MasterEquationsForT2

We shall simply apply the definition implicit in Bloch equations (as written in the rotating frame), and calculate the relaxation rate as the ratio between the average time derivative of the nuclear magnetization and the average magnetization at time {$t$}, starting from the transverse relaxation rate, or

{$$ \tag{1}\begin{equation} \frac 1 {T_2} = - \frac {\left\langle {dM_x(t)}/{dt}\right\rangle} {\langle M_x(t)\rangle} \end{equation}$$}

Here we have omitted the primes on the coordinates, but {$x$} and {$y$} refer to the rotating frame, so that {$\langle M_x(t) \rangle = \langle M_x(0) \rangle $} is constant. The averages may be computed with the density matrix formalism, i.e.

{$$ \tag{2}\begin{equation} \langle M_x(t)\rangle = n \gamma \hbar\quad Tr \rho(t) I_x \end{equation}$$}

and

{$$ \tag{3}\begin{equation} \left\langle \frac {dM_x(t)}{dt}\right\rangle = n \gamma \hbar\quad \overline {\frac {d}{dt} Tr \rho(t) I_x} = n \gamma \hbar\quad \overline {Tr \frac {d\rho}{dt} I_x} \end{equation}$$}

In the last equation the overline recalls that an average must be computed over the electron dynamics, which is irrelevant for the purely nuclear quantity {$I_x$} in the previous equation. Summarizing, the transverse relaxation rate is simply given by the ratio

{$$ \tag{4}\begin{equation}\frac 1 {T_2} = - \overline{Tr \frac {d\rho}{dt} I_x}/ (Tr \rho I_x)\end{equation}$$}.

We know that the density matrix satisfies the Heisenberg equation

{$$ \frac {d\rho} {dt} = \frac i {\hbar} \left[ {\cal H}_0+ {\cal H}_1(t),\rho\right]$$}

where {$ {\cal H}_0=-\hbar\gamma B I_z$} and {$ {\cal H_1}(t)=\hbar\gamma \mathbf{B_e}(t) \cdot \mathbf{I} $} may be treated as a small perturbation, not in view of its instantaneous value (very large), but thanks to its vanishing time average. In the Appendices? we also show that, moving to the interaction representation by means of the unitary transformation

{$$ \rho_I(t)=e^{i\frac {\cal H_0} \hbar t}\rho(t)e^{-i\frac {\cal H_0} \hbar t}$$}

we obtain a simpler dynamical equation for the Interaction density matrix

{$$ \frac {d\rho_I} {dt} = \frac i {\hbar} \left[ {\cal H}_{1I}(t),\rho_I(t)\right]$$}

where the Hamiltonian is just the perturbation, {$ {\cal H}_{1I}(t)=e^{i\frac {\cal H_0} \hbar t}{\cal H}_{1}(t)e^{-i\frac {\cal H_0} \hbar t}$}, once again in the Interaction representation. This may be thought as the quantum equivalent of the rotating frame, where time evolution is dictated by the perturbation, {$ \cal H_1$}, since the effect of {$ \cal H_0$} coincides with the motion of the reference frame. We do not expect, therefore, to obtain precessions from the averages computed with {$\rho_I$}, but this is irrelevant when we are interested in the relaxation.

We further show in the Appendices, Eq(X)? that a first order approximation for {$\rho_I(t)$} on the right hand side leads to the approximate expression

{$$ \frac {d\rho_I} {dt} \approx \frac i {\hbar} \left[ {\cal H}_{1I}(t),\rho_I(0)\right] - \frac 1 {\hbar^2} \int_0^t d\tau \left[ {\cal H}_{1I}(t),\left[ {\cal H}_{1I}(t-\tau), \rho_I(0)\right] \right] $$}

We recall that the much faster electron spin dynamics allows us to treat it as a random statistical process, which we take into account by the overline average

{$$ \overline{\frac {d\rho_I} {dt}} \approx -\frac 1 {\hbar^2} \int_0^t d\tau \overline {\left[ {\cal H}_{1I}(t),\left[ {\cal H}_{1I}(t-\tau), \rho_I(0)\right] \right]} $$}

We have considered that the electron average of the first term vanishes because {$ \overline {{\cal H}_{1I}} $} does (since it is the vanishing average Zeeman interaction with the fast-fluctuating electron field {$\mathbf{B_e}(t)$}). More accurately this average should be accomplished by the use of a density matrix which includes electron degrees of freedom, in a larger Hilbert space which is the tensor product of those of the electrons and the nuclei. The more formal treatment would allow for correlation effects which we are here forcedly neglecting.

The last step before we can actually use this formalism is to assume the simplifying hypothesis that the electron dynamics has just one intrinsic characteristic time {$\tau_c$}, such that for {$t\gg \tau_c$} the electron self-correlations are lost (the electron system has lost memory of its previous history). In this case we may suppose that the right-hand-side integral vanishes for {$t\gg \tau_c$}, and we may as well replace the upper limit with {$+\infty$}

{$$ \tag{5}\begin{equation} \overline{\frac {d\rho_I} {dt}} \approx -\frac 1 {\hbar^2} \int_0^\infty d\tau \overline {\left[ {\cal H}_{1I}(t),\left[ {\cal H}_{1I}(t-\tau), \rho_I(0)\right] \right]} \end{equation}$$}

We now recall that, after a {$\frac \pi 2$} pulse, the density matrix for a spin {$I=\frac 1 2$} is:

{$$ \rho_I(0)=\rho(0)=\frac 1 2 \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} = \frac 1 2 (\mathbf{1} + \sigma_x) $$}

If we compute average of traceless observables {$O$}, as it is often the case with spin operators, the density matrix may be simply written {$\rho=I_x$}, since {$Tr \mathbb{1}O=TrO=0$}. In order to calculate Eq. (4) we may start considering that

{$${\cal H}_{1I}(t)= =- \hbar \left(\omega_{ex}(t)I_{xI}(t) +\omega_{ey}(t)I_{yI}(t)+\omega_{ez}(t)I_{zI}(t)\right)$$}

where, for each Cartesian component {$\alpha=x,y,z$}, we have written {$\gamma B_{e\alpha}=\omega_{e\alpha}$} and he three spin operators in the interaction representation are {$I_{\alpha I}(t)= e^{-i{\cal H}_0 t/\hbar}\,I_\alpha\,e^{i{\cal H}_0 t/\hbar}$}. It is convenient to rewrite {${\cal H}_{1I}$} using {$I_x=\frac {I_++I_-} 2$}, {$I_y=\frac {I_+-I_-} {2i}$}. The coefficients of the operators {$I_{\pm}$} are

{$$ \tag{6}\begin{equation} \omega_{e\pm}=\frac 1 2 (\omega_{ex}\mp i\omega_{ey})\end{equation}$$}

and the Interaction hamiltonian is

{$$ \tag{7}\begin{equation}{\cal H}_{1I}(t)= =- \hbar \left(\omega_{e+}(t)I_{+I}(t) +\omega_{e-}(t)I_{-I}(t)+\omega_{ez}(t)I_{Iz}(t)\right)\end{equation}$$}

Since {$ \left[ {\cal H}_0 ,I_z \right] = 0 $} we have {$I_{zI}=I_z$}. For the other two it is sufficient to consider their action on the eigenstates of {$\cal H_0$} to see that

{$$ \tag{8}\begin{equation}I_{\pm I}(t)=e^{\mp i \gamma B t} I_{\pm} \end{equation}$$}

As we anticipated above Eq. (5), we are going to consider the NMR signal at times {$t\gg \tau$}, since the latter is the much shorter time characteristic of electronic dynamics. Substituting Eq. (8) and (7) in Eq. (5) the time dependence of the spin operators will give rise to combinations of factors {$e^{\pm i\gamma B t}$} -- from the operator {${\cal H}_I(t)$} -- and {$e^{\pm i\gamma B (t-\tau)}$} -- from the operator {${\cal H}_I(t-\tau)$}. They naturally split into time independent and {$e^{\pm i\gamma B \tau}\approx 1$} factors, on one side, and the {$t$} dependent factors, that are high frequency oscillating terms with zero average.

Only the first kind leads to a measurable relaxation. Let us identify them and limit our calculation to them only. There are nine terms in Eq. (5) arising from the products of {${\cal H}_I(t){\cal H}_I(t-\tau)$}. They are due to the following combinations of spin operators:

{$$ \begin{matrix} I_\pm(t) I_\pm(t-\tau) & & e^{\mp i\gamma B(t + (t-\tau))} =&e^{\mp i(2t - \tau)}\\ \color {red} {I_\pm(t) I_\mp(t-\tau)} & & e^{\mp i\gamma B(t - (t-\tau))} =& \color {red}{e^{\mp i \gamma B\tau}}\\ I_\pm(t),I_z(t-\tau) &\mbox{leading to oscillating factors}& & e^{\mp i\gamma Bt} \\ I_z(t) I_\pm(t-\tau) & & & e^{\mp i\gamma B (t-\tau)}\\ \color {red} {I_z(t)I_z(t-\tau)} & & & \color {red} 1 \\ \end{matrix} $$}

In conclusion we must consider only the red terms, the so called secular combinations {$I_\pm(t)I_\mp(t-\tau)$} and {$I_z(t)I_z(t-\tau)$}.

Let us insert the secular contributions in Eq. (5) (in the absence of a strong external field, the full list of terms must be considered, as it is shown in the zero field case). The minus signs of the two {${\cal H}_{1I}$} cancel and so do their two factor {$\hbar$} with {$-1/\hbar^2$} leaving a minus sign. The right hand side of Eq. (5) with its double commutator may then be spelled out into nine terms:

{$$ \begin{align} - \overline{\omega_{e+}(t)\omega_{e-}(t-\tau)} e^{-i\gamma B \tau} [I_+,[I_-,I_x]] & = - \overline{\omega_{e+}(0)\omega_{e-}(\tau)} e^{-i\gamma B \tau} I_+ \\ - \overline{\omega_{e-}(t)\omega_{e+}(t-\tau)} e^{+i\gamma B \tau} [I_-,[I_+,I_x]] & = - \overline{\omega_{e-}(0)\omega_{e+}(\tau)} e^{+i\gamma B \tau} I_- \\ - \overline{\omega_{ez}(t)\omega_{ez}(t-\tau)} [I_z,[I_z,I_x]] & = - \overline{\omega_{ez}(0)\omega_{ez}(\tau)} I_x \\ \end{align}$$}

where we have applied translational invariance in time and time inversion symmetry: {$\overline{\omega(t)\omega(t-\tau)}=\overline{\omega(0)\omega(\tau)}$}.

We are now in the position to evaluate the numerator in Eq. (4). The trace has three contributions, from the operators {$I_+I_x$}, {$I_-I_x$} and {$I_x^2$}. From the definition of {$I_\pm=I_x\pm I_y$}, and the fact that {$Tr I_x I_y =0$} they all reduce to {$Tr I_x^2 = \frac 1 3 Tr I^2 = \frac {I(I+1)} 3 $}. We can then write

{$$ \overline{Tr \frac {d\rho} {dt}} I_x = -\frac {I(I+1)} 3 \int_0^\infty d\tau \left\{ \overline{\omega_{e+}(0)\omega_{e-}(\tau)} e^{-i\gamma B \tau} + \overline{\omega_{e-}(0)\omega_{e+}(\tau)} e^{i\gamma B \tau} + \overline{\omega_{ez}(0)\omega_{ez}(\tau)} \right\} $$}

Substituting back the definitions of {$\omega_{e\pm}$} Eq. (6), the ensemble averages of the frequency correlations are

{$$ \overline{\omega_{e+}(0)\omega_{e-}(\tau)} e^{-i\gamma B \tau} + \overline{\omega_{e-}(0)\omega_{e+}(\tau)} e^{i\gamma B \tau} = \frac 1 2 [ \overline{\omega_{ex}(0)\omega_{ex}(\tau)}+ \overline{\omega_{ey}(0)\omega_{ey}(\tau)}] \cos \gamma B \tau + \frac 1 2 [ \overline{\omega_{ex}(0)\omega_{ey}(\tau)} - \overline{\omega_{ey}(0)\omega_{ex}(\tau)}] \sin \gamma B \tau$$}

The terms proportional to {$\sin \gamma B t$} represent a build-up of out of transverse coherence with a {$\pi/2$} shift with respect to the initial phase. The build-up competes with the relaxation process and since the electron correlation is much faster then the nuclear Larmor period {$\tau\ll 1/\gamma B$} we shall neglect the {$\sin \gamma B t$} here. The relaxation rate from Eq. (4) then is:

{$$ \begin{equation} \frac 1 {T_2} = \int_0^\infty d\tau \left\{ \frac 1 2 \left[\overline{\omega_{ex}(0)\omega_{ex}(\tau)} + \overline{\omega_{ey}(0)\omega_{ey}(\tau)}\right] \cos\gamma B \tau + \overline{\omega_{ez}(0)\omega_{ez}(\tau)} \right\} \end{equation}$$}

This may be read as the sum of three terms, coinciding with the Fourier transform of the self-correlation functions of the frequencies corresponding to the electron hyperfine field components, namely at the Larmor frequency {$\omega=\gamma B$} for the transverse components and at {$\omega=0$} for the longitudinal component.

We have supposed for simplicity that a single correlation time {$\tau$} governs the electron spin dynamics, such that {$\overline{\omega_{e\alpha}(0)\omega_{e\alpha}(\tau)} = \omega_0^2 e^{-t/\tau}~\alpha=x,y,z$}. In this case the integrals are straightforward and give

{$$\begin{align} \omega_0^2 \int_0^\infty d\tau e^{-t/\tau} \cos \gamma B \tau &= \frac {\omega_0^2\tau} {1+\gamma^2 B^2 \tau^2}\\ \omega_0^2 \int_0^\infty d\tau e^{-t/\tau} &= \omega_0^2\tau \end{align}$$}

hence {$T_2^{-1}=\omega_0^2\tau [1+1/(1+\gamma^2B^2\tau^2)]$}. This expression may be read in the following way: there are two contributions, from longitudinal and transverse components with respect to the initial nuclear spin direction, just after the pulse sequence, say {$z$}. The longitudinal component, that comes from the {$\omega=0$} correlations of the {$z$} components of the local magnetic field, has a weight equal to that of the {$\omega=\omega_L$} correlations of the transverse components of the field. That is because there are two transverse components, but only one of them is secular, i.e. along the total local field.

Zero field case

If there is no {${\cal H}_0$}, as e.g in zero magnetic field, with no quadrupolar interactions, then no term can be discarded in Eq.(5) and the whole list reads

{$$ \begin{align} - \overline{\omega_{e+}(t)\omega_{e+}(t-\tau)} e^{-i\gamma B (2t-\tau)} [I_+,[I_+,I_x]] & = - \overline{\omega_{e+}(0)\omega_{e+}(\tau)} e^{-i\gamma B (2t-\tau)} -I_+ & \\ {\color{red} {- \overline{\omega_{e+}(t)\omega_{e-}(t-\tau)} e^{-i\gamma B \tau} [I_+,[I_-,I_x]] } } & = {\color{red}{- \overline{\omega_{e+}(0)\omega_{e-}(\tau)} e^{-i\gamma B \tau} I_+ } } & {\color{red}\longleftarrow}\\ & = {\color{red}{- \overline{\omega_{e-}(0)\omega_{e+}(\tau)} e^{i\gamma B \tau} I_- } } & {\color{red}\longleftarrow} \\ - \overline{\omega_{e-}(t)\omega_{e-}(t-\tau)} e^{i\gamma B (2t-\tau)} [I_-,[I_-,I_x]] & = \overline{\omega_{e-}(0)\omega_{e-}(\tau)} e^{i\gamma B (2t-\tau)} -I_- \\ - \overline{\omega_{e+}(t)\omega_{ez}(t-\tau)} e^{-i\gamma B t} [I_+,[I_z,I_x]] & = - \overline{\omega_{e+}(0)\omega_{ez}(\tau)} e^{-i\gamma B t} -I_z \\ - \overline{\omega_{e-}(t)\omega_{ez}(t-\tau)} e^{i\gamma B t} [I_-,[I_z,I_x]] & = \overline{\omega_{e-}(0)\omega_{ez}(\tau)} e^{i\gamma B t} -I_z\\ - \overline{\omega_{ez}(t)\omega_{e+}(t-\tau)} e^{-i\gamma B (t-\tau)} [I_z,[I_+,I_x]] & = 0\\ - \overline{\omega_{ez}(t)\omega_{e-}(t-\tau)} e^{i\gamma B (t-\tau)} [I_z,[I_-,I_x]] & = 0 \\ {\color{red} {- \overline{\omega_{ez}(t)\omega_{ez}(t-\tau)} [I_z,[I_z,I_x]] }} &= {\color{red} {- \overline{\omega_{ez}(0)\omega_{ez}(\tau)} I_x } } & {\color{red} {\longleftarrow }} \end{align}$$}