Continuum Limits

First of all, it is necessary to say that, regardless of the spacetime dimension and of the symmetry group which are chosen, the dimensionless parameters $\alpha$ and $\lambda$ are the true free parameters of the model. Besides the requirements of stability, there is no reason to limit their range a priori. Limitations may arise, however, from the discussion of physically meaningful observables, expressed as expectation values, specially in the continuum limit. We start therefore with no more than the stability conditions that $\lambda\geq 0$, and that $\alpha\geq 0$ if $\lambda=0$, as the limitations for $\alpha$ and $\lambda$.

In the continuum limit, when we make $N\to\infty$ and $a\to 0$, most dimensionless renormalized quantities we calculated here go to zero. In order to recover the physically meaningful results in the limit, before we take the limit we must rewrite these dimensionless quantities in terms of the corresponding dimensionfull quantities, using the scalings listed in Section 2, Equation (1). Since in the continuum limit $\sigma_{0}$ and $\sigma_{\mathfrak{N}}$ become identical, in all cases where this is possible we will write the formulas in terms of $\sigma_{0}$ only, producing in this way equations which are equivalent to the original ones for the purposes of that limit.

Starting with the expectation value of the field, in the case in which there is no external source $j_{0}$, in which case the limit must be taken within the broken-symmetric phase of the model if we are to have the possibility of a non-zero result, from Equation (7) we have

$$\begin{eqnarray*} \left\langle\phi_{\mathfrak{N}}(x_{\mu})\right\rangle & = & ... ...rak{N}+2)\sigma_{0}^{2}\right]}} {\sqrt{\lambda}\,a^{(d-2)/2}}. \end{eqnarray*}$$

Since for $d\geq 3$ the denominator goes to zero in the continuum limit, if the field $\phi_{\mathfrak{N}}(x_{\mu})$ is to have a finite expectation value, then it is necessary that $v_{0}$ approach zero in the limit, which forcefully takes us to points over the critical line, which is characterized by $v_{0}=0$ and by the equation that states that the quantity within the square root above is zero.

Since the critical line starts at the Gaussian point and extends to the quadrant where $\lambda>0$ and $\alpha\leq 0$, it follows that all possible continuum limits originating from the broken-symmetric phase must go to points in the parameter plane where $\alpha\leq 0$, the case $\alpha=0$ being the Gaussian point and corresponding to the Gaussian sector of the model. In $d=4$, in particular, all possible non-trivial continuum limits necessarily correspond to strictly negative values of $\alpha$. A particular sequence of values of $(\alpha,\lambda)$ approaching the critical line defines both a path in the parameter space of the model and a rate of progress along that path, leading to that particular continuum limit, and is called a continuum limit flow. A continuum limit is completely characterized by its flow, and is not characterized completely just by a point $(\alpha,\lambda)$ in the parameter plane.

Going back to the case in which we have an external source present, we may now rewrite Equation (9) in terms of the renormalized dimensionfull quantities, thus obtaining

$$a^{d-4} V_{0}^{3} - \left( \frac{m_{\mathfrak{N}}^{2}}{... ...ght) V_{0} + \left( \frac{J_{0}}{2\lambda} \right) = 0.$$

In the case $d=3$ we see that, if $\lambda$ is not zero, then the first term dominates over the others, and therefore we conclude simply that $V_{0}^{3}=0$. It follows that in this case there is no spontaneous symmetry breaking and no effect of the external source over $V_{0}$ in the continuum limit. If we wish to have any interesting structure in the model in this case, we are forced to make $\lambda=0$ in the limit. If we do that at the appropriate rate, there may be interesting continuum limits sitting right over the Gaussian point. In the case $d\geq 5$, on the other hand, we see that the first term vanishes, and we are left with $J_{0}=m_{\mathfrak{N}}^{2}V_{0}$, which is characteristic of a free, or trivial theory. In the case $d=4$ we get the equation

$$V_{0}^{3} - \left( \frac{m_{\mathfrak{N}}^{2}}{2\lambda}... ...ght) V_{0} + \left( \frac{J_{0}}{2\lambda} \right) = 0.$$

It is interesting to calculate the discriminant $\Delta$ of this cubic equation, which turns out to be

$$\Delta = \frac{3^{3}}{2^{2}\lambda^{2}} \left( \sqrt{\f... ...{\frac{2}{3^{3}\lambda}}\,m_{\mathfrak{N}}^{3}-J_{0} \right).$$

We can see now that the number of roots of the equation depends on the value of $J_{0}$ in a simple way. If we have

$$J_{0} < \sqrt{\frac{2}{3^{3}\lambda}}\,m_{\mathfrak{N}}^{3},$$

then $\Delta>0$ and therefore there are three distinct simple real roots. If we have

$$J_{0} = \sqrt{\frac{2}{3^{3}\lambda}}\,m_{\mathfrak{N}}^{3},$$

then $\Delta=0$ and the three roots merge into one triple real root. Finally, if

$$J_{0} > \sqrt{\frac{2}{3^{3}\lambda}}\,m_{\mathfrak{N}}^{3},$$

then $\Delta<0$ and there is a single real root, the other two having non-zero imaginary parts. This supports the idea that as $J_{0}$ increases along positive values, the left well of the potential becomes shallower and eventually there is no possibility for the local distribution of the field $\varphi_{\mathfrak{N}}$ to fit within it, even to form a meta-stable state. One of the roots relates to the third extremum of the potential, the local maximum between the two minima. It is clear that, when there is more than one solution to the equation, only the largest solution corresponds to a stable state and is therefore relevant in the context of the symmetry breaking driven by a positive $J_{0}$.

The same analysis regarding critical behavior and the critical line is valid for the renormalized masses. Considering first the limits from the symmetric phase, with no external source $j_{0}$, we have $\alpha_{0}=\alpha_{\mathfrak{N}}$ with $m_{0}^{2}=\alpha_{0}/a^{2}$, and therefore using Equation (10) we have

$$\begin{eqnarray*} \alpha_{0} & = & \alpha+\lambda(\mathfrak{N}+2)\sigma_{0}^{... ...rac {\sqrt{\alpha+\lambda(\mathfrak{N}+2)\sigma_{0}^{2}}} {a}. \end{eqnarray*}$$

We can see that, regardless of how we take the limit, we will necessarily have $m_{0}=m_{\mathfrak{N}}$ in this case. Observe that the numerator on the right-hand side is the quantity which, according to the equation of the critical line, is zero over that line, and hence approaches zero when $(\alpha,\lambda)$ tends to a point on the critical line. Once more we see that, if we are to have a finite value for $m_{0}$, we must approach the critical line on the continuum limit, in such a way that the quantity $\alpha+\lambda(\mathfrak{N}+2)\sigma_{0}^{2}$ approaches zero as $a^{2}$ or faster. If the approach is such that the quantity in the numerator behaves exactly as $a^{2}$, then we have a finite and non-zero value of $m_{0}$. If the approach is faster than that, then we will have $m_{0}=0$. On the other hand, if it is too slow, then we may end up with an infinite $m_{0}$ in the limit.

The same type of mechanism works for limits from the broken-symmetric phase, except that in that case we will always have $m_{0}=0$ in the limit, as we will now demonstrate. As we saw before in Equation (11), we have for $\alpha_{0}$

$$\alpha_{0} = 2\lambda \left( \sigma_{0}^{2} - \sigma_{\mathfrak{N}}^{2} \right),$$

which indeed goes to zero in the limit. However, the analysis of the limit is not so simple, due to the fact that on finite lattices $\alpha_{0}$ appears in the right-hand side of the equation as well. If we write it explicitly, using Equations (B.6) and (B.7) of Appendix B, we get an equation involving $\alpha_{0}$ and $\alpha_{\mathfrak{N}}$,

$$\begin{eqnarray*} \alpha_{0} & = & 2\lambda\, \frac{1}{N^{d}} \sum_{k_{\mu}... ...right] \left[\rho^{2}(k_{\mu})+\alpha_{\mathfrak{N}}\right] }. \end{eqnarray*}$$

Now, if $\alpha_{\mathfrak{N}}=\alpha_{0}$, which implies that $m_{\mathfrak{N}}=m_{0}$, then the right-hand side is zero, and therefore so is $\alpha_{0}$. This in turn implies that $m_{0}=0$, as expected. This is in fact one possibility, we may indeed have both $m_{0}$ and $m_{\mathfrak{N}}$ zero in the limit. If, on the other hand, we have $\alpha_{\mathfrak{N}}\neq\alpha_{0}$, then we may write the equation as

$$\frac{m_{0}^{2}}{m_{\mathfrak{N}}^{2}-m_{0}^{2}} = 2\lamb... ...ght] \left[\rho^{2}(k_{\mu})+\alpha_{\mathfrak{N}}\right] },$$

where we wrote the left-hand side in terms of dimensionfull quantities. Obviously, because both $\lambda$ and the sum are necessarily positive quantities, it is impossible to have $m_{\mathfrak{N}}<m_{0}$. Here we see that, if we have both $m_{0}$ and $m_{\mathfrak{N}}>m_{0}$ different from zero in the limit, then the left-hand side has a non-zero limit and therefore the normalized sum on the right-hand side must be non-zero in the limit.

However, one can check numerically that, for $d\geq 4$ and in the type of continuum limit that we consider here, the normalized sum does indeed go to zero in the limit. This implies that in these dimensions, which include $d=4$, we cannot have both $m_{0}$ and $m_{\mathfrak{N}}$ different from zero in the limit. Since $m_{\mathfrak{N}}>m_{0}$, this implies that we must always have $m_{0}=0$ in the limit. What we have here, as one should expect, are the Goldstone bosons brought about by the process of spontaneous symmetry breaking.

For the longitudinal mass parameter $m_{\mathfrak{N}}$ we have, using Equation (13),

$$\begin{eqnarray*} \alpha_{\mathfrak{N}} & = & -2 \left[\alpha+\lambda(\mathf... ...left[\alpha+\lambda(\mathfrak{N}+2)\sigma_{0}^{2}\right]}} {a}, \end{eqnarray*}$$

so that exactly the same argument that was used for $\alpha_{0}$ in the symmetric phase applies. We see therefore that the need to approach the critical line when one takes continuum limits in this model is a rather general characteristic of the model. This makes the critical line the locus of all physically possible continuum limits of the model. This means that making $\alpha<0$ is not a choice that we have, since it is forced upon us by the need to obtain physically meaningful continuum limits.

Let us now discuss the continuum limits of the transversal renormalized mass in the presence of an external source. We have the result in Equation (5), valid in either phase,

$$\alpha_{0}(j_{0}) = \lambda v_{0}^{2} + \left[ \alpha + \lambda (\mathfrak{N}+2) \sigma_{0}^{2} \right].$$

Observe that this equation implies that it is still necessary to approach the critical line in the continuum limit, and in the same ways as before. In the symmetric phase, if $\alpha_{0}(0)$ is the corresponding result in the absence of external sources, which corresponds to $v_{0}=0$, we may write

$$\alpha_{0}(j_{0}) = \alpha_{0}(0) + \lambda v_{0}^{2}.$$

Rewriting all quantities in terms of the corresponding dimensionfull ones we have

$$m_{0}^{2}(J_{0}) = m_{0}^{2}(0) + \lambda a^{d-4} V_{0}^{2}.$$

We see that for $d=3$ we are forced to make $\lambda\to 0$ in the limit. In the case $d=4$ no additional constraints on $\lambda$ arise, and we get the relation

$$m_{0}(J_{0}) = \sqrt { m_{0}^{2}(0) + \lambda V_{0}^{2} },$$

describing indirectly how $m_{0}(J_{0})$ increases with $J_{0}$ through the variation of $V_{0}$. In the case $d=5$ the term containing $\lambda$ vanishes in the limit, and we get simply that $m_{0}(J_{0})=m_{0}(0)$, meaning that in this case $m_{0}(J_{0})$ does not really depend on $J_{0}$ in the continuum limit.

In the broken-symmetric phase we may start with Equation (12) for the transversal mass parameter. If we recall that we have already shown that in this phase we must have $m_{0}(0)=0$ in the limit, we may make $\sigma_{\mathfrak{N}}=\sigma_{0}$ in this formula and thus obtain

$$\alpha_{0} = \lambda \left( v_{0}^{2} - v_{0,{\rm SSB}}^{2} \right).$$

In terms of dimensionfull quantities we have therefore

$$m_{0}^{2}(J_{0}) = \lambda a^{d-4} \left( V_{0}^{2} - V_{0,{\rm SSB}}^{2} \right),$$

which gives us back $m_{0}(0)=0$ in the absence of external sources. Not much changes in the discussion of the various possible dimensions. We may restrict our comments to the case $d=4$, in which we get a fairly simple relation giving $m_{0}(J_{0})$ in the presence of the external source,

$$m_{0}(J_{0}) = \sqrt{\lambda} \sqrt { V_{0}^{2} - V_{0,{\rm SSB}}^{2} }.$$

The same analysis can be made for the longitudinal mass in the presence of an external source. In this case we have the result in Equation (6), valid in either phase,

$$\alpha_{\mathfrak{N}}(j_{0}) = 3\lambda v_{0}^{2} + \l... ...\alpha + \lambda (\mathfrak{N}+2) \sigma_{0}^{2} \right].$$

The necessity to approach the critical line remains in force here. In the symmetric phase, if $\alpha_{\mathfrak{N}}(0)$ is the corresponding result in the absence of external sources, which corresponds to $v_{0}=0$, we may write

$$\alpha_{\mathfrak{N}}(j_{0}) = \alpha_{\mathfrak{N}}(0) + 3\lambda v_{0}^{2}.$$

Rewriting all quantities in terms of the corresponding dimensionfull ones we have

$$m_{\mathfrak{N}}^{2}(J_{0}) = m_{\mathfrak{N}}^{2}(0) + 3\lambda a^{d-4} V_{0}^{2}.$$

Once more we see that for $d=3$ we are forced to make $\lambda\to 0$ in the limit. In the case $d=4$ we get simply the relation

$$m_{\mathfrak{N}}(J_{0}) = \sqrt { m_{\mathfrak{N}}^{2}(0) + 3\lambda V_{0}^{2} },$$

describing indirectly how $m_{\mathfrak{N}}(J_{0})$ increases with $J_{0}$ through the variation of $V_{0}$. In the case $d=5$ the term containing $\lambda$ vanishes in the limit, and we get simply that $m_{\mathfrak{N}}(J_{0})=m_{\mathfrak{N}}(0)$, meaning that in this case $m_{\mathfrak{N}}(J_{0})$ also does not depend on $J_{0}$ in the continuum limit.

In the broken-symmetric phase we may start with Equation (14) for the longitudinal mass parameter

$$\alpha_{\mathfrak{N}} - \alpha_{\mathfrak{N},{\rm SSB}} = 3\lambda \left( v_{0}^{2} - v_{0,{\rm SSB}}^{2} \right).$$

where $\alpha_{\mathfrak{N},{\rm SSB}}$ is the value of the parameter in the absence of external sources. In terms of dimensionfull quantities we have therefore

$$m_{\mathfrak{N}}^{2}(J_{0}) - m_{\mathfrak{N}}^{2}(0) = ... ... a^{d-4} \left( V_{0}^{2} - V_{0,{\rm SSB}}^{2} \right),$$

Once again not much changes in the discussion of the various possible dimensions. In the case $d=4$ we get

$$m_{\mathfrak{N}}(J_{0}) = \sqrt { m_{\mathfrak{N}}^{2}(... ...lambda \left( V_{0}^{2} - V_{0,{\rm SSB}}^{2} \right) }.$$

It is interesting to note that, both for the transversal and longitudinal masses, the dependence of the renormalized masses on the external source $J_{0}$ seems to be a peculiar feature of the case $d=4$, which is absent for $d\geq 5$.