Calculating nuclear shieldings

From a response theory of view, we can calculate the nuclear shielding for normalized wave functions in the Rayleigh-Ritz quotient as \[\begin{equation} \require{physics} \sigma_{Ki, m} % = \sum_{PQ} % \eval{ % \pdv{ % \tilde{h}_{PQ} ( % \vb{B}_{\text{ext}}, % \qty{ \vb{M}_K } % ) % }{ M_{Ki} }% { B_{\text{ext}, \thinspace m} } % }_{ % \vb{B}_{\text{ext}, \thinspace 0} % \qty{ \vb{M}_{K, \thinspace 0} } % } % D_{PQ} % + \qty( \vb{x}^{\text{T}} \vb{F}_{\vb{p}} )_{Ki, \thinspace m} % \thinspace , \end{equation}\] in which \(\vb{x}\) is the wave function parameter response due to the nuclear permanent magnetic dipole moments: \[\begin{equation} (\vb{x})_{j, Ki} % = \eval{ % \pdv{ % p^\star_j ( % \vb{B}_{\text{ext}, \thinspace 0}, % \qty{ \vb{M}_K } % ) % }{ M_{Ki} } % }_{ \qty{ \vb{M}_{K, \thinspace 0} } } % \end{equation}\] and \(\vb{F}_{\vb{p}}\) is the parameter response force due to the magnetic field: \[\begin{equation} \qty( % \vb{F}_{\vb{p}} % )_{jm} % = % \eval{ % \pdv{ % E ( % \vb{B}_{\text{ext}, \thinspace 0}, % \qty{ \vb{M}_K }, % \vb{p} % ) % }{p_j}{ B_{\text{ext}, \thinspace m} } % }_{ \vb{p}^\star, \vb{B}_{\text{ext}, \thinspace 0}, \qty{ \vb{M}_{K, \thinspace 0} } } % \thinspace . \end{equation}\]