# 7.7: Appendix - Magnetic Dipole and Electric Quadrupole Transitions

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The second term in the expansion in eq. (6.39) leads to magnetic dipole and electric quadrupole transitions, which we will describe here. The interaction potential is

$$V^{(2)}(t)=-\frac{q}{m}\left[i A_{0}(\hat{\varepsilon} \cdot \bar{p})(\bar{k} \cdot \bar{r}) e^{-i \omega t}-i A_{0}^{*}(\hat{\varepsilon} \cdot \bar{p})(\bar{k} \cdot \bar{r}) e^{i \omega t}\right]$$

We can use the identity

\begin{aligned} (\hat{\varepsilon} \cdot \bar{p})(\bar{k} \cdot \bar{r}) &=\hat{\varepsilon} \cdot(\overline{p r}) \cdot \bar{k} \\ &=\frac{1}{2} \hat{\varepsilon}(\overline{p r}-\overline{r p}) \bar{k}+\frac{1}{2} \hat{\varepsilon}(\overline{p r}+\overline{r p}) \bar{k} \end{aligned}

to separate V(t) into two distinct light–matter interaction terms:

$V^{(2)}(t)=V_{m a g}^{(2)}(t)+V_{Q}^{(2)}(t) \label{6.112}$

$V_{m a g}^{(2)}(t)=\frac{-i q}{2 m} \hat{\varepsilon} \cdot(\overline{p r}-\overline{r p}) \cdot \bar{k}\left(A_{0} e^{-i \omega t}+A_{0}^{*} e^{i \omega t}\right) \label{6.113}$

$V_{Q}^{(2)}(t)=\frac{-i q}{2 m} \hat{\varepsilon} \cdot(\overline{p r}+\overline{r p}) \cdot \bar{k}\left(A_{0} e^{-i \omega t}+A_{0}^{*} e^{i \omega t}\right) \label{6.114}$

where the first $$V_{\operatorname{mag}}^{(2)}$$ gives rise to magnetic dipole transitions, and the second $$V_{Q}^{(2)}$$ leads to electric quadrupole transitions.

For the notation above, $$\overline{p r}$$ represents an outer product (tensor product $$\bar{p}: \bar{r}$$), so that

$\hat{\varepsilon} \cdot(\overline{p r}) \cdot \bar{k}=\left(\begin{array}{lll} \varepsilon_{x} & \varepsilon_{y} & \varepsilon_{z} \end{array}\right)\left(\begin{array}{ccc} p_{x} r_{x} & p_{x} r_{y} & p_{x} r_{z} \\ p_{y} r_{x} & p_{y} r_{y} & p_{y} r_{z} \\ p_{z} r_{x} & p_{z} r_{y} & p_{z} r_{z} \end{array}\right)\left(\begin{array}{c} k_{x} \\ k_{y} \\ k_{z} \end{array}\right)$

This expression is meant to imply that the component of r that lies along k can influence the magnitude of p along $$\varepsilon$$. Alternatively this term could be written $$\sum_{a, b=x, y, z} \varepsilon_{a}\left(p_{a} r_{b}\right) k_{b}$$. These interaction potentials can be simplified and made more intuitive. Considering first eq. \ref{6.113}, we can use the vector identity $$(\bar{A} \times \bar{B}) \cdot(\bar{C} \times \bar{D})=(\bar{A} \times \bar{C})(\bar{B} \times \bar{D})-(\bar{A} \times \bar{D})(\bar{B} \times \bar{C})$$ to show

\begin{aligned} \frac{1}{2} \hat{\varepsilon} \cdot(\overline{p r}-\overline{r p}) \cdot \bar{k} &=\frac{1}{2}[(\hat{\varepsilon} \cdot \bar{p})(\bar{r} \cdot \bar{k})-(\hat{\varepsilon} \cdot \bar{r})(\bar{p} \cdot \bar{k})]=\frac{1}{2}[(\bar{k} \cdot \hat{\varepsilon}) \cdot(\bar{r} \times \bar{p})] \\ &=\frac{1}{2}(\bar{k} \times \hat{\varepsilon}) \cdot \bar{L} \end{aligned} \label{6.116}

For electronic spectroscopy, $$\bar{L}$$ is the orbital angular momentum. Since the vector $$\bar{k} \times \hat{\varepsilon}$$ describes the direction of the magnetic field $$\bar{B}, \text {and since} A_{0}=B_{0} / 2 i k$$

$V_{m d}^{(2)}(t)=\frac{-q}{2 m} \bar{B}(t) \cdot \bar{L} \quad \bar{B}(t)=\bar{B}_{0} \cos \omega t$

$$\bar{B} \cdot \bar{L} \text {more generally is} \bar{B} \cdot(\bar{L}+2 \bar{S})$$ when considering the spin degrees of freedom. In the case of an electron,

$\frac{q \bar{L}}{m}=\frac{2 c}{\hbar} \beta \bar{L}=\frac{2 c}{\hbar} \bar{\mu}_{m a g}$

where the Bohr magneton $\beta=\sum_{i} e \hbar / 2 m_{i} c, \text {and} \bar{\mu}_{m a g}=\beta \bar{L}$ is the magnetic dipole operator. So we have the form for the magnetic dipole interaction

$V_{m a g}^{(2)}(t)=-\frac{c}{\hbar} \bar{B}(t) \cdot \bar{\mu}_{m a g}$

For electric quadrupole transitions, once can simplify eq. \ref{6.114} by evaluating matrix elements for the operator $$(\overline{p r}+\overline{r p})$$.

$\overline{p r}+\overline{r p}=\frac{i m}{\hbar}\left[\left[H_{0}, \bar{r}\right] \bar{r}-\bar{r}\left[\bar{r}, H_{0}\right]\right]=\frac{-i m}{\hbar}\left[\bar{r} \bar{r}, H_{0}\right]$

and

$V_{Q}^{(2)}(t)=\frac{-q}{2 \hbar} \hat{\varepsilon} \cdot\left[\bar{r} \bar{r}, H_{0}\right] \cdot \bar{k}\left(A_{0} e^{-i \omega t}+A_{0}^{*} e^{i \omega t}\right) \label{6.121}$

Here $$\bar{r} \bar{r}$$ is an outer product of vectors. For a system of many charges (i), we define the quadrupole moment, a traceless second rank tensor

$\begin{array}{l} \bar{Q}=\sum_{i} q_{i} \bar{r} \otimes \bar{r} \\ Q_{m n}=\sum_{i} q_{i}\left(3 r_{m i} \cdot r_{n i}-r_{i}^{2} \delta_{m n}\right) \quad m, n=x, y, z \end{array}$

Now, using $$A_{0}=E_{0} / 2 i \omega$$ eq. (\ref{6.121}) becomes

$V(t)=-\frac{1}{2 i \hbar \omega} \bar{E}(t) \cdot\left[\overline{\bar{Q}}, H_{0}\right] \cdot \hat{k} \quad \bar{E}(t)=\bar{E}_{0} \cos \omega t \label{6.123}$

Since the matrix element $$\left\langle k\left|\left[Q, H_{0}\right]\right| \ell\right\rangle=\hbar \omega_{k \ell} \overline{\bar{Q}}_{k \ell}$$, we can write the electric quadrupole transition moment as

\begin{aligned} V_{k \ell} &=\frac{i E_{0} \omega_{k \ell}}{2 \omega}\langle k|\hat{\varepsilon} \cdot \overline{\bar{Q}} \cdot \hat{k}| \ell\rangle \\ &=\frac{i E_{0} \omega_{k \ell}}{2 \omega} \overline{\bar{Q}}_{k \ell} \end{aligned}

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