Critical Line:

Here we discuss the physical significance of our result for $j_{0}$ as a function of $v_{0}$. As we shall see, this bears on the critical behavior of the model. First of all, let us discuss the case $j_{0}=0$, that is without external sources at all, which from Equation (4) results in the equation

$$v_{0} \left\{ \lambda v_{0}^{2} + \alpha + \lambda ... ...{2} + 3 \sigma_{\mathfrak{N}}^{2} \right] \right\} = 0.$$

Observe that we do not assume that $v_{0}$ is automatically zero. Since we must have $\lambda\geq 0$, in the part of the parameter space of the model in which the quantity shown below is positive,

$$\alpha + \lambda \left[ (\mathfrak{N}-1) \sigma_{0}^{2} + 3 \sigma_{\mathfrak{N}}^{2} \right] > 0,$$

the only possible solution to the equation is in fact $v_{0}=0$. This is the symmetric phase of the model. On the other hand, in the complementary region of the parameter space of the model, in which that same quantity is negative,

$$\alpha + \lambda \left[ (\mathfrak{N}-1) \sigma_{0}^{2} + 3 \sigma_{\mathfrak{N}}^{2} \right] < 0,$$

and once again because we must have $\lambda\geq 0$, there are two other solutions besides the $v_{0}=0$ solution, given by

$$\lambda v_{0}^{2} = - \left\{ \alpha + \lambda \lef... ...a_{0}^{2} + 3 \sigma_{\mathfrak{N}}^{2} \right] \right\}.$$ (D.7)

Let us observe that since $\lambda>0$ we must have $\alpha<0$ here. This is the broken-symmetric phase of the model, where these solutions corresponds to the local minima of the potential, while $v_{0}=0$ corresponds to the local maximum. If we look for the locus in the $(\alpha,\lambda)$ parameter plane of the model in which the $v_{0}=0$ solution becomes the only possible solution, we arrive at the equation

$$\alpha + \lambda \left[ (\mathfrak{N}-1) \sigma_{0}^{2} + 3 \sigma_{\mathfrak{N}}^{2} \right] = 0.$$

This is an equation giving $\lambda$ in terms of $\alpha$, which thus determines a certain curve in the parameter plane of the model, in this case a straight line. This is the critical line, which separates the two phases of the model. An example of the parameter space of the model, showing the critical line, can be seen in Figure 1, on page . To the right of this line the model is symmetric and we have $\left\langle\vec{\varphi}(n_{\mu})\right\rangle=0$. To the left, the symmetry is broken and we have $\left\langle\varphi_{\mathfrak{N}}(n_{\mu})\right\rangle\neq 0$.

It might seem odd that we find here what looks like a complete phase transition even on finite lattices. In fact, it is a known fact that there are no real phase transitions on finite lattices with periodical boundary conditions, a situation in which all one can hope to get are approximations of this behavior. However, it is in fact possible to get complete phase transitions on finite lattices if one uses other boundary conditions or changes other aspects of the system, such as the imposition of self-consistency conditions [2], just as we do in the Gaussian-Perturbative technique. Strictly speaking, however, the position of the critical line that we find here is not completely well-defined on finite lattices, because there is a slight dependence on $\alpha_{0}$ and $\alpha_{\mathfrak{N}}$ through $\sigma_{0}$ and $\sigma_{\mathfrak{N}}$. This small dependence vanishes in the continuum limit, of course.

Since $\sigma_{0}$ and $\sigma_{\mathfrak{N}}$ are strictly positive, and $(\mathfrak{N}-1)\geq 0$, we can see that this critical line starts at the Gaussian point $(\alpha,\lambda)=(0,0)$, and extends to the quadrant where $\alpha<0$ and $\lambda>0$. Besides, since in the continuum limit $\sigma_{0}$ and $\sigma_{\mathfrak{N}}$ become identical, we may write the following equivalent equation for purposes of that limit,

$$\alpha + (\mathfrak{N}+2) \lambda \sigma_{0}^{2} = 0.$$ (D.8)

This is the known result for the critical line, obtained previously without the introduction of any external sources at all [1].

Going back to the general case, when the external source $j_{0}$ is not zero, then the equation of the critical line determines it in terms of $v_{0}$, in either phase of the model, for as we saw before in Equation (4) we have

$$j_{0} = v_{0} \left\{ \lambda v_{0}^{2} + \alpha + ... ...a_{0}^{2} + 3 \sigma_{\mathfrak{N}}^{2} \right] \right\}.$$

Given a point $(\alpha,\lambda)$ in the parameter space of the model, this clearly and directly determines $j_{0}$ in terms of $v_{0}$. Conversely, given $j_{0}$ one may determine the corresponding $v_{0}$ by solving this simple algebraic cubic equation. Using the result for $\alpha_{\mathfrak{N}}$ in Equation (6), we may write this cubic equation in a simpler and more explicit form, in terms of the renormalized mass,

$$v_{0}^{3} - \left( \frac{\alpha_{\mathfrak{N}}}{2\lambda... ...ght) v_{0} + \left( \frac{j_{0}}{2\lambda} \right) = 0.$$ (D.9)

In the simple case in which we make $\lambda=0$, returning to the free-field theory, we at once have that $\alpha_{\mathfrak{N}}=\alpha_{0}=\alpha$, and the result reduces to

$$j_{0} = \alpha\, v_{0},$$

which is the familiar result for the free theory.