Spin-independent Hamiltonian

We assume a Hamiltonian for two identical electrons of the form

$$\hat{H}=-\frac{\hbar^2}{2m}\Delta_1+ V({\bf r}_1) -\frac{\hbar^2}{2m}\Delta_2 + V({\bf r}_2) + U\left(\vert{\bf r}_1 -{\bf r}_2\vert\right)$$ (2.17)

which does not depend on the spin. The Hamiltonian is symmetric with respect to the particle indices 1 and 2. The solutions of the stationary Schrödinger equation $\hat{H}\psi({\bf r}_1,{\bf r}_2)= E \psi({\bf r}_1,{\bf r}_2)$ for the orbital parts of the wave function can be classified into symmetric and anti-symmetric with respect to swapping ${\bf r}_1$ and ${\bf r}_2$: this is because we have

$$\hat{H} \psi({\bf r}_1,{\bf r}_2) = E \psi({\bf r}_1,{\bf r}_2) \leftrightarrow \hat{H} \psi({\bf r}_2,{\bf r}_1) = E \psi({\bf r}_2,{\bf r}_1)$$

$$\leftrightarrow \underline{\hat{H} \hat{\Pi}_{12}}\psi({\bf r}_1,{\bf r}_2) = E \... ...r}_1,{\bf r}_2) = \underline{ \hat{\Pi}_{12} \hat{H} }\psi({\bf r}_1,{\bf r}_2)$$

$$\leftrightarrow [\hat{H}, \hat{\Pi}_{12}]=0,$$ (2.18)

which means that the permutation operator $\hat{\Pi}_{12}$ commutes with the Hamiltonian. The eigenstates of $\hat{H}$ can therefore be chosen such they are also simultaneous eigenstates of $\hat{\Pi}_{12}$ which are symmetric and antisymmetric wave functions with respect to swapping ${\bf r}_1$ and ${\bf r}_2$.

Since the total wave function (orbital times spin) must be antisymmetric, this means that for energy levels corresponding to symmetric orbital wave functions lead to spin singlets with total spin $S=0$. Energy levels corresponding to anti-symmetric orbital wave functions lead to spin triplets with total spin $S=1$. Even though there is no spin-dependent interaction term in the Hamiltonian, the spin and the possible energy values are not independent of each other!