2) \pm \Psi_{n_1,n_2}(x_2, x_1)\big) \tag{6}$$

The "spin-part" of the wave equation can be written as follows. The symmetric functions:

$$\Psi^{\text{Spin}}_{1,1} = |\alpha(1)\alpha(2)\rangle \tag{7.1}$$ $$\Psi^{\text{Spin}}_{1,0} = \frac{1}{\sqrt{2}}(|\alpha(1)\alpha(2)\rangle + |\beta(1)\alpha(2)\rangle \tag{7.2}$$ $$\Psi^{\text{Spin}}_{1,-1} = |\beta(1)\beta(2)\rangle \tag{7.3}$$ $$\Psi^{\text{Spin}}_{0,0} = \frac{1}{\sqrt{2}}(|\alpha(1)\alpha(2)\rangle - |\beta(1)\alpha(2)\rangle \tag{7.4}$$

For the wave equation to be anti-symmetric, we need

$$\Psi(x_1,x_2)=\Psi_{s}^{\text{Space}}(x_1,x_2)\cdot\Psi_{0,0}^{ \text{Spin}} \tag{8}$$ resp. $$\Psi(x_1,x_2)=\Psi_{a}^{\text{Space}}(x_1,x_2)\cdot\Psi_{1,M}^{ \text{Spin}}, \quad M=0,\pm 1 \tag{8}$$

Questions: Now I'm a bit confused about what's going on here. Why do we first calculate the wave equation for neutral particles and then somehow change it? What I don't get is: What is there to change? The wave equation is explicitly given in 3.2. There isn't anything we can change? Do we maybe use this wave equation in 3.2 as a basis for the one we are looking for the Fermions. Furthermore I don't fully get 7.* - what kind of basis do we use here and why is it two dimensional? We have two particles, each particle is a 1/2-spin and so we have a 2-dimensional space for each particle resulting in a 4-dimensional space, so what exactly is $\alpha, \beta$?

In general, the whole process is way too fast and it just lacks details for me to really get it. What's the main idea behind what we do here?

Answer 1

First I note that your $V(x_1,x_2)=V(x_1)+V(x_2)$. Thus there are in fact two independent equations. This leads to the Hartree wave function 3.2. Note that the spin degree of freedom is not explicit here, so 3.2 actually describes 4 degenerate wave functions.

Two indistinguishable fermions however require a wave function that is antisymmetric under particle exchange. This requirement leads to the wave functions 7. These wave functions have the same energy, momentum etc as 3.2 because there are still no two particle interactions and no Pauli interactions in the Hamiltonian. Only two particle properties are affected. In this case you can think of $\vec r_1 - \vec r_2$ and $\vec s_1 \cdot \vec s_2$. These operators do not appear in the Hamiltonian so the energy degeneracy remains.