3
$\begingroup$

For reference I am trying to work out the derivation in this paper, in which the partition function for an Ising model is approximated by replacing the Ising variables $\sigma_i$ with $N$ component vectors $\mathbf{s}_i$ subject to the condition that $\left| \mathbf{s}_i\right|^2 = N$, and taking the limit $N \rightarrow \infty$ (why this produces accurate results is a mystery as far as I can tell, if anyone has insight on this I would greatly appreciate it). I summarize the relevant math below:

Setup

The Hamiltonian is $$H = -\frac{J}{2}\sum_{ij} V_{ij} \sigma_i \sigma_j$$ where $V_{ij}$ is an adjacency matrix ($V_{ij} = 1$ if $i$ and $j$ are nearest neighbors, and 0 otherwise) and $J$ is the interaction strength, and the $\sigma_i = \pm 1$. The partition function reads

$$Z = \sum_{\left\{\sigma_i = \pm 1\right\}} \exp\left(-\frac{\beta J}{2} \sum_{ij} V_{ij} \sigma_i \sigma_j\right) \tag{3}$$

We can also define continuous variables $s_i$ and define the partition function as

$$Z = \int \prod_j \left(ds_j \, \delta(\left|s_j\right|^2 - 1)\right) \exp\left(-\frac{\beta J}{2} \sum_{ij} V_{ij} s_i s_j\right) \tag{4}$$

Now we define a new $O(N)$ model where the $s_i$ are $N$ component vectors with norm $\left|\mathbf{s}_i\right|^2 = N$. The partition function is

$$Z_N = \int \prod_j \left(d\mathbf{s}_j \, \delta(\left|\mathbf{s}_j\right|^2 - N)\right) \exp\left(-\frac{\beta J}{2} \sum_{ij} V_{ij} \mathbf{s}_i\cdot \mathbf{s}_j\right)\tag{6}$$

Clearly $Z_1 \equiv Z$. We implement the delta function by introducing a constraint field at each site, $\mu_i$, using

$$\delta(x) = \int_{-\infty}^{\infty} d\mu\, e^{i\mu x}$$

to write

$$Z_N = \int \prod_j d\mathbf{s}_j d\mu_j \exp\left(-\frac{1}{2}\sum_j i\mu_j\left(\left|\mathbf{s}_j\right|^2 - N)\right)\right) \exp\left(-\frac{\beta J}{2} \sum_{ij} V_{ij} \mathbf{s}_i\cdot \mathbf{s}_j\right)\tag{6b}$$

The factor $1/2$ will give us an irrelevant overall factor of 2 which I ignore (since $\delta(ax) = \delta(x)/\left|a\right|$). Write the integration measure as $\mathcal{D}\mathbf{s} \mathcal{D}\mu$ for simplicity. This is where the paper skips some steps.

My attempt to complete the derivation

Here is my attempt to continue: rewrite this as

$$Z_N = \int\mathcal{D}\mathbf{s} \mathcal{D}\mu \exp\left(\frac{N}{2}\sum_j i\mu_j\right) \exp\left(-\frac{1}{2} \sum_{ij} \left( \delta_{ij} (i\mu_j)\left|\mathbf{s_j}\right|^2 + \beta J V_{ij} \mathbf{s}_i\cdot \mathbf{s}_j\right)\right)$$

$$Z_N = \int \mathcal{D}\mu \exp\left(\frac{N}{2}\sum_j i\mu_j\right) \int\mathcal{D}\mathbf{s}\,\exp\left(-\frac{1}{2} \sum_{ij} \left( \delta_{ij} (i\mu_j)+ \beta J V_{ij}\right) \mathbf{s}_i\cdot \mathbf{s}_j\right)$$

Define the matrix $\mu_{ij} = \delta_{ij} \mu_j$. Let $s_i^a$ be the $a$'th component of $\mathbf{s}_i$. Then we write this as

$$Z_N = \int \mathcal{D}\mu \exp\left(\frac{N}{2}\mathrm{Tr}[i\mu]\right) \int\mathcal{D}\mathbf{s}\,\exp\left(-\frac{1}{2} \sum_{a=1}^N\sum_{ij} s_i^a\left( i\mu+ \beta J V\right)_{ij} s_j^a\right)$$

Let $M_{ij} = (i\mu + \beta J V)_{ij}$, and write $d\mathbf{s}_i = \prod_{a=1}^N ds_i^a$, then we have

$$Z_N = \int \mathcal{D}\mu \exp\left(\frac{N}{2}\mathrm{Tr}[i\mu]\right) \prod_{a=1}^N\int \prod_{j}ds_j^a\,\exp\left(-\frac{1}{2} \sum_{ij} s_i^aM_{ij} s_j^a\right)$$

My question

We can perform the $N$ identical multivariate gaussian integrals, but I don't see how this will lead us to equation 7 in the paper:

$$Z_N = \int \mathcal{D}\mu \exp\left(-\frac{N}{2}\Big(-\mathrm{Tr}[i\mu] + \mathrm{Tr[log}(M)]\Big)\right) \tag{7}$$

Supplemental question

I would also like to understand more precisely how we obtain the saddle point condition, $$M^{-1}_{ii} = 1 \qquad (\text{no sum over }i),\tag{8}$$ which comes from finding the maximum of the exponent, something like $$\frac{d}{d(i\mu_j)}\Big(-\mathrm{Tr}[i\mu] + \mathrm{Tr[log}(i\mu + \beta J V)]\Big) = 0 $$ but how do we do this more rigorously since we are dealing with the logarithm of a matrix? And why do we expect $\mu$ to be purely imaginary?

$\endgroup$
2
$\begingroup$

Hints:

  1. The trace of the logarithm in eq. (7) is the logarithm of the determinant from the Gaussian integration over the ${\bf s}$-variables, cf. the identity $$\ln \det (M) ~=~ {\rm tr}\ln(M).\tag{A}$$

  2. The action $$S(\lambda):={\rm tr}(-\lambda + \ln M ),\tag{B}$$ with $$M~:=~\lambda +\beta J V,\tag{C}$$ has infinitesimal variation $$ \delta S(\lambda)~=~-\sum_i\delta \lambda_{ii} + \sum_{ij} M^{-1}_{ij} \delta \lambda_{ji},\tag{D}$$ whose stationary condition leads to OP's sought-for eq. (8). (Recall that $\lambda_{ij}$ is a diagonal matrix.)

  3. Here we have used the identity $$ \delta {\rm tr}\ln(M) ~=~{\rm tr}(M^{-1}\delta M) ,\tag{E}$$ or equivalently, $$ \delta {\rm tr}(A) ~=~{\rm tr}(e^{-A}\delta e^{A}),\tag{F} $$ if we write $M=e^A$. The identity (F) follows e.g. from cyclicity of the trace and the identity $$ e^{-A}\delta e^{A}~=~\int_0^1\! ds~ e^{-sA} (\delta A) e^{sA}, \tag{G} $$ which is proven in my Phys.SE answer here.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.