Categories: Physics, Quantum mechanics.

Selection rules

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

⟨f∣O^∣i⟩=⟨ℓfmf∣O^∣ℓimi⟩\begin{aligned} \matrixel{f}{\hat{O}}{i} = \matrixel{\ell_f m_f}{\hat{O}}{\ell_i m_i} \end{aligned}

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

Parity rules

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

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

Where Π^\hat{\Pi} is the parity operator. We wrap this property of O^\hat{O} in the states ∣ℓfmf⟩\Ket{\ell_f m_f} and ∣ℓimi⟩\Ket{\ell_i m_i}:

⟨ℓfmf∣O^∣ℓimi⟩=−⟨ℓfmf∣Π^†O^Π^∣ℓimi⟩=−⟨ℓfmf∣(−1)ℓfO^(−1)ℓi∣ℓimi⟩=(−1)ℓf+ℓi+1⟨ℓfmf∣O^∣ℓimi⟩\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 Δℓ≡ℓf−ℓi\Delta \ell \equiv \ell_f - \ell_i must be odd. This leads to the following selection rule, often referred to as Laporte’s rule:

Δℓ  is odd\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 ⟨ℓfmf∣O^∣ℓimi⟩=0\matrixel{\ell_f m_f}{\hat{O}}{\ell_i m_i} = 0. We can derive an analogous rule for any operator E^\hat{E} which is even under parity:

Π^†E^Π^=E^  ⟹  Δℓ  is even\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 r^\vu{r} or x^\hat{x}, and the associated selection rules are known as dipole rules.

For the zz-component of angular momentum mm we have the following:

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

We know that the angular momentum zz-component operator L^z\hat{L}_z satisfies:

[L^z,x^]=iℏy^[L^z,y^]=−iℏx^[L^z,z^]=0\begin{aligned} \comm{\hat{L}_z}{\hat{x}} = i \hbar \hat{y} \qquad \comm{\hat{L}_z}{\hat{y}} = - i \hbar \hat{x} \qquad \comm{\hat{L}_z}{\hat{z}} = 0 \end{aligned}

We take the first relation, and wrap it in ⟨ℓfmf∣\Bra{\ell_f m_f} and ∣ℓimi⟩\Ket{\ell_i m_i}, giving:

iℏ⟨ℓfmf∣y^∣ℓimi⟩=⟨ℓfmf∣L^zx^∣ℓimi⟩−⟨ℓfmf∣x^L^z∣ℓimi⟩=ℏmf⟨ℓfmf∣x^∣ℓimi⟩−ℏmi⟨ℓfmf∣x^∣ℓimi⟩=ℏ(mf−mi)⟨ℓfmf∣x^∣ℓimi⟩\begin{aligned} i \hbar \matrixel{\ell_f m_f}{\hat{y}}{\ell_i m_i} &= \matrixel{\ell_f m_f}{\hat{L}_z \hat{x}}{\ell_i m_i} - \matrixel{\ell_f m_f}{\hat{x} \hat{L}_z}{\ell_i m_i} \\ &= \hbar m_f \matrixel{\ell_f m_f}{\hat{x}}{\ell_i m_i} - \hbar m_i \matrixel{\ell_f m_f}{\hat{x}}{\ell_i m_i} \\ &= \hbar (m_f - m_i) \matrixel{\ell_f m_f}{\hat{x}}{\ell_i m_i} \end{aligned}

Next, we do the same thing with the second relation, for [L^z,y^][\hat{L}_z, \hat{y}], giving:

−iℏ⟨ℓfmf∣x^∣ℓimi⟩=⟨ℓfmf∣L^zy^∣ℓimi⟩−⟨ℓfmf∣y^L^z∣ℓimi⟩=ℏmf⟨ℓfmf∣y^∣ℓimi⟩−ℏmi⟨ℓfmf∣y^∣ℓimi⟩=ℏ(mf−mi)⟨ℓfmf∣y^∣ℓimi⟩\begin{aligned} - i \hbar \matrixel{\ell_f m_f}{\hat{x}}{\ell_i m_i} &= \matrixel{\ell_f m_f}{\hat{L}_z \hat{y}}{\ell_i m_i} - \matrixel{\ell_f m_f}{\hat{y} \hat{L}_z}{\ell_i m_i} \\ &= \hbar m_f \matrixel{\ell_f m_f}{\hat{y}}{\ell_i m_i} - \hbar m_i \matrixel{\ell_f m_f}{\hat{y}}{\ell_i m_i} \\ &= \hbar (m_f - m_i) \matrixel{\ell_f m_f}{\hat{y}}{\ell_i m_i} \end{aligned}

Respectively isolating the two above results for x^\hat{x} and y^\hat{y}, we arrive at these equations:

⟨ℓfmf∣x^∣ℓimi⟩=i(mf−mi)⟨ℓfmf∣y^∣ℓimi⟩⟨ℓfmf∣y^∣ℓimi⟩=−i(mf−mi)⟨ℓfmf∣x^∣ℓimi⟩\begin{aligned} \matrixel{\ell_f m_f}{\hat{x}}{\ell_i m_i} &= i (m_f - m_i) \matrixel{\ell_f m_f}{\hat{y}}{\ell_i m_i} \\ \matrixel{\ell_f m_f}{\hat{y}}{\ell_i m_i} &= - i (m_f - m_i) \matrixel{\ell_f m_f}{\hat{x}}{\ell_i m_i} \end{aligned}

By inserting the first into the second, we find (part of) the selection rule:

⟨ℓfmf∣y^∣ℓimi⟩=(mf−mi)2⟨ℓfmf∣y^∣ℓimi⟩\begin{aligned} \matrixel{\ell_f m_f}{\hat{y}}{\ell_i m_i} &= (m_f - m_i)^2 \matrixel{\ell_f m_f}{\hat{y}}{\ell_i m_i} \end{aligned}

This can only be true if Δm=±1\Delta m = \pm 1, unless the inner products of x^\hat{x} and y^\hat{y} are zero, in which case we cannot say anything about Δm\Delta m yet. Assuming the latter, we take the inner product of the commutator [L^z,z^]=0\comm{\hat{L}_z}{\hat{z}} = 0, and find:

0=⟨ℓfmf∣L^zz^∣ℓimi⟩−⟨ℓfmf∣z^L^z∣ℓimi⟩=ℏmf⟨ℓfmf∣z^∣ℓimi⟩−ℏmi⟨ℓfmf∣z^∣ℓimi⟩=ℏ(mf−mi)⟨ℓfmf∣z^∣ℓimi⟩\begin{aligned} 0 &= \matrixel{\ell_f m_f}{\hat{L}_z \hat{z}}{\ell_i m_i} - \matrixel{\ell_f m_f}{\hat{z} \hat{L}_z}{\ell_i m_i} \\ &= \hbar m_f \matrixel{\ell_f m_f}{\hat{z}}{\ell_i m_i} - \hbar m_i \matrixel{\ell_f m_f}{\hat{z}}{\ell_i m_i} \\ &= \hbar (m_f - m_i) \matrixel{\ell_f m_f}{\hat{z}}{\ell_i m_i} \end{aligned}

If ⟨f∣z^∣i⟩≠0\matrixel{f}{\hat{z}}{i} \neq 0, we require Δm=0\Delta m = 0. The previous requirement was Δm=±1\Delta m = \pm 1, implying that ⟨f∣x^∣i⟩=⟨f∣y^∣i⟩=0\matrixel{f}{\hat{x}}{i} = \matrixel{f}{\hat{y}}{i} = 0 whenever ⟨f∣z^∣i⟩≠0\matrixel{f}{\hat{z}}{i} \neq 0. Only if ⟨f∣z^∣i⟩=0\matrixel{f}{\hat{z}}{i} = 0 does the previous rule Δm=±1\Delta m = \pm 1 hold, in which case the inner products of x^\hat{x} and y^\hat{y} are nonzero.

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

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

We start from the following relation (which is already quite a chore to prove):

[L^2,[L^2,r^]]=2ℏ2(r^L^2+L^2r^)\begin{aligned} \Comm{\hat{L}^2}{\comm{\hat{L}^2}{\vu{r}}} = 2 \hbar^2 (\vu{r} \hat{L}^2 + \hat{L}^2 \vu{r}) \end{aligned}

To begin with, we want to find the commutator of L^2\hat{L}^2 and x^\hat{x}:

[L^2,x^]=[L^x2,x^]+[L^y2,x^]+[L^z2,x^]=[L^y2,x^]+[L^z2,x^]=L^y[L^y,x^]+[L^y,x^]L^y+L^z[L^z,x^]+[L^z,x^]L^z\begin{aligned} \comm{\hat{L}^2}{\hat{x}} &= \comm{\hat{L}_x^2}{\hat{x}} + \comm{\hat{L}_y^2}{\hat{x}} + \comm{\hat{L}_z^2}{\hat{x}} = \comm{\hat{L}_y^2}{\hat{x}} + \comm{\hat{L}_z^2}{\hat{x}} \\ &= \hat{L}_y \comm{\hat{L}_y}{\hat{x}} + \comm{\hat{L}_y}{\hat{x}} \hat{L}_y + \hat{L}_z \comm{\hat{L}_z}{\hat{x}} + \comm{\hat{L}_z}{\hat{x}} \hat{L}_z \end{aligned}

Evaluating these commutators gives us:

[L^y,x^]=[z^p^x,x^]−[x^p^z,x^]=z^[p^x,x^]+[z^,x^]p^x−x^[p^z,x^]−[x^,x^]p^z=−iℏz^[L^z,x^]=[x^p^y,x^]−[y^p^x,x^]=x^[p^y,x^]+[x^,x^]p^y−y^[p^x,x^]−[y^,x^]p^x=iℏy^\begin{aligned} \comm{\hat{L}_y}{\hat{x}} &= \comm{\hat{z} \hat{p}_x}{\hat{x}} - \comm{\hat{x} \hat{p}_z}{\hat{x}} = \hat{z} \comm{\hat{p}_x}{\hat{x}} + \comm{\hat{z}}{\hat{x}} \hat{p}_x - \hat{x} \comm{\hat{p}_z}{\hat{x}} - \comm{\hat{x}}{\hat{x}} \hat{p}_z = - i \hbar \hat{z} \\ \comm{\hat{L}_z}{\hat{x}} &= \comm{\hat{x} \hat{p}_y}{\hat{x}} - \comm{\hat{y} \hat{p}_x}{\hat{x}} = \hat{x} \comm{\hat{p}_y}{\hat{x}} + \comm{\hat{x}}{\hat{x}} \hat{p}_y - \hat{y} \comm{\hat{p}_x}{\hat{x}} - \comm{\hat{y}}{\hat{x}} \hat{p}_x = i \hbar \hat{y} \end{aligned}

Which we then insert back into the original equation, yielding:

[L^2,x^]=iℏ(−L^yz^−z^L^y+L^zy^+y^L^z)\begin{aligned} \comm{\hat{L}^2}{\hat{x}} &= i \hbar (- \hat{L}_y \hat{z} - \hat{z} \hat{L}_y + \hat{L}_z \hat{y} + \hat{y} \hat{L}_z) \end{aligned}

This can be simplified by introducing some more commutators:

[L^2,x^]=iℏ( ⁣− ⁣([L^y,z^]+z^L^y)−z^L^y+([L^z,y^]+y^L^z)+y^L^z)\begin{aligned} \comm{\hat{L}^2}{\hat{x}} &= i \hbar \big( \!-\! ( \comm{\hat{L}_y}{\hat{z}} + \hat{z} \hat{L}_y ) - \hat{z} \hat{L}_y + ( \comm{\hat{L}_z}{\hat{y}} + \hat{y} \hat{L}_z ) + \hat{y} \hat{L}_z \big) \end{aligned}

Evaluating these commutators gives us:

[L^y,z^]=[z^p^x,z^]−[x^p^z,z^]=z^[p^x,z^]+[z^,z^]p^x−x^[p^z,z^]−[x^,z^]p^z=iℏx^[L^z,y^]=[x^p^y,y^]−[y^p^x,y^]=x^[p^y,y^]+[x^,y^]p^y−y^[p^x,y^]−[y^,y^]p^x=−iℏx^\begin{aligned} \comm{\hat{L}_y}{\hat{z}} &= \comm{\hat{z} \hat{p}_x}{\hat{z}} - \comm{\hat{x} \hat{p}_z}{\hat{z}} = \hat{z} \comm{\hat{p}_x}{\hat{z}} + \comm{\hat{z}}{\hat{z}} \hat{p}_x - \hat{x} \comm{\hat{p}_z}{\hat{z}} - \comm{\hat{x}}{\hat{z}} \hat{p}_z = i \hbar \hat{x} \\ \comm{\hat{L}_z}{\hat{y}} &= \comm{\hat{x} \hat{p}_y}{\hat{y}} - \comm{\hat{y} \hat{p}_x}{\hat{y}} = \hat{x} \comm{\hat{p}_y}{\hat{y}} + \comm{\hat{x}}{\hat{y}} \hat{p}_y - \hat{y} \comm{\hat{p}_x}{\hat{y}} - \comm{\hat{y}}{\hat{y}} \hat{p}_x = - i \hbar \hat{x} \end{aligned}

Substituting these then leads us to the first milestone of this proof:

[L^2,x^]=iℏ( ⁣− ⁣iℏx^−z^L^y−z^L^y−iℏx^+y^L^z+y^L^z)=2iℏ(y^L^z−z^L^y−iℏx^)\begin{aligned} \comm{\hat{L}^2}{\hat{x}} &= i \hbar \big( \!-\! i \hbar \hat{x} - \hat{z} \hat{L}_y - \hat{z} \hat{L}_y - i \hbar \hat{x} + \hat{y} \hat{L}_z + \hat{y} \hat{L}_z \big) \\ &= 2 i \hbar (\hat{y} \hat{L}_z - \hat{z} \hat{L}_y - i \hbar \hat{x}) \end{aligned}

Repeating this process for [L^2,y^]\comm{\hat{L}^2}{\hat{y}} and [L^2,z^]\comm{\hat{L}^2}{\hat{z}}, we find analogous expressions:

[L^2,y^]=2iℏ(z^L^x−x^L^z−iℏy^)[L^2,z^]=2iℏ(x^L^y−y^L^x−iℏz^)\begin{aligned} \comm{\hat{L}^2}{\hat{y}} &= 2 i \hbar (\hat{z} \hat{L}_x - \hat{x} \hat{L}_z - i \hbar \hat{y}) \\ \comm{\hat{L}^2}{\hat{z}} &= 2 i \hbar (\hat{x} \hat{L}_y - \hat{y} \hat{L}_x - i \hbar \hat{z}) \end{aligned}

Next, we take the commutator with L^2\hat{L}^2 of the commutator we just found:

[L^2,[L^2,x^]]=2iℏ([L^2,y^L^z]−[L^2,z^L^y]−iℏ[L^2,x^])=2iℏ(y^[L^2,L^z]+[L^2,y^]L^z−z^[L^2,L^y]−[L^2,z^]L^y−iℏ[L^2,x^])\begin{aligned} \comm{\hat{L}^2}{\comm{\hat{L}^2}{\hat{x}}} &= 2 i \hbar \big(\comm{\hat{L}^2}{\hat{y} \hat{L}_z} - \comm{\hat{L}^2}{\hat{z} \hat{L}_y} - i \hbar \comm{\hat{L}^2}{\hat{x}}\big) \\ &= 2 i \hbar \big( \hat{y} \comm{\hat{L}^2}{\hat{L}_z} + \comm{\hat{L}^2}{\hat{y}} \hat{L}_z - \hat{z} \comm{\hat{L}^2}{\hat{L}_y} - \comm{\hat{L}^2}{\hat{z}} \hat{L}_y - i \hbar \comm{\hat{L}^2}{\hat{x}} \big) \end{aligned}

Where we used that [L^2,L^y]=[L^2,L^z]=0\comm{\hat{L}^2}{\hat{L}_y} = \comm{\hat{L}^2}{\hat{L}_z} = 0. The other commutators look familiar:

[L^2,[L^2,x^]]=2iℏ([L^2,y^]L^z−[L^2,z^]L^y−iℏ[L^2,x^])\begin{aligned} \comm{\hat{L}^2}{\comm{\hat{L}^2}{\hat{x}}} &= 2 i \hbar \big( \comm{\hat{L}^2}{\hat{y}} \hat{L}_z - \comm{\hat{L}^2}{\hat{z}} \hat{L}_y - i \hbar \comm{\hat{L}^2}{\hat{x}} \big) \end{aligned}

By inserting the expressions we found earlier for these commutators, we get:

[L^2,[L^2,x^]]=−4ℏ2(z^L^xL^z−x^L^z2−iℏy^L^z+y^L^xL^y−x^L^y2+iℏz^L^y)+2ℏ2(L^2x^−x^L^2)\begin{aligned} \comm{\hat{L}^2}{\comm{\hat{L}^2}{\hat{x}}} &= - 4 \hbar^2 \big( \hat{z} \hat{L}_x \hat{L}_z - \hat{x} \hat{L}_z^2 - i \hbar \hat{y} \hat{L}_z + \hat{y} \hat{L}_x \hat{L}_y - \hat{x} \hat{L}_y^2 + i \hbar \hat{z} \hat{L}_y \big) \\ &\qquad\qquad + 2 \hbar^2 \big( \hat{L}^2 \hat{x} - \hat{x} \hat{L}^2 \big) \end{aligned}

Substituting the well-known commutators iℏL^y=[L^z,L^x]i \hbar \hat{L}_y = \comm{\hat{L}_z}{\hat{L}_x} and iℏL^z=[L^x,L^y]i \hbar \hat{L}_z = \comm{\hat{L}_x}{\hat{L}_y}:

[L^2,[L^2,x^]]=−4ℏ2(z^L^xL^z+y^L^xL^y−x^L^y2−x^L^z2+z^[L^z,L^x]−y^[L^x,L^y])+2ℏ2(L^2x^−x^L^2)=−4ℏ2(z^L^zL^x+y^L^yL^x−x^L^y2−x^L^z2)+2ℏ2(L^2x^−x^L^2)\begin{aligned} \comm{\hat{L}^2}{\comm{\hat{L}^2}{\hat{x}}} &= - 4 \hbar^2 \big( \hat{z} \hat{L}_x \hat{L}_z + \hat{y} \hat{L}_x \hat{L}_y - \hat{x} \hat{L}_y^2 - \hat{x} \hat{L}_z^2 + \hat{z} \comm{\hat{L}_z}{\hat{L}_x} - \hat{y} \comm{\hat{L}_x}{\hat{L}_y} \big) \\ &\qquad\qquad + 2 \hbar^2 \big( \hat{L}^2 \hat{x} - \hat{x} \hat{L}^2 \big) \\ &= - 4 \hbar^2 \big( \hat{z} \hat{L}_z \hat{L}_x + \hat{y} \hat{L}_y \hat{L}_x - \hat{x} \hat{L}_y^2 - \hat{x} \hat{L}_z^2 \big) + 2 \hbar^2 \big( \hat{L}^2 \hat{x} - \hat{x} \hat{L}^2 \big) \end{aligned}

By definition, L^x2+L^y2+L^z2=L^2\hat{L}_x^2 + \hat{L}_y^2 + \hat{L}_z^2 = \hat{L}^2, which we use to arrive at:

[L^2,[L^2,x^]]=−4ℏ2(z^L^zL^x+y^L^yL^x+x^L^x2−x^L^2)+2ℏ2(L^2x^−x^L^2)=−4ℏ2(z^L^zL^x+y^L^yL^x+x^L^x2)+2ℏ2(L^2x^+x^L^2)\begin{aligned} \comm{\hat{L}^2}{\comm{\hat{L}^2}{\hat{x}}} &= - 4 \hbar^2 \big( \hat{z} \hat{L}_z \hat{L}_x + \hat{y} \hat{L}_y \hat{L}_x + \hat{x} \hat{L}_x^2 - \hat{x} \hat{L}^2 \big) + 2 \hbar^2 \big( \hat{L}^2 \hat{x} - \hat{x} \hat{L}^2 \big) \\ &= - 4 \hbar^2 \big( \hat{z} \hat{L}_z \hat{L}_x + \hat{y} \hat{L}_y \hat{L}_x + \hat{x} \hat{L}_x^2 \big) + 2 \hbar^2 \big( \hat{L}^2 \hat{x} + \hat{x} \hat{L}^2 \big) \end{aligned}

The second term is what we want to prove, so the first term must vanish:

z^L^zL^x+y^L^yL^x+x^L^x2=(r^⋅L^)L^x=(r^⋅(r^×p^))L^x=(p^⋅(r^×r^))L^x=0\begin{aligned} \hat{z} \hat{L}_z \hat{L}_x + \hat{y} \hat{L}_y \hat{L}_x + \hat{x} \hat{L}_x^2 = (\vu{r} \cdot \vu{L}) \hat{L}_x = (\vu{r} \cdot (\vu{r} \cross \vu{p})) \hat{L}_x = (\vu{p} \cdot (\vu{r} \cross \vu{r})) \hat{L}_x = 0 \end{aligned}

Where L^=r^×p^\vu{L} = \vu{r} \cross \vu{p} by definition, and the cross product of a vector with itself is zero.

This process can be repeated for [L^2,[L^2,y^]]\comm{\hat{L}^2}{\comm{\hat{L}^2}{\hat{y}}} and [L^2,[L^2,z^]]\comm{\hat{L}^2}{\comm{\hat{L}^2}{\hat{z}}}, leading us to:

[L^2,[L^2,x^]]=2ℏ2(x^L^2+L^2x^)[L^2,[L^2,y^]]=2ℏ2(y^L^2+L^2y^)[L^2,[L^2,z^]]=2ℏ2(z^L^2+L^2z^)\begin{aligned} \comm{\hat{L}^2}{\comm{\hat{L}^2}{\hat{x}}} &= 2 \hbar^2 (\hat{x} \hat{L}^2 + \hat{L}^2 \hat{x}) \\ \comm{\hat{L}^2}{\comm{\hat{L}^2}{\hat{y}}} &= 2 \hbar^2 (\hat{y} \hat{L}^2 + \hat{L}^2 \hat{y}) \\ \comm{\hat{L}^2}{\comm{\hat{L}^2}{\hat{z}}} &= 2 \hbar^2 (\hat{z} \hat{L}^2 + \hat{L}^2 \hat{z}) \end{aligned}

At last, this brings us to the desired equation for [L^2,[L^2,r^]]\comm{\hat{L}^2}{\comm{\hat{L}^2}{\vu{r}}}, with r^=(x^,y^,z^)\vu{r} = (\hat{x}, \hat{y}, \hat{z}).

We then multiply this relation by ⟨f∣=⟨ℓfmf∣\Bra{f} = \Bra{\ell_f m_f} on the left and ∣i⟩=∣ℓimi⟩\Ket{i} = \Ket{\ell_i m_i} on the right, so the right-hand side becomes:

2ℏ2⟨f∣r^L^2 ⁣ ⁣+ ⁣L^2 ⁣r^∣i⟩=2ℏ2(⟨f∣r^L^2∣i⟩+⟨f∣L^2r^∣i⟩)=2ℏ2(ℏ2ℓi(ℓi ⁣+ ⁣1)⟨f∣r^∣i⟩+ℏ2ℓf(ℓf ⁣+ ⁣1)⟨f∣r^∣i⟩)=2ℏ4(ℓf(ℓf ⁣+ ⁣1)+ℓi(ℓi ⁣+ ⁣1))⟨f∣r^∣i⟩\begin{aligned} 2 \hbar^2 \matrixel{f}{\vu{r} \hat{L}^2 \!\!+\! \hat{L}^2 \!\vu{r}}{i} &= 2 \hbar^2 \big( \matrixel{f}{\vu{r} \hat{L}^2}{i} + \matrixel{f}{\hat{L}^2 \vu{r}}{i} \big) \\ &= 2 \hbar^2 \big( \hbar^2 \ell_i (\ell_i \!+\! 1) \matrixel{f}{\vu{r}}{i} + \hbar^2 \ell_f (\ell_f \!+\! 1) \matrixel{f}{\vu{r}}{i} \big) \\ &= 2 \hbar^4 \big(\ell_f (\ell_f \!+\! 1) + \ell_i (\ell_i \!+\! 1)\big) \matrixel{f}{\vu{r}}{i} \end{aligned}

And, likewise, the left-hand side becomes:

⟨f∣[L^2,[L^2,r^]]∣i⟩=⟨f∣L^2[L^2,r^]∣i⟩−⟨f∣[L^2,r^]L^2∣i⟩=ℏ2ℓf(ℓf ⁣+ ⁣1)⟨f∣[L^2,r^]∣i⟩−ℏ2ℓi(ℓi ⁣+ ⁣1)⟨f∣[L^2,r^]∣i⟩=ℏ2(ℓf(ℓf ⁣+ ⁣1)−ℓi(ℓi ⁣+ ⁣1))⟨f∣[L^2,r^]∣i⟩=ℏ2(ℓf(ℓf ⁣+ ⁣1)−ℓi(ℓi ⁣+ ⁣1))(⟨f∣L^2r^∣i⟩−⟨f∣r^L^2∣i⟩)=ℏ4(ℓf(ℓf ⁣+ ⁣1)−ℓi(ℓi ⁣+ ⁣1))2⟨f∣r^∣i⟩\begin{aligned} \matrixel{f}{\comm{\hat{L}^2}{\comm{\hat{L}^2}{\vu{r}}}}{i} &= \matrixel{f}{\hat{L}^2 \comm{\hat{L}^2}{\vu{r}}}{i} - \matrixel{f}{\comm{\hat{L}^2}{\vu{r}} \hat{L}^2}{i} \\ &= \hbar^2 \ell_f (\ell_f \!+\! 1) \matrixel{f}{\comm{\hat{L}^2}{\vu{r}}}{i} - \hbar^2 \ell_i (\ell_i \!+\! 1) \matrixel{f}{\comm{\hat{L}^2}{\vu{r}}}{i} \\ &= \hbar^2 \big( \ell_f (\ell_f \!+\! 1) - \ell_i (\ell_i \!+\! 1) \big) \matrixel{f}{\comm{\hat{L}^2}{\vu{r}}}{i} \\ &= \hbar^2 \big( \ell_f (\ell_f \!+\! 1) - \ell_i (\ell_i \!+\! 1) \big) \big( \matrixel{f}{\hat{L}^2 \vu{r}}{i} - \matrixel{f}{\vu{r} \hat{L}^2}{i} \big) \\ &= \hbar^4 \big( \ell_f (\ell_f \!+\! 1) - \ell_i (\ell_i \!+\! 1) \big)^2 \matrixel{f}{\vu{r}}{i} \end{aligned}

Obviously, both sides are equal to each other, leading to the following equation:

2ℓf(ℓf ⁣+ ⁣1)+2ℓi(ℓi ⁣+ ⁣1)=(ℓf(ℓf ⁣+ ⁣1)−ℓi(ℓi ⁣+ ⁣1))2\begin{aligned} 2 \ell_f (\ell_f \!+\! 1) + 2 \ell_i (\ell_i \!+\! 1) &= \big( \ell_f (\ell_f \!+\! 1) - \ell_i (\ell_i \!+\! 1) \big)^2 \end{aligned}

To proceed, we rewrite the right-hand side like so:

(ℓf(ℓf ⁣+ ⁣1)−ℓi(ℓi ⁣+ ⁣1))2=(ℓf2−ℓi2+ℓf−ℓi)2=((ℓf+ℓi)(ℓf−ℓi)+(ℓf−ℓi))2=(ℓf+ℓi)2(ℓf−ℓi)2+2(ℓf+ℓi)(ℓf−ℓi)2+(ℓf−ℓi)2=((ℓf+ℓi)2+2(ℓf+ℓi)+1)(ℓf−ℓi)2=(ℓf+ℓi+1)2(ℓf−ℓi)2\begin{aligned} \big( \ell_f (\ell_f \!+\! 1) - \ell_i (\ell_i \!+\! 1) \big)^2 &= \big( \ell_f^2 - \ell_i^2 + \ell_f - \ell_i \big)^2 \\ &= \big( (\ell_f + \ell_i) (\ell_f - \ell_i) + (\ell_f - \ell_i) \big)^2 \\ &= (\ell_f + \ell_i)^2 (\ell_f - \ell_i)^2 + 2 (\ell_f + \ell_i) (\ell_f - \ell_i)^2 + (\ell_f - \ell_i)^2 \\ &= \big( (\ell_f + \ell_i)^2 + 2 (\ell_f + \ell_i) + 1 \big) (\ell_f - \ell_i)^2 \\ &= (\ell_f + \ell_i + 1)^2 (\ell_f - \ell_i)^2 \end{aligned}

And then we do the same to the left-hand side, yielding:

2(ℓf2+ℓi2+ℓf+ℓi)=2ℓf2+2ℓi2+2ℓfℓi−2ℓfℓi+2ℓf+2ℓi+1−1=(ℓf+ℓi+1)2+ℓf2+ℓi2−2ℓfℓi−1=(ℓf+ℓi+1)2+(ℓf−ℓi)2−1\begin{aligned} 2 (\ell_f^2 + \ell_i^2 + \ell_f + \ell_i) &= 2 \ell_f^2 + 2 \ell_i^2 + 2 \ell_f \ell_i - 2 \ell_f \ell_i + 2 \ell_f + 2 \ell_i + 1 - 1 \\ &= (\ell_f + \ell_i + 1)^2 + \ell_f^2 + \ell_i^2 - 2 \ell_f \ell_i - 1 \\ &= (\ell_f + \ell_i + 1)^2 + (\ell_f - \ell_i)^2 - 1 \end{aligned}

The equation above has thus been simplified to the following form:

(ℓf+ℓi+1)2+(ℓf−ℓi)2−1=(ℓf+ℓi+1)2(ℓf−ℓi)2\begin{aligned} (\ell_f + \ell_i + 1)^2 + (\ell_f - \ell_i)^2 - 1 &= (\ell_f + \ell_i + 1)^2 (\ell_f - \ell_i)^2 \end{aligned}

Rearranging yields a product equal to zero, so one or both of the factors must vanish:

0=(ℓf+ℓi+1)2(ℓf−ℓi)2−(ℓf+ℓi+1)2−(ℓf−ℓi)2+1=((ℓf+ℓi+1)2−1)((ℓf−ℓi)2−1)\begin{aligned} 0 &= (\ell_f + \ell_i + 1)^2 (\ell_f - \ell_i)^2 - (\ell_f + \ell_i + 1)^2 - (\ell_f - \ell_i)^2 + 1 \\ &= \big( (\ell_f + \ell_i + 1)^2 - 1 \big) \big( (\ell_f - \ell_i)^2 - 1 \big) \end{aligned}

The first factor is zero if ℓf=ℓi=0\ell_f = \ell_i = 0, in which case the matrix element ⟨f∣r^∣i⟩=0\matrixel{f}{\vu{r}}{i} = 0 anyway. The other, non-trivial option is therefore:

(ℓf−ℓi)2=1\begin{aligned} (\ell_f - \ell_i)^2 = 1 \end{aligned}

Rotational rules

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

[L^2,s^]=0[L^z,s^]=0[L^±,s^]=0\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 L^±≡L^x±iL^y\hat{L}_\pm \equiv \hat{L}_x \pm i \hat{L}_y. The inner product of any such s^\hat{s} must obey these selection rules:

Δℓ=0Δm=0\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 ⟨ℓf∣∣s^∣∣ℓi⟩\matrixel{\ell_f}{|\hat{s}|}{\ell_i} is the reduced matrix element, which is identical to ⟨ℓfmf∣s^∣ℓimi⟩\matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i}, but with a different notation to say that it does not depend on mfm_f or mim_i:

⟨ℓfmf∣s^∣ℓimi⟩=δℓfℓiδmfmi⟨ℓf∣∣s^∣∣ℓi⟩\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}

Firstly, we look at the commutator of s^\hat{s} with the zz-component L^z\hat{L}_z: 0=⟨ℓfmf∣[L^z,s^]∣ℓimi⟩=⟨ℓfmf∣L^zs^∣ℓimi⟩−⟨ℓfmf∣s^L^z∣ℓimi⟩=ℏ(mf−mi)⟨ℓfmf∣s^∣ℓimi⟩\begin{aligned} 0 = \matrixel{\ell_f m_f}{\comm{\hat{L}_z}{\hat{s}}}{\ell_i m_i} &= \matrixel{\ell_f m_f}{\hat{L}_z \hat{s}}{\ell_i m_i} - \matrixel{\ell_f m_f}{\hat{s} \hat{L}_z}{\ell_i m_i} \\ &= \hbar (m_f - m_i) \matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i} \end{aligned}

Which can only be true if mf ⁣− ⁣mi=0m_f \!-\! m_i = 0, unless, of course, ⟨ℓfmf∣s^∣ℓimi⟩=0\matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i} = 0 by itself.

Secondly, we look at the commutator of s^\hat{s} with the total angular momentum L^2\hat{L}^2:

0=⟨ℓfmf∣[L^2,s^]∣ℓimi⟩=⟨ℓfmf∣L^2s^∣ℓimi⟩−⟨ℓfmf∣s^L^2∣ℓimi⟩=ℏ2(ℓf(ℓf ⁣+ ⁣1)−ℓi(ℓi ⁣+ ⁣1))⟨ℓfmf∣s^∣ℓimi⟩\begin{aligned} 0 = \matrixel{\ell_f m_f}{\comm{\hat{L}^2}{\hat{s}}}{\ell_i m_i} &= \matrixel{\ell_f m_f}{\hat{L}^2 \hat{s}}{\ell_i m_i} - \matrixel{\ell_f m_f}{\hat{s} \hat{L}^2}{\ell_i m_i} \\ &= \hbar^2 \big( \ell_f (\ell_f \!+\! 1) - \ell_i (\ell_i \!+\! 1) \big) \matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i} \end{aligned}

Assuming ⟨ℓfmf∣s^∣ℓimi⟩≠0\matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i} \neq 0, this can only be satisfied if the following holds:

0=ℓf2+ℓf−ℓi2−ℓi=(ℓf+ℓi)(ℓf−ℓi)+(ℓf−ℓi)\begin{aligned} 0 = \ell_f^2 + \ell_f - \ell_i^2 - \ell_i = (\ell_f + \ell_i) (\ell_f - \ell_i) + (\ell_f - \ell_i) \end{aligned}

If ℓf=ℓi=0\ell_f = \ell_i = 0 this equation is trivially satisfied. Otherwise, the only option is ℓf ⁣− ⁣ℓi=0\ell_f \!-\! \ell_i = 0, which is another part of the selection rule.

Thirdly, we look at the commutator of s^\hat{s} with the ladder operators L^±\hat{L}_\pm:

0=⟨ℓfmf∣[L^±,s^]∣ℓimi⟩=⟨ℓfmf∣L^±s^∣ℓimi⟩−⟨ℓfmf∣s^L^±∣ℓimi⟩=Cf⟨ℓf(mf ⁣∓ ⁣1)∣s^∣ℓimi⟩−Ci⟨ℓfmf∣s^∣ℓi(mi ⁣± ⁣1)⟩\begin{aligned} 0 = \matrixel{\ell_f m_f}{\comm{\hat{L}_\pm}{\hat{s}}}{\ell_i m_i} &= \matrixel{\ell_f m_f}{\hat{L}_\pm \hat{s}}{\ell_i m_i} - \matrixel{\ell_f m_f}{\hat{s} \hat{L}_\pm}{\ell_i m_i} \\ &= C_f \matrixel{\ell_f (m_f\!\mp\!1)}{\hat{s}}{\ell_i m_i} - C_i \matrixel{\ell_f m_f}{\hat{s}}{\ell_i (m_i\!\pm\!1)} \end{aligned}

Where CfC_f and CiC_i are constants given below. We already know that Δℓ=0\Delta \ell = 0 and Δm=0\Delta m = 0, so the above matrix elements are only nonzero if mf=mi±1m_f = m_i \pm 1. Therefore:

Ci=ℏℓi(ℓi+1)−mi(mi±1)Cf=ℏℓf(ℓf ⁣+ ⁣1)−mf(mf ⁣∓ ⁣1)=ℏℓf(ℓf ⁣+ ⁣1)−(mi ⁣± ⁣1)(mi ⁣± ⁣1 ⁣∓ ⁣1)=ℏℓf(ℓf ⁣+ ⁣1)−mi(mi ⁣± ⁣1)\begin{aligned} C_i &= \hbar \sqrt{\ell_i (\ell_i + 1) - m_i (m_i \pm 1)} \\ C_f &= \hbar \sqrt{\ell_f (\ell_f \!+\! 1) - m_f (m_f \!\mp\! 1)} \\ &= \hbar \sqrt{\ell_f (\ell_f \!+\! 1) - (m_i \!\pm\! 1) (m_i \!\pm\! 1 \!\mp\! 1)} \\ &= \hbar \sqrt{\ell_f (\ell_f \!+\! 1) - m_i (m_i \!\pm\! 1)} \end{aligned}

In other words, Cf=CiC_f = C_i. The above equation therefore reduces to:

⟨ℓfmi∣s^∣ℓimi⟩=⟨ℓf(mi ⁣± ⁣1)∣s^∣ℓi(mi ⁣± ⁣1)⟩\begin{aligned} \matrixel{\ell_f m_i}{\hat{s}}{\ell_i m_i} &= \matrixel{\ell_f (m_i \!\pm\! 1)}{\hat{s}}{\ell_i (m_i\!\pm\!1)} \end{aligned}

Which means that the value of the matrix element does not depend on mim_i (or mfm_f) at all.

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

[L^z,V^z]=0[L^z,V^±]=±ℏV^±[L^±,V^z]=∓ℏV^±[L^±,V^±]=0[L^±,V^∓]=±2ℏV^z\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 V^\vu{V} must obey the following selection rules:

Δℓ=0  or ±1Δm=0  or ±1\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):

⟨ℓfmf∣V^z∣ℓimi⟩=Cmi 0 mfℓi 1 ℓf⟨ℓf∣∣V^∣∣ℓi⟩⟨ℓfmf∣V^+∣ℓimi⟩=−2Cmi 1 mfℓi 1 ℓf⟨ℓf∣∣V^∣∣ℓi⟩⟨ℓfmf∣V^−∣ℓimi⟩=2Cmi −1 mfℓi 1 ℓf⟨ℓf∣∣V^∣∣ℓi⟩\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 O^\hat{O} and multi-particle state ∣Ψ⟩\Ket{\Psi}, we can say:

⟨Ψ∣O^∣Ψ⟩=⟨P^Ψ∣O^∣P^Ψ⟩\begin{aligned} \matrixel{\Psi}{\hat{O}}{\Psi} = \matrixel{\hat{P} \Psi}{\hat{O}}{\hat{P} \Psi} \end{aligned}

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

⟨P^Ψ∣O^∣P^Ψ⟩=⟨Ψ∣P^−1O^P^∣Ψ⟩=⟨Ψ∣P^−1P^O^∣Ψ⟩=⟨Ψ∣O^∣Ψ⟩\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 ∣s⟩\Ket{s} and an antisymmetric state ∣a⟩\Ket{a} (see Pauli exclusion principle), which obey the following for a permutation P^\hat{P}:

P^∣s⟩=∣s⟩P^∣a⟩=−∣a⟩\begin{aligned} \hat{P} \Ket{s} = \Ket{s} \qquad \hat{P} \Ket{a} = - \Ket{a} \end{aligned}

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

⟨s∣O^∣a⟩=⟨P^s∣O^∣a⟩=⟨s∣P^−1O^∣a⟩=⟨s∣O^P^∣a⟩=⟨s∣O^∣P^a⟩=−⟨s∣O^∣a⟩\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:

⟨s∣O^∣a⟩=0\begin{aligned} \boxed{ \matrixel{s}{\hat{O}}{a} = 0 } \end{aligned}

References

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