Consider two non-interacting Fermions (half-integer spin) confined in a 'box'. Construct the anti-symmetric wavefunctions and compare the corresponding ground-state energies of the two systems; one with particles of identical spin and the other with particles of opposite spin.

Now, the wavefunction with opposite spin is incredibly easy, One can just duplicate the spin wavefunction for helium, right?

What confuses me is the "ground state" energy of a wavefunction with two fermions of identical spin. Isn't that impossible? Wouldn't this look like non-excited helium with the same spin electrons in the 1S orbital (which can't be done)?

  • $\begingroup$ You can't put two same-spin electrons onto 1s orbital. Your spin-polarized ground state is the "best available option" of 1s2s $\endgroup$
    – Slaviks
    Nov 4, 2011 at 21:37
  • $\begingroup$ im aware - hence my original question and confusion. $\endgroup$
    – drjrm3
    Nov 4, 2011 at 21:47
  • $\begingroup$ Sorry, then I don't really get your question (and the confusion). Maybe it's terminology? 1s2s with restricted total spin projection is the lowest energy state under this extra constraint, not really the "ground state". Or is it something else which causes confusion? $\endgroup$
    – Slaviks
    Nov 4, 2011 at 21:50
  • $\begingroup$ You're confusing the ground state of the system with the ground state of the individual particles. The ground state of the system is the state with the lowest total energy. You have to take the required symmetries into account. $\endgroup$ Nov 6, 2011 at 23:17
  • $\begingroup$ Think of it this way: Solving the Schrödinger equation gives you a bunch of (single-particle) eigenstates. When building (non-interacting) fermionic wavefunctions, you just put fermions into the states, starting with the lowest available one and following the Pauli exclusion principle. So if the spins are different, both can go into the 1S orbital. If the spins are the same, one goes in the lowest available one, 1S, the other one goes in the next-lowest available one, 2S. $\endgroup$
    – Lagerbaer
    May 3, 2013 at 4:29

1 Answer 1


The particle of mass $m$ in the box of length $L$ in 1D is solved by wavefunctions $$ \begin{align} \psi_{n\alpha}&=A\sin (k_n x) e^{-\omega_n t}|\alpha \rangle\;, \\ k_n&=\frac{n\pi}{L}\;,\\ E_n&=\hbar \omega_n\;,\\ \omega_n&=\frac{\pi h n^2}{4L^2m}\;. \end{align} $$ Here, $|\alpha \rangle$ represents the spin state.

The global fermionic wavefunction for two particles is constructed from all pairs by anti-symmetrization, as $$ \Psi_{n\alpha m\beta}(x_1,x_2,t)=\psi_{n\alpha}(x_1,t)\psi_{m\beta}(x_2,t) - \psi_{m\beta}(x_1,t)\psi_{n\alpha}(x_2,t)\;. $$ Energy of state $\Psi_{n\alpha m\beta}(x_1,x_2,t)$ can be calculated as $$ (H_1+H_2)\Psi_{n\alpha m\beta}(x_1,x_2,t)=(E_n+E_m)\Psi_{n\alpha m\beta}(x_1,x_2,t)\;, $$ since each of the one-particle Hamiltonians acts on the respective one-particle wavefunction $\psi_{n\alpha}(x_1,t)$, which yields its eigenenergy $E_n$.

For identical spins, we are interested only in solutions for which $\alpha=\uparrow$ and $\beta=\uparrow$. The ground state is the lowest lying energy state of the system. In this case, it would correspond to $\Psi_{1\uparrow 1\uparrow}$, but this function is identically zero. Then next two lowest lying states are $\Psi_{1\uparrow 2\uparrow}$ and $\Psi_{2\uparrow 1\uparrow}$. Thanks to the antisymmetrization, $\Psi_{1\uparrow 2\uparrow} = -\Psi_{2\uparrow 1\uparrow}$ and it represents the ground state of the system with energy $E_1+E_2$. For opposite spins, we choose $\alpha=\uparrow$ and $\beta=\downarrow$. Here, the lowest lying energy state is $\Psi_{1\uparrow 1\downarrow}$ and it has energy $2E_1$.


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