An article serves to reorganize the frame and structure of the group theory course in the autumn semester at PKU. I try to devote myself into the picture instead of pure symbol manipulating, which I think the lecture note mainly does.

The group here is the symmetry group, say \(G\), of the system. It should be a subgroup of \(SO(3)\) under which the Hamiltonian is invariant \(P_g H = H P_g\). To every \(g\) in \(G\) we assign an operator \(P_g\). \(P_g\)s should furnish an infinite-dimensional, projective representation of \(G\). Yet we only consider scalar particles and define \(P_g: \psi (x) \mapsto \psi (g^{-1} x)\). They thus furnish an authentic infinite-dimensional representation.

The states of particles are represented by a (multiplets of) function(s), i. e. the wave function \(\psi(x)\).

The transformation of the spacetime coordinates naturally induces a transformation of functions.

\[ x \mapsto gx \implies f \mapsto f', \]

with \(f'\) satisfying \(f'(x)=f(g^{-1}x)\).

This can be viewed as the "biggest" representation of all. Note it is also faithful.

Symmetry arises when two observers have the same description of the system. Because the change of observer (thus the change of description) is often viewed as the transformation of physical entities(?) itself, symmetry can also be called invariance under certain transformation.

Example: Two observers describe the same event as \(x\) and \(gx\). The wave functions then \(\psi (x)\) and \(\psi (g^{-1} x)=P_g \psi (x)\). Hamiltonian is a linear map

\[ H: \psi \mapsto \psi'. \]

So, in another description the same transformation will be

\[ H': P_g \psi \mapsto P_g \psi'. \]

Namely \(H' = P_g H P_g^{-1}\), and if \(H' = H\), this "symmetry" will let the two descriptions evolve in time in the same manner.

\[ P_g (e^{-iHt/\hbar}\psi) = e^{-iHt/\hbar}(P_g \psi). \]

So, under a symmetry transformation the description of the physcal state will change, but the way of evolving is invariant.

The main confusion here is that the lectureer has abused the symbol of "irreducible representations" to an unbearable extent. Representations are not a series of matrices. They are all the data including vector spaces and linear transformation on it.

The picture is simple: we try to "classify" the vectors in the total Hilbert space, namely we want to devide the space into a direct sum of all (minimum) invariant subspaces (Yes, we're assuming an infinite-dimensional representation to be completely reducible. Apparently physicists' brains aren't well-functional enough to deal with math rigorousness.) under \(G\)'s action.

In the "finest" division one subspace must be in a eigenspace of \(H\). In other words an invariant subspace including vectors of two different eigenvalues of \(H\) must be futher devided, till every subspace furnishes an irreducible representation (This is exactly the definition of "irreducible".) Every of these representations belongs to an equivalence class of "unequal and irreducible representation". That's the true meaning of such nonsense as "This system has two \(\Gamma _1\) representations, one \(\Gamma _2\) representation, blah blah."

This classification can partly be done by the "projecting operators". But they can only project a vector into the direct sum of all subspaces that belongs to the same unequal and irreducible representation.

• Selection Rule

We try to figure out if \((\psi _1, H' \psi _2)\) vanishes, where \(H'\) is the pertubative Hamiltonian. The straightforward way of calculating is usually impossible. A coarser way of asking is if it's "prohibited by symmetry". If \(\psi _1\) belongs to "the subspace of one representation", and \(H' \psi _2\) belongs to another, a theorem assures that these two vectors are orthognal.

The representation mentioned above doesn't have to be irreducible. We just do the direct sum decomposition. If they do not share a common irreducible representation, they correspond to orthognal subspaces.

The subspace we chose for \(\psi _1\) is just the irreducible subspace. For \(H' \psi _2\) we consider the direct product representation of irreducible representations corresponding to \(H'(x)\) and \(\psi _2\).

The spatial tranlation group is Abelian, it has n one-dimensional representations. So we classify the states into n invariant subspaces, each corresponding to a representation, or a \(k\).

But as mentioned before, this classification is not the "finest". It contains states from different energy eigenvalues. Thus it can be devided further into lines in Hilbert space.

In a word, every wave vector in the Brillouin zone corresponds to infinitely many states, all with different energy.