Nuclear spin-rotation constants

Introduction

In this tutorial we introduce the calculation of nuclear spin-rotation tensors as given in the DIRAC code, based on the theoretical developments by I. Agustín Aucar et al. For details, you are welcome to consult [Aucar_JCP2012].

The nuclear spin-rotation (SR) tensor elements of a nucleus \(N\) are given by

\[M_{N,\alpha\beta} = M_{N,\alpha\beta}^{NU} + M_{N,\alpha\beta}^{EV} + M_{N,\alpha\beta}^{LR}\]

They have three terms: the first of them is independent of the electronic variables, whereas the second and third are given by an expectation value (EV) and a linear response (LR) function, respectively.

In tensorial notation, the nuclear spin-rotation contributions of a nucleus \(N\) are given, in SI units, by:

\[\begin{split}{\bf M}_N^{NU} =& \; \frac{1}{4\pi\epsilon_0c^2} \, \frac{e^2 \hslash^2}{2 m_p h} \, g_N \sum_{M \neq N} \frac{Z_M}{|{\bf R}_{MN}|^3} \\ & \hspace{2cm} \left( \left\{ \left[ {\bf R}_{M,CM} - \left( 1 - \frac{Z_N m_p}{m_N g_N} \right) {\bf R}_{N,CM} \right] \cdot {\bf R}_{MN} \right\} {\bf 1} \right. \\ & \hspace{4cm} \left. - \left[ {\bf R}_{M,CM} - \left( 1 - \frac{Z_N m_p}{m_N g_N} \right) {\bf R}_{N,CM} \right] {\bf R}_{MN} \right) \; \cdot \; {\bf I}^{-1} \\ & \\ & \\ {\bf M}_N^{EV} =& \; \frac{1}{4\pi\epsilon_0c^2} \, \frac{e^2 \hslash^2}{2 m_p h} \, g_N \left( 1 - \frac{Z_N m_p}{m_N g_N} \right) \\ & \hspace{2cm} \left[ \left( \langle 0 | \frac{{\bf r}-{\bf r}_N}{|{\bf r} -{\bf r}_N|^3 } | 0 \rangle \cdot {\bf R}_{N,CM} \right) {\bf 1} - \langle 0 | \frac{{\bf r}-{\bf r}_N}{|{\bf r} -{\bf r}_N|^3 } | 0 \rangle \; \; {\bf R}_{N,CM}\right] \; \cdot \; {\bf I}^{-1} \\ & \\ & \\ {\bf M}_N^{LR} =& \; \frac{1}{4\pi\epsilon_0c^2} \, \frac{e^2 \hslash^2}{2 m_p h} \, g_N \; \langle\langle \; \left (\frac{{\bf r}-{\bf r}_N}{|{\bf r} -{\bf r}_N|^3 }\times c \, {\bf \alpha}\right) \; ; \; {\bf J}_e \; \rangle\rangle \; \cdot \; {\bf I}^{-1}\end{split}\]

where \({\bf R}_{MN}\) is the position of nucleus \(M\) with respect to the position of nucleus \(N\); \({\bf R}_{N,CM}\) is the position of nucleus \(N\) with respect to the molecular center of mass; \({\bf I}\) is the inertia tensor of the molecule and \({\bf J}_e = \left({\bf r} - {\bf R}_{CM} \right) \times {\bf p}+{\bf S}_e\) is the electronic total angular momentum.

For molecules at their equilibrium geometry, it is found that the sum of the first two terms, \({\bf M}_N^{NU} (eq) + {\bf M}_N^{EV} (eq)\), is equal to a new tensor which is independent of the electronic variables (see Eq. 60 of [Aucar_JCP2012]),

\[\begin{split}{\bf M}_N^{nuc} =& \; {\bf M}_N^{NU} (eq) \; + \; {\bf M}_N^{EV} (eq) \\ =& \; \frac{1}{4\pi\epsilon_0c^2} \, \frac{e^2 \hslash^2}{2 m_p h} \, g_N \sum_{M \neq N} Z_M \left[ \left( {\bf R}_{M,CM} \cdot \frac{{\bf R}_{MN}}{|{\bf R}_{MN}|^3} \right) {\bf 1} - {\bf R}_{M,CM} \frac{{\bf R}_{MN}}{|{\bf R}_{MN}|^3} \right] \cdot {\bf I}^{-1}\end{split}\]

Therefore, the electronic dependence of the nuclear spin-rotation tensor of a nucleus \(N\) in a molecule in equilibrium is completely given by its linear response term, \({\bf M}_N^{LR}\) (see Eq. 59 of [Aucar_JCP2012]).

Note

The current implementation gives by default (.PRINT values up to 3) results only for molecules in equilibrium.

Application to the HF molecule

As an example, we show a calculation of the SR constant of the fluorine nucleus at the Hydrogen fluoride molecule. The input file spinrot.inp is given by

**DIRAC
.TITLE
 Spin-rotation constant
.WAVE FUNCTION
.PROPERTIES
**HAMILTONIAN
.URKBAL
**INTEGRALS
*READIN
.UNCONTRACT
**WAVE FUNCTIONS
.SCF
*SCF
.EVCCNV
1.0E-7
**PROPERTIES
.SPIN-R
.PRINT
4
*LINEAR RESPONSE
.THRESH
1.0D-7
*END OF

whereas the molecular input file HF_cv3z.mol is

INTGRL
Hydrogen fluoride. Experimental bond length: 0.917 A
dyall.cv3z basis set
C   2              A .10D-15
       9.0    1
F       0.00000000000    0.00000000000     0.00000000000       Isotope=19
LARGE BASIS dyall.cv3z
       1.0    1
H       0.00000000000    0.00000000000     0.91700000000       Isotope=1
LARGE BASIS dyall.cv3z
FINISH

The calculation is run using:

pam --inp=spinrot --mol=HF_cv3z

As a result, SR constants are obtained at the coupled Hartree-Fock level. The code also works at the DFT level.

It is also interesting to note that as .URKBAL is requested in the present calculation, the results will be very close to those obtained using the RKB prescription, because the diamagnetic-like (e-p) contributions to \(M_{N,\alpha\beta}^{LR}\) are almost zero (see Eq. 60 of [Aucar_JCP2012]).

Reading the output file

As the .PRINT flag in the input file is set to 4, the results are fully detailed and given in both kHz and ppm units.

The SR constant of the fluorine nucleus, in kHz, will look like:

                       Spin-rotation constants (kHz) for F
                       -----------------------------------

  Nuclear g-value:     5.257736

  Total spin-rotation constant (SRC)      :   -317.17730994

    Nuclear contribution to SRC (M^nuc)   :     52.47420202
    Electronic contribution to SRC (M^LR) :   -369.65151197
            M^LR-L(e-e) :   -381.52309467
            M^LR-S(e-e) :     11.85860365
            M^LR-L(e-p) :      4.27896875
            M^LR-S(e-p) :     -4.26598970

******************************************************
 ********      M^NU         ********:     55.00521470
 ********      M^EV         ********:     -2.51499774
 ********      M^LR         ********:   -369.65151197
 ********      M^total      ********:   -317.16129500
******************************************************

One should recall that in the case of linear molecules, as the present one, only one tensor element is printed out (the nuclear spin-rotation constant) because for these molecules the SR tensor has only two equal and non-zero diagonal elements.

As it can be seen, the total SR constant of the fluorine nucleus is given, and then separated in its two terms (nuc and LR). The latter contribution, the linear response function, is further separated in their (e-e) and (e-p) parts, as well as in their \(\mathbf{L}\) and \(\mathbf{S}\) parts. It is seen how the (e-p) contribution to the linear response term of the SR constants is almost zero.

In addition, the results are fully detailed, showing the three terms mentioned above (NU, EV and LR).

Finally, if one is interested in the relationship between the SR constants and .SHIELDING (see [Aucar_JPCL2016] and [AucarChap2019]), which is given, in SI units, by

\[{\bf \sigma}_N = \; \frac{1}{4\pi\epsilon_0c^2} \, \frac{e^2}{2} \; \langle\langle \; \left (\frac{{\bf r}-{\bf r}_N}{|{\bf r} -{\bf r}_N|^3 }\times c \, {\bf \alpha}\right) \; ; \; \left( {\bf r}-{\bf r}_{GO} \times c \, {\bf \alpha}\right) \; \rangle\rangle\]

then we print also the SR constants in ppm, by multiplying each SR tensor element \(M_{N,\alpha\beta}\) (given in kHz) by \(\frac{m_p I_{\beta \gamma}}{g_N}\,\frac{2\, \pi \, 10^9}{m_e \, \hslash}\), where \(m_p\) and \(m_e\) are the proton and electron masses, respectively, \(\hslash\) is the reduced Planck’s constant, and \(g_N\) is the g-factor of nucleus N.

In this way, one obtains the following output:

                      Spin-rotation constants (in ppm) for F
                      --------------------------------------

  Total spin-rotation constant (SRC)      :    -88.19431889

    Nuclear contribution to SRC (M^nuc)   :     14.59097597
    Electronic contribution to SRC (M^LR) :   -102.78529486
            M^LR-L(e-e) :   -106.08630700
            M^LR-S(e-e) :      3.29740318
            M^LR-L(e-p) :      1.18981000
            M^LR-S(e-p) :     -1.18620104

******************************************************
 ********      M^NU         ********:     15.29474932
 ********      M^EV         ********:     -0.69932024
 ********      M^LR         ********:   -102.78529486
 ********      M^total      ********:    -88.18986578
******************************************************