Nuclear shielding
Suppose we bring a molecule and one of its nuclear permanent magnetic dipole moment \(\require{physics} \vb{M}_K\) into an external magnetic field \(\vb{B}_{\text{ext}}\). (Jameson and Buckingham 1980) The surrounding electrons may indirectly interact with that magnetic moment, so the (interaction) energy of that nuclear permanent magnetic dipole moment with the external magnetic field is then given by: \[\begin{equation} \label{eq:E_int_B_eff_M_K} E_{\text{int}} ( % \vb{B}_{\text{ext}}, % \vb{M}_K % ) % = - \vb{M}_K \vdot \vb{B}_{\text{eff}}(\vb{R}_K; \vb{B}_{\text{ext}}) % \thinspace , \end{equation}\] where \(\vb{B}_{\text{eff}}(\vb{R}_K; \vb{B}_{\text{ext}})\) is the effective magnetic field at the location \(\vb{R}_K\) of the nuclear moment. It is the sum of the external magnetic field and an induced magnetic field due to the electrons: \[\begin{equation} \vb{B}_{\text{eff}} ( % \vb{r}; % \vb{B}_{\text{ext}} % ) = % \vb{B}_{\text{ext}} % + \vb{B}_{\text{ind}} ( % \vb{r}; % \vb{B}_{\text{ext}} % ) % \thinspace , \end{equation}\] where the induced magnetic field \(\vb{B}_{\text{ind}}\) can be assumed to be dependent on the external field strength. Up to second order, we can expand the induced magnetic field around the field-free case as: \[\begin{equation} \label{eq:B_ind_expansion} \begin{split} B_{\text{ind}, \thinspace i} ( % \vb{r}; % \vb{B}_{\text{ext}} % ) % = & \sum_m % \eval{ % \dv{ % B_{\text{ind}, \thinspace i} ( % \vb{r}; % \vb{B}_{\text{ext}} % ) % }{ % B_{\text{ext}, \thinspace m} % } }_{ \vb{B}_{\text{ext}, \thinspace 0} } % B_{\text{ext}, \thinspace m} \\ & + \frac{1}{2!} \sum_{mn} % \eval{ % \frac{ % \dd{^2}{ % B_{\text{ind}, \thinspace i} ( % \vb{r}; % \vb{B}_{\text{ext}} % ) % } }{ % \dd{ B_{\text{ext}, \thinspace m} } % \dd{ B_{\text{ext}, \thinspace n} } % } % }_{ \vb{B}_{\text{ext}, \thinspace 0} } % B_{\text{ext}, \thinspace m} % B_{\text{ext}, \thinspace n} % \thinspace , \end{split} \end{equation}\] in which the zero-th order term is omitted. I still have to clear up why:Analogous to the discussion of dipole moment, we will continue by writing the effective magnetic field as: \[\begin{equation} \label{eq:B_eff_sigma} \vb{B}_{\text{eff}} ( % \vb{r}; % \vb{B}_{\text{ext}} % ) = % ( % \vb{I} % - \boldsymbol{\sigma}(\vb{r}) % ) \vb{B}_{\text{ext}} % \thinspace , \end{equation}\] where the induced magnetic field is given by \[\begin{equation} \label{eq:B_ind_shielding} \vb{B}_{\text{ind}} ( % \vb{r}; % \vb{B}_{\text{ext}} % ) % = - \boldsymbol{\sigma}(\vb{r}) \vb{B}_{\text{ext}} % \thinspace , \end{equation}\] such that we can see that \(\boldsymbol{\sigma}(\vb{r})\) relates the induced magnetic field with the externally applied magnetic field.
By extension, \(\boldsymbol{\sigma}(\vb{R}_K)\) expresses how much the nuclear permanent magnetic moment is `shielded’ by the electrons from the influence of the external magnetic field. Accordingly, \(\boldsymbol{\sigma}(\vb{r})\) is called the nuclear shielding (tensor field). The origin of the minus sign in equation \(\eqref{eq:B_eff_sigma}\) is then also clear: if (one of the tensor components of) \(\boldsymbol{\sigma}(\vb{r})\) is positive, the induced magnetic field (in the corresponding direction) will be counteracting the external magnetic field, a situation that would qualitatively correspond to Lenz’ law.
The linear coefficient is called the (permanent) nuclear shielding (tensor field) \(\boldsymbol{\sigma}^{(0)}\): \[\begin{equation} \sigma^{(0)}_{im}(\vb{r}) % = \eval{ % \dv{ % B_{\text{ind}, \thinspace i} ( % \vb{r}; % \vb{B}_{\text{ext}} % ) % }{ % B_{\text{ext}, \thinspace m} % } }_{ \vb{B}_{\text{ext}, \thinspace 0} } % \end{equation}\] and we will call the quadratic coefficient the nuclear shielding (tensor field) magnetizability: \[\begin{equation} \tau_{imn}(\vb{r}) % = \eval{ % \frac{ % \dd{^2}{ % B_{\text{ind}, \thinspace i} ( % \vb{r}; % \vb{B}_{\text{ext}} % ) % } }{ % \dd{ B_{\text{ext}, \thinspace m} } % \dd{ B_{\text{ext}, \thinspace n} } % } % }_{ \vb{B}_{\text{ext}, \thinspace 0} } % \thinspace . \end{equation}\] Similarly to the relation between polarizability and electrical dipole moment, and magnetizability and magnetic dipole moment, the nuclear shielding magnetizability expresses, up to first order, how much the nuclear shielding deviates from the (permanent) nuclear shielding due to second-order effects of the external magnetic field: \[\begin{equation} \tau_{imn}(\vb{r}) % = \eval{ % \dv{ % \sigma_{im}(\vb{r}) }{ % B_{\text{ext}, \thinspace n} % } }_{ \vb{B}_{\text{ext}, \thinspace 0} } % \thinspace . \end{equation}\]
Up to first order in the magnetic field, the effective magnetic field at the nucleus \(\vb{R}_K\) is then given by: \[\begin{equation} \vb{B}_{\text{eff}} ( % \vb{R}_K; % \vb{B}_{\text{ext}} % ) = % (\vb{I} - \boldsymbol{\sigma}^{(0)}(\vb{R}_K)) % \thinspace \vb{B}_{\text{ext}} % \thinspace . \end{equation}\] We then indeed find that the (permanent) nuclear shielding (tensor field) is related to the mixed second derivative of the interaction energy (cfr. equation \(\eqref{eq:E_int_B_eff_M_K}\)): \[\begin{equation} \frac{ % \dd{^2}{ % E_{\text{int}} ( % \vb{B}_{\text{ext}}, % \vb{M}_K % ) % } % }{ % \dd{M_{Ki}} \dd{B_{\text{ext}, \thinspace m}} % } % = \sigma^{(0)}_{im}(\vb{R}_K) % \thinspace . \end{equation}\]