Discrete Symmetry Groups: Rotations, Mirrors, and Vanishing Matrix Elements

Symmetry is one of the most powerful tools in physics — not just for aesthetics, but for hard computational results. If a Hamiltonian has a symmetry, you can often conclude that entire blocks of its matrix representation vanish without doing any integrals. This problem works through a concrete discrete symmetry group to illustrate how.

Setup: the group $\mathcal{G}$ #

Consider the set of rotations about the $z$ -axis by multiples of $\pi/2$ :

$\mathcal{G} = \{D_z(0),\, D_z(\pi/2),\, D_z(\pi),\, D_z(3\pi/2)\}$

where $D_z(a)D_z(b) = D_z(a+b)$ (angles add under composition, modulo $2\pi$ ).

Is $\mathcal{G}$ a group?#

Writing $D_z(n) \equiv D_z(n\pi/2)$ for short, the multiplication table is:

$*$	0	1	2	3
0	$D_z(0)$	$D_z(1)$	$D_z(2)$	$D_z(3)$
1	$D_z(1)$	$D_z(2)$	$D_z(3)$	$D_z(0)$
2	$D_z(2)$	$D_z(3)$	$D_z(0)$	$D_z(1)$
3	$D_z(3)$	$D_z(0)$	$D_z(1)$	$D_z(2)$

Checking the four group axioms:

Identity: $D_z(0)$ — visible from the first row and column
Closure: every entry in the table is in $\mathcal{G}$ ✓
Inverses: $D_z(N)^{-1} = D_z(|N-4|)$ , i.e. $D_z(1)^{-1} = D_z(3)$ , $D_z(2)^{-1} = D_z(2)$
Associativity: follows directly from angle addition: $D(F)D(G)D(H) = D(F+G+H)$ , which is manifestly associative

So $\mathcal{G} \cong \mathbb{Z}_4$ (the cyclic group of order 4).

Is it abelian?#

Yes — the multiplication table is symmetric across the diagonal, meaning $D_z(a)D_z(b) = D_z(b)D_z(a)$ for all elements. This is obvious from angle addition: $a+b = b+a$ .

Subgroups#

$\mathcal{G}$ has a subgroup $\mathbb{Z}_2 = \{D_z(0), D_z(2)\} = \{D_z(0), D_z(\pi)\}$ :

$*$	0	2
0	$D_z(0)$	$D_z(2)$
2	$D_z(2)$	$D_z(0)$

This is closed, and $D_z(2)$ is its own inverse since $D_z(\pi)$ applied twice gives $D_z(2\pi) = D_z(0)$ .

Eigenvalues of symmetry operators#

Mirror operator $M_x$ #

The mirror $M_x$ reflects $(x,y) \to (-x, y)$ . Applying it twice returns the original state:

$M_x^2 |x,y\rangle = M_x|-x,y\rangle = |x,y\rangle \implies M_x^2 = \mathbf{1}$

So if $M_x|\psi\rangle = \lambda|\psi\rangle$ , then $\lambda^2 = 1$ , giving:

$\lambda(M_x) = \pm 1$

States with $\lambda = +1$ are even (symmetric) under reflection; $\lambda = -1$ are odd (antisymmetric).

Rotation operators $D_z(n\pi/2)$ #

The eigenvalues depend on the order of the element — how many times it must be applied to return to the identity:

$D_z(0)$ : order 1, so $\lambda^1 = 1 \implies \lambda = +1$
$D_z(\pi/2)$ : order 4, so $\lambda^4 = 1 \implies \lambda = \pm 1, \pm i$
$D_z(\pi)$ : order 2, so $\lambda^2 = 1 \implies \lambda = \pm 1$
$D_z(3\pi/2)$ : order 4, so $\lambda^4 = 1 \implies \lambda = \pm 1, \pm i$

Do rotations and mirrors commute?#

In general, no. Consider four atoms sitting in the four quadrants, labelled by which quadrant they occupy: $|1\rangle, |2\rangle, |3\rangle, |4\rangle$ .

$M_x D_z(1)|1\rangle = M_x|2\rangle = |1\rangle$

$D_z(1) M_x|1\rangle = D_z(1)|2\rangle = |3\rangle$

Since $|1\rangle \neq |3\rangle$ , the operators do not commute: $[M_x, D_z(\pi/2)] \neq 0$ .

But $M_x$ and $D_z(\pi)$ do commute#

We can verify this by tracking how each operator permutes the four quadrant states. Define the action on an ordered tuple $(a,b,c,d)$ representing which atom is in quadrants 1,2,3,4:

$M_x\{a,b,c,d\} = \{b,a,d,c\} \qquad D_z(2)\{a,b,c,d\} = \{c,d,a,b\}$

Then:

$D_z(2)\,M_x\{1,2,3,4\} = D_z(2)\{2,1,4,3\} = \{4,3,2,1\}$

$M_x\,D_z(2)\{1,2,3,4\} = M_x\{3,4,1,2\} = \{4,3,2,1\}$

Same result — so $[M_x, D_z(\pi)] = 0$ , meaning they share simultaneous eigenstates.

Using symmetry to kill matrix elements#

Here is the payoff. Suppose a potential $V = aXY$ acts on the system. How does it transform?

$M_x V = M_x(aXY) = a(-X)Y = -aXY = -V \quad \Rightarrow \quad \{M_x, V\} = 0 \text{ (anticommutes)}$

$D_z(2) V = D_z(2)(aXY) = a(-X)(-Y) = aXY = V \quad \Rightarrow \quad [D_z(2), V] = 0 \text{ (commutes)}$

Now label eigenstates by their parities: $|\epsilon_{D}, \epsilon_{M}\rangle$ where $\epsilon_D = \pm 1$ is the $D_z(2)$ eigenvalue and $\epsilon_M = \pm 1$ is the $M_x$ eigenvalue. Consider the matrix element $\langle +,+|V|+,+\rangle$ :

$\langle +,+|V|+,+\rangle = \langle +,+|D_z(2)M_x\,V\,M_x D_z(2)|+,+\rangle$

Since $M_x$ anticommutes with $V$ and commutes with $D_z(2)$ :

$= (-1)\langle +,+|D_z(2)\,V\,\underbrace{M_x M_x}_{1}\,D_z(2)|+,+\rangle = -\langle +,+|V|+,+\rangle$

The only number equal to its own negative is zero:

$\boxed{\langle +,+|V|+,+\rangle = 0}$

Notice that the argument only ever used the $M_x$ label: whatever $\epsilon_M$ is, it appears twice and squares to 1, while the anticommutation supplies the minus sign. So every diagonal element vanishes, including the mixed-parity ones such as $\langle +,-|V|+,-\rangle$ . The expectation value of $V$ in any $M_x$ eigenstate is zero.

The general rule follows from applying the same trick to an off-diagonal element $\langle \epsilon_D, \epsilon_M|V|\epsilon_D', \epsilon_M'\rangle$ :

$M_x V M_x = -V$ gives $\epsilon_M\epsilon_M' \langle\cdots\rangle = -\langle\cdots\rangle$ , so the element vanishes unless $\epsilon_M' = -\epsilon_M$ ;
$D_z(2) V D_z(2) = V$ gives $\epsilon_D\epsilon_D' \langle\cdots\rangle = \langle\cdots\rangle$ , so the element vanishes unless $\epsilon_D' = \epsilon_D$ .

So $V = aXY$ can only connect states with the same $D_z(\pi)$ parity and opposite mirror parity, e.g. $\langle +,+|V|+,-\rangle$ . Everything else is zero.

The lesson: knowing how a perturbation transforms under the symmetry group of the Hamiltonian immediately tells you which matrix elements vanish — no integration required. This is the essence of selection rules in atomic physics, and of the Wigner–Eckart theorem in group theory.

Setup: the group G\mathcal{G}G#

Is G\mathcal{G}G a group?#