One Yb ion in a trigonal environment

Splitting of the Yb states

Yb ions crystallizing in the delafossite structure are embedded in an (almost) octahedral environment of chalconides like oxygen, sulfur, or selenium. Counting electrons gives a threefold oxidation state such that the f electrons are in a $4f^{13}$ configuration. Strong spin-orbit coupling splits the $(2s+1)(2\ell+1)=14$ states of the free ion into two groups of states, a $j=\ell+s=7/2$ octet and a typically electronvolts higher $j=\ell-s=5/2$ sextett. Put into an ideal octet, the Yb ions are subject to a cubic crystal field. this splits the $j=7/2$ octet into two doublets $\Gamma_6$ and $\Gamma_7$ plus a $\Gamma_8$ quartet, the $j=5/2$ sextett accordingly into a $\Gamma_7$ and a $\Gamma_8$ multiplet. In any crystal field with less than cubic symmetry (delafossites: trigonal distortion of the octahedral environment) the quartets split further into two doublets. The typical splittings between the doublets are of oder $10\,\rm meV$. This is great for experimental investigations: while at temperatures above of order $100\,\text K$, the full $j=7/2$ moment determines magnetic properties, there is a crossover to the low-temperature $\Gamma_6^T$ doublet generating all the physics we see.

Crystal-field Hamiltonian

The crystal-field Hamiltonian in a trigonal environment can be written as

\[{\cal H}_\text{CEF}=\sum_{l=2,4,6}\sum_{m=0}^lB_l^mO_l^m\]

where the $O_l^m$ are linear combinations of products of the angular-momentum operators and are called Stevens operator equivalents, the $B_l^m$ are numerical factors called crystal-field parameters. Here, $l$ labels the number of components of the angular-momentum operator in an operator product, and $m$ labels the number of ladder operators in that product, i.e., an operator $O_l^m$ maps a $\left|j,m_j\right\rangle$ state onto a linear combination of $\left|j,m_j\pm m\right\rangle$ states.

The summation is restricted to even $l$ because we are discussing time-reversal invariant systems only and $l\le6$ since a $4f$ electron with $\ell=3$ cannot have multipoles larger than $2\ell=6$. The $l=0$ term is omitted because that’s just a constant. If the point group of the Yb ion contains a $p$-fold symmetry axis, then only terms with $m$ being a multiple of $p$ are nonzero. The $B_l^m$ in principle can be calculated, however are usually determined by experiment.

Ideal octahedron – cubic symmetry

Although the delafossites all have trigonally distorted octahedra, we can gain some insight by first looking at the ideal case. Choosing a threefold axis (normal to the center of two opposite equilateral triangles) as the quantization axis, the crystal-field Hamiltonian is

\[{\cal H}_\text{CEF}^{(3)} = B_4^{(3)}\left(O_4^0-20\sqrt2O_4^3\right) +B_6^{(3)}\left(O_6^0+\frac{35\sqrt2}4O_6^3+\frac{77}8O_6^6\right).\]

Essentially, apart from a global energy scale, we have a one-parameter Hamiltonian. The eigensystem is comprised of two doublets and a quartet,

\[\begin{aligned} \Gamma_6 &:\quad \mp\frac13\sqrt{\frac{35}6}\left|\frac72,\pm\frac52\right\rangle -\frac13\sqrt{\frac73}\left|\frac72,\mp\frac12\right\rangle \pm\frac13\sqrt{\frac56}\left|\frac72,\mp\frac72\right\rangle, &\quad \left\langle\Gamma_6\right|J_z\left|\Gamma_6\right\rangle =\mathop{\rm diag}\left(\pm\frac76\right), \nonumber\\ \Gamma_7 &:\quad \mp\frac13\sqrt{\frac72}\left|\frac72,\pm\frac72\right\rangle +\frac{\sqrt5}3\left|\frac72,\pm\frac12\right\rangle \pm\frac1{3\sqrt2}\left|\frac72,\mp\frac52\right\rangle, &\quad \left\langle\Gamma_7\right|J_z\left|\Gamma_7\right\rangle =\mathop{\rm diag}\left(\pm\frac32\right), \nonumber\\ \Gamma_8 &:\quad \left\{\begin{aligned} &\frac13\sqrt{\frac{14}3}\left|\frac72,\pm\frac72\right\rangle \pm\frac13\sqrt{\frac53}\left|\frac72,\pm\frac12\right\rangle +\frac23\sqrt{\frac23}\left|\frac72,\mp\frac52\right\rangle\\ &\left|\frac72,\pm\frac32\right\rangle \end{aligned}\right., &\quad \left\langle\Gamma_8\right|J_z\left|\Gamma_8\right\rangle = \mathop{\rm diag}\left(\pm\frac76,\pm\frac32\right) \end{aligned}\]

determined by symmetry only, independent of the crystal-field parameters. Its eigenvalues are

\[\begin{aligned} E_{\Gamma_6}&=-315\left(4B_4^{(3)}+45B_6^{(3)}\right), \nonumber\\ E_{\Gamma_7}&=\phantom{-}405\left(4B_4^{(3)}-21B_6^{(3)}\right), \nonumber\\ E_{\Gamma_8}&=-180\left(B_4^{(3)}-63B_6^{(3)}\right). \end{aligned}\]

Structure of the crystal-field states

Given the structure of the crystal-field Hamiltonian, we can read off that three of the four doublets, in particular the ground-state doublet, have the form

\[\left|\psi^+\right\rangle = -\alpha{\rm e}^{\text i\phi_\alpha}\left|\frac72,\frac72\right\rangle +\beta\left|\frac72,\frac12\right\rangle +\gamma{\rm e}^{-\text i\phi_\gamma}\left|\frac72,-\frac52\right\rangle,\] \[\left|\psi^-\right\rangle = \alpha{\rm e}^{-\text i\phi_\alpha}\left|\frac72,-\frac72\right\rangle +\beta\left|\frac72,-\frac12\right\rangle -\gamma{\rm e}^{\text i\phi_\gamma}\left|\frac72,\frac52\right\rangle\]

with real coefficients, and the fourth are the pure $\left|7/2,\pm3/2\right\rangle$ states. We note that the sign of $\left|7/2,-1/2\right\rangle=T\left|7/2,1/2\right\rangle$ doesn’t change under time reversal $T$. This is a general feature of the time-reversal operator, giving $T\left|j,\pm1/2\right\rangle=(-)^{j\pm1/2}\left|j,\mp1/2\right\rangle$ with $T^2=-1$ for half-integer total momentum $j$.

Pseudospin

Using the standard definition of the components of the angular momentum operator, the matrix elements of $J$ within the ground-state doublet can be read off:

\[\begin{aligned} \left\langle\psi^\pm\left|J_z\right|\psi^\pm\right\rangle &= \pm\frac12\left(7\alpha^2+\beta^2-5\gamma^2\right), \\ \left\langle\psi^-\left|J_x\right|\psi^+\right\rangle &= \sqrt7\alpha\gamma{\rm e}^{\text i(\phi_\alpha-\phi_\gamma)}+2\beta^2 =\left\langle\psi^+\left|J_x\right|\psi^-\right\rangle^*, \\ \left\langle\psi^-\left|J_y\right|\psi^+\right\rangle &= \text i\left(\sqrt7\alpha\gamma{\rm e}^{\text i(\phi_\alpha-\phi_\gamma)}+2\beta^2\right) \\ &=\left\langle\psi^+\left|J_y\right|\psi^-\right\rangle^*.\end{aligned}\]

All other matrix elements vanish.

The total momentum operator $J$ transforms like $J=-TJT^{-1}$ under time reversal $T$, see below. With $T=UK$, $U=\exp(\text i\pi J_y)$ in standard representation and $K$ the complex conjugation, this requires the matrix elements of $J_z$ and $J_x$ to be real, those of $J_y$ to be purely imaginary, which is equivalent to $\phi_\alpha=\phi_\gamma+2\pi n$, $n\in\mathbb Z$.

This indeed allows us to introduce a pseudospin $S$ for the ground-state doublet by mapping

\[\begin{aligned} g_jJ_z &\to g_\parallel S_z, \\ g_jJ_x &\to g_\perp S_x, \\ g_jJ_y &\to g_\perp S_y, \\ g_\parallel &:= g_j\left(7\alpha^2+\beta^2-5\gamma^2\right), \\ g_\perp &:= g_j\left(2\sqrt7\alpha\gamma+4\beta^2\right), \\ g_j &= 1+\frac{j(j+1)+s(s+1)-\ell(\ell+1)}{2j(j+1)} \nonumber\\ &=\frac87\approx1.14 \quad\mbox{(Landé factor)}.\end{aligned}\]

The Zeeman splitting then is given by

\[{\cal H}_\text{Zeeman} = -\mu_0g_j\mu_\text B\sum_\alpha J_\alpha H_\alpha \to -\mu_0\mu_\text B\left[ g_\parallel S_zH_z +g_\perp\left(S_xH_x+S_yH_y\right) \right],\]

where $\mu_0$ is the magnetic permeability constant and $\mu_\text B$ the Bohr magneton.

The above equations give an argument why $\left|7/2,\pm3/2\right\rangle$ cannot be the ground state here: We would have $\left\langle7/2,\pm3/2\left|J_{x,y}\right|7/2,\mp3/2\right\rangle=0$, i. e. $g_\perp\equiv0$, no coupling to a magnetic field applied perpendicular to the threefold axis enforced by trigonal symmetry alone. As we will see, this is in contradiction to experiment. To the best of our knowledge, no Yb compound is known to have the $\left|7/2,\pm3/2\right\rangle$ doublet as ground state.