Absence of Symmetry Breaking in 1D Ising Model--Continuum Version I have seen arguments for why there is no symmetry breaking in the 1D Ising model--for example, using the transfer matrix method to explicitly solve the model, and another of energy-entropy arguments showing that a uniformly spin up or down state is unstable to the creation of domain walls.
However, I was wondering if there are explicit arguments like this except using the continuum version of the Ising model, given by the action:
$$S = \int d^dx \; [c_1 (\partial\phi)^2 + c_2\phi^2 + c_4\phi^4]$$
From reading Altland and Simons, I believe the argument involves showing that this theory allows for instantons, and in $d=1$, one can make a similar energy-entropy argument. However I am not sure how to do this explicitly using the functional formalism. Does anyone have any source recommendations on how to go through this argument, or some pointers on how to perform the argument?
 A: Hm, why do not to compute 2-point correlation function? I am not sure that I understand you correctly but Merwin-Wagner theorem in many textbooks is illustrated by this simple computation.
The idea is following: suppose that SSB is possible, then you can consider fluctuations that are represented by massless scalar field.
First, consider that this field is massive. So, I deal with massive scalar field in $d$-dimensions. My bare propagator in looks like
$$D(r) = \int\frac{d^dk}{(2\pi)^d}\frac{e^{ikr}}{k^2+m^2}.$$
This integral can be evaluated in many different ways, one of them is to use Schwinger parametrization for denominator $(k^2+m^2)$ which helps to perform integration over angles. Finally, I deal with integral
$$ D(r)=\frac{1}{2(2\pi)^{d/2}}\int_0^{\infty}d\alpha\,\alpha^{-d/2}\exp\left(-\frac{\alpha m^2}{2}-\frac{r^2}{2\alpha}\right).$$
Then set $m=0$ and check that integral becomes divergent for $d\leq 2$. This integral is convergent for $d>2$.
It is quite rough example how to check that SSB is not possble in $d\leq 2$.
