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For a free theory in two dimensions with Lagrangian

$$\mathcal{L}=\int d^2 x \phi\nabla\phi$$

where $\nabla$ is the 2D Laplacian/D'Lambertian for Euclidean/Minkowski signature we have that the correlation function goes as

$$\langle\phi(x)\phi(y)\rangle=\ln\frac{|x-y|^2}{\mu^2}$$

where you insert $\mu$ in order to make the argument of the logarithm dimensionless. This correlation function grows with separation distance?! This is counter to my intuition that far separated excitations should be decorrelated. What is some good intuition for why the further separated particles are in $2$D the greater their correlation function?

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    $\begingroup$ If you want a quantum field theory that satisfies the wightmann axioms, you would be forced to conclude that the state corresponding to $\phi$ needs to be excluded from the Hilbert space. i.e, the theory is made up of operators that are functions of $\phi$ (example $\partial \phi$), that satisfy the cluster property, but $\phi$ itself is NOT an operator in this theory. $\endgroup$
    – Anonjohn
    Commented Mar 9, 2021 at 17:19
  • $\begingroup$ @Anonjohn I suppose the OP is interested, why there is a growth, instead of decrease, the logarithmic law itself, which is valid even at classical level, without referring to quantization $\endgroup$ Commented Mar 9, 2021 at 18:10
  • $\begingroup$ @Anonjohn thanks! Both classical and quantum interpretations are interesting. It would be nice to have intuition for why if I simulated the system on a computer using the path integral as a statistical weighting, that in the case of 2D case the path integral gives a growing correlation function. So regardless of physical relevance, why in the path integral sense do these correlation functions grow. $\endgroup$
    – Luke
    Commented Mar 9, 2021 at 18:17
  • $\begingroup$ Also would be interesting if there are actual $2D$ condensed matter systems where the system is accurately modelled by the free boson. $\endgroup$
    – Luke
    Commented Mar 9, 2021 at 18:19

1 Answer 1

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Let's see what happens when we try to define the model unambiguously. We can do this in several different ways. I'll consider one way.

Defining the model

I'll treat $\phi(x)$ as being real-valued field (non-compact target space). I'll work in 2d euclidean space with finite volume, saving the infinite-volume limit for later. I'll use periodic boundary conditions in both of the two dimensions, with period $L$. Thanks to the periodic boundary conditions, the action shown in the question might as well be $S\equiv\int d^2x\ (\partial\phi)^2$, where $\partial\phi$ is the gradient of $\phi$.

The correlation function should be something like $$ \newcommand{\pl}{\partial} \newcommand{\la}{\langle} \newcommand{\ra}{\rangle} \la\phi(x)\phi(y)\ra \sim \int [d\phi]\ e^{-S[\phi]}\phi(x)\phi(y), \tag{1} $$ but this is ill-defined because $S$ is invariant under the change of variable $\phi(x)\to\phi(x)+c$ for any constant $c$. In other words, $S$ is independent of the zero mode of $\phi(x)$. The naïve functional integral includes an integral over the zero mode, and since the integrand is not a bounded function of the zero mode, the integral does not converge.

Thanks to the finite-volume condition, this problem has an easy solution. In finite volume, the set of possible wavenumbers is discrete, so we can write $$ \phi(x)=\frac{1}{L^2}\sum_p e^{ip\cdot x}\tilde\phi(p). \tag{2} $$ The zero mode (the mode with wavenumber $p=(0,0)$) is $$ \tilde\phi(0)=\int d^2x\ \phi(x). \tag{3} $$ We can define (1) just by omitting the zero mode in (2), so that the original $\phi(x)$ is replaced with $$ \phi(x)=\frac{1}{L^2}\sum_{p\neq (0,0)} e^{ip\cdot x}\tilde\phi(p). \tag{4} $$ Now we can define (1) by $$ \la\phi(x)\phi(y)\ra \propto \int \left(\prod_{p\neq (0,0)}d\tilde\phi(p)\right)\ e^{-S[\phi]}\phi(x)\phi(y), \tag{5} $$ with $\phi$ given by (4). Omitting the zero mode doesn't affect the action $S$, which was already independent of the zero mode. However, omitting the zero mode does affect the rest of the integrand, and that's why we did it: now the integral converges.

A side-effect of using (4) is that $\phi(x)$ is no longer a local operator, because (4) and (2) differ by the nonlocal quantity (3), and "local" is defined by the original version (2). Keep this in mind when considering the infinite-volume limit, below.

The infinite-volume limit

To understand how the definition (5) leads to the expression $\ln|x-y|^2/\mu^2$ shown in the question, consider the infinite-volume limit. Before taking the infinite-volume limit, the correlation function (5) is a periodic function of both $x$ and $y$, with period $L$ in both dimensions. Explicitly, $$ \la\phi(x)\phi(y)\ra \propto \frac{1}{L^2}\sum_{p\neq (0,0)}\frac{e^{ip\cdot(x-y)}}{p^2}, \tag{6} $$ where the components of $p$ are integers times $2\pi/L$. We can write this as $$ \la\phi(x)\phi(y)\ra \propto \lim_{m\to 0} \left(\frac{1}{L^2}\sum_{p}\frac{e^{ip\cdot(x-y)}}{p^2+m^2} -\frac{1}{L^2m^2}\right) \tag{7} $$ where now the sum is over all wavenumbers $p$ because the zero-wavenumber term is explicitly subtracted. In the limit $L\to\infty$, the quantity in large parentheses becomes $$ \left(\frac{1}{L^2}\sum_{p}\frac{e^{ip\cdot(x-y)}}{p^2+m^2} -\frac{1}{L^2m^2}\right) \to \int\frac{d^2p}{(2\pi)^2}\ \frac{e^{ip\cdot(x-y)}}{p^2+m^2} \propto \int\frac{d^2p}{(2\pi)^2}\ \frac{e^{ip\cdot(x-y)m}}{p^2+1}. \tag{8} $$ Notice that the limit $m\to 0$ is no longer defined: the limits $L\to\infty$ and $m\to 0$ don't commute. At this point, we have two options:

  • We can change the rules in the middle of the game and decide that we don't really need to take the limit $m\to 0$ after all. For $|x-y|m\ll 1$, this leads to the result shown in the question with $\mu\propto 1/m$. (Also see this post).

  • We can accept that $\la\phi(x)\phi(y)\ra$ doesn't have an infinite-volume limit. That shouldn't bother us, because $\phi(x)$ isn't a local operator anyway. Correlation functions of local operators are the things we really care about.

The gradient $\pl\phi(x)$ is a local operator: like the action, it was already independent of the zero mode. In the infinite-volume limit, the correlation function $\la\pl\phi(x)\pl\phi(y)\ra$ has a monotonically decreasing magnitude as a function of $|x-y|$, as expected for local operators, and it's independent of the ad-hoc regulator $\mu\propto 1/m$.

The question is why $\la\phi(x)\phi(y)\ra$ doesn't have that property. At least in the approach I used here, the answer is that $\la\phi(x)\phi(y)\ra$ is undefined in the infinite-volume limit. We define it by arbitrarily keeping $m=1/\mu$ finite, and then the result shown in the question is valid for $|x-y|/\mu\ll 1$. For such values of $|x-y|$, the result shown in the question has a monotonically decreasing magnitude as a function of distance, as expected.

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  • $\begingroup$ I do not see where the dimensionality plays a role in the argument? $\endgroup$
    – Luke
    Commented Mar 10, 2021 at 15:26
  • $\begingroup$ @Luke Most of the answer applies for any number $N$ of dimensions, after replacing factors of $L^2$ with $L^N$. The part specific to 2d is going from the left-hand side of equation (8) to $\ln|x-y|^2/\mu^2$. With $N>2$, the right-hand of (8) would be different, the regulator could be removed, and the function would approach zero as $|x-y|\to \infty$, like we would normally expect a correlation function to do. $\endgroup$ Commented Mar 10, 2021 at 23:58

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