# A Ising model and Hamilton function

1. Nov 2, 2016

### LagrangeEuler

In case of Ising model we are working with effective Hamiltonian. So lets look to spins which interact. In a case of feromagnet energy function is defined by
$H=-JS_1S_2$
We have two possibilities. $S_1$ and $S_2$ has different values. And $S_1$ and $S_2$ has the same value. In first case
$H=-J \cdot 1 \cdot (-1)=J$
and in the second case
$H=-J \cdot 1 \cdot 1=-J$
Because $-J<J$ spins like in this case to have parallel orientation. It is clear for me and easy to understand. However I have problem with form of Hamiltonian. Is this Hamiltonian potential or kinetial energy? How energy could be negative $-J$?

2. Nov 2, 2016

### Staff: Mentor

It is potential energy from spin-spin interaction. What is the problem with having a negative energy?

3. Nov 2, 2016

### LagrangeEuler

And why we do not have kinetic energy? It is not a problem. But I never saw that someone calls this energy potential. Why we do not call it potential energy for Ising spin system?

4. Dec 17, 2016

### bolbteppa

Spin satisfies
\begin{align}
\mathbf{S}_1 \cdot \mathbf{S}_2 &= \frac{1}{2}(\mathbf{S}^2 - \mathbf{S}_1^2 - \mathbf{S}_2^2 ) \\
&= \frac{1}{2}[S(S+1) - S_1(S_1 + 1) - S_2(S_2 + 1)].
\end{align}
If the total spin is $\frac{1}{2} - \frac{1}{2} = 0$ then
\begin{align}
\mathbf{S}_1 \cdot \mathbf{S}_2 &= \frac{1}{2}[0 - \frac{1}{2}(\frac{1}{2} + 1) - \frac{1}{2}(\frac{1}{2} + 1)] \\
&= (1/2)(-(3/4) - (3/4)) \\
&= (1/2)(-6/4) \\
&= - 3/4
\end{align}
If the total spin is $\frac{1}{2} + \frac{1}{2} = 1$ then
\begin{align}
\mathbf{S}_1 \cdot \mathbf{S}_2 &= \frac{1}{2}[2 - \frac{1}{2}(\frac{1}{2} + 1) - \frac{1}{2}(\frac{1}{2} + 1)] \\
&= \frac{1}{2}(2 - 3/4 - 3/4) \\
&= (1/2)(2/4) \\
&= 1/4
\end{align}

Treating the potential $U = U(\mathbf{r}_1 - \mathbf{r}_2)$ as a perturbation, the average interaction energy is
\begin{align*}
<U> &= \int dV_1 dV_2 \psi^* U \psi \\
&= \int dV_1 dV_2 \frac{1}{\sqrt{2}}[\psi_1^*(\mathbf{r}_1) \psi_2^* (\mathbf{r}_2) \pm \psi_2^*(\mathbf{r}_1) \psi_1^*(\mathbf{r}_2)]U\frac{1}{\sqrt{2}}[\psi_1(\mathbf{r}_1) \psi_2 (\mathbf{r}_2) \pm \psi_2(\mathbf{r}_1) \psi_1(\mathbf{r}_2)] \\
&= \int dV_1 dV_2 \frac{1}{2}[\psi_1^*(\mathbf{r}_1) \psi_2^* (\mathbf{r}_2) \pm \psi_2^*(\mathbf{r}_1) \psi_1^*(\mathbf{r}_2)]U[\psi_1(\mathbf{r}_1) \psi_2 (\mathbf{r}_2) \pm \psi_2(\mathbf{r}_1) \psi_1(\mathbf{r}_2)] \\
&= \int dV_1 dV_2 \frac{1}{2} \{ [U\psi_1^*(\mathbf{r}_1) \psi_2^* (\mathbf{r}_2)\psi_1(\mathbf{r}_1) \psi_2 (\mathbf{r}_2) + U \psi_2^*(\mathbf{r}_1) \psi_1^*(\mathbf{r}_2)\psi_2(\mathbf{r}_1) \psi_1(\mathbf{r}_2) ] \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \pm [U \psi_1^*(\mathbf{r}_1) \psi_2^* (\mathbf{r}_2) \psi_2(\mathbf{r}_1) \psi_1(\mathbf{r}_2) + U \psi_2^*(\mathbf{r}_1) \psi_1^*(\mathbf{r}_2)\psi_1(\mathbf{r}_1) \psi_2 (\mathbf{r}_2) ] \} \\
&= \frac{1}{2} \{ \int dV_1 dV_2 U\psi_1^*(\mathbf{r}_1) \psi_2^* (\mathbf{r}_2)\psi_1(\mathbf{r}_1) \psi_2 (\mathbf{r}_2) + \int dV_1 dV_2 U \psi_2^*(\mathbf{r}_1) \psi_1^*(\mathbf{r}_2)\psi_2(\mathbf{r}_1) \psi_1(\mathbf{r}_2) ] \\
& \ \ \ \pm [ \int dV_1 dV_2 U \psi_1^*(\mathbf{r}_1) \psi_2^* (\mathbf{r}_2) \psi_2(\mathbf{r}_1) \psi_1(\mathbf{r}_2) + \int dV_1 dV_2 U \psi_2^*(\mathbf{r}_1) \psi_1^*(\mathbf{r}_2)\psi_1(\mathbf{r}_1) \psi_2 (\mathbf{r}_2) ] \} \\
&= \frac{1}{2} \{ \int dV_1 dV_2 U\psi_1^*(\mathbf{r}_1) \psi_2^* (\mathbf{r}_2)\psi_1(\mathbf{r}_1) \psi_2 (\mathbf{r}_2) + \int dV_2 dV_1 U \psi_2^*(\mathbf{r}_2) \psi_1^*(\mathbf{r}_1)\psi_2(\mathbf{r}_2) \psi_1(\mathbf{r}_1) ] \\
& \ \ \ \pm [ \int dV_1 dV_2 U \psi_1^*(\mathbf{r}_1) \psi_2^* (\mathbf{r}_2) \psi_2(\mathbf{r}_1) \psi_1(\mathbf{r}_2) + \int dV_2 dV_1 U \psi_2^*(\mathbf{r}_2) \psi_1^*(\mathbf{r}_1)\psi_1(\mathbf{r}_2) \psi_2 (\mathbf{r}_1) ] \} \\
&= \int dV_1 dV_2 U\psi_1^*(\mathbf{r}_1) \psi_2^* (\mathbf{r}_2)\psi_1(\mathbf{r}_1) \psi_2 (\mathbf{r}_2) \pm \int dV_1 dV_2 U \psi_1^*(\mathbf{r}_1) \psi_2^* (\mathbf{r}_2) \psi_2(\mathbf{r}_1) \psi_1(\mathbf{r}_2)
\end{align*}
so that $<U> = A \pm J$, where the $\textit{exchange integral}$
$$\pm J = \pm \int dV_1 dV_2 U \psi_1^*(\mathbf{r}_1) \psi_2^* (\mathbf{r}_2) \psi_2(\mathbf{r}_1) \psi_1(\mathbf{r}_2)$$
depends on the spin and so can be characterized, using $1 = 4 \mathbf{S}_1 \cdot \mathbf{S}_2$, spin 1, and $- 3 = 4 \mathbf{S}_1 \cdot \mathbf{S}_2$, spin 0, we can express, in terms of the spin, using $1 = 4 \mathbf{S}_1 \cdot \mathbf{S}_2$ to find $- J$ via
$$-J = - J \frac{1}{2}(1 + 1) = -J\frac{1}{2}(1 + 4 \mathbf{S}_1 \cdot \mathbf{S}_2)$$
or using $- 3 = 4 \mathbf{S}_1 \cdot \mathbf{S}_2$ to find $+J$ via
$$J = - J \frac{1}{2}(1 - 3) = - J \frac{1}{2}(1 + 4 \mathbf{S}_1 \cdot \mathbf{S}_2)$$
implying that the $\textit{spin exchange operator}$ is
$$\hat{V}_{ex} = - \frac{1}{2}J(1 +4 \mathbf{S}_1 \cdot \mathbf{S}_2).$$
In general,
$$\hat{V}_{ex} = - \frac{1}{2}\sum_{i<j} J_{ij}(1 + 4 \mathbf{S}_i \cdot \mathbf{S}_j) = - \sum_{i<j} J_{ij}(\frac{1}{2} + 2 \mathbf{S}_i \cdot \mathbf{S}_j) = - \sum_{i,j} J_{ij}(\frac{1}{4} + \mathbf{S}_i \cdot \mathbf{S}_j).$$

The Ising model considers the partition function
\begin{align}
Z' &= \sum e^{-H/T} = \sum_{\mathbf{S}} e^{-\beta (- \sum_{i,j} J_{ij}(\frac{1}{4} - \mathbf{S}_i \cdot \mathbf{S}_j)} = \sum_{\mathbf{S}} e^{\beta \sum_{i,j} J_{ij}\frac{1}{4}} e^{ \beta \sum_{i,j} J_{ij} \mathbf{S}_i \cdot \mathbf{S}_j} \\
&= e^{\beta \sum_{i,j} J_{ij}\frac{1}{4}} \sum_{\mathbf{S}} e^{ \beta \sum_{i,j} J_{ij} \mathbf{S}_i \cdot \mathbf{S}_j} = A \sum_{\mathbf{S}} e^{ \beta \sum_{i,j} J_{ij} \mathbf{S}_i \cdot \mathbf{S}_j}
\end{align}
or rather $Z = Z'/A$, namely
$$Z = \sum_{\mathbf{S}} e^{ \beta \sum_{i,j} J_{ij} \mathbf{S}_i \cdot \mathbf{S}_j}$$
in the case where neighbors have the same interaction strength, $J_{ij} = J$,
$$Z = \sum_{\mathbf{S}} e^{ \beta J \sum_{i,j} \mathbf{S}_i \cdot \mathbf{S}_j}$$
where you assume spin is non-zero in the z direction
$$Z = \sum_{s^z} e^{ \beta J \sum_{i,j} s^z_i s^z_j}$$
and then magically assume $s^z_i = s_i = \pm 1$ to find the Ising model:
$$Z = \sum_{s} e^{ \beta J \sum_{i,j} s_i s_j}$$
Can you state the last four lines in a better way?

Hopefully the meaning of the Hamiltonian, the meaning of $J$ and the minus sign is clearer.

5. Dec 17, 2016

### nrqed

There is no kinetic energy because the particles are at fixed positions. As to why it is not usually called a potential energy, I would say it is more a matter of convention. It is often referred to as an interaction energy and rarely as potential energy, this is true. My guess is that "potential energy" is often associated to an energy that is a function of the position of particles and here the energy does not change by changing the position of the particles but by changing their spin states.