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: 3486575627 ISBN: 3486575627 ISBN: 3486575627 ISBN: 3486575627

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Next: Magnetization Recovery in a Up: Nuclear Relaxation Previous: Nuclear Relaxation

Magnus Expansion and Relaxation

Let us assume the Hamiltonian describing the spin system to be a sum of a large time-independent term $\mathcal{H}_0$ and a small time-dependent term $\mathcal{H}_1(t)$ . Having the section in mind we can directly go to the interaction representation:

$\begin{displaymath}i \hbar \frac{d \widetilde{\rho}}{dt}=\left[ \widetilde{\mathcal{H}}_1(t), \widetilde{\rho} \right], \end{displaymath}$

(3.34)

with

$\displaystyle \widetilde{\rho}$	:=	$\displaystyle e^{\frac{i}{\hbar}\mathcal H_0 t} \rho e^{\frac{-i}{\hbar}\mathcal H_0 t}$	(3.35)
$\displaystyle \widetilde{\mathcal H}_1(t)$	:=	$\displaystyle e^{\frac{i}{\hbar}\mathcal H_0 t} \mathcal{H}_1(t) e^{\frac{-i}{\hbar}\mathcal H_0 t}.$	(3.36)

For small perturbations $\widetilde{\mathcal H}_1(t)$ or for small t one can expand $\mathrm{exp} \left( \frac{-i}{\hbar} \widetilde{\mathcal H}_1(t) t \right)$ . Together with the first three terms of the expansion of $\widetilde{\rho}$ one obtains -- after some algebra -- the expression

$\displaystyle \widetilde{\rho} (t)$	$\textstyle \approx$	$\displaystyle \rho(0) + \frac{-i}{\hbar} \int_0^t\left[ \widetilde{\mathcal H}_1(t'), \rho(0) \right] dt' \notag$	(3.37)
	-	$\displaystyle \frac{1}{\hbar^2} \int_0^t \int_0^{t'} \left[\widetilde{\mathcal ... ...\left[ \widetilde{\mathcal H}_1(t''), \rho(0) \right]\right] dt' dt'' + ... \ .$	(3.38)

If we further assume that the spins reach equilibrium among themselves much faster than they relax with the lattice; or in other words if we assume that the spin system can be described by one single temperature T_S, and that the system is basically described by the Zeeman Hamiltonian $\mathcal H_0 \approx -\gamma_\mathrm{N} \hbar H_0 I_z$ , the unperturbet density matrix $\rho(0)$ is given by:

$\begin{displaymath}\rho (0) \approx \frac{ \mathbf{1} + \beta \gamma \hbar H_0} {\mathrm{Tr} \mathbf{1}}, \end{displaymath}$

(3.39)

with $\beta =k_\mathrm{B} T_S$ This equation holds in the high temperature limit where the energy of the lattice is much larger than the difference between the energy levels.

Since we are interested in a measurable quantity, which the magnetisation recovery for example is, it is sufficient to know only $\left<I_z\right> = \mathrm{ Tr} \widetilde{\rho}(t) I_z$ . Together with equation we obtain the result

$\begin{displaymath}\frac{1}{T_1} = \frac{1}{\hbar^2 \mathrm{Tr}I_z^2} \int_0^\i... ...athcal H}_1(0)] [\widetilde{\mathcal H}_1(-\tau),I_z] d\tau . \end{displaymath}$

(3.40)

It is interesting to notice, that the second term in the expansion (

) does not contribute to the recovery and that we therefore had to go to second order. It is suggested to read the application of this equation in the case of a fluctuating transversal field, which can be found in [#!kind!#]. The result will be used in section

Equation () can be written in a not obviously equivalent way [#!slichter!#]:

$\begin{displaymath}\frac{1}{T_1} = \frac{1}{2}\frac{\sum_{m,n}W_{mn}(E_m-E_n)^2} {\sum_{m}E_m^2}, \end{displaymath}$

(3.41)

with the transition probability W_mn given by Fermi's golden rule

$\begin{displaymath}W_{mn} = \frac{2 \pi}{\hbar}\sum_{i,f} \left\vert\left<m,i\ve... ...{int}} \vert f,n\right> \right\vert^2 \delta(E_f+E_m-E_i-E_n), \end{displaymath}$

(3.42)

with $\mathcal{H}_{\mathrm{int}}$ denoting the interaction of the spin with the lattice due to what ever process^3.3, and $\left\vert i,m\right>$ denoting the initial and $\left\vert f,n\right>$ the final state. Which of the two equations is more useful depends on whether the perturbation or the interaction is given.

Next: Magnetization Recovery in a Up: Nuclear Relaxation Previous: Nuclear Relaxation




Festkörperphysik (Broschiert) von Siegfried Hunklinger

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