Quantum Mechanics, symmetry, and the consequence of the symmetry's irreducible representations for the eigenfunctions

Question

I am reading Anthony Zee's textbook on group theory for physicists.

I am looking for explanation or further reading for the following point:

The author is showing the consequences an invariance of the Hamiltonian has on the solutions to the wave equation:

The transformations T that leave H invariant ($T^\dagger H T = H$) form a group.
Since T is unitary (I assume by Maschke's theorem), $$ HT = TH $$ Denote the eigenvectors of H to the eigenvalue E by $\psi^a$ $$H(T\psi^a) = (HT)\psi^a = (TH)\psi^a = ET\psi^a $$ and so $\psi'^a = T \psi^a$ is also an eigenvector to E and must be a linear combination of $\psi^a$s. We can therefore write the transformations of the $\psi^a$s by the symmetry operator T: $$\psi'^a = (D(T))^{ab}\psi^b \tag{$\star$} $$

The author has suggestively written the matrix expressing the linear combination as $D(T)$ to identify with an irreducible representation of T.

He uses the above for an interesting proof of how the solutions to the $\text{Schrodinger}$ equation for a Hamiltonian left invariant by $\mathbb{Z}_2$ can be chosen to be either even or odd:
Taking $r\in \mathbb{Z}_2$ to be reflection: $$r\psi(x) = \psi(-x) = \pm 1 * \psi(x) $$ where $\pm1$ are the 2 possible irreducible representations of r.

My main question is:
How does one identify $(D(T))$ with an irreducible representation of T ? (in ($\star$) ?) Should this be obvious ?

I was also wondering whether the choice of representation in ($\star$) must be the same for eigenvectors corresponding to all eigenvalues. i.e. in the parity example, must a choice be made for all eigenvalues to be either even or odd ? Or can the representation be chosen on an individual basis ?

Answer 1

A choice be made for all eigenvalues to be either even or odd ?

The analysis in group theory can classify how many different sysmmetry styles will be in the eigen functions according to the symmeytry group of the Hamiltonian. It can tell from the given basis functions to determine what irreducible representations (symmetric styles) will show up in the resultant eigenvectors.

For example, if your assume a $\cos ax$ function, it must be a even solution; and $\sin ax$ forms an odd solution.

can the representation be chosen on an individual basis?

You should say: "can the representation(s) be chosen on a set of bases?

Yes. This is one of the main tasks for group analysis. For example, in molecule $NH_3$, providing the bases $N(2P_x, 2P_y, 2P_z)$ and $H(1s)$, these bases will render solutions only of symmetric styles (irreducible representations) $A_1$ and $E$ (a 2-dim representation).

How to identify (D(T)) with an irreducible representation of T ?

The question should be stated as How to determine the (matrix) representation, $D_T(g)$, with given bases functions?

Here $T$ is only a notation to identify the representation. The notation system in solid state is complete different from that of molecule study. The matrix representation for each element g, $D(g)$ is depend on the basis chosen, but the trace of matrix does not depend on the bases functions. The trace is a same value for all elements in a class, know as the character of a class $\chi(C)$. Therefore, we are seldom using the whole matrix representation, $D(g)$ for each element, instead working with charater $\chi(C)$ in class-space. There are several nice orthogonal relations about the characters. These can be use to decompose a given set of bases into several irreducible representations.

This is certainly not obvious. It is a major achievement of group theory to construct a systemmatic way of decomposing from the given bases into serval irreducible (matrix) representations by computing the character (trace) of each class. A class is a sub-set of all the conjugate elements in the group. All the elements in a class have same character (trace of the matrix representation). For a finit group, the number of different classes is equal to the number of irreducible representation.

The character tables for various classes of finite groups had been well documented. Therefore, the end of a group course is learning how to use the character tables.

For example $NH_3$ with bases $N(P_x, P_y)$ and $H(1s)$, total number of bases is 5. The sysmmetry group of $NH_3$ is $c3v$ - two 3-fold rotationals and 3 mirror reflections, together with identity, number group element is $n_g= 6$:

$$ G(c_3v) =\left\{I, C_1, C_2, \sigma_1, \sigma_2, \sigma_3\right\} $$ It contains all the sysmetry operations of a normal triangle.

This group can be divided into 3 classes: $$ \{I\},\,\, 2C=\{ C_1, C_2\}, \,\,\text{ and } \,3\sigma=\{\sigma_1, \sigma_2, \sigma_3\} $$

And the character table:

\begin{align} & & I & & 2C & & 3\sigma \\ A_1 & & 1 & & 1 & & 1 \\ A_2 & & 1 & & 1 & & -1 \tag{1}\\ E & & 2 & & -1 & & 0 \end{align} $A_1$ and $A_2$ are 1-d representations, $A_1$ is invariant under all system operations (character all $1$s), and $A_2$ has a symmetry that is invariant under $120^o$ rotations, but changes sign under reflections. $E_2$ is a 2-d represnetation which means the eigen values has a 2-fold degeneracy.

How to determine the 5 bases $$ \phi_1=N(p_x), \, \phi_1=N(p_y),\,\phi_3=H_1(S),\,\phi_4=H_2(S),\,\phi_5=H_3(S), $$ will render what types of representations?

First, you construct a $5 \times 5$ matrix (called nature representation), and calculate the trace of this matrix for each class sysmetry operation.

The character values of this $5\times 5$ matrix for the three classes is : $$\tag{2} \chi =\left( 5, -1, 1\right) $$ For class $I$, all $\phi$s do not moves, therefore, the diagonal elements are all $1$, lead to a trace $= 5$. For rotation $120^o$ the H(s) all moved to other site, and the rotation of $P_x$ and $P_y$ give the two diagonal elements $\cos 120^o = -\frac{1}{2}$, therefore trace $=-\frac{1}{2}-\frac{1}{2}+0+0+0= -1$. For reflection about $y$-axis, $P_x = -P_x$, $P_y = P_y$, 2 of H(1s)s exchanged, the other one remains invariant. The trace $=-1 + 1 + 0 +0 +1 = 1$.

The we decompose Eq.(2) in the character table in Eq.(1) using the orthgonal relation between the columns of Eq.(1) $\sum_{class} n_c \chi_{r_1} \chi_{r_2} = n_g \delta_{r_1,r_2}$ where $n_c$ is the number of element in class $C$, and $n_g$ the number of element in the group $G$: $$ \chi = 1 \chi_{A_1} + 0 \chi_{A_2} + 2 \chi_{E} $$

Thus, for the bases $\{N(P_x, P_y),\,3H(S)\}$ will render eigenvectors of $1 \, A_1$ sysmmetry type and $2 \, E$ symmetry types. Representation $E$ is 2-dimesional, each eigen values is double generacy. Therefore total is 5 state.

The group theory analysis provide no infomation about the energy. Therefore, the arrangement of these 3 eigen values cannot be determined from group theory.