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: 3486582739 ISBN: 3486582739 ISBN: 3486582739 ISBN: 3486582739

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Next: The Knight Shift Up: The Spin Hamiltonian Previous: The Zeeman Interaction

The Quadrupolar Interaction

Besides the Zeeman interaction of the spin of the nucleaus with the external field, one would expect also the electrostatic interaction of the charge distribution of the nucleus with the charge distribution due to other nuclei in its surroundings to be a significant term in the total Hamiltonian. In this section we will discuss this term and its implication for NMR measurements. The classical electrostatic energy of a nucleus at the origin surrounded by a charge distribution is given by

$\begin{displaymath}E = \int \rho({\bf {r}}) \cdot V ({\bf {r}}) d^3r, \end{displaymath}$

(3.7)

where $\rho(\mathrm{\bf {r}})$ denotes the charge distribution of the nucleus and $V (\bf {r})$ the potential due to the environement. We expand the potential $V (\bf {r})$ in a Taylor's series about the origin and rewrite (

$\begin{displaymath}E = V(0) \int \rho d^3x + \sum_{\alpha} V_{\alpha} \int x_{\a... ...V_{\alpha \beta} \int x_{\alpha} x_{\beta} \rho d^3x + . . ., \end{displaymath}$

(3.8)

defining

$\displaystyle V_{\alpha}$	:=	$\displaystyle \frac{\partial V}{\partial x_{\alpha}}\Big\vert _{r=0}$	(3.9)
$\displaystyle V_{\alpha \beta}$	:=	$\displaystyle \frac{\partial^2 V}{\partial x_{\alpha} \partial x_{\beta}} \Big\vert _{r=0}.$	(3.10)

The first term in () is the electrostatic energy of the nucles treated as a point charge. It is a constant offset and of no interest for NMR because it does not affect any transitions we are measuring. The second term in () vanishes -- as every even term in this expansion does -- because of parity considerations (see [#!benno!#] and [#!domi!#]) The third term is the only term of interest as far as NMR is concerned. It is called the electrical quadrupole term. It shows the electrostatic energy of a non-spherical nuclear charge distribution varying with its orientation relative to the orientation of the field gradient. Turning the nucleus end by end does not affect its energy. Consequently the quadrupolar interaction does not affect the single transition of a spin $\frac{1}{2}$ nucleus. Because V is a real symmetric square matrix there is always a coordinate system of principal axes of the potential V such that (see [#!berberian!#]):

$\begin{displaymath}V_{\alpha \beta}=\delta_{\alpha \beta} \cdot V_{\alpha \beta} \end{displaymath}$

(3.11)

and with the Laplace's equation we see that

$\begin{displaymath}\mathrm{\bf {Tr}} \ V_{\alpha \beta} = \sum_{\alpha} V_{\alpha \alpha} = 0. \end{displaymath}$

(3.12)

In cubic symmetry the equation

V_xx = V_yy = V_zz = 0

(3.13)

holds.

It is very convenient to introduce the quandrupolar tensor

$\begin{displaymath}Q_{\alpha \beta} := \int(3 x_{\alpha} x_{\beta} - \delta_{\alpha \beta} r^2) \rho \cdot d^3x. \end{displaymath}$

(3.14)

Because of equation (

) and (

) the third term of (

) reduces to

$\begin{displaymath}E^{(2)} = \frac{1}{6} \sum_{\alpha \beta} V_{\alpha \beta} Q_{\alpha \beta}. \end{displaymath}$

(3.15)

To obtain the Hamiltonian operator we have to find a quantum mechanical expression for the quadrupole operator. This is easily done by replacing the classical charge distribution by a sum over all protons. The Hamiltonian then is

$\begin{displaymath}\mathcal{H}_Q = \frac{1}{6} \sum_{\alpha \beta} V_{\alpha \beta} Q^{(op)}_{\alpha \beta}, \end{displaymath}$

(3.16)

with

$\begin{displaymath}Q_{\alpha \beta}^{(op)} = e \sum_{\mathrm{protons} \ k}(3 x_{k \alpha} x_{k \beta} - \delta_{\alpha \beta} r_k^2). \end{displaymath}$

(3.17)

To deal with () involves to deal with the quadrupole operator () which is a many particle operator. Since the form () is a sum over all protons in the nucleus it would enable us to treat also transitions to excited core states which will never happen in NMR experiments. For our purposes it is sufficient to deal with a formalism where the nucleus remains in a sufficently stable state. It is therefore adequate to characterize the eigenstates of a nucleus by the quantum number of the total angular momentum, denoted by I, the z-component of it, denoted by m and a set of other quantum numbers we are not interested in, denoted by $\xi$ . As already mentioned, in NMR there will be only transitions between subspaces with the same I and $\xi$ : $\vert I,m,\xi> \longleftrightarrow \vert I,m',\xi>$ . Thus our only interest applies to the matrix elements of the quadrupole operator which are diagonal in I and $\xi$ , such as

$\begin{displaymath}\left< I m \xi \left \vert Q_{\alpha \beta}^{(op)} \right\vert I m' \xi \right>. \end{displaymath}$

(3.18)

By means of the Wigner-Eckart theorem (conf. [#!theis!#]) which applies to irreducible tensor operators such as our quadrupole operator (see [#!slichter!#]), these matrix elements can be written as

$\begin{displaymath}\left< I m \xi \left \vert Q_{\alpha \beta}^{(op)} \right\ver... ...a}) - \delta_{\alpha \beta}I^2 \right\vert I m' \xi \right>, \end{displaymath}$

(3.19)

with C a constant, independent of m, $\alpha$ and $\beta$ . It can be obtained from the evaluation of the matrix elements (

) with $\alpha=\beta=x$ and m=m'=I:

$\begin{displaymath}C = \frac{eQ}{I(2I-1)}, \end{displaymath}$

(3.20)

with eQ the quadrupole coupling constant

$\begin{displaymath}eQ = \left< I I \xi \left\vert \sum_{\mathrm{protons} \ k } (3 z_k^2 - r_k^2) \right\vert I I \xi \right>, \end{displaymath}$

(3.21)

which one usually obtains from measurements or simply from tabulars. The fact that $V_{\alpha \beta}$ is symmetric leads to further simplifications of the Hamiltonian. Using Laplace's equation after having choosen a set of principal axes x, y, z relative to which $V_{\alpha \beta}$ is diagonal, $\mathcal H _{Q}$ gives

$\begin{displaymath}\mathcal{H}_Q = \frac{e^2qQ}{4I(2I-1)}\left[ 3I_z^2 -I(I+1)+\frac{\eta}{2} (I_+^2+I_-^2) \right], \end{displaymath}$

(3.22)

with eq = V_zz, $\eta=\frac{V_{xx} - V_{yy}}{V_{zz}}$ and $\left\vert V_{zz} \right\vert \geq \left\vert V_{xx} \right\vert \geq \left\vert V_{yy} \right\vert$ . It is due to the Wigner-Eckart theorem, that all the nuclear properties only enter into the equation through eQ and that the electrostatic field due to the environement enters only through $\eta$ and eq. $\eta$ is called the asymmetry paramter and reflects the site symmetry of the nuclear site under investigation. In the case of having the nucleus at a site of the symmetry c4 with respect to the z-axis, V_xx=V_yy and therefore $\eta = 0$ . In these cases it is useful to define the quadrupole frequency by

$\begin{displaymath}\nu_Q := \frac{3e^2qQ}{h2I(2I-1)}. \end{displaymath}$

(3.23)

It denotes the distance in frequency between the transitions $1/2 \longleftrightarrow 3/2$ and $-3/2 \longleftrightarrow -1/2$ . In cubic symmetry eq vanishes due to the Laplace's equation. In lower symmetries $\nu_Q$ has to be obtained by diagonalizing $V_{\alpha \beta}$ which can be found with a simple numerical program such as described in [#!domi!#].

The total Hamiltonian of a nuclear spin with a quadrupole moment in the presence of an external field can now be written:

$\begin{displaymath}\mathcal{H} = - \gamma_\mathrm{N} \hbar \mathrm{\bf {H \cdot ... ...} \left[ 3I_z^2 -I(I+1)+\frac{\eta}{2} (I_+^2+I_-^2) \right]. \end{displaymath}$

(3.24)

Our interest is now focussed on finding the eigenvalues of (). In the most general case this implies that we have to solve a secular equation of the order of 2I+1. Here we only want to discuss the special case wich occurred in our measurements, namely the case of a comparably high field and a field gradient with c4 symmetry ( $\eta = 0$ ). For all further considerations we choose the Z-axis along the magnetic field and write $I_Z = I_z \mathrm{cos} \theta + I_x \mathrm{sin} \theta$ , where $\theta$ denotes the angle between the z-axis (principal axis of $V_{\alpha \beta}$ , given by the crystal orientation) and the Z-axis (axis of the external magnetic field). In that frame all the off-diagonal elements of the hamiltonian are due to the quadrupole interaction, which is, in our special case, assumed to be weaker than the Zeeman interaction. It therefore can be treated as a perturbation. After a little algebra (see [#!abragam!#]) one obtains

E_m⁽0)	=	$\displaystyle -\gamma \hbar H m := h \nu_L m,$	(3.25)
E_m⁽1)	=	$\displaystyle \frac{1}{4} h \nu_Q(3\mu^2-1)(m^2-\frac{1}{3}a),$	(3.26)
E_m⁽2)	=	$\displaystyle -h \left( \frac{\nu^2_Q}{12\nu_L}\right)m \left[ \begin{matrix} \... ...mu^2)(8m^2-4a+1) \\ + \frac{3}{8}(1-\mu^2)^2(-2m^2+2a-1) \end{matrix} \right],$	(3.27)

with a:=I(I+1) and $\mu := \mathrm{cos} \theta$ .

The frequency at which the transition $m \longleftrightarrow m-1$ appears in the spectrum now is:

$\begin{displaymath}\nu_m := \frac{E_{m-1} - E_{m}}{h} = \nu_L + \nu_m^{(1)} +\nu_m^{(2)} + ... \ . \end{displaymath}$

(3.28)

Since our experiments were all performed in ¹¹B which has spin 3/2 we want to calculate a spectrum for I=3/2 explicitly up to second order in $\nu_Q$ . With

$\begin{displaymath}\nu_m^{(1)} = -\nu_Q (m-\frac{1}{2}) \frac{3 \mu^2 -1}{2} \end{displaymath}$

(3.29)

we see that in general for half integer spins the central line is not shifted in first order. Due to equation (

) there is no first order correction in the central line. The two other transitions of our example are shifted by an amount $\nu_m(\theta)$ depending on the orientation of the lattice relative to the external field such that finally the situation arises which is shown in figure

**Figure:** With only Zeeman interaction the energy levels of a spin are equidistant such that we would see only one line at $\nu_L$ . A small perturbation due to the quadrupole interaction shifts the energy levels depending on the orientation of the crystal relative to the external field such that the situation arises which is plotted on the right hand side. In that picture there are shown both, the spectrum of a crystal oriented with $\theta = \pi \text{\ or \ } 0$ and with $\theta = \pi /2$ respectively.
$\includegraphics[width=12cm]{quad_splitt.eps}$

If we go on to second order also the central line is slightly shifted. From equation () one obtains for the central line:

$\begin{displaymath}\nu_{1/2}^{(2)}=\frac{-\nu_Q^2}{16\nu_L} (a-\frac{3}{4})(1-\mu^2)(9\mu^2-1). \end{displaymath}$

(3.30)

Since E_m⁽²⁾ is an even function the distance between the wings is not affected by the second order correction and the first-order formula

$\begin{displaymath}\Delta \nu = \nu_Q (m-\frac{1}{2})(3\mu^2-1) \end{displaymath}$

(3.31)

can be used to determine $\nu_Q$ from the distance between the two wings in the spectrum.

Since we are very often measuring materials which are good conductors we have to be aware of the skin effect, which prevents the high frequency fields needed for our measurements from entering into the sample. To avoid this one usually powders the sample into little grains which are smaller than the skin depth. In the same time one obtains all possible orientations of the crystal relative to the field equally probable. But some samples tend to orient in the external field -- for example due to magnetic moments -- and the spatial distribution is no more homogenous. We just want to show you some powder spectra under different conditions which we obtained with a little numerical simulation (see figure ). Parts of the C-program can be found in the appendix .

**Figure:** The first viewgraph shows two simulated frequency spectra with $\nu_L = 70.961$ MHz and $\nu_Q = 622$ kHz. The line is of a Lorentzian shape and the line width is assumed to be completely given by the conditions of irradiation. The second viewgraph shows a sharp lined spectrum for pulses longer than we can do. Due to the limited computational facilities it was to time-expensive to increase the over-all resolution. But an enlarged view of the central line is given in the small box.
$\includegraphics[width=11cm]{powder_spectra.eps}$

Next: The Knight Shift Up: The Spin Hamiltonian Previous: The Zeeman Interaction




Festkörperphysik (Gebundene Ausgabe) von Neil W. Ashcroft, David N. Mermin

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