Dipole moments

The electrical dipole moment \(\require{physics} \boldsymbol{\mu}\) governs the interaction energy of a system with an external (uniform) electric field \(\vb{F}_{\text{ext}}\): \[\begin{equation} E_{\text{int}}(\vb{F}_{\text{ext}}) % = - \boldsymbol{\mu} \vdot \vb{F}_{\text{ext}} % \thinspace , \end{equation}\] in which the minus sign makes sure that the energy lowers when the electrical dipole moment vector is aligned with the electric field vector. Expanding the energy in orders of the external field about the field-free case \(\vb{F}_{\text{ext}, \thinspace 0}\) up to second order: \[\begin{equation} \begin{split} E(\vb{F}_{\text{ext}}) % = &E (\vb{F}_{\text{ext}, \thinspace 0}) % + \sum_m % \eval{ % \dv{ % E(\vb{F}_{\text{ext}}) % }{ F_{\text{ext}, \thinspace m} } % }_{ \vb{F}_{\text{ext}, \thinspace 0} } % F_{\text{ext}, \thinspace m} \\ &+ \frac{1}{2!} \sum_{mn} % \eval{ % \frac{ % \dd{E(\vb{F}_{\text{ext}})} % }{ % \dd{F_{\text{ext}, \thinspace m}} % \dd{F_{\text{ext}, \thinspace n}} } % }_{ \vb{F}_{\text{ext}, \thinspace 0} } % F_{\text{ext}, \thinspace m} % F_{\text{ext}, \thinspace n} % \thinspace , \end{split} \end{equation}\] we must make the connection that the perturbational terms are actually the interaction energy: \[\begin{equation} - \boldsymbol{\mu} \vdot \vb{F}_{\text{ext}} = \sum_m \eval{ \dv{ E(\vb{F}_{\text{ext}}) }{ F_{\text{ext}, \thinspace m} } }_{ \vb{F}_{\text{ext}, \thinspace 0} } F_{\text{ext}, \thinspace m} \\ + \frac{1}{2!} \sum_{mn} \eval{ \frac{ \dd{^2}{E(\vb{F}_{\text{ext}})} }{ \dd{F_{\text{ext}, \thinspace m}} \dd{F_{\text{ext}, \thinspace n}} } }_{ \vb{F}_{\text{ext}, \thinspace 0} } F_{\text{ext}, \thinspace m} F_{\text{ext}, \thinspace n} \thinspace . \end{equation}\] The electrical dipole moment can also be expanded around the field-free case \(\vb{F}_{\text{ext}, \thinspace 0}\) (up to first order): \[\begin{equation} \mu_m(\vb{F}_{\text{ext}}) % = \mu_m^{(0)} % + \sum_n % \eval{ % \dv{ % \mu_m(\vb{F}_{\text{ext}}) % }{ F_{\text{ext}, \thinspace n}} % }_{ \vb{F}_{\text{ext}, \thinspace 0} } % F_{\text{ext}, \thinspace n} % \thinspace , \end{equation}\] in which \[\begin{equation} \boldsymbol{\mu}^{(0)} = \boldsymbol{\mu}(\vb{F}_{\text{ext}, \thinspace 0}) \end{equation}\] is called the permanent electrical dipole moment. The linear response coefficient of the electrical dipole moment is then introduced as the electrical polarizability: \[\begin{equation} \alpha_{mn} = \eval{ \dv{ \mu_m(\vb{F}_{\text{ext}}) }{ F_{\text{ext}, \thinspace n}} }_{ \vb{F}_{\text{ext}, \thinspace 0} } \thinspace . \end{equation}\] Combining these insights, we can calculate the permanent electrical dipole moment of a system as \[\begin{equation} \mu^{(0)}_m % = - \eval{ % \dv{ % E(\vb{F}_{\text{ext}}) % }{ F_{\text{ext}, \thinspace m} } % }_{ \vb{F}_{\text{ext}, \thinspace 0} } % \end{equation}\] and the electrical polarizability as \[\begin{equation} \alpha_{mn} % = - \sum_{mn} % \eval{ % \frac{ % \dd{^2}{E(\vb{F}_{\text{ext}})} % }{ % \dd{F_{\text{ext}, \thinspace m}} % \dd{F_{\text{ext}, \thinspace n}} } % }_{ \vb{F}_{\text{ext}, \thinspace 0} } % \thinspace . \end{equation}\]

From classical electrostatics, we furthermore know that we can calculate the dipole moment from the charge distribution \(\rho(\vb{r})\) of the system: \[\begin{equation} \boldsymbol{\mu} % = \int \dd{\vb{r}} % \rho(\vb{r}) % \thinspace % \vb{r} % \thinspace . \end{equation}\]

Analogously, we can expand the magnetic dipole moment \(\vb{m}\) around the field-free case \(\vb{B}_{\text{ext}, \thinspace 0}\) as: \[\begin{equation} m_m(\vb{B}_{\text{ext}}) % = m^{(0)}_m % + \sum_n % \eval{ % \dv{ % m_m(\vb{B}_{\text{ext}}) % }{ B_{\text{ext}, \thinspace n}} % }_{ \vb{B}_{\text{ext}, \thinspace 0} } % B_{\text{ext}, \thinspace n} % \thinspace , \end{equation}\] in which \[\begin{equation} \vb{m}^{(0)} = \vb{m}(\vb{B}_{\text{ext}, \thinspace 0}) \end{equation}\] is accordingly called the permanent magnetic dipole moment and can be calculated as a response of the energy with respect to the uniform magnetic field \(\vb{B}_{\text{ext}}\): \[\begin{equation} m_m^{(0)} % = - \eval{ % \dv{ % E(\vb{B}_{\text{ext}}) % }{ B_{\text{ext}, \thinspace m} } % }_{ \vb{B}_{\text{ext}, \thinspace 0} } % \thinspace . \end{equation}\] The magnetizability \(\xi_{mn}\) is then defined and can be calculated as: \[\begin{align} \xi_{mn} % &= \eval{ % \dv{ % m_m(\vb{B}_{\text{ext}}) % }{ B_{\text{ext}, \thinspace n}} % }_{ \vb{B}_{\text{ext}, \thinspace 0} } \\ &= - \sum_{mn} % \eval{ % \frac{ % \dd{^2}{E(\vb{B}_{\text{ext}})} % }{ % \dd{B_{\text{ext}, \thinspace m}} % \dd{B_{\text{ext}, \thinspace n}} } % }_{ \vb{B}_{\text{ext}, \thinspace 0} } % \thinspace . \end{align}\]

From electromagnetic theory, the magnetic dipole moment of a system can be calculated from the current density \(\vb{j}\) as \[\begin{equation} \vb{m} % = \frac{1}{2} \int \dd{\vb{r}} \thinspace % \vb{r} \cross \vb{j}(\vb{r}) % \thinspace . \end{equation}\]