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MasterEquationsForT1

< The master equations for 1/T2 | Index | Master Equations: quadrupolar case >

Following the same lines the longitudinal relaxation rate is :

{$$ \frac 1 {T_1} = - \frac 1 { M_{eq}} \left\langle \frac {dM_z(t)} {dt} \right\rangle = - \frac 1 {I_{z}(0)} \left\langle \frac {dI_z(t)} {dt} \right\rangle $$}

The simplest experiment to measure {$T_1^{-1}$} is the inversion recovery, whereby the initial state is obtained by a {$\pi$} nutation, i.e. an inversion of the spin populations. The density matrix in the initial state, just after the inversion, is in this case

{$$\rho = \frac 1 2 (1-\sigma_z)\sim I_z$$}

when evaluating the average of traceless tensors. Hence the calculation of the density matrix derivative {$d\rho(t)/dt$} is the same up to Eqs. (5)-(13) in the previous page. In particular the terms that give rise to non oscillating {$e^\pm \gamma B \tau,\, 1$} factors are those that contain the {$[I_\pm,[I_\mp,I_\xi]]$} and {$[I_z,[I_z,I_\xi]]$} double commutators, respectively. Here {$\xi=x,z$} represents the density matrix, depending on whether we compute {$T_{1,2}^{-1}$}.

Therefore the three relevant terms in the present case are

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

and consequently

{$$\begin{equation} \frac 1 {T_1} = \int_0^\infty d\tau \left[\overline{\omega_{ex}(0)\omega_{ex}(\tau)} + \overline{\omega_{ey}(0)\omega_{ey}(\tau)} \right]\cos \gamma B \tau = 2\frac {\omega_0^2\tau}{1+\gamma^2B^2\tau^2}\end{equation}$$}

where we have supposed the same electronic spin correlation functions {$\overline{\omega_{ex}(0)\omega_{ex}(\tau)}=\overline{\omega_{ey}(0)\omega_{ey}(\tau)}=\omega_0^2 e^{-t/\tau}$} as for the {$T_2^{-1}$} case.

Comparing Eq. (4) with 1/T₂, Eq. (10) we may say that 1/T₁ is due exclusively to transverse components, but both transverse components count now, hence their weight is double that of the {$1/T_2$} case. That is because, when the local field is along {$z$} bot {$x$} and {$y$} fluctuating fields can nutate the nuclear spin, acting like a random radio frequency-

< The master equations for 1/T2 | Index | Master Equations: quadrupolar case >