# Crystal Field Theory I

School is back in session, so I figure now is as good a time as any to start writing again!  In this series of posts I will discuss a subject that is a perennial favorite: the physical basis of crystal field theory.  We will consider the origin of the infamous crystal field parameter 10Dq, and later dwell on the Slater/Racah parameters (F0/F2/F4 and A/B/C, respectively).

### Introduction

Crystal field theory (CFT) was first developed by Hans Bethe and John Van Vleck – two physicists who made significant contributions to the modern quantum-mechanical theory of magnetism.  Beth and Van Vleck wanted to understand the paramagnetic behavior of simple transition metal salts (e.g. ferrous chloride). Their ideas grew from the theory of atomic electronic structure, which was relatively well developed at the time; and they elegantly showcase the power of group theory without requiring too much mathematical virtuosity on the part of the reader – a good combination, I think!

Before moving forward, we need to remind ourselves of a few things.  First of all, we need to understand free ion terms.  There are two ways we can think about them.  For first row transition metal ions (TMI), the most commonly-used description comes from the Russell-Saunders (LS) coupling scheme introduced in most undergraduate inorganic chemistry classes.  In the LS coupling  scheme, the spin-orbit splitting of a given free ion term is taken as a perturbation on the relatively large crystal field splitting.  This is in contrast to the splitting of lanthanide free ion terms, whose components’ characteristics are largely determined by spin-orbit coupling.  The validity of the LS coupling scheme for first row TMIs follows from the fact that 1) the 3d orbitals are relatively small, leading to large coulomb/exchange integrals and 2) the magnitude of spin-orbit coupling has a quartic dependence on nuclear charge, and is therefore is small for first row metals.

Implicit in our choice of the LS coupling scheme is the notion that an ‘uncoupled’ basis of direct product microstates

$\left| {d_n ,\chi } \right\rangle \in \left\{ {\left| {d_n } \right\rangle \otimes \left| \chi \right\rangle \left| {n = 0, \pm 1, \pm 2,\chi = \alpha ,\beta } \right.} \right\}$

can be used to construct good approximate descriptions of the components of first row TMI free ion terms.  For example: a d1 electronic configuration yields a 2D free ion term which is split into 2D5/2 and 2D3/2 levels by S.O. coupling.  The magnitude of the S.O. coupling is taken to be relatively small, however, so we turn our attention to the parent 2D term and its 10 constituent microstates (we will consider S.O. coupling later as a perturbation to the crystal field splitting).

CFT attempts to explain, in a perturbation-theoretic sense, how the components of a parent free ion term (2D in this example) are mixed by an external potential due to a symmetrical collection of point charges (i.e. ligands).  In other words, we will be studying a sort of intramolecular Stark effect. This point of view is admittedly simplistic;  but by throwing covalent contributions to bonding under the bus, we are free to focus our attention on the symmetry-determined aspects of electronic structure that I find most interesting.

### The Crystal Field Hamiltonian as a Multipole Expansion

To understand how a 2D free ion term is split by a crystal field, we need to phrase the question in a way that is amenable to calculation.  In other words, the way we represent the crystal field perturbation Hamiltonian $\hat H_{cf}$ should take advantage of whatever symmetry the problem has.  We will consider an octahedral crystal field for the sake of simplicity, but what follows may be generalized to other cases without substantial effort.

Moving forward, the general strategy will be to expand $\hat H_{cf}$ in an appropriate (infinite) basis, and then use symmetry considerations to determine which terms in the expansion contribute to the physics at hand.  Once we have chosen a suitably truncated representation of $\hat H_{cf}$, we will study how it mixes the 10 components of a 2D free ion term into Eg and T2g levels split by 10Dq.

The angular components of the 2D basis states may be represented by spherical harmonics, i.e.

$\left| {d_m } \right\rangle = \left| {R_{n,2} } \right\rangle \otimes \left| {Y_{2,m} } \right\rangle$

Spherical harmonics have a number of nice algebraic properties which we can utilize if we expand $\hat H_{cf}$ as a series of spherical harmonics sharing the same coordinate origin as the metal-centered basis states (i.e. as a multipole expansion):

$\hat H_{cf} = \alpha \sum\limits_{l = 0}^\infty {\sum\limits_{m = - l}^{m = l} {c_{lm} \hat Y_{lm} } }$

where scalar properties of the potential have been subsumed into the constant $\alpha$ (which I will promptly drop and forget about until later posts).

As written, this expansion isn’t terribly useful.  The rules of vector coupling allow us to perform a massive truncation, however:  two states for which l=2 can be only be coupled by components of $\hat H_{cf}$ having l$\le$ =2+2=4.

$\left\langle {d_a ,\chi _b } \right|\hat Y_{l,m} \left| {d_c ,\chi _d } \right\rangle = \left\langle {Y_{2a} } \right|\hat Y_{lm} \left| {Y_{2b} } \right\rangle \times \delta _{bd} = 0$.

Next, we take advantage of the octahedral symmetry of $\hat H_{cf}$: the potential must remain invariant under Oh symmetries, i.e. it must transform as the totally symmetric (a1g) representation of Oh.  Right off the bat, then, we can toss out the ${\hat Y_{1m} }$ and ${\hat Y_{3m} }$ terms, as they are ungerade and therefore cannot contribute to the totally symmetric potential.  The contribution from the  ${\hat Y_{00} }$ component affects all d orbitals equivalently (since ${\hat Y_{00} }$ transforms as a1g), so it is uninteresting to us.  The ${\hat Y_{2m} }$ components, which transform like the 5 d orbitals, are mixed into eg and t2g sets (in accord with the symmetry of the potential); no linear combination of the ${\hat Y_{2m} }$ spans a1g, so these components are also uninteresting to us.

This leaves the ${\hat Y_{4m} }$ set, which we consider a bit more carefully. First, we formalize our statement about invariance of $\hat H_{cf}$ under Oh symmetries by insisting $\hat \Lambda \hat H_{cf} = \hat H_{cf}$ $\forall \hat \Lambda \in O_h$. Now suppose $\hat \Lambda = \hat C_4$. Then

$\hat H_{cf} = \hat C_4 \hat H_{cf} = \hat C_4 \sum\limits_{m = - 4}^{m = 4} {c_{4m} \hat Y_{4m} } = \sum\limits_{m = - 4}^{m = 4} {e^{im\pi /2} \delta _{nm} c_{4m} \hat Y_{4m} } = \sum\limits_{n = - 4}^{n = 4} {e^{in\pi /2} c_{4n} \hat Y_{4n} }$

The coefficients of the ${\hat Y_{4-4} }$, ${\hat Y_{40} }$, and ${\hat Y_{44} }$ components remain invariant, so some linear combination of these components transforms as a1g. To understand the nature of this linear combination, we investigate the action of a few more symmetry operations on $\hat H_{cf}$. Suppose $\hat \Lambda = \hat C'_2$. Then

$\hat C'_2 \hat H_{cf} = \hat C'_2 \left( {c_{44} \hat Y_{44} + c_{40} \hat Y_{40} + c_{4 - 4} \hat Y_{4 - 4} } \right) = c_{44} \hat Y_{4 - 4} + c_{40} \hat Y_{40} + c_{4 - 4} \hat Y_{44}$

From this manipulation, we see $c_{44} = c_{4 - 4}$, i.e. (ignoring normalization) $\hat H_{cf} = \hat Y_{40} + c\left( {\hat Y_{4 - 4} + \hat Y_{44} } \right)$. To find the value of the remaining coefficient c, we consider one final equation. Suppose $\hat \Lambda = \hat C_3$. In this case, it is helpful to write out the spherical harmonics in their cartesian form (modulo a multiplicative factor of $\sqrt {9/4\pi }$:

$\hat H_{cf} = \sqrt {\frac{1}{{64}}} \left( {35\frac{{z^4 }}{{r^4 }} + 30\frac{{z^2 }}{{r^2 }} + 3} \right) + c\sqrt {\frac{{35}}{{128}}} \left[ {\left( {\frac{{x + iy}}{r}} \right)^4 + \left( {\frac{{x - iy}}{r}} \right)^4 } \right]$

It follows that

$\hat C_3 \hat H_{cf} = \sqrt {\frac{1}{{64}}} \left( {35\frac{{x^4 }}{{r^4 }} + 30\frac{{x^2 }}{{r^2 }} + 3} \right) + c\sqrt {\frac{{35}}{{128}}} \left[ {\left( {\frac{{y + iz}}{r}} \right)^4 + \left( {\frac{{y - iz}}{r}} \right)^4 } \right]$

Since the two previous expressions must be equivalent, we can solve for c by collecting terms in z4, and ultimately arrive at

$3/8 + 2c\sqrt {35/128} = 1 \Rightarrow c = \sqrt {5/14}$

which leaves us with a suitable representation of $\hat H_{cf}$ (modulo the constant describing its magnitude):

$\hat H_{cf} = \hat Y_{40} + \sqrt {5/14} \left( {\hat Y_{4 - 4} + \hat Y_{44} } \right)$

In the next post, we will apply this perturbation Hamiltonian to a 2D free ion term, characterize the resulting states, and hopefully gain a detailed understanding of how the d orbitals mix and split in the presence of an octahedral crystal field!