MasterEquationsQuadrupolar

This page addresses relaxation in perturbed high field NMR, i.e. in a large external or internal magnetic field {$B$}. We loosely follow Mehring and the nuclear spin Hamiltonian is {$$ \tag{1} \begin{equation} {\cal H} = {\cal H}_Z + {\cal H}' \end{equation}$$} with two inequivalent spin, {$I$}, detected, and {$S$}, undetected, possibly the electron, so that {$ {\cal H}_z = \omega_I I_z + \omega_S S_z$}, where {$\omega_{I,S}=^{I,S}\gamma B$} is just the reference frequency for each species (no shift). We skip here a term {${\cal H}_1$} for the radiofrequency field {$B_1$} applied along {$\hat x$} and a lattice term {${\cal H}_L$}. Dipolar, quadrupolar and hyperfine (even direct) spin interactions are in {${\cal H}'$}.

The full hierarchy of energy scales is, from largest to smallest, Zeeman interactions by far, then radiofrequency and spin interactions, with spin lattice as smallest. Therefore the system is analysed in the basis set of the Zeeman interactio, considering an appropriate rotating frame (notably, {$ [{\cal H}_Z,{\cal H}'] \ne 0$}) and in first approximation the other terms are restricted to their secular components (in the rotating frame).

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Interaction representation

The time-dependent spin Hamiltonian in this representation, equivalent to the classical system seen from the rotating frame at frequency {$\omega_I$}, is

{$$ {\cal H}^i(t) = e^{i{\cal H}_Zt/\hbar} {\cal H} e^{-i{\cal H}_Zt/\hbar} = (\omega-\omega_I)I_z + \underbrace{ e^{i(\omega_I I_z + \omega_S S_z)t} {\cal H}' e^{-i(\omega_I I_z + \omega_S S_z)t} }_{{\cal H}'^ì(t)} $$}

For the duration of the radiofrequency pulse the first term on the left includes also the secular part of the radiofrequency in the rotating frame, i.e. {$^I\gamma B_1 I_x$}. For short pulses the time dependent relaxation {${\cal H}'^i(t)$} can be neglected and on resonance, {$\omega =\omega_I$}, the dynamics is the pure nutation in the radiofrequency field {$B_1$}.

In the following all {$t$}-dependent operators are in the interaction representation, hence all z components of the spin operators have no explicit time dependence.

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Isotropic case: irreducible representations of tensor operators

We are considering for instance powders of a crystal with some isotropic lattice dynamics that modulates the interactions described by Eq. (1). We use the irreducible spherical rank-2 tensor representation of the spin operators in {${\cal H}'$} and hence in

{$$ \tag{2} \begin{equation}{\cal H}'^i(t) = \sum_{k=0}^2 \sum_{q=-k}^k (-1)^q A_{kq} {\cal T}_{k\,q}^i(t) = \sum_{k=0}^2 \sum_{q=-k}^k (-1)^q A_{kq} \, e^{i(\omega_IIz+\omega_sS_z)t} {\cal T}_{k\, -q}e^{-i(\omega_IIz+\omega_sS_z)t}\end{equation}$$}

Any rank-2 tensor can be obtained, in terms of its Cartesian components, {${\cal T}_{ij},\,(i,j=x,y,z)$}, as the sum of a scalar {${\cal T}_0= Tr({\cal T})/3$}, a traceless anti-symmetric rank-1 tensor {${\cal T}_1 = ({\cal T}_{ij}-{\cal T}_{ji})/2$} and a traceless rank-2 tensor {${\cal T}_2 = ({\cal T}_{ij}+{\cal T}_{ji})/2 -{\cal T}_0$} that transform as spherical harmonics of order 0,1 and 2, respectively. This means that in a new reference frame identified by Euler angles {$\alpha,\beta,\gamma$} the tensor is

{$$ {\cal T}'_{kq} = \sum_{p=-k}^k {\cal T}_{kp}D^k_{pq}(\alpha,\beta,\gamma)$$}

where {$D$} are Wigner matrices. It is convenient to project the tensor along the tensor product of three complex unit vectors {$(\hat x + i\hat y)\sqrt 2, (\hat x -i\hat y)\sqrt 2,\hat z$} of the Gauss (or Bloch) sphere. The irreducible spherical tensor components in this reference frame are

{$$ \tag{3}\begin{align} {\cal T}_{00} &= -\sqrt3 {\cal T}_0\nonumber\\ {\cal T}_{10} &= -\frac i {\sqrt2}({\cal T}_{xy}-{\cal T}_{yx})\nonumber\\ {\cal T}_{1\pm1} &= - \frac 1 2 [{\cal T}_{zx}\pm {\cal T}_{xz}-i({\cal T}_{zy}-{\cal T}_{yz})]\nonumber\\ {\cal T}_{20} &= \frac 3 {\sqrt6}({\cal T}_{zz}-{\cal T}_0)\nonumber\\ {\cal T}_{2\pm1} &= \mp\frac 1 2 [{\cal T}_{zx}+{\cal T}_{xz} \pm({\cal T}_{zy}+{\cal T}_{yz})]\nonumber\\ {\cal T}_{2\pm2} &= \frac 1 2[{\cal T}_{xx}-{\cal T}_{yy}\pm i ({\cal T}_{xy}+{\cal T}_{yx})] \end{align}$$}

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Weak collision master equations

Relaxation is due to the fact that {$ \cal{H}'$} does not commute with the secular spin lattice interaction. So when a collision between spin and lattice excitations takes place, it projects the nuclear spin state onto different states of the same chosen base (that of {${\cal H}_Z$}). We assume the weak collision limit, where collision rate {$\tau^{-1}$} is much faster than both the nuclear spin precession frequency {$\omega_I$} and the resulting relaxation rates {$T_{1,2}^{-1}$}. The interaction {$ {\cal H}'^i(t)$} is that of Eq. (2).

We start from a density matrix that has been rotated by the application of a pulse, hence its statistical average is {$\langle I_\xi(t) \rangle$} is {$I_\xi=e^{-i\beta I_y}I_z e^{i\beta I_y}$} at {$t=0$} and, under appropriate conditions for single exponential relaxation (e.g. the whole {$I$} spectrum is irradiated by strong, short pulses and the whole spectrum is detected) obeys

{$$ \frac d {dt} \overline{\langle I_\xi(t) \rangle} = -\frac 1 {T_{1\rho}} \overline{\langle I_\xi(t) \rangle} $$}

in the rotating frame. As in the previous pages {$\langle A(t) \rangle = Tr A\rho(t)$} while {$\overline A$} is the average over the lattice dynamics. The {$T_1^{-1}$} rate is obtained with {$\beta=0$}, in resonant conditions {$\omega=\omega_I$}, so that {$\hat \xi = \hat z$}, whereas {$\beta=\pi/2$}, hence {$\hat \xi = \hat x$} and {$\omega\rightarrow\omega_I$} provide {$T_2^{-1}$}. By second-order time-dependent perturbation theory in the former case we get

{$$ \frac d {dt} \overline{\langle I_z(t) \rangle} = - \int_0^t dt' \overline{Tr [{\cal H}'(t),[{\cal H}'(t'-t),I_x]]} $$}

and we actually get the equation we expects if we assume the interaction-representation time evolution operator {$U(t,t')=U(t-t')$}, to be time invariant. We recognize that in the weak collision limit only the real part ({$\Re$}) of this integral, corresponding to diagonal relaxation, is non vanishing, since coherent off-diagonal terms cannot build up, and in these conditions the integral can be extended to {$t\rightarrow \infty$}. We further assume that {$\overline{\langle I_z(t')\rangle}$} in the right-hand-side can be factored out of the integral as {$\overline{\langle I_z(t)\rangle}$}. This is not strictly true for {$I>\frac 1 2$} and a set of master equations are needed in the general case, giving rise to multi-exponential relaxation (see Bloch-Redfield equations in that case). It is still true in notable special experimental conditions mentioned above (whole spectrum) the whole spectrum. We assume this to happen and by substituting Eq. (2) into the above expression we get

{$$ \Re \int_0^\infty d\tau \sum_{k=0}^2 \sum_{q=-k}^k (-1)^q \overline{A_{kq}(0)A_{k-q}(\tau)} \frac {Tr\,[{\cal T}_{kq}(0),[{\cal T}_{k-q}(\tau),I_z]]\rho(t)} {\langle I_z(t)\rangle} $$}

This is what is called {$T_{1\rho}$} rate in the NMR jargon, where {$\rho$} stands here for rotating frame. Notice that the denominator of the last term

{$$\overline{\langle I_z(t)\rangle} = Tr I_z(t)\rho(t) = Tr I_z^2 = \frac {I(I+1)} 3$$}

since both the lattice average and the time evolution of a constant of motion are irrelevant, and we get

{$$ \tag{4} \begin{equation} \frac 1 {T_1^ì} = \frac 3 {I(I+1)} \Re \int_0^\infty d\tau \sum_{k=0}^2 \sum_{q=-k}^k (-1)^q \overline{A_{kq}(0)A_{k-q}(\tau)} Tr\,[{\cal T}_{kq}(0),[{\cal T}_{k-q}(\tau),I_z]]]\rho(t) \end{equation}$$}

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Dipolar

The dipolar interaction can be written as

{$$ \tag{5} \begin{equation} {\cal H}_D = \frac {\mu_0}{4\pi} \frac {^I\gamma^S\gamma \hbar^2}{r^3} \sum_{i,j =x,y,z} I_i D_{ij} S_j \end{equation}$$}

where {$r, \hat r$} are the distance and the unit vector joining the two spin, and the (traceless) dipolar tensor is {$D_{ij}=\delta_{ij}-3\hat r_i \hat r_j\,(i,j=x,y,z)$}. Identifying the spherical spin operator components is like writing the dipolar alphabet (Abragam p. 103, Slichter 3rd edition p. 65). One must spell out Eq. (5), separate the six (A-E) different functions of {$\theta,\phi$}, recognize them as spherical harmonics and rewrite their spin operators using {$I_z,I_\pm,Sz,S_\pm$}. Factoring out a coupling coefficient {$\lambda=\mu_0\,^I\gamma\,^S\gamma\,\hbar^2/4\pi r^3$} this yields spherical-harmonics-projected functions

{$$ \begin{align*} A_{20} &= \sqrt{\frac 3 2} D_{zz} = \sqrt{\frac 3 2}(3\cos^2\theta-1) \propto Y^0_2(\theta,\phi) \\ A_{2\pm 1} &=\mp (D_{xz}-iD_{yz}) = 3\sin\theta\cos\theta e^{\mp i \phi} \propto Y^{\mp 1}_2(\theta,\phi) \\ A_{2\pm 2} &= \frac 1 2 (D_{xx} - D_{yy} \pm 2i D_{xy}) = \frac 1 2 \sin^2\theta e^{\pm 2i\phi} \propto Y^{\mp 2}_2(\theta,\phi) \end{align*} $$}

and their spin operator components

{$$\begin{align*} {\cal T}_{20} &= \frac 1 {\sqrt 6} (3I_zS_z -\mathbf I \cdot \mathbf S)\\ {\cal T}_{2\pm1} &= \mp\frac 1 2 (I_zS_\pm + I_\pm S_z)\\ {\cal T}_{2\pm2} &= \frac 1 2 I_\pm S_\pm \end{align*}$$}

In order to compute Eq. (4) we need to evaluate the interaction time evolution of the relevant spin operator components, {$$e^{i(\omega_I I_z+\omega_SS_z)t}{\cal T}_{2 q}e^{-i(\omega_I I_z+\omega_SS_z)t} = \begin{cases}{\cal T}_{2 0}, & q=0\\ \mp\frac 1 2 (e^{\pm i\omega_St}I_zS_\pm + e^{\pm i\omega_It}I_\pm S_z), &q=\pm1\\e^{\pm i(\omega_I+\omega_S)t}{\cal T}_{2 \pm2}& q=\pm 2\end{cases}$$}

The numerator of the last term in Eq. (4) reads

{$$ \tag{6} \begin{align} Tr\,[{\cal T}_{kq}(0),[{\cal T}_{k-q}(\tau),I_z]]\, \rho(t) &= \begin{cases}Tr\,[{\cal T}_{2 0},[{\cal T}_{2 0},I_x]]\rho(0) & q=0\\ \frac 1 4 Tr\,[(I_zS_\pm + I_\pm S_z)[(e^{\mp i\omega_S\tau}I_zS_\mp + e^{\mp i\omega_I\tau}I_\mp S_z), I_z]]\,\rho(0)&q=\pm1\\ e^{\mp i(\omega_I+\omega_S)\tau}\,Tr\,[{\cal T}_{2 \pm2},[{\cal T}_{2 \mp2},I_z]]\,\rho(0)\,\rho(0)& q=\pm 2 \end{cases} \\ &= \begin{cases}Tr\,[3I_zS_z-\mathbf I\cdot \mathbf S,[3I_zS_z-\mathbf I\cdot \mathbf S,I_z]]\,I_z= 0, & q=0\\ \frac 1 4 Tr\, [(I_zS_\pm + I_\pm S_z)[(e^{\mp i\omega_S\tau}I_zS_\mp + e^{\mp i\omega_I\tau}I_\mp S_z), I_z]]\,I_z=\frac 1 4 e^{\mp i\omega_I\tau}\,\frac {I(I+1)} 3 \,\frac {S(S+1)} 3&q=\pm1\\ e^{\mp i(\omega_I+\omega_S)\tau}\,\frac 1 4 Tr\,[I_\pm S_\pm,[I_\mp S_\mp,I_z]]\,I_z=e^{\mp i(\omega_I+\omega_S)\tau} \,\frac {I(I+1)} 3\, \frac {S(S+1)} 3 & q=\pm 2 \end{cases} \end{align} $$}

where we used the fact that several commutators either vanish or produce traceless operators. Numerator and denominator in Eq. (4) yield

{$$ \tag{7} \begin{equation} \frac 1 {T_1^i} = \frac {\mu_0\,^I\gamma\,^S\gamma\,\hbar^2} {16\pi r^3} \, \frac {S(S+1)} 3 \left[ \, j_{21}(\omega_I) + 4j_{22}(2\omega_I) \,\right] \end{equation}$$}

where the spectral densities {$j_{q2}(\omega)$}, the Fourier transform of the time self correlation of the corresponding spherical harmonics, represent the component of the power spectrum of the lattice angular fluctuations at frequency {$\omega$}

{$$j_{2q}(\omega) = \Re\int_0^\infty d\tau (-1)^q \, \left[ \overline{A_{2q}(0)A_{2-q}(\tau)}+\overline{A_{2-q}(0)A_{2q}(\tau)} \right]\,e^{\mp i \omega\tau }$$} Top

Quadrupolar

The quadrupole electric coupling in the reference frame of the principal axes of the tensor is

{$$ {\cal H}_Q = Q\frac {V_{zz}}{4I(2I-1)}[3I_z^2-I(I+1)+\frac \eta 2 (I^2_+ + I^2_-)]$$}

where {$Q$} is the nuclear quadrupole moment, {$V_{zz}$} is the principal value of the electric field gradient (a traceless tensor) and {$\eta =(V_{xx}-V_{yy})/V_{zz}$} is the asymmetry parameter. Since {$I(I+1)=I_x^2+I_y^2+I_z^2$} and {$V_{xx}/V_{zz}=(\eta-1)/2$}, {$V_{yy}/V_{zz}=(-\eta-1)/2$} we can also write

{$${\cal H}_Q = \frac {Q}{2I(2I-1)}[V_{zz}I^2_z +V_{xx} I^2_x + V_{yy}I_y^2]$$}

It is clear from this last expression that the traceless quadrupole Hamiltonian has the same spherical representation of the traceless dipolar Hamiltonian, with the constraint {$S=I$} and {$Q\lambda = QV_{zz}/(2I(2I-1))$}. We write the spherical tensor operator components ignoring constant terms

{$$\begin{align*} {\cal T}_{20} &= \frac 1 {\sqrt 6} 3I_z^2 \\ {\cal T}_{2\pm1} &= \mp\frac 1 2 (I_zI_\pm + I_\pm I_z)\\ {\cal T}_{2\pm2} &= \frac 1 2 I_\pm^2 \end{align*}$$}

with the same spherical-harmonics components as in the dipolar case

{$$\begin{align*} A_{20} &= \frac {3V_{zz}}{4I(2I-1)}\\ A_{2\pm2} &= \frac {\eta V_{zz}}{4I(2I-1)} \end{align*}$$}

We can borrow the result of Eq. (6) and impose the constraints to obtain

{$$ \tag{8} \begin{align} Tr\,[{\cal T}_{kq}(0),[{\cal T}_{k-q}(\tau),I_z]]\, \rho(t) &= \begin{cases}Tr\,[{\cal T}_{2 0},[{\cal T}_{2 0},I_x]]\rho(0) & q=0\\ \frac 1 4(e^{\mp i\omega_S\tau}\, Tr\,[(I_zI_\pm + I_\pm I_z)[I_zI_\mp + I_\mp I_z), I_z]]\,\rho(0)&q=\pm1\\ e^{\mp 2i\omega_I\tau}\,Tr\,[{\cal T}_{2 \pm2},[{\cal T}_{2 \mp2},I_z]]\,\rho(0)\,\rho(0)& q=\pm 2 \end{cases} \\ &= \begin{cases}Tr\,[3I_z^2-I(I+1),[3I_z^2-I(I+1),I_z]]\,I_z= 0, & q=0\\ \frac 1 4 e^{\mp i\omega_I\tau}\, Tr\, [(I_zI_\pm + I_\pm I_z)[(I_z I_\mp + I_\mp I_z), I_z]]\,I_z=\frac 1 4 e^{\mp i\omega_I\tau}\,\left(\frac {I(I+1)} 3 \right)^2&q=\pm1\\ \frac 1 4 e^{\mp 2i\omega_I\tau}\, Tr\,[I_\pm^2,[I_\mp^2,I_z]]\,I_z=e^{\mp 2i\omega_I\tau} \,\left(\frac {I(I+1)} 3\right)^2 & q=\pm 2 \end{cases} \end{align} $$}

Equation (7) becomes

{$$ \tag{9}\begin{equation}\frac 1 {T_1^i} = \frac {QV_{zz}} {2I(2I-1)}\, \frac {I(I+1)} 3 \left[j_{2 1}(\omega_I) + 4 j_{2 2}(2\omega_I)\right]\end{equation}$$}

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Hyperfine

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References

M. Mehring 1983 Springer Ch. 9 Appendix Relaxations

A. Tunstall Proc. Royal Society 1961 78 1

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