Appendix - The effect of geometric frustration on some correlated electron systems

1Tight binding systems¶

1.1Dispersion for ordered phases¶

Here we show how to calculate dispersion for selected ordered phases, which have relatively small unit cells. We define the unit cell for each phase, and go to $k-$ space where the Hamiltonian becomes block diagonal.

1.1.1Spectrum for collinear phases¶

The Hamiltonian can be diagonalized by Fourier transformation. We write the Hamiltonian $H$ as $H = H_{0}+H_{J}$ , where, $H_0$ is given by,

\begin{align} \nonumber H_{0} &=&\sum_{\vec{X}\sigma} [ \epsilon_{1}f^{\dagger}_{\vec{X},\sigma}f_{\vec{X},\sigma} +\epsilon_{2}m^{\dagger}_{\vec{X}+\vec{a_1},\sigma}m^{\dagger}_{\vec{X}+\vec{a_1},\sigma}] -t\sum_{\vec{X},\sigma,\vec{\delta} \in \textrm{NN}}(f^{\dagger}_{\vec{X},\sigma}m_{\vec{X}+\vec{\delta},\sigma}+\textrm{h.c.}) \\ &~&-t'\sum_{\vec{X},\sigma,\vec{\delta}\in \textrm{NNN}}(m^{\dagger}_{\vec{X}+\vec{a_1},\sigma}m_{\vec{X}+\vec{a_1}+\vec{\delta},\sigma}+\textrm{h.c.})\\ \end{align}

(1)

and $H_J$ is given by

H_{J} = J \sum_{\vec{X}}\vec{S}(\vec{X}) \cdot \vec{\sigma}_{\alpha,\beta}f^{\dagger}_{\vec{X},\alpha}f_{\vec{X},\beta}

(2)

The lattice vector $\vec{X}$ is defined as $\vec{X} = n_{1}\vec{A_{1}} + n_{2}\vec{A_{2}} + n_{3}\vec{A_{3}}$ with $A_i,i=1,2,3$ as the primitive lattice vectors ( $A_1=(2,0,0)$ , $A_2=(1,1,0)$ , $A_3=(0,1,1)$ ), defining the periodicity of lattice with the 2 site unit cell. With this periodicity, the unit cell has one ‘ $f$ ’ and one ‘ $m$ ’ site at $(0,0,0)$ and $(1,0,0)$ respectively. Now doing a Fourier transform on ‘ $f$ ’ operators (similarly for ' $m$ ’s)

\begin{align*} f^{\dagger}_{\vec{X},\sigma} = \frac{1}{\sqrt{N}}\sum_{\vec{k}}f^{\dagger}_{\vec{k},\sigma}\exp(i\vec{k}\cdot\vec{X}) \end{align*}

(3)

This simplifies the non-magnetic part $H_{0}$ as follows,

\begin{align*}\nonumber H_{0} = \sum_{\vec{k},\sigma} \left[ (\epsilon_{2}+A'_{\vec{k}})m^{\dagger}_{\vec{k},\sigma}m_{\vec{k},\sigma} +\epsilon_{1}f^{\dagger}_{\vec{k},\sigma}f_{\vec{k},\sigma}+(A_{\vec{k}}f^{\dagger}_{\vec{k},\sigma}m_{\vec{k},\sigma}+\textrm{h.c.})\right] \end{align*}

(4)

=\sum_{\vec{k},\sigma}\left(f^{\dagger}_{\vec{k},\sigma} m^{\dagger}_{\vec{k},\sigma} \right) \left(\begin{array}{cc} \epsilon_1 & A_{\vec{k}}\\A_{\vec{k}} & \epsilon_2+A^{'}_{\vec{k}} \end{array}\right) \left(\begin{array}{c} f_{\vec{k},\sigma} \\ m_{\vec{k},\sigma} \end{array}\right)

(5)

Which is reduced to $2\times 2$ block. the amplitudes $A_{\vec{k}}=-2t(\cos{k_{x}}+\cos{k_{y}}+\cos{k_{z}})$ and $A'_{\vec{k}}=-4t'(\cos{k_{x}}\cos{k_{y}}+\cos{k_{y}}\cos{k_{z}}+\cos{k_{z}}\cos{k_{x}})$ are just the cubic and FCC dispersions.

Next, we have to simplify the $H_J$ part. For the collinear phases, $\vec{S}(\vec{X})$ can be expressed as $\vec{S}(\vec{X}) = (0,0,e^{i\vec{q}\cdot\vec{X}})$ . For FM, $\vec{q}$ is trivially $(0,0,0)$ . For A-type, $\vec{q} = (\frac{\pi}{2},-\frac{\pi}{2},\frac{\pi}{2})$ , while for C-type $\vec{q} = (0,\pi,-\pi)$ . Now, plugging this value of $\vec{S}(\vec{X})$ in $H_J$ and doing the Fourier transform for the $H_J$ , we get,

H_J = J\sum_{\vec{x}}\sigma f^{\dagger}_{\vec{k},\sigma}f_{\vec{k}+\vec{q},\sigma} \quad ;\sigma= \pm 1

(6)

Now $\vec{q}=0$ for FM, so $H_J$ becomes diagonal. Thus total Hamiltonian $H$ still remains $2\times 2$ block, and the eigenvalues for the FM are solutions of the following $2\times 2$ block

H_{2X2}(\vec{k},\sigma) = \left( \begin{array}{cc} \epsilon_1 + J\sigma& A_{\vec{k}} \\ A_{\vec{k}} & \epsilon_{2} + A'_{\vec{k}} \end{array} \right)

(7)

For A-type and C-type phases, we get matrix elements connecting $|\vec{k},\sigma\rangle \to |\vec{k}+\vec{q},\sigma\rangle \to |\vec{k},\sigma\rangle$ , so that now we get to solve following $4\times4$ block

H_{4\times4}(\vec{k},\sigma) = \left( \begin{array}{cccc} \epsilon_1 & J\sigma & A_{\vec{k}} & 0 \\ J\sigma & \epsilon_1 & 0 & A_{\vec{k}+\vec{q}} \\ A_{\vec{k}} & 0 & \epsilon_2+ A'_{\vec{k}} & 0 \\ 0 & A_{\vec{k}+\vec{q}} & 0 & \epsilon_2+ A'_{\vec{k}} \end{array} \right)

(8)

From these we obtain the spectrum for F,A,C phases on large ( $\sim 100^3-500^3$ ) lattices, which can be used to calculate the density of states, phase diagram, phase separation windows etc.

1.2Spectrum for the ‘flux’ phase¶

The unit cell for the ‘flux’ phase has 4 $B$ , and 4 $B'$ atoms lying on the corners of the cube. The primitive lattice vectors become $A_i=\{(2,0,0),(0,2,0),(0,0,2)\}$ At finite $J$ , the same procedure (as for collinear phases) will reduce the Hamiltonian into $16\times 16$ block. To make life a bit simple, we use the J $\rightarrow \infty$ limit on the Hamiltonian for the ‘flux’ phase, which is same as used in (Sanyal & Majumdar (2009)) except its the 3D version.

Figure 1:Unit cell structure for the Flux phase.

This gives us 4 spin-less $f_i$ levels and 8 $m_{i,\sigma}$ levels in the unit cell, which upon simplification reduces to $12\times 12$ block. With the basis $\left(\begin{array}{cc} f_i(k) & m_{i\uparrow}(k) \end{array}\right)_{i\in \{1,2,3,4\}}$ The Hamiltonian breaks into $12\times 12$ block given as follows

H = \left( \begin{array}{cccccccccccc} \Delta & 0 & 0 & 0 & t_{1\uparrow}a_1 & t_{1\downarrow}a_1 & t_{1\uparrow}a_2 & t_{1\downarrow}a_2 & 0 & 0 & t_{1\uparrow}a_3 & t_{1\downarrow}a_3 \\ 0 & \Delta & 0 & 0 & t_{2\uparrow}a_2 & t_{2\downarrow}a_2 & t_{2\uparrow}a_1 & t_{2\downarrow}a_1 & t_{2\uparrow}a_3 & t_{2\downarrow}a_3 & 0 & 0 \\ 0 & 0 & \Delta & 0 & t_{3\uparrow}a_3 & t_{3\downarrow}a_3 & 0 & 0 & t_{3\uparrow}a_2 & t_{3\downarrow}a_2 & t_{3\uparrow}a_1 & t_{3\downarrow}a_1 \\ 0 & 0 & 0 & \Delta & 0 & 0 & t_{4\uparrow}a_3 & t_{4\downarrow}a_3 & t_{4\uparrow}a_1 & t_{4\downarrow}a_1 & t_{4\uparrow}a_2 & t_{4\downarrow}a_2 \\ t_{1\uparrow}^{*}a_1 & t_{2\uparrow}^{*}a_2 & t_{3\uparrow}^{*}a_3 & 0 & 0 & 0 & t_{12} & 0 & t_{23} & 0 & t_{13} & 0 \\ t_{1\downarrow}^{*}a_1 & t_{2\downarrow}^{*}a_2 & t_{3\downarrow}^{*}a_3 & 0 & 0 & 0 & 0 & t_{12} & 0 & t_{23} & 0 & t_{13} \\ t_{1\uparrow}^{*}a_2 & t_{2\uparrow}^{*}a_1 & 0 & t_{4\uparrow}^{*}a_3 & t_{12} & 0 & 0 & 0 & t_{13} & 0 & t_{23} & 0 \\ t_{1\downarrow}^{*}a_2 & t_{2\downarrow}^{*}a_1 & 0 & t_{4\downarrow}^{*}a_3 & 0 & t_{12} & 0 & 0 & 0 & t_{13} & 0 & t_{23} \\ 0 & t_{2\uparrow}^{*}a_3 & t_{3\uparrow}^{*}a_2 & t_{4\uparrow}^{*}a_1 & t_{23} & 0 & t_{13} & 0 & 0 & 0 & t_{12} & 0 \\ 0 & t_{2\downarrow}^{*}a_3 & t_{3\downarrow}^{*}a_2 & t_{4\downarrow}^{*}a_1 & 0 & t_{23} & 0 & t_{13} & 0 & 0 & 0 & t_{12} \\ t_{1\uparrow}^{*}a_3 & 0 & t_{3\uparrow}^{*}a_1 & t_{4\uparrow}^{*}a_2 & t_{13} & 0 & t_{23} & 0 & t_{12} & 0 & 0 & 0 \\ t_{1\downarrow}^{*}a_3 & 0 & t_{3\downarrow}^{*}a_1 & t_{4\downarrow}^{*}a_2 & 0 & t_{13} & 0 & t_{23} & 0 & t_{12} & 0 & 0 \end{array} \right)

(9)

Where the symbols in the above are defined as

\begin{align*}\nonumber \alpha = \sqrt{\frac{\sqrt{3}+1}{2\sqrt{3}}};~~\beta = \sqrt{\frac{\sqrt{3}-1}{2\sqrt{3}}}; ~~ z = \frac{1-i}{\sqrt{2}};~~ a_1 = 2\cos{k_1}; a_2 = 2\cos{k_2}; a_3 = 2\cos{k_3}\\\nonumber t_{12} = -4t'\cos{k_1}\cos{k_2}; t_{23} = -4t'\cos{k_2}\cos{k_3}; t_{13} = -4t'\cos{k_1}\cos{k_3};\\\nonumber t_{1\uparrow} = t_{2\uparrow} = -t\alpha; t_{3\uparrow} = t_{4\uparrow} = -t\beta; t_{1\downarrow} = -t_{2\downarrow} = tz\beta; t_{3\downarrow} = -t_{4\downarrow} = -tz^{*}\alpha \end{align*}

(10)

This gives us $H({\bf k})_{12\times 12}$ , which is very difficult to diagonalize analytically, but still saves us from diagonalizing full real-space matrix of $\cal{O}(N)$ size, and reduces the problem to $\cal{O}(N)$ number of diagonalizations of 12 sized matrix.

The comparison of energies using the $\epsilon_{\bf k}$ obtained for F,A,C and flux phases, one can draw the magnetic phase diagram for large lattice size ( $N\sim 400^3$ ). The Figure 2, shows the phase diagram, which can be compared to that obtained through real space calculations done on 20³ size.

Phase diagram based on k space based diagonalisation for (a) t'=0, (b) t'=0.3, (c) t'=-0.3. System size N=160^{3}. Here we can only use F, A, C and ‘flux’ phase as candidate states but some of the complexity of more elaborate phase diagrams of , are already present. — (a)

2FCC lattice setup¶

On the FCC lattice, the primitive lattice translation vectors are

\vec{A}_1~=~(0,1,1),\qquad \vec{A}_2~=~(1,0,1),\qquad \vec{A}_3~=~(1,1,0)

(11)

All the points on FCC lattice are expressed in integer units of these, i.e., $\vec{X} = n_1\vec{A}_1+n_2\vec{A}_2+n_3\vec{A}_3$ = $(n_2+n_3,n_3+n_1,n_1+n_2)$ = $\vec{X}(n_1,n_2,n_3)$ . Each site has 12 neighbours $X+\delta$ , where

\begin{align*} \nonumber \delta =& (~1,~1,~0),(-1,-1,~0),\\\nonumber &(~1,~0,~1),(-1,~0,-1),\\\nonumber &(~0,~1,~1),(~0,-1,-1)\\\nonumber &(~1,-1,~0),(-1,~1,~0),\\\nonumber &(~1,~0,-1),(-1,~0,~1),\\\nonumber &(~0,~1,-1),(~0,-1,~1) \end{align*}

(12)

Now lets see how the neighbours are labeled

\begin{align*} \nonumber N_1 &= X(n_1,n_2,n_3) + (1,1,0) = (n_2+n_3,n_3+n_1,n_1+n_2) + (1,1,0)\\\nonumber &= (n_2+(n_3+1),(n_3+1)+n_1,n_1+n_2) = X(n_1,n_2,n_3+1) \end{align*}

(13)

Or,

\begin{align*} \nonumber N_1 &= X(n_1,n_2,n_3) + (1,1,0) = X(n_1,n_2,n_3+1) \end{align*}

(14)

Similarly, one gets the other five as

\begin{align*} \nonumber N_2 &= X(n_1,n_2,n_3) + (-1,-1,0) = X(n_1,n_2,n_3-1)\\\nonumber N_3 &= X(n_1,n_2,n_3) + (1,0,1) = X(n_1,n_2+1,n_3)\\\nonumber N_4 &= X(n_1,n_2,n_3) + (-1,0,-1) = X(n_1,n_2-1,n_3)\\\nonumber N_5 &= X(n_1,n_2,n_3) + (0,1,1) = X(n_1+1,n_2,n_3)\\\nonumber N_6 &= X(n_1,n_2,n_3) + (0,-1,-1) = X(n_1-1,n_2,n_3) \end{align*}

(15)

Next, we have,

\begin{align*} \nonumber N_7 &= X(n_1,n_2,n_3) + (1,-1,0) = (n_2+n_3,n_3+n_1,n_1+n_2) + (1,-1,0)\\\nonumber &= ((n_2+1)+n_3,n_3+(n_1-1),(n_1-1)+(n_2+1)) = X(n_1-1,n_2+1,n_3) \end{align*}

(16)

Or,

\begin{align*} \nonumber N_7 &= X(n_1,n_2,n_3) + (1,-1,0) = X(n_1-1,n_2+1,n_3) \end{align*}

(17)

Similarly the other neighbours are labeled as

\begin{align*} \nonumber N_8 &= X(n_1,n_2,n_3) + (-1,1,0) = X(n_1+1,n_2-1,n_3)\\\nonumber N_9 &= X(n_1,n_2,n_3) + (1,0,-1) = X(n_1-1,n_2,n_3+1)\\\nonumber N_{10} &= X(n_1,n_2,n_3) + (-1,0,1) = X(n_1+1,n_2,n_3-1)\\\nonumber N_{11} &= X(n_1,n_2,n_3) + (0,1,-1) = X(n_1,n_2-1,n_3+1)\\\nonumber N_{12} &= X(n_1,n_2,n_3) + (0,-1,1) = X(n_1,n_2+1,n_3-1) \end{align*}

(18)

Thus if we label the sites $(n_1,n_2,n_3)$ , the first six neighbours have labels, $(n_1\pm 1,n_2,n_3)$ , $(n_1,n_2\pm 1,n_3)$ , and $(n_1,n_2,n_3\pm 1)$ , and the second six neighbours have labels $(n_1\pm 1,n_2\mp 1,n_3)$ , $(n_1\pm 1,n_2,n_3\mp 1)$ , and $(n_1,n_2\pm 1,n_3\mp 1)$ .

Thus, the 12 neighbours of the FCC lattice correspond to 12 ‘neighbours’ in the ‘label’ lattice, first six of which are nearest neighbours, and second six of which are 6 selected next-nearest neighbours. So, though we need to define actual spin configurations on the FCC lattice, the corresponding Hamiltonian of FCC lattice is equivalent to that defined on the ‘label’ lattice.

Table 1:The neighbours on fcc lattice, (a) nearest neighbours (NN) and (b) next nearest neighbours (NNN).

$\delta$	$\delta'$	J $_{x}$	J $_{y}$	J $_{z}$
( 1, 1, 0)	( 0, 0, 1)	- $t$	- $t$	0
(-1,-1, 0)	( 0, 0,-1)	$t$	$t$	0
( 1, 0, 1)	( 0, 1, 0)	- $t$	0	- $t$
(-1, 0,-1)	( 0,-1, 0)	$t$	0	$t$
( 0, 1, 1)	( 1, 0, 0)	0	- $t$	- $t$
( 0,-1,-1)	(-1, 0, 0)	0	$t$	$t$
( 1,-1, 0)	(-1, 1, 0)	- $t$	$t$	0
(-1, 1, 0)	( 1,-1, 0)	$t$	- $t$	0
( 1, 0,-1)	(-1, 0, 1)	- $t$	0	$t$
(-1, 0, 1)	( 1, 0,-1)	$t$	0	- $t$
( 0, 1,-1)	( 0,-1, 1)	0	- $t$	$t$
( 0,-1, 1)	( 0, 1,-1)	0	$t$	- $t$

(a)

$\delta$	$\delta'$	J $_{x}$	J $_{y}$	J $_{z}$
( 2, 0, 0)	(-1, 1, 1)	- $t'$	0	0
(-2, 0, 0)	( 1,-1,-1)	$t'$	0	0
( 0, 2, 0)	( 1,-1, 1)	0	- $t'$	0
( 0,-2, 0)	(-1, 1,-1)	0	$t'$	0
( 0, 0, 2)	( 1, 1,-1)	0	0	- $t'$
( 0, 0,-2)	(-1,-1, 1)	0	0	$t'$

(b)

References¶

Sanyal, P., & Majumdar, P. (2009). Phys. Rev. B, 80, 054411.

The effect of geometric frustration on some correlated electron systems

Summary of the thesis

The effect of geometric frustration on some correlated electron systems

References