10.4: Appendix - Rotational Transition Matrix Elements

In calculating matrix elements involving rotation or orbital angular momentum, one often comes across integrals over a product of three spherical harmonics:

\[\int d\Omega \left[ Y_{J_3}^{M_3} ( \theta ,\phi ) \right]^{*} Y_{J_1}^{M_1 } ( \theta ,\phi ) Y_{J_2 }^{M_2 } ( \theta ,\phi ) \label{eq10.4.01}\]

where the integral over spherical angles has been abbreviated as

\[\int d\Omega = \frac{1}{4\pi } \int_0 ^{2\pi } d\phi \int_0 ^{\pi } \sin \theta \, d\theta \label{eq10.4.02}\]

To evaluate such integrals, one can take advantage of a reduction formula that relates the product of two spherical harmonics to an expansion in single harmonics, where the expansion factors are related to Clebsch Gordon Coefficients (CGCs):

\[Y_{J_1}^{M_1} Y_{J_2}^{M_2}=\sum_{J_3, M_3} \sqrt{\frac{\left(2 J_1+1\right)\left(2 J_2+1\right)}{4 \pi\left(2 J_3+1\right)}}\left\langle J_1 0 ; J_2 0| L 0\right\rangle\left\langle J_1 M_1 ; J_2 M_2 | J_3 M_3\right\rangle Y_{J_3}^{M_3} \label{eq10.4.03}\]

With this reduction formula and the orthonormality of spherical harmonics, one obtains the general solution:

\[\int d \Omega \left[ Y_{J_3 }^{M_3 } \right]^{*} Y_{J_1 }^{M_1 }Y_{J_2 }^{M_2 } = \sqrt {\frac{\left( 2J_1 + 1 \right)\left( 2J_2 + 1 \right)}{4\pi \left( 2J_3 + 1 \right)}} \left\langle {J_1 0;J_2 0} | {L0} \right\rangle \left\langle {J_1 M_1 ;J_2 M_2 } | {J_3 M_3 } \right\rangle \label{eq10.4.04}\]

These expressions can be used to calculate the rotational matrix elements for electric dipole transitions. The transition dipole orientation for interactions with linearly polarized light is calculated by decomposing the dipole orientation into the laboratory frame \(x,y,z\) directions as:

\[\begin{aligned}
(\hat{\boldsymbol{\mu}} \cdot \hat{\epsilon}) & =\hat{z} \cos \theta+\hat{y} \sin \theta \sin \varphi+\hat{x} \sin \theta \cos \varphi \\
& =\hat{z} \sqrt{\frac{4 \pi}{3}} Y_1^0+\hat{x} \sqrt{\frac{2 \pi}{3}}\left(Y_1^{-1}-Y_1^1\right) + \mathrm{i} \hat{y} \sqrt{\frac{2 \pi}{3}}\left(Y_1^{-1}+Y_1^1\right)
\end{aligned} \label{eq10.4.05}\]

Looking at the leading term, the rotational matrix element for the \(z\)-projection of the dipole orientation can be evaluated by setting \(\left( \hat \mu \cdot \hat z \right) = \cos \theta = \sqrt {\tfrac{4\pi }{3}} Y_1^{0}\), resulting in the integral

\[\left\langle J'M' \right|\cos \theta \left| JM \right\rangle = \sqrt {\tfrac{4\pi }{3}} \int d\Omega \, Y_{J' \! ,M'}^{*}Y_{1,0} Y_{J,M} \label{eq10.4.06}\]

Using eq. (\ref{eq10.4.04}) with \(J_1 = 1;M_1 = 0\) gives

\[\mu _{z} = \left\langle J' \! ,M'\left| \cos \theta \right|J,M \right\rangle = \left( \frac{2J + 1}{2J' + 1} \right)^{\frac{1}{2}}\left\langle {J0;10} | {J'0} \right\rangle \left\langle {JM;10} | {J'M'} \right\rangle \label{eq10.4.07}\]

We can now simplify this using the CGCs for \(J_1 = 1,M_1 = 0\). The non-zero coefficients are:

\[\begin{aligned}
&\left\langle {J_2 M;1,0} | {\left( J_2 + 1 \right)M} \right\rangle = \sqrt {\dfrac{\left( J_2 - M + 1 \right)\left( J_2 + M + 1 \right)}{\left( 2J_2 + 1 \right)\left( J_2 + 1 \right)}} \quad & & \left( \Delta J = + 1,\Delta M = 0 \right) \\
&\left\langle {J_2 M;1,0} | {J_2 M} \right\rangle = \dfrac{M}{\sqrt {J_2 \left( J_2 + 1 \right)} }\quad & & \left( \Delta J = 0,\Delta M = 0 \right) \\
&\left\langle {J_2 M;1,0} | {\left( J_2 - 1 \right)M} \right\rangle = - \sqrt {d\frac{\left( J_2 - M \right)\left( J_2 + M \right)}{J_2 \left( 2J_2 + 1 \right)}} \quad & & \left( \Delta J = - 1,\Delta M = 0 \right)
\end{aligned} \label{eq10.4.08}\]

Individually, these CGCs are non-zero for \(\Delta J = 0, \pm 1\) and \(\Delta M = 0\), but \(\Delta J = 0\), \(\Delta M = 0\) doesn’t involve any change of state. So, we find \(\Delta J = \pm 1, \Delta M = 0\) are allowed transitions. For these two terms, the rotational transition dipole matrix elements obtained by evaluating eq. (\ref{eq10.4.07}) are

\[\mu _{z} = \left\langle J' \! ,M'\left| \hat r \cdot \hat z \right|J,M \right\rangle = c_{J + ,M}^{z}\delta _{J' \! ,J + 1}\delta _{M,M'} + c_{J - ,M}^{z}\delta _{J' \! ,J - 1}\delta _{M,M'} \label{eq10.4.09}\]

where the coefficients for \(\Delta J = \pm 1\) transitions are:

\[\begin{aligned}
c_{J + ,M}^{z} & = \left\langle J + 1,M\left| \hat r \cdot \hat z \right|J,M \right\rangle = \sqrt {\frac{\left( J + M + 1 \right)\left( J - M + 1 \right)}{\left( 2J + 1 \right)\left( 2J + 3 \right)}} = \sqrt {\frac{(J + 1)^{2} - M^{2}}{4(J + 1)^{2} - 1}} \\
c_{J - ,M}^{z} &= \left\langle J - 1,M\left| \hat r \cdot \hat z \right|J,M \right\rangle = \sqrt {\frac{\left( J + M \right)\left( J - M \right)}{\left( 2J - 1 \right)\left( 2J + 1 \right)}} = \sqrt {\frac{J^{2} - M^{2}}{4J^{2} - 1}} \\
\end{aligned} \label{eq10.4.10}\]

A similar process can be applied to the components of the dipole vector oriented along the \(x\) and \(y\) directions. The resulting matrix elements are non-zero for \(J = J' \pm 1\) and \(M = M' \pm 1\). Looking at \(x\):

\[\begin{aligned}
\mu _{x} = \left\langle J' \! ,M'\right| \hat r \cdot \hat x \left|J,M \right\rangle = \frac{1}{\sqrt {2} }& \left(\delta _{M' \! ,M - 1}\left( c_{J + ,M -}^{x}\delta _{J' \! ,J + 1} + c_{J - ,M -}^{x}\delta _{J' \! ,J - 1} \right)\right. \\
& \kern.4em \left. - \delta _{M' \! ,M + 1}\left( c_{J + ,M +}^{x}\delta _{J' \! ,J + 1} + c_{J - ,M +}^{x}\delta _{J' \! ,J - 1} \right) \right)\end{aligned}\label{eq10.4.11}\]\[\begin{aligned}
\mu _{y} = \left\langle J' \! ,M' \right| \hat r \cdot \hat y \left|J,M \right\rangle = \frac{1}{\sqrt {2} }&\left( \delta _{M' \! ,M - 1}\left( c_{J + ,M -}^{y}\delta _{J' \! ,J + 1} + c_{J - ,M -}^{y}\delta _{J' \! ,J - 1} \right)\right. \\
& \kern.4em \left. - \delta _{M' \! ,M + 1}\left( c_{J + ,M +}^{y}\delta _{J' \! ,J + 1} + c_{J - ,M +}^{y}\delta _{J' \! ,J - 1} \right) \right)\end{aligned}\label{eq10.4.12}\]

The coefficients for \(\mu_x\) are given by:

\[\begin{aligned}
& c_{J + ,M -}^{x} = \left\langle J + 1,M - 1\left| \hat r \cdot \hat x \right|J,M \right\rangle = \sqrt {\frac{\left( J - M + 1 \right)\left( J - M + 2 \right)}{2\left( 2J + 1 \right)\left( 2J + 3 \right)}} \\
& c_{J - ,M -}^{x} = \left\langle J - 1,M - 1\left| \hat r \cdot \hat x \right|J,M \right\rangle = \sqrt {\frac{\left( J + M \right)\left( J + M - 1 \right)}{2\left( 2J - 1 \right)\left( 2J + 1 \right)}} \\
& c_{J + ,M +}^{x} = \left\langle J + 1,M + 1\left| \hat r \cdot \hat x \right|J,M \right\rangle = \sqrt {\frac{\left( J + M + 2 \right)\left( J + M + 1 \right)}{2\left( 2J + 1 \right)\left( 2J + 3 \right)}} \\
& c_{J - ,M +}^{x} = \left\langle J - 1,M + 1\left| \hat r \cdot \hat x \right|J,M \right\rangle = \sqrt {\frac{\left( J - M \right)\left( J - M + 1 \right)}{2\left( 2J - 1 \right)\left( 2J + 1 \right)}} \\
\end{aligned} \label{eq10.4.13}\]

The coefficients for \(\hat y\) are closely related those for \(\hat x\):

\[\begin{aligned}
& c_{J + ,M +}^{y} = \left\langle J + 1,M + 1\right| \hat r \cdot \hat y \left|J,M \right\rangle = \mathrm{i}c_{J + ,M +}^{x} \\
& c_{J + ,M -}^{y} = \left\langle J + 1,M - 1\right| \hat r \cdot \hat y \left|J,M \right\rangle = - \mathrm{i}c_{J + ,M -}^{x} \\
& c_{J - ,M +}^{y} = \left\langle J - 1,M + 1\right| \hat r \cdot \hat y \left|J,M \right\rangle = \mathrm{i}c_{J - ,M +}^{x} \\
& c_{J - ,M -}^{y} = \left\langle J - 1,M - 1\right| \hat r \cdot \hat y \left|J,M \right\rangle = - \mathrm{i}c_{J - ,M -}^{x} \\
\end{aligned} \label{eq10.4.14}\]

Or:

\[\begin{gathered}
c_{J + ,M \pm}^{x} = - \mathrm{i} c_{J + ,M \pm}^{y} \\
c_{J - ,M -}^{x} = \mathrm{i} c_{J - ,M -}^{y} \\
\end{gathered} \label{eq10.4.15}\]

Returning to eq. (\ref{eq10.4.05}), we see that the same expression can be regrouped as

\[(\hat{\boldsymbol{\mu}} \cdot \hat{\epsilon})=\hat{z} \sqrt{\frac{4 \pi}{3}} Y_1^0-\sqrt{\frac{2 \pi}{3}} Y_1^1(\hat{x}-\mathrm{i}\hat{y})+\sqrt{\frac{2 \pi}{3}} Y_1^{-1}(\hat{x}+\mathrm{i}\hat{y}) \label{eq10.4.16}\]

Grouping by the angular momentum eigenstate illustrates why left and right circularly polarized waves, \(\left| \pm \right\rangle = \tfrac{1}{\sqrt {2} }\left( \hat x \pm \mathrm{i} \hat y \right)\), are used to induce selective changes in the spin angular momentum of electrons and a quantum electromagnetic field. Photons are spin-\(1\) bosons that have a spin angular momentum projected in the direction of propagation: \( \pm \hbar \) for left and right circularly polarized fields. Using the relationships in eq. (\ref{eq10.4.15}) to evaluate the matrix elements for these polarization states by summing the coefficients, one finds that the left circularly polarized fields are responsible for \(\Delta M = - 1\) transitions whereas the right circularly polarized fields are responsible for \(\Delta M = + 1\). No spin changes (\(\Delta M = 0\)) are induced by \(z\)-polarized light. All matrix elements require a \( \pm 1\) change of \(J\) because any change in the matter's angular momentum is matched by an equal and opposite change in the photon's orbital angular momentum.

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