Tuesday, September 1, 2009

Frustrated spin triangles

Frustrated Ising spin triangles


When we talk about spin liquid, one often invoke the frustrated spin system, for which the classical example is the Ising spin triangle with antiferromagnetic interaction.

Let's then look at such a model system see how it behaves, which will be very revealing for understanding more complicated system.

1. Hamiltonian and basis

Suppose there is a (localized) spin triangle of sites A,B,C, in the language of second quantization, there are 8 possible states, which we take as the basis:

$\phi_1 =|\uparrow_A\uparrow_B\uparrow_C>$;
$\phi_2 =|\uparrow_A\uparrow_B\downarrow_C>$;
$\phi_3 =|\uparrow_A\downarrow_B\uparrow_C>$;
$\phi_4 =|\uparrow_A\downarrow_B\downarrow_C>$;
$\phi_5 =|\downarrow_A\uparrow_B\uparrow_C>$;
$\phi_6 =|\downarrow_A\uparrow_B\downarrow_C>$;
$\phi_7 =|\downarrow_A\downarrow_B\uparrow_C>$;
$\phi_8 =|\downarrow_A\downarrow_B\downarrow_C>$;



eq=
\phi_1 =|\uparrow_A\uparrow_B\uparrow_C>\\
\phi_2 =|\uparrow_A\uparrow_B\downarrow_C>\\
\phi_3 =|\uparrow_A\downarrow_B\uparrow_C>\\
\phi_4 =|\uparrow_A\downarrow_B\downarrow_C>\\
\phi_5 =|\downarrow_A\uparrow_B\uparrow_C>\\
\phi_6 =|\downarrow_A\uparrow_B\downarrow_C>\\
\phi_7 =|\downarrow_A\downarrow_B\uparrow_C>\\
\phi_8 =|\downarrow_A\downarrow_B\downarrow_C>


The spin Hamiltonian will be:
$H=\frac{1}{2}[\sum_{i \ne j}^{}J_{ij} S^z_iS^z_j+(S^+_iS^-j+S^-_iS^+_j)/2]$

2. Eigenstates

We can diagonalize the Hamitonian and get the eigenstates and eigenenergies.

$\xi _i=\sum_{j}^{}c_{ij}\phi_{j}$


Table:

E=-3/4J


E=3/4J



ξ1ξ2ξ3ξ4ξ5ξ6ξ7ξ8
ci10.0000.0000.0000.0001.0000.0000.0000.000
ci2-0.8160.0000.0000.0000.0000.5770.0000.000
ci30.4080.7070.0000.0000.0000.5770.0000.000
ci40.0000.0000.0000.8160.0000.000-0.5770.000
ci50.408-0.7070.0000.0000.0000.5770.0000.000
ci60.0000.0000.707-0.4080.0000.000-0.5770.000
ci70.0000.000-0.707-0.4080.0000.000-0.5770.000
ci80.0000.0000.0000.0000.0000.0000.0001.000






eq=\xi _i=\sum_{j}^{}c_{ij}\phi_{j}

It turns out that the system become two energy subspaces.

The first subspace corresponds to degenerate ground states with energy -3/4J.
One can see that |Sz|= 0 for those states.

The second subspace corresponds to a spin quartet with S=3/2.



3. Field and temperature dependence:  magnetization plateau




By varying temperature and magnetic field, one can study the magnetization change.
Above is the result we found for different J value. Starting from left are J=0,1,2,3,4,5,6,7,9,10,11

1) J=0 corresponds to isolated spins, which is actually Brolluvin function
2) J>=9: there is a clear plateau at low magnetization which corresponds to the saturation magnetization of the first subspace of the eigenstates.
3) J=1-7: intermediate cases

4. Field and temperature dependence.


In real experiment, one can not vary J unfortunately. Instead, we only have control over B and T. Next, we show the B,T dependence of the magnetization with fixed J.



In this picture, we can see that at very low temperature, the two plateau is obvious.

One can show that a system with 6 Ising spin which form triangular lattice also has similar behavior, in the sense that there is a low magnetization plateau of 1/3 of the magnitude. The 10 Ising spin system is not calculable for me however due to the computational difficulty but one can imaging the similarity. The key is that for the frustrate the spin system, there is a ground state with huge degeneracy which can behave like a paramagnetic system with reduced spin magnitude.

5. specific heat




As shown in the above figure (J=1,2,3 from the left), the specific heat has a peak at a energy scale proportional to the exchange interaction.

This is actually very typical behavior of two-level system.

6. specific field in magnetic fields



From left to right, the curves correspond to different magnetic fields. One can see that at intermediate fields, there are two peaks in the specific heat.



1 comment:

Tony said...

Hi, Xiao-Shan, this is cool, the transition plateau of the magnetism in spin lattice is nicely displayed. Good job. I've always wanted to learn this, ha