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# 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

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

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.

$(2) \qquad\qquad \langle M_x(t)\rangle = n \gamma \hbar\quad Tr \rho(t) I_x$

and

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

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 ralaxation rate is simply given by the ratio $\frac 1 {T_2} = - \overline{Tr \frac {d\rho}{dt} I_x}/ (Tr \rho I_x)$.

We know that the density matrix satisfies the Heisenberg equation

$\frac {d\rho} {dt} = \frac i {\hbar} \left[ {\cal H_0+ H}_1(t),\rho\right]$

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 left 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 $+\infty$

$(4) \qquad\qquad \overline{\frac {d\rho_I} {dt}} \approx -\frac 1 {\hbar^2} \int_0^\infty d\tau \overline {\left[ {\cal H}_{1I}(t),\left[ {\cal H}_{1I}(t-\tau), \rho_I(0)\right] \right]}$

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

$\rho_I(0)=\rho(0)=\frac 1 2 \left[ \begin{array} 1 & 1 \\ 1 & 1 \end{array}\right] = \frac 1 2 (\mathbb{1} + \sigma_x)$

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 $I_{\alpha I}(t)=e^{i\frac {\cal H_0} \hbar t}I_\alpha e^{-i\frac {\cal H_0} \hbar t}$. It is convenient to rewrite ${\cal H}_{1I}$ using $I_x=\frac {I_++I_-} 2$, $I_y=\frac {I_+-I_-} {2i}$ and $\omega_{e\pm}=\frac12 (\omega_{ex}\mp i\omega_{ey})$

${\cal H}_{1I}(t)= =- \hbar \left(\omega_{e+}(t)I_{+I}(t) +\omega_{e-}(t)I_{-I}(t)+\omega_{ez}(t)I_{z}\right)$

where we have exploited the fact that $\left[ {\cal H}_0 ,I_z \right] = 0$. It is easy to show by inspection on the basis of the eigenstates of ${\cal H}_0$ that the other two operators appearing on the right-hand-side may be rewritten as

$I_{\pm I}(t)=e^{\mp i \gamma B t} I_{\pm}$

The double commutator in Eq. (4) may then be spelled out into nine terms:

$\hbar^2 \overline{\omega_{e+}(t)\omega_{e+}(t-\tau)} e^{-i\gamma B (2t-\tau)} [I_+,[I_+,I_x]] = -\hbar^2 \overline{\omega_{e+}(0)\omega_{e+}(\tau)} e^{-i\gamma B (2t-\tau)} I_+$

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

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

$\red \hbar^2 \overline{\omega_{ez}(t)\omega_{ez}(t-\tau)} [I_z,[I_z,I_x]] = \hbar^2 \overline{\omega_{ez}(0)\omega_{ez}(\tau)} I_x \qquad \qquad \longleftarrow$

where we have applied translational invariance in time.

It is important to recognize the physical origin of the relaxation mechanism: the averaged products of two electron frequencies at different times represent the different components of the time self-correlation of the electronic field (i.e. the self-correlation of the electron spin $s$). With the simple assumption made so far these functions will decay exponentially with a characteristic time $\tau_c$. Within this approximation all terms containing factors $e^{i \gamma B t}$ are actually oscillating rapidly in time in the rotaing frame, i.e they will not contribute effectively to the relaxation and we may disregard them in comparison with the second, the third and the last term, indicated by the red arrows.

We are now in the position to evaluate the numerator of the ratio $\frac 1 {T_2} = - \overline{Tr \frac {d\rho}{dt} I_x}/ (Tr \rho I_x)$. The trace has three contributions, from the operators $I_+I_x$, $I_-I_x$ and $I_x^2$, and the first two traces are equal to the third (try it out!). Therefore we can write

$\overline{Tr \frac {d\rho} {dt}} I_x = - \int_0^\infty d\tau \left{\overline{\omega_{e+}(0)\omega_{e-}(\tau)} e^{-i\gamma B \tau} + \overline{\omega_{e-}(0)\omega_{e+}(\tau)} e^{i\gamma B \tau} + \overline{\omega_{ez}(0)\omega_{ez}(\tau)} \right} (Tr I_x^2)$

where the trace on the right is actually equal to $Tr \rho Ix$, hence the integral is term in front of it is, by Eq. (1) is the relaxation rate:

$\frac 1 {T_2} = \int_0^\infty d\tau \left{ \overline{\omega_{e+}(0)\omega_{e-}(\tau)} e^{-i\gamma B \tau} + \overline{\omega_{e-}(0)\omega_{e+}(\tau)} e^{i\gamma B \tau} + \overline{\omega_{ez}(0)\omega_{ez}(\tau)} \right}$

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.

Page last modified on February 06, 2013, at 11:00 PM