Categories: Physics, Quantum mechanics.

# Selection rules

In quantum mechanics, it is often necessary to evaluate matrix elements of the following form, where $$\ell$$ and $$m$$ respectively represent the total angular momentum and its $$z$$-component:

\begin{aligned} \matrixel{f}{\hat{O}}{i} = \matrixel{\ell_f m_f}{\hat{O}}{\ell_i m_i} \end{aligned}

Where $$\hat{O}$$ is an operator, $$\ket{i}$$ is an initial state, and $$\ket{f}$$ is a final state (usually at least; $$\ket{i}$$ and $$\ket{f}$$ can be any states). Selection rules are requirements on the relations between $$\ell_i$$, $$\ell_f$$, $$m_i$$ and $$m_f$$, which, if not met, guarantee that the above matrix element is zero.

## Parity rules

Let $$\hat{O}$$ denote any operator which is odd under spatial inversion (parity):

\begin{aligned} \hat{\Pi}^\dagger \hat{O} \hat{\Pi} = - \hat{O} \end{aligned}

Where $$\hat{\Pi}$$ is the parity operator. We wrap this property of $$\hat{O}$$ in the states $$\ket{\ell_f m_f}$$ and $$\ket{\ell_i m_i}$$:

\begin{aligned} \matrixel{\ell_f m_f}{\hat{O}}{\ell_i m_i} &= - \matrixel{\ell_f m_f}{\hat{\Pi}^\dagger \hat{O} \hat{\Pi}}{\ell_i m_i} \\ &= - \matrixel{\ell_f m_f}{(-1)^{\ell_f} \hat{O} (-1)^{\ell_i}}{\ell_i m_i} \\ &= (-1)^{\ell_f + \ell_i + 1} \matrixel{\ell_f m_f}{\hat{O}}{\ell_i m_i} \end{aligned}

Which clearly can only be true if the exponent is even, so $$\Delta \ell \equiv \ell_f - \ell_i$$ must be odd. This leads to the following selection rule, often referred to as Laporte’s rule:

\begin{aligned} \boxed{ \Delta \ell \:\:\text{is odd} } \end{aligned}

If this is not the case, then the only possible way that the above equation can be satisfied is if the matrix element vanishes $$\matrixel{\ell_f m_f}{\hat{O}}{\ell_i m_i} = 0$$. We can derive an analogous rule for any operator $$\hat{E}$$ which is even under parity:

\begin{aligned} \hat{\Pi}^\dagger \hat{E} \hat{\Pi} = \hat{E} \quad \implies \quad \boxed{ \Delta \ell \:\:\text{is even} } \end{aligned}

## Dipole rules

Arguably the most common operator found in such matrix elements is a position vector operator, like $$\vu{r}$$ or $$\hat{x}$$, and the associated selection rules are known as dipole rules.

For the $$z$$-component of angular momentum $$m$$ we have the following:

\begin{aligned} \boxed{ \Delta m = 0 \:\:\mathrm{or}\: \pm 1 } \end{aligned}

Meanwhile, for the total angular momentum $$\ell$$ we have the following:

\begin{aligned} \boxed{ \Delta \ell = \pm 1 } \end{aligned}

## Rotational rules

Given a general (pseudo)scalar operator $$\hat{s}$$, which, by nature, must satisfy the following relations with the angular momentum operators:

\begin{aligned} \comm*{\hat{L}^2}{\hat{s}} = 0 \qquad \comm*{\hat{L}_z}{\hat{s}} = 0 \qquad \comm*{\hat{L}_{\pm}}{\hat{s}} = 0 \end{aligned}

Where $$\hat{L}_\pm \equiv \hat{L}_x \pm i \hat{L}_y$$. The inner product of any such $$\hat{s}$$ must obey these selection rules:

\begin{aligned} \boxed{ \Delta \ell = 0 } \qquad \quad \boxed{ \Delta m = 0 } \end{aligned}

It is common to write this in the following more complete way, where $$\matrixel{\ell_f}{|\hat{s}|}{\ell_i}$$ is the reduced matrix element, which is identical to $$\matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i}$$, but with a different notation to say that it does not depend on $$m_f$$ or $$m_i$$:

\begin{aligned} \boxed{ \matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i} = \delta_{\ell_f \ell_i} \delta_{m_f m_i} \matrixel{\ell_f}{|\hat{s}|}{\ell_i} } \end{aligned}

Similarly, given a general (pseudo)vector operator $$\vu{V}$$, which, by nature, must satisfy the following commutation relations, where $$\hat{V}_\pm \equiv \hat{V}_x \pm i \hat{V}_y$$:

$\begin{gathered} \comm*{\hat{L}_z}{\hat{V}_z} = 0 \qquad \comm*{\hat{L}_z}{\hat{V}_{\pm}} = \pm \hbar \hat{V}_{\pm} \qquad \comm*{\hat{L}_{\pm}}{\hat{V}_z} = \mp \hbar \hat{V}_{\pm} \\ \comm*{\hat{L}_{\pm}}{\hat{V}_{\pm}} = 0 \qquad \comm*{\hat{L}_{\pm}}{\hat{V}_{\mp}} = \pm 2 \hbar \hat{V}_z \end{gathered}$

The inner product of any such $$\vu{V}$$ must obey the following selection rules:

\begin{aligned} \boxed{ \Delta \ell = 0 \:\:\mathrm{or}\: \pm 1 } \qquad \boxed{ \Delta m = 0 \:\:\mathrm{or}\: \pm 1 } \end{aligned}

In fact, the complete result involves the Clebsch-Gordan coefficients (from spin addition):

$\begin{gathered} \boxed{ \matrixel{\ell_f m_f}{\hat{V}_{z}}{\ell_i m_i} = C^{\ell_i \: 1 \: \ell_f}_{m_i \: 0 \:m_f} \matrixel{\ell_f}{|\hat{V}|}{\ell_i} } \\ \boxed{ \matrixel{\ell_f m_f}{\hat{V}_{+}}{\ell_i m_i} = - \sqrt{2} C^{\ell_i \: 1 \: \ell_f}_{m_i \: 1 \:m_f} \matrixel{\ell_f}{|\hat{V}|}{\ell_i} } \\ \boxed{ \matrixel{\ell_f m_f}{\hat{V}_{-}}{\ell_i m_i} = \sqrt{2} C^{\ell_i \: 1 \: \ell_f}_{m_i \: -1 \:m_f} \matrixel{\ell_f}{|\hat{V}}{|\ell_i} } \end{gathered}$

## Superselection rule

Selection rules are not always about atomic electron transitions, or angular momenta even.

According to the principle of indistinguishability, permuting identical particles never leads to an observable difference. In other words, the particles are fundamentally indistinguishable, so for any observable $$\hat{O}$$ and multi-particle state $$\ket{\Psi}$$, we can say:

\begin{aligned} \matrixel{\Psi}{\hat{O}}{\Psi} = \matrixel*{\hat{P} \Psi}{\hat{O}}{\hat{P} \Psi} \end{aligned}

Where $$\hat{P}$$ is an arbitrary permutation operator. Indistinguishability implies that $$\comm*{\hat{P}}{\hat{O}} = 0$$ for all $$\hat{O}$$ and $$\hat{P}$$, which lets us prove the above equation, using that $$\hat{P}$$ is unitary:

\begin{aligned} \matrixel*{\hat{P} \Psi}{\hat{O}}{\hat{P} \Psi} = \matrixel{\Psi}{\hat{P}^{-1} \hat{O} \hat{P}}{\Psi} = \matrixel{\Psi}{\hat{P}^{-1} \hat{P} \hat{O}}{\Psi} = \matrixel{\Psi}{\hat{O}}{\Psi} \end{aligned}

Consider a symmetric state $$\ket{s}$$ and an antisymmetric state $$\ket{a}$$ (see Pauli exclusion principle), which obey the following for a permutation $$\hat{P}$$:

\begin{aligned} \hat{P} \ket{s} = \ket{s} \qquad \hat{P} \ket{a} = - \ket{a} \end{aligned}

Any obervable $$\hat{O}$$ then satisfies the equation below, again thanks to the fact that $$\hat{P} = \hat{P}^{-1}$$:

\begin{aligned} \matrixel{s}{\hat{O}}{a} = \matrixel*{\hat{P} s}{\hat{O}}{a} = \matrixel{s}{\hat{P}^{-1} \hat{O}}{a} = \matrixel{s}{\hat{O} \hat{P}}{a} = \matrixel*{s}{\hat{O}}{\hat{P} a} = - \matrixel{s}{\hat{O}}{a} \end{aligned}

This leads us to the superselection rule, which states that there can never be any interference between states of different permutation symmetry:

\begin{aligned} \boxed{ \matrixel{s}{\hat{O}}{a} = 0 } \end{aligned}

1. D.J. Griffiths, D.F. Schroeter, Introduction to quantum mechanics, 3rd edition, Cambridge.