Set of supplementary notes.

58 CHAPTER 4. ELECTRONIC STRUCTURE
Now, we are faced with two possibilities:
〉
∣
Ĥ ∣ˆTψ = ˆT
〉 〉
∣
∣Ĥψ ∣
= E ∣ˆTψ
1. If the Hamiltonian has no degenerate eigenstates, then there is only one eigenstate with
this energy, and ˆT maps it onto itself (with a complex pre-factor). So the non-degenerate
eigenstates of Ĥ are always also eigenstates of ˆT.
2. If the Hamiltonian has degenerate eigenstates, then these are not necessarily also eigenstates
of ˆT. The set of degenerate eigenstates with a particular eigenvalue forms a subspace.
Ĥ maps a state within this subspace onto itself, but ˆT maps states within the
subspace onto other states within the subspace. Because ˆT is a unitary operator (symmetry
operation leaves 〉 ’length’ of ψ unchanged), we can diagonalise the associated matrix
T mn =
〈ψ m ˆTψn . That means that we can find basis states within the degenerate subspace,
which are eigenstates of ˆT. In other words, we can choose a set of states which are
simultaneously eigenstates of Ĥ and of ˆT.
We find, then, that a complete set of eigenstates
same time eigenstates of the symmetry operator ˆT.
Ĥ can always be found which are at the
For two commuting operators Ĥ, ˆT, we can always choose
simultaneous eigenstates of both Ĥ and ˆT.
This motivates us to use the eigenvalue of ˆT to give an eigenstate of Ĥ a meaningful label.
In the lattice, Ĥ commutes with translation operator ˆT a , where a is a Bravais lattice vector. To
find the possible eigenvalues of translation operator ˆT a , let it operate on plane wave states |k〉,
which after all are also eigenstates of ˆT (not necessarily of Ĥ, though!) and form a complete
basis set. This gives us the set of all possible eigenvalues of ˆT:
ˆT a |k〉 = e ika |k〉
If we now choose Ĥ eigenstates |ψ〉 which are also eigenstates of ˆT: Ĥ |ψ〉 = E |ψ〉 =⇒
ˆT a |ψ〉 = c a |ψ〉, where c a is an eigenvalue of ˆT a , then we know that the ˆT a -eigenvale c a must
be of the form e ika , because these form a complete set of eigenvalues for ˆT.
This realisation leads directly to one form of Bloch’s theorem:
ˆT a |ψ〉 = e ika |ψ〉 (4.5)
From now on, we use the k from the exponent in (4.5) to label the energy eigenstate:
ψ → ψ k .
4.2.2 Bloch’s theorem
We now join everything up and apply it specifically to electrons subject to a periodic potential.
The energy eigenstates of electrons in a lattice: