My problem is to take the $d$-dimensional Ising Hamiltonian, $$H = -\sum_{i,j}\sigma_i J_{i,j} \sigma_j - \sum_{i} \tilde{h}_i \sigma_i$$ where $J_{ij}$ is a matrix describing the couplings between sites $i$ and $j$. Applying a Hubbard-Stratonovich transformation, rewrite the partition function as $$Z = N_0 \int d^N \psi \exp\left\{-\left[\frac{1}{4}\sum_{i,j} \psi_i K_{ij} \psi_j - \sum_{i} \ln[\cosh(h_i+\psi_i)]\right]\right\}$$ where $N_0$ is an overall normalization constant, $K_{ij} = (\beta J_{ij})^{-1}$, and $h_i = \beta\tilde{h}_i$. This much is relatively straightforward. We write the field as $\psi_i = \phi_i - h_i$, and we can show that $\left<\phi_i\right> \propto J_{ij} \left<\sigma_j\right>$, i.e. it can be interpreted as a "mean field" at site $i$ due to the interaction with all other sites.

Next we assume that the variation in the field is small, $\left|\phi_i\right|<<1$, we set $h_i = 0$, and expand $\ln \cosh(x) \approx \frac{1}{2}x^2 - \frac{1}{12}x^4$ to get $$Z \approx N_0\int d^N\psi \exp\left\{-\left[\frac{1}{4}\sum_{i,j}\phi_i K_{ij} \phi_j - \sum_i \left[\frac{\phi_i^2}{2} - \frac{\phi_i^4}{12}\right]\right]\right\}$$

Now we take the continuum limit, in units where the lattice spacing is unity, labeling each site by its position $\mathbf{r}$, which gives $$Z\rightarrow \mathcal{N} \int \mathcal{D}\phi\, \exp\left\{-\frac{1}{2}\left[\frac{1}{2}\int d\mathbf{r}\,d\mathbf{r}'\,\phi(\mathbf{r}) K(\mathbf{r}-\mathbf{r}') \phi(\mathbf{r}') - \int d\mathbf{r}\,\left[\phi(\mathbf{r})^2 - \frac{\phi(\mathbf{r})^4}{6}\right]\right]\right\}$$

This is where I am not sure how to proceed. I am told to expand $\phi(\mathbf{r}')$ as a small variation from the value at $\mathbf{r}$, i.e. $$\phi(\mathbf{r}') \approx \phi(\mathbf{r}) + (x_\mu'-x_\mu)\partial_\mu \phi(\mathbf{r}) + \frac{1}{2}(x_\mu' - x_\mu)(x_{\nu}'-x_\nu)\partial_\mu \partial_\nu \phi(\mathbf{r}) + \cdots$$ and introduce the Fourier transform $\tilde{K}(\mathbf{q}) = \int d\mathbf{r} K(\mathbf{r}) e^{-i\mathbf{q}\cdot\mathbf{r}}$ and write the continuum action as $$S = \int d^d\mathbf{r} \left[c_1 \left(\partial \phi\right)^2 + c_2 \phi^2 + c_4 \phi^4\right]$$ and find the coefficients in terms of $\tilde{K}(0)$ and $\tilde{K}''(0)$.

I believe that I can argue that $K$ is only a function of $\left|\mathbf{r}-\mathbf{r}'\right|$, in which case $K(\mathbf{r}-\mathbf{r}')(x_\mu'-x_\mu)$ is odd about the point $\mathbf{r}$, and so integrating over $d\mathbf{r}'$ (treating $\mathbf{r}$ as constant) will kill any term except those which depend on the square of the difference, leaving me with $$\int d\mathbf{r}\,d\mathbf{r}'\,\phi(\mathbf{r}) K(\mathbf{r}-\mathbf{r}') \phi(\mathbf{r}') = \int d\mathbf{r}\,d\mathbf{r}' K(\mathbf{r}-\mathbf{r}')\left(\phi(\mathbf{r})^2 + \frac{1}{2}(x_\mu'-x_\mu)^2\phi(\mathbf{r})\partial_\mu^2 \phi(\mathbf{r})\right) $$

The first term I can deal with, but it's the second term that I don't know how to deal with.

  • $\begingroup$ In general $K_{ij}$ is the adjiacency matrix of the nearest neighbor graph. The difference between your variables $x’-x=a$ is the lattice constant. Hence that term gives rise to the “kinetic” term of the field theory $\endgroup$
    – lcv
    Commented Mar 14, 2018 at 8:46
  • $\begingroup$ I am not sure that I can simply say that $(x_\mu' - x_\mu)$ is the lattice constant (which I have set to 1), for the purposes of the integration. I am assuming that somehow I will be able to convert $\int d\mathbf{r}' K(\mathbf{r}-\mathbf{r}') (x_\mu'-x_\mu)^2 \sim\tilde{K}''(\mathbf{0})$, and integrating by parts I can move one of the $\partial_\mu$ onto the other $\phi(\mathbf{r})$ to get a $(\partial_\mu\phi)^2$ term. I'm just not sure how to do it. $\endgroup$
    – Kai
    Commented Mar 14, 2018 at 18:22
  • $\begingroup$ What I've included above is the way the problem was presented to me, I think that it would be better to leave the lattice spacing in explicitly and take the limit that $a\rightarrow 0$, since this is supposed to be a "coarse grained" mean field theory $\endgroup$
    – Kai
    Commented Mar 14, 2018 at 18:26
  • $\begingroup$ I'm late to the party, but if the variation in the field is small you can just drop all higher derivatives, which kills the second term. $\endgroup$ Commented Mar 15, 2018 at 17:29

1 Answer 1


You can write the (scaled) interaction part of the action as: $$S_I \equiv \int_{\mathbb R^d}d^d \mathbf r \ \phi(\mathbf r)\int_{\mathbb R^d}d^d \mathbf r' \ K(\mathbf r-\mathbf r') \ \phi(\mathbf r')$$ Let's take the inner integral over $\mathbf r'$ first (I will call it $\mathcal I$ to make things easier). Expanding $\phi(\mathbf r')$ around $\mathbf r$ gives : $$\mathcal I \equiv\int_{\mathbb R^d}d^d \mathbf r' \ K(\mathbf r-\mathbf r') \ \phi(\mathbf r') \approx \int_{\mathbb R^d}d^d \mathbf r' \ K(\mathbf r-\mathbf r') \ \bigg(\phi(\mathbf r)+ \sum_{i=1}^d (x_i'-x_i)\partial_i \phi(\mathbf r) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\frac 12\sum_{i=1}^d\sum_{j=1}^d (x_i'-x_i)(x_j'-x_j)\partial_i \partial_j \phi(\mathbf r) \bigg)$$ Now take the integral inside to get: $$\mathcal I\approx \phi(\mathbf r) \int_{\mathbb R^d}d^d \mathbf r' \ K(\mathbf r-\mathbf r') \ + \sum_{i=1}^d \partial_i \phi(\mathbf r)\int_{\mathbb R^d}d^d \mathbf r' \ (x_i'-x_i) K(\mathbf r-\mathbf r') +\frac 12\sum_{i=1}^d\sum_{j=1}^d \partial_i \partial_j \phi(\mathbf r) \times\int_{\mathbb R^d}d^d \mathbf r'(x_i'-x_i)(x_j'-x_j)K(\mathbf r-\mathbf r') \bigg)$$ Now assuming that the coupling is homogenous, $K(\mathbf r-\mathbf r')\equiv K(\mathbf r'-\mathbf r)$. With that in mind, and also changing variables $\mathbf R \equiv \mathbf r'-\mathbf r$, we get:

$$\mathcal I \approx \phi(\mathbf r) \int_{\mathbb R^d}d^d \mathbf R \ K(\mathbf R) \ + \sum_{i=1}^d \partial_i \phi(\mathbf r)\int_{\mathbb R^d}d^d \mathbf R \ R_i K(\mathbf R) +\frac 12\sum_{i=1}^d\sum_{j=1}^d \partial_i \partial_j \phi(\mathbf r) \ \ \ \ \times\int_{\mathbb R^d}d^d \mathbf R \ R_iR_jK(\mathbf R) \bigg)$$ You can relate each of the integrals over $\mathbf R$ to the Fourier transform of $K(\mathbf R)$ defined as $\tilde K(\mathbf q) \equiv \int_{\mathbb R^d} d^d \mathbf R \ K(\mathbf R)\exp(-i \mathbf{q} . \mathbf R)$:

- First integral: $$\int_{\mathbb R^d}d^d \mathbf R \ K(\mathbf R) = \int_{\mathbb R^d}d^d \mathbf R \ K(\mathbf R) \ e^{-i \mathbf q. \mathbf R} |_{\mathbf q =0} = \tilde K(\mathbf 0)$$ - Second integral: $$\int_{\mathbb R^d}d^d \mathbf R \ R_i K(\mathbf R) =0$$ Because of the integrand being odd as you mentioned.

- Third integral:
For this one we first note that as you mentioned the integral is zero for all different $i,j$. For $i=j$, first note that: $$\frac{\partial^2}{\partial q_i^2} \int_{\mathbb R^d}d^d \mathbf R \ K(\mathbf R) \ e^{-i \mathbf q. \mathbf R} = \int_{\mathbb R^d}d^d \mathbf R \ (-i)(-i) R_i R_i K(\mathbf R) \ e^{-i \mathbf q. \mathbf R} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ = - \int_{\mathbb R^d}d^d \mathbf R \ R_i^2 K(\mathbf R) \ e^{-i \mathbf q. \mathbf R} $$ Which implies: $$\int_{\mathbb R^d}d^d \mathbf R \ R_i^2 K(\mathbf R)=-\frac{\partial^2}{\partial q_i^2} \int_{\mathbb R^d}d^d \mathbf R \ K(\mathbf R) \ e^{-i \mathbf q. \mathbf R}|_{\mathbf q=0}=-\frac{\partial^2}{\partial q_i^2}\tilde K(\mathbf q) |_{\mathbf q=0} $$ Now if you assume that the coupling is also isotropic, i.e. $\exists \mathcal K : K(\mathbf R) \equiv \mathcal K(|\mathbf R|)$, the Fourier transform of $K$ will become a single variable function, meaning that the third integral is just $-\tilde K''(0)$.

In summary, $\mathcal I$ is: $$\mathcal I \approx \phi(\mathbf r) \tilde K(0) \ - \frac 12\sum_{i=1}^d \partial_i^2 \phi(\mathbf r) \tilde K''(0)$$ Thus, the interaction term in the action is: $$S_I = \int_{\mathbb R^d}d^d \mathbf r \ \phi(\mathbf r) \mathcal I = \int_{\mathbb R^d}d^d \mathbf r \ \phi(\mathbf r)\bigg(\phi(\mathbf r) \tilde K(0) \ - \frac 12\sum_{i=1}^d\partial_i^2 \phi(\mathbf r) \tilde K''(0)\bigg)$$

$$=\tilde K(0)\int_{\mathbb R^d}d^d \mathbf r \ \phi^2(\mathbf r) - \frac {\tilde K''(0)}2\sum_{i=1}^d \int_{\mathbb R^d}d^d \mathbf r \ \phi(\mathbf r) \ \partial_i^2 \phi(\mathbf r) $$ Integrating by parts in the second term results in (boundary terms vanish because $\phi(\mathbf r) \to 0$ as $|\mathbf r| \to \infty$ so that the integrals converge): $$S_I =\tilde K(0)\int_{\mathbb R^d}d^d \mathbf r \ \phi^2(\mathbf r) + \frac {\tilde K''(0)}2\sum_{i=1}^d\int_{\mathbb R^d}d^d \mathbf r \ \partial_i \phi(\mathbf r) \ \partial_i \phi(\mathbf r)$$ $$=\tilde K(0)\int_{\mathbb R^d}d^d \mathbf r \ \phi^2(\mathbf r) + \frac {\tilde K''(0)}2 \int_{\mathbb R^d}d^d \mathbf r \ \sum_{i=1}^d (\partial_i \phi(\mathbf r))^2$$

$$=\int_{\mathbb R^d}d^d \mathbf r \ \bigg( \tilde K(0) \phi^2(\mathbf r) + \frac {\tilde K''(0)}2 \ \big( \partial \phi(\mathbf r) \big)^2 \bigg)$$ Plugging this in the full action finally gives: $$S[\phi] =\int_{\mathbb R^d}d^d \mathbf r \ \bigg( \frac {\tilde K''(0)}8 \ \big( \partial \phi(\mathbf r) \big)^2 + \left(\frac {\tilde K(0)}{4}- \frac 12\right) \phi^2(\mathbf r) + \frac 1{12} \phi^4(\mathbf r) \bigg) $$ Notice that the coefficient of the quadratic term can change sign with temperature (through $\tilde K$), which is a sign of a phase transition.

  • $\begingroup$ Thanks so much! I was able to get the quadratic term correct and I did note that since $\tilde{K} \propto \beta^{-1}$ that at low temperature the sign will change and there will be a phase transition, I just wasn't able to get the first term. I have not gone through all of your solution yet but thank you. $\endgroup$
    – Kai
    Commented Mar 14, 2018 at 23:13
  • $\begingroup$ You're more than welcome! $\endgroup$ Commented Mar 14, 2018 at 23:17
  • $\begingroup$ I had been trying to do something similar to what you did for the third integral but I was inserting the inverse transform first, after which I couldn't figure out how to apply the derivatives, since I had both $K$ and the phase factor depending on $\mathbf{q}$. I hadn't thought of it this way. Thank you again this has been quite instructive and helpful. $\endgroup$
    – Kai
    Commented Mar 14, 2018 at 23:19
  • $\begingroup$ I'm happy I could help. $\endgroup$ Commented Mar 14, 2018 at 23:20
  • 1
    $\begingroup$ I was just looking over this again, I believe you meant to say that the boundary term vanishes because $\phi(\mathbf r)\to 0$ as $|\mathbf r| \to \infty$. $\endgroup$
    – Kai
    Commented Mar 15, 2018 at 2:50

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