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I am considering a single-electron system with two possible sites. In the tight-binding model, I have computed that the energy of each site is $E_1=1$ and $E_2=-1$.

I have then been told that this means that the wavefunction is $$\Psi_1 = \frac{1}{\sqrt{2}}(\phi_1 - \phi_2)$$ $$\Psi_2 = \frac{1}{\sqrt{2}}(\phi_1+\phi_2).$$ So this would mean that $E_1$ corresponds to an antisymmetric superposition of states, and $E_2$ corresponds to a symmetric superposition of states.

However, I am confused about the concept of symmetric and antisymmetric for a single electron. I was always under the impression that symmetrisation of wavefunctions always referred to multiple electrons with the aim of determining whether or not they could exist together in the same state. (I.e. if two electrons had an antisymmetric wavefunction, they cannot exist in the same state, whereas they can if they have a symmetric wavefunction.)

Can someone please explain.

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In essence, the (anti)symmetry in this case refers to the parity of the wavefunction with respect to inversion of the two sites; the mental model behind this is the ground and excited states of a double well where the middle barrier is relatively large (so the ground state is an even function with presence in both wells, and the first excited state is similar but odd).

In a model as abstract as yours, speaking of 'symmetric' and 'antisymmetric' requires a bit more structure, since you could happily change $\phi_2$ for $-\phi_2$ without affecting the physics. To use the terms, then, you normally presuppose the existence of a parity operator $P$ which takes $P\phi_1=\phi_2$ and $P\phi_2=\phi_1$, and which commutes with the hamiltonian. With that structure in place, then, $\psi_e=(1+P)\phi_1$ and $\psi_o=(1-P)\phi_2$ are even and odd eigenfunctions of $P$, and they're rightfully called symmetric and antisymmetric.

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