NMR Studies in Hexaborides Diplomarbeit in experimenteller Festkörperphysik.

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NMR Studies in Hexaborides Diplomarbeit in experimenteller Festkörperphysik.

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ISBN: 3055014006 ISBN: 3055014006 ISBN: 3055014006 ISBN: 3055014006

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Next: Relaxation Mechanisms Up: Nuclear Relaxation Previous: Magnus Expansion and Relaxation

Magnetization Recovery in a System described by more than one Temperature

In the case mentioned above we considered only Zeeman interaction and therefore one single line in the spectrum. In this case the dipole-dipole interaction can establish a single spin temperature. In the case of a non-vanishing quadrupole splitting -- as we have it in the Hexaborides -- the spin-spin interaction does no more establish a fast thermal equilibrium between the different states. This causes the magnetization recovery to be no more describable by a single exponential with a characteristic time scale T₁. In this section we will give an outline of the derivation of an expression for the magnetization recovery following the arguments of [#!rhodes!#]. A detailed calculation of the results we are going to quote here is given in [#!benno!#].

Let us start with a time-independent Hamiltonian in the eigenstate representation consisting of the Zeeman and the quadrupole interaction:

$\begin{displaymath}\mathcal{H}_0 = \sum_{n}E_n \left\vert n\right<\left>n\right\vert. \end{displaymath}$

(3.43)

With the further assumption that $T_2 \ll T_1$ the off-diagonal matrix elements of $\rho$ vanish long before the spin system relaxes with the lattice and we can again write:

$\begin{displaymath}\rho = \frac{\mathbf{1}-\sum_n \beta_n E_n P_n}{\mathrm{Tr} \mathbf{1}}, \end{displaymath}$

(3.44)

with $\beta_n$ representing the spin temperature of the state n and $P_n := \left\vert n\right>\left<n\right\vert$ .

The time-dependent perturbation we need to obtain relaxation is of the form:

$\begin{displaymath}\mathcal{H}_1(t) = \sum_{m,n}\mathcal{H}_{mn} \left\vert m\right>\left<n\right\vert. \end{displaymath}$

(3.45)

With the same series expansion we introduced in the last section and the substitution $\alpha_m := E_m \beta_m$ we obtain an expression similar to equation ():

$\begin{displaymath}\frac{d \alpha_m}{dt} = - \frac{1}{\hbar^2} \sum_n \alpha_n ... ...athcal H}_1(0)] [\widetilde{\mathcal H}_1(-\tau),P_n] d\tau . \end{displaymath}$

(3.46)

The evaluation of this expression is straight forward but a bit lengthy. It has been done for a general case in [#!kind!#] and for our case I=3/2 in [#!benno!#]. Here we only quote the result:

$\begin{displaymath}\alpha_m(t) = \sum_n c_n \nu_{mn}e^{-\lambda_n t}, \end{displaymath}$

(3.47)

with

$\displaystyle \bf {\boldsymbol{\lambda}}$	=	$\displaystyle \left( \begin{matrix} 2W \\ 0 \\ 6W \\ 12W \\ \end{matrix}\right),$	(3.48)
$\displaystyle \left[ \nu_{mn} \right]$	=	$\displaystyle \left( \begin{matrix} 0.67 & -0.5 & -0.5 & -0.22 \\ 0.22 & -0.5 ... ...2 & -0.5 & 0.5 & -0.67 \\ -0.67 & -0.5 & -0.5 & 0.22 \\ \end{matrix} \right),$	(3.49)

and W a common constant of the transition probabilities. The constants c_n are given by the initial conditions, which are displayed by the deviation of the situation after the irradiation from the situatuation in equlibrium with the lattice $\bf {\alpha} _\mathrm{equil.} - \bf {\alpha}_0$ . We are going to quote the results due to three different initial conditions.

If we assume the comb irradiation to be short enough that it does only affect the central transition we are speaking of the so called fast irradiation. The initial conditions then are

$\displaystyle \alpha_{3/2}(0)$	=	$\displaystyle \alpha_{-3/2}(0) = 0,$	(3.50)
$\displaystyle \alpha_{-1/2}(0)$	=	$\displaystyle - \alpha_{1/2}(0) \approx \frac{\gamma \beta \hbar H_0}{2},$	(3.51)

where $\alpha_m$ denotes the deviations of the m'th level from equilibrium. In this case the magnetization recovery is

$\begin{displaymath}m(\infty)-m(t) = C \left(0.1 e^{-t/T_1} + 0.9 e^{-6t/T_1}\right). \end{displaymath}$

(3.52)

If we assume the comb irradiation to be long enough that the levels we did not irradiate come to thermal equlibrium with the transition we did irradiate, we are speaking of the so called slow irradiation. The magnetization recovery then is

$\begin{displaymath}m(\infty)-m(t) = C \left(0.4 e^{-t/T_1} + 0.6 e^{-6t/T_1}\right). \end{displaymath}$

(3.53)

If we finally assume that the pulses in the comb are short enough (broad enough in frequency) that all transitions in the spectrum are irradiated more or less equally, we obtain again a magnetization recovery which is describable by a single exponential:

$\begin{displaymath}m(\infty)-m(t) = C e^{-t/T_1}. \end{displaymath}$

(3.54)

Next: Relaxation Mechanisms Up: Nuclear Relaxation Previous: Magnus Expansion and Relaxation




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