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:


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?



  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.


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