Some Mean Field Results

We will perform in this section a few analytical calculations, in the usual mean-field approximation with $N=1$, of some observables of the scalar field models, using the ideas presented in section 2.2. The approximations with $N\geq 2$, involving the extension of the method which was also discussed in that section, are usually too complex for an analytical treatment and are better characterized, therefore, as material for performing stochastic simulations, which will be discussed in a future volume. The calculations we will present here can be understood as exact analytical solutions for the lattice systems with a single active site. We will start with the Ising models, in which the calculations are somewhat simpler.

A quantity of particular interest in the Ising models is the magnetization, which is the order parameter of these models. It can be defined as the expectation value $v_{R}$ of the field, which will be pointing predominantly in the direction that we choose arbitrarily for the symmetry breaking. This is the direction in which we will keep oriented the fields at the border. We will calculate $v_{R}$ in the mean-field approximation as a function of the parameter $\beta$ and of the dimension $d$ of space-time. As we saw before in section 2.2, the parameter $\beta$ can be understood as the inverse of a fictitious temperature. For a given value of $d$, the critical point $\beta_{c}$ is the value of the inverse temperature for which the magnetization $v_{R}(d,\beta)$ ceases to be zero, when we ``cool'' the system and therefore increase $\beta$. For the calculations it is convenient to use the dimensionless field $\psi$ that was introduced in section 2.2, so that for the purposes of this section we will use, rather than $v_{R}=\langle\varphi\rangle$, the quantity $\nu_{R}=\langle\psi\rangle$, where $v_{R}=\sqrt{\beta}\nu_{R}$. It is $\nu_{R}$ rather than $v_{R}$ that is more closely related to the standard definition of the magnetization of statistical mechanics. As was derived in that section, the action can be written in terms of $\psi$ as

$$S[\psi]=-\beta\sum_{\ell}\psi(\ell_{-})\psi(\ell_{+}),$$

while the constraint is written as

$$\psi^{2}=1,$$

where the field $\psi$ is, therefore, either $+1$ or $-1$. We now freeze all the sites except one at the value $\psi=\nu_{R}$, arbitrarily choosing the positive direction as the direction of orientation of the fields. Therefore in our $N=1$ mean-field approximation the relation $\nu_{R}=\langle\psi\rangle$ can be written as

$$\nu_{R}=\frac{\displaystyle \sum[\psi=\pm 1]\;\psi\;e^{2d \... ...i}}{\displaystyle \sum[\psi=\pm 1]\;e^{2d \beta\nu_{R}\psi}}.$$

Since the sum is over a single site, in this simple $O(1)$ case we can immediately write the result as

$$\nu_{R}=\frac{\displaystyle \sinh(2d \beta\nu_{R})}{\displaystyle \cosh(2d \beta\nu_{R})},$$ (2.3.1)

thus obtaining an algebraic equation that in principle gives us the complete solution for $\nu_{R}$ in the $N=1$ system. This equation can be solved numerically in order to provide us with graphs of $\nu_{R}$ as a function of $\beta$ (problem 2.3.2). It can also be used to determine the value of $\beta$ at which $\nu_{R}=0$ becomes the only possible solution (problem 2.3.1), by the use of series expansions. However, the simplest way to obtain the critical points is to simply expand its right-hand side around $\nu_{R}=0$. This is sufficient to determine the critical points because the phase transition is of second order in these models and, therefore, we have that $\nu_{R}\approx 0$ for $\beta\approx\beta_{c}$. Expanding and keeping only up to first-order terms (problem 2.3.3) we get

$$\nu_{R}=2d \beta\nu_{R}.$$

A factor of $\nu_{R}$ cancels off and we are left with

$$1=2d \beta.$$

After the cancellation of the trivial solution $\nu_{R}=0$ we may impose the condition $\nu_{R}=0$ in order to find the region where only this zero solution is possible. Observe that the equation above does not depend on $\nu_{R}$. Had we kept the higher-order terms of the expansion, the condition $\nu_{R}=0$ would eliminate them at this point, leaving only the equation above. Hence we obtain the mean-field approximation for the critical points

$$\beta_{c}(d)=\frac{1}{2d}.$$ (2.3.2)

This technique of expansion around the critical point is useful in cases in which the mean-field results cannot be obtained explicitly, or in which the exact solutions of the resulting equations are difficult to determine analytically.

In four dimensions this $N=1$ result is in fairly good agreement with the numerical results obtained for larger values of $N$. The agreement is poorer in three dimensions, very poor is two dimensions, and the result fails completely in one dimension, in which there is no critical point at all. Observe that the success or lack thereof on the $N=1$ approximation in reproducing well the results for large values of $N$ is not a diagnostic about the validity of the method itself, bur rather an indication of whether or not the results for increasing values of $N$ accumulate around some finite value of $\beta$. For $d=1$ the values $\beta_{c}(N)$ for each lattice size diverge to $+\infty$ when $N\rightarrow\infty$, so that no finite lattice can represent well the limit. In larger dimensions the critical points $\beta_{c}(N)$ of the finite lattices converge to finite values in the $N\rightarrow\infty$ limit, so that in these cases the approximation of the limiting result by the results for finite lattices makes some sense, being better or worse depending on the speed of convergence.

Table 2.3.1 contains results for the critical points of the Ising models for dimensions from $d=1$ up to $d=4$. For $d=4$ the result was obtained by means of extrapolations to the $N\rightarrow\infty$ limit of the results of stochastic simulations with periodical boundary conditions on lattices with $N$ from $4$ to $10$. For $d=3$ we quote the most precise result that we know about [36]. For $d=2$ we quote the well-known result for the two-dimensional Ising model [37]. Since the estimate from periodical simulations is obtained from the continuous and differentiable curves of the magnetization as a function of $\beta$, the error bars indicated are only approximate estimates. The entry $\infty$ indicates the case in which the theorems about long-distance order [31] imply that there is no phase transition, which is the same in which $\beta_{c}(N)\rightarrow\infty$ when $N\rightarrow\infty$ in the simulations with self-consistent fixed boundary conditions. The results of equation (2.3.2) are also included, for comparison. As one can see, in $d=4$ the mean-field result deviates about 17% from the numerical results, while for $d=3$ it deviates by about 25%.

Table 2.3.1: Critical points of the Ising models in dimensions from $d=1$ to $d=4$.

$d$	$\mathfrak{N}$	$\beta_{c}$
		periodical	mean-field
$1$	$1$	$\infty$	$0.500$
$2$	$1$	$0.4406868$	$0.250$
$3$	$1$	$0.22165(4)$	$0.16\bar{6}$
$4$	$1$	$0.15\pm 0.02$	$0.125$

Calculating the magnetization in an infinitesimal neighborhood of $\beta_{c}$ one can also obtain the mean-field approximation for the corresponding critical exponent. The definition of this critical exponent is as follows. If we have, close to the critical point in the broken-symmetrical phase,

$$\nu_{R}\approx C\left(\beta-\beta_{c}\right)^{\gamma}$$ (2.3.3)

for some non-vanishing constant $C$, then $\gamma$ is the critical exponent of $\nu_{R}$. Once more, this can be done using the complete solution in terms of the hyperbolic functions (problem 2.3.4), but the simplest way to obtain the result is to expand equation (2.3.1) to third order in $\nu_{R}$ (problem 2.3.3), thus obtaining

$$\nu_{R}=2d \beta\nu_{R}-\frac{8}{3}d^{3}\beta^{3}\nu_{R}^{3}.$$

Once more a factor of $\nu_{R}$ cancels out and we obtain

$$1=2d \beta\left[1-\frac{4}{3}d^{2}\beta^{2}\nu_{R}^{2}\right].$$

Remembering that $\beta_{c}=1/(2d)$ and considering that we are in the vicinity of the critical point, with $\beta\approx\beta_{c}$, we may write this as

$$\nu_{R}=\sqrt{3}\;\frac{\beta_{c}}{\beta} \;\sqrt{\frac{\bet... ...{\frac{\beta-\beta_{c}}{\beta_{c}}} =C(\beta-\beta_{c})^{1/2}.$$ (2.3.4)

This shows that the mean-field approximation for the critical exponent is $\gamma=1/2$, and determines the value of the constant $C=\sqrt{3/\beta_{c}}$. This value for the critical exponent of the magnetization is characteristic of the mean-field calculations with $N=1$.

Another mean-field calculation of interest is that of the critical curves of the $\lambda \varphi ^{4}$ polynomial models, which are given by equation of the type $\lambda=f(\alpha)$. In this case we will use the usual dimensionless field $\varphi$ as our variable. If we define the magnetization for this case as $v_{R}=\langle\varphi\rangle$ we may write, in the $N=1$ mean-field approximation,

$$v_{R}=\frac{\displaystyle \int_{-\infty}^{\infty}{\rm d}\var... ...\left[(d+\alpha/2)\varphi^{2}+(\lambda/4)\varphi^{4}\right]}}.$$

If we use here our ``radial-angular'' decomposition of $\varphi$ into its absolute value and its sign, $\varphi=\psi\vert\varphi\vert$, with $\psi=\pm 1$, as was done in section 2.1, we may write this as

$$v_{R}=\frac{\displaystyle \sum[\psi=\pm 1]\int_{0}^{\infty}{... ...\left[(d+\alpha/2)\varphi^{2}+(\lambda/4)\varphi^{4}\right]}}.$$

We may now execute the sum over the single active site and hence get

$$v_{R}=\frac{\displaystyle \int_{0}^{\infty}{\rm d}\varphi\;\... ...\left[(d+\alpha/2)\varphi^{2}+(\lambda/4)\varphi^{4}\right]}}.$$ (2.3.5)

In this case it is also possible to perform the integrations analytically in terms of special functions (problem 2.3.5) but, close to the critical curve, it suffices to calculate the right-hand side of this equation for small values of $v_{R}$. Therefore, we expand equation (2.3.5) to first order in $v_{R}$ and obtain

$$v_{R}=2d v_{R}\;\frac{\displaystyle \int_{0}^{\infty}{\rm d... ...\left[(d+\alpha/2)\varphi^{2}+(\lambda/4)\varphi^{4}\right]}}.$$

Just like before, a factor of $v_{R}$ cancels out and we obtain as our mean-field result an equation giving implicitly the critical curve $\lambda=f(\alpha)$,

$$\frac{1}{2d}=\frac{\displaystyle \int_{0}^{\infty}{\rm d}\va... ...\left[(d+\alpha/2)\varphi^{2}+(\lambda/4)\varphi^{4}\right]}}.$$ (2.3.6)

Once more the integrations can be done, this time in terms of the parabolic cylinder functions ${\bf D}_{\nu}$ (problem 2.3.6), but this does not help us to solve this equation in order to write the equation of the critical curve in explicit form. The fact that the left-hand side of this equation is equal to the mean-field value of the critical point $\beta_{c}$ of the corresponding Ising model is not an accident, it is clearly related to the fact that the $\lambda\rightarrow\infty$ limits of the $\lambda \varphi ^{4}$ models converge to the Ising models (problem 2.3.7).

Although we are not able to solve equation (2.3.6) analytically in order to write the equation $\lambda=f(\alpha)$ of the critical curve in explicit form, it is possible to solve the asymptotic form of the equation, for large values of $\lambda$ (problem 2.3.8). Doing this we discover that in this limit the critical curve is asymptotic to the straight line defined by the equation

$$\lambda(\alpha)=-\frac{1}{\beta_{c}}(\alpha+2d),$$

where $\beta_{c}=1/(2d)$ are the critical points of the corresponding Ising models, so that the asymptote cuts the $\lambda$ axis at the point $-4d^{2}$ and forms with the negative $\alpha$ semi-axis an angle $\theta$ such that $\tan(\theta)=1/\beta_{c}$. Expanding equation (2.3.6) for small values of $-\alpha$ and $\lambda$ one can also obtain the slope of the tangent line to the critical curve at the Gaussian point (problem 2.3.9). Doing this we obtain for this tangent line the equation

$$\lambda(\alpha)=-\frac{1}{\beta_{c}}\;\frac{\alpha}{3}.$$

We see that the asymptotic slope is $-1/\beta_{c}$, larger therefore than the slope at the Gaussian point, which is $-1/(3\beta_{c})$, by a factor of $3$, thus showing that the critical curve has its concavity turned mostly upwards.

Figure 2.3.1: The mean-field critical curve of the $\lambda \varphi ^{4}$ model with $O(1)$ symmetry, in $d=3$, for the smaller values of the parameters.

Figure 2.3.2: The mean-field critical curve of the $\lambda \varphi ^{4}$ model with $O(1)$ symmetry, in $d=3$, for the larger values of the parameters.

Figure 2.3.3: The mean-field critical curve of the $\lambda \varphi ^{4}$ model with $O(1)$ symmetry, in $d=4$, for the smaller values of the parameters.

Figure 2.3.4: The mean-field critical curve of the $\lambda \varphi ^{4}$ model with $O(1)$ symmetry, in $d=4$, for the larger values of the parameters.

In order to get the graphs of the critical curve we are compelled to solve the equation by numerical means. In fact, this can be done not only for our $O(1)$ case here, but for the $SO(\mathfrak{N})$ generalizations as well. A curious fact is that it is in fact easier to do this for $\mathfrak{N}=2$ and the other even-$\mathfrak{N}$ cases than for $\mathfrak{N}=1$ and the other odd-$\mathfrak{N}$ cases, because in the even-$\mathfrak{N}$ cases it turns out that the integrals can be written in terms of the error function. For the odd-$\mathfrak{N}$ cases we must use direct numerical integration in order to solve the equation, which is a technique that can be used in all cases [38]. The graphs in figures from 2.3.1 to 2.3.4 show the critical curves obtained by such numerical means in a few cases. Each graph shows also the tangent line at the Gaussian point and the asymptotic line for large values of $\lambda$.

Figure 2.3.1 shows the solution in the case $d=3$, for fairly small values of the parameters $-\alpha$ and $\lambda$, while figure 2.3.2 shows the same solution for larger values of the parameters. Figures 2.3.3 and 2.3.4 show the corresponding data for the case $d=4$. It should be noted that, as one can see in the two graphs with the larger values of the parameters, the asymptotic lines actually cross the critical curves. This implies that at some location for even larger values of the parameters the critical curves must have inflection points and reverse their concavities, presumably approaching their asymptotes from below rather than from above. One can see, looking directly at the data shown in the graphs, that the critical curves do in fact slowly approach the asymptotes, but it seems that the location of the inflection point is at very large values of the parameters, that so far have not been probed numerically. It is interesting to observe that this behavior seems to be characteristic of the $O(1)$ models. In the $SO(\mathfrak{N})$ models with $\mathfrak{N}>1$ one does not see this type of crossing, and the critical curves seem to approach their asymptotes from above.

We are led to conclude that the mean-field approximation confirms the basic properties of the critical curves that we discussed before in sections 1.1 and 1.3. One can improve on the comparison with the perturbative results by calculating the mean-field approximation for the quantity $\sigma^{2}=\langle\varphi^{2}\rangle$ over the critical curve, where $v_{R}=\langle\varphi\rangle$ is zero. The mean-field approximation for $\sigma^{2}$ is given by

$$\sigma_{\rm MF}^{2}=\frac{\displaystyle \sum[\psi=\pm 1]\int... ...\left[(d+\alpha/2)\varphi^{2}+(\lambda/4)\varphi^{4}\right]}},$$

and if we execute the sum over the signs at the single active site we get

$$\sigma_{\rm MF}^{2}=\frac{\displaystyle \int_{0}^{\infty}{\r... ...\left[(d+\alpha/2)\varphi^{2}+(\lambda/4)\varphi^{4}\right]}}.$$

At the critical curve, where $v_{R}=0$, this reduces to

$$\sigma_{\rm MF}^{2}=\frac{\displaystyle \int_{0}^{\infty}{\r... ...\left[(d+\alpha/2)\varphi^{2}+(\lambda/4)\varphi^{4}\right]}}.$$

Note that the right-hand side of this equation is exactly equal to the right-hand side of equation (2.3.6), which determines the critical line, and that therefore we have

$$\sigma_{\rm MF}^{2}=\frac{1}{2d}=\beta_{c},$$

so that the equation of the tangent line to the mean-field critical curve at the Gaussian point can be written as

$$\lambda(\alpha)=-\frac{1}{\sigma_{\rm MF}^{2}}\;\frac{\alpha}{3}.$$

If we now recall the perturbative result for the equation of the tangent line to the critical curve at the Gaussian point, which can be obtained from equation (1.3.7), and can be written as

$$\lambda(\alpha)=-\frac{1}{\sigma_{0}^{2}}\;\frac{\alpha}{3},$$

we see that the two results are identical except for the exchange of $\sigma_{0}^{2}$ and $\sigma_{\rm MF}^{2}$. Therefore, the two results for the slope of the critical curve at the Gaussian point coincide within the expected level of precision of the mean-field approximation, since the mean-field result is just an approximation, while the perturbative result of the Gaussian approximation is, for this particular quantity, presumably exact.

We can also calculate the critical exponent of $v_{R}$ in the case of the polynomial models. In order to do this it is necessary to expand equation (2.3.5) up to a higher order in $v_{R}$, so as to allow us to write the differential of $v_{R}^{2}(\alpha,\lambda)$ as a function of ${\rm d}\alpha$ e ${\rm d}\lambda$. This is the work proposed in problem 2.3.10, and it can be shown that it is possible to write the differential of $v_{R}^{2}$ as

$${\rm d}(v_{R}^{2})=C_{\alpha}\;{\rm d}\alpha+C_{\lambda}\;{\rm d}\lambda,$$

where the coefficients $C_{\alpha}$ e $C_{\lambda}$ are finite and non-vanishing expressions in the vicinity of the critical curve. The fact that we are able to write the differential of $v_{R}^{2}$ directly in terms of the differentials of $\alpha$ and $\lambda$ with coefficients that are finite and not zero over the critical curve is sufficient to show that the critical exponent of $v_{R}$ is, once more, $\gamma=1/2$.

In conclusion, we have discovered that the mean-field method is related, through its generalizations, to systems on finite lattices with fixed and self-consistent boundary conditions. These systems define a second family of continuum limits for the models, which does not necessarily have to be identical to the one defined by the systems with periodical boundary conditions. Although the two families of limits have, by and large, similar properties, many of the details are not identical and some quantities of interest may, in fact, depend on the boundary conditions. For example, it may be that only part of the discrepancies shown in table 2.3.1 for the critical points of the Ising models is due to the fact that we are using the $N=1$ mean-field approximation in the right-hand column. Another part of the discrepancies may be due to the fact that we are comparing results for systems with very different boundary conditions, since the left-hand column refers to systems with periodical boundary conditions. A situation like this is probably more likely to be realized for the finer, more delicate observables, such as the critical exponents and correlation functions, than for the more basic objects such as the critical points.

The choice of boundary conditions is an important subject within the structure of the quantum theory, just as it is in the classical theory. It should be noted that there are other ways, besides the one that we examined here, to implement fixed boundary conditions, some of which may be physically more natural and compelling from the point of view of quantum field theory. One example of this kind of thing was proposed in the last chapter of a previous volume of this series of books [35]. Other ideas related to that one may be discussed in future volumes.

Problems

1. Verify that $\nu_{R}=0$ always satisfies equation (2.3.1). In order to discover whether there are values of $\beta$ for which it is possible to have a solution $\nu_{R}\neq 0$, use the series expansions of the hyperbolic functions to obtain the relation

   
   

   $$\sum_{k=0}^{\infty}\frac{(2d \beta\nu_{R})^{2k+1}}{(2k+1)!} \left[(2k+1)-2d \beta\right]=0.$$
   
   

   Observe that all the factors in each term of this series are always positive except for the last one on the right. Use this fact to determine the interval of values of $\beta$ for which it is possible to have $2d \beta\nu_{R}$ and therefore $\nu_{R}$ different from zero as a solution of this equation, and thus determine the mean-field critical points $\beta_{c}$ of the Ising models.

2. Write a program to solve numerically equation (2.3.1) and plot graphs of $\nu_{R}$ as a function of $\beta$ for some values of $d\geq 3$. Consider using the exponential bisection method and consider the results of problem 2.3.1. Verify how close to the exact result in the broken-symmetrical phase is the ansatz

   
   

   $$\nu_{R}=\sqrt{\frac{3(\beta-\beta_{c})}{3\beta-2\beta_{c}}},$$
   
   

   which has the correct behavior for $\beta$ close to $\beta_{c}$ and that tends to $1$ for $\beta\rightarrow\infty$.

3. Expand the right-hand side of equation (2.3.1) up to the third order on the variable $A=2d \beta\nu_{R}$ and show that the terms with even powers of the variable vanish, thus obtaining the result

   
   

   $$\nu_{R}=A-\frac{A^{3}}{3}.$$
   
   

4. Starting from the relation $\nu_{R}\approx C\left(\beta-\beta_{c}\right)^{\gamma}$ that defines the critical exponent $\gamma$ of $\nu_{R}$, write the corresponding relation for the parameter $A=2d \beta\nu_{R}$ and differentiate the resulting equation with respect to $\beta$, thus obtaining an expression for the critical exponent,

   
   

   $$\gamma=\frac{\beta-\beta_{c}}{A}\;\frac{{\rm d}A}{{\rm d}\beta} -\frac{\beta-\beta_{c}}{\beta}.$$
   
   

   In the vicinity of the critical point we have $\beta\approx\beta_{c}$ and $A\approx 0$, so that we may write for $\gamma$

   
   

   $$\gamma=\frac{\beta-\beta_{c}}{A}\;\frac{{\rm d}A}{{\rm d}\beta}.$$ (2.3.7)

   
   

   In order to calculate ${\rm d}A/{\rm d}\beta$, differentiate equation (2.3.1) with respect to $A$, and thus obtain the result

   
   

   $$\frac{{\rm d}A}{{\rm d}\beta}=\frac{2d\;\sinh^{2}(A)}{\sinh(A)\cosh(A)-A}.$$
   
   

   Use this in the expression for $\gamma$ and expand it to second order around the critical point, then use the result $\beta_{c}=1/(2d)$, in order to obtain for $\gamma$ the expression

   
   

   $$\gamma=\frac{3}{2C^{2}\beta_{c}}(\beta-\beta_{c})^{1-2\gamma}.$$
   
   

   Note that, given the second-order nature of the phase transition in these models, we know that $\gamma$ must be within the interval $(0,1]$. Examine the behavior of this expression in the cases $1/2<\gamma\leq 1$ and $0<\gamma<1/2$, and show by reduction to absurd that the only possible value for the critical exponent is $\gamma=1/2$. Substitute this value in the equation above and obtain the value of the constant $C$, thus reproducing equation (2.3.4) which was derived in the text.

5. Use the series expansions of the hyperbolic functions $\sinh(B)$ and $\cosh(B)$, with $B=2d v_{R}\varphi$, that appear in equation (2.3.5), in order to rewrite that equation in the form

   
   

   $$\frac{1}{2d}=\frac{\displaystyle \sum_{k=0}^{\infty}\frac{(2... ...\left[(d+\alpha/2)\varphi^{2}+(\lambda/4)\varphi^{4}\right]}}.$$
   
   

   The integrals may now be written in terms of the parabolic cylinder functions ${\bf D}_{\nu}$ [15], so use them to write the final form of the equation

   
   

   $$\frac{\sqrt{2\lambda}}{2d}=\frac {\sum_{k=0}^{\infty}\frac{(... ...1}{2}\right)} \left(\frac{2d+\alpha}{\sqrt{2\lambda}}\right)}.$$
   
   

6. Use the same techniques of problem 2.3.5 in order to calculate the integrals that appear in equation (2.3.6), and thus obtain

   
   

   $$\frac{\sqrt{2\lambda}}{2d}=\frac{{\bf D}_{-\left(\frac{3}{2}... ...1}{2}\right)} \left(\frac{2d+\alpha}{\sqrt{2\lambda}}\right)}.$$ (2.3.8)

   
   

   Observe that this result is the same of problem 2.3.5 if we truncate the two series that appear there, leaving only their first terms, those with $k=0$.

7. Take explicitly the Ising-model limit of equation (2.3.6), making $\lambda\rightarrow\infty$ and $\alpha\rightarrow-\infty$ with $\alpha=-\beta\lambda$, and show that it reduces to the known result for the value of the critical point of the Ising model.

8. Obtain the asymptotic form of the critical curve, for large values of $-\alpha$ and $\lambda$, using in equation (2.3.8) the asymptotic expansion of the parabolic cylinder functions ${\bf D}_{\nu}$ [25], to the lowest non-vanishing order, thus obtaining the result

   
   

   $$\lambda\left(\alpha\right)=-\frac{1}{\beta_{c}}\left(\alpha+2d\right).$$
   
   

9. In order to obtain the behavior of the critical curve for small values of $-\alpha$ and $\lambda$, write equation (2.3.6) in the form

   
   

   $$\int_{0}^{\infty}{\rm d}\varphi\;\varphi^{2} \;e^{-\left[(d+... ...{-\left[(d+\alpha/2)\varphi^{2}+(\lambda/4)\varphi^{4}\right]}$$
   
   

   and differentiate implicitly in terms of ${\rm d}\alpha$ e ${\rm d}\lambda$, applying the resulting coefficients at the Gaussian point. The integrals that appear in these coefficients are expressible in terms of the $\Gamma$ function [14], so use them to obtain

   
   

   $$\frac{{\rm d}\lambda}{{\rm d}\alpha}=-\frac{2d}{3}.$$
   
   

   Integrate this first-order differential equation for $\lambda(\alpha)$ with the boundary condition $\lambda(0)=0$, thus obtaining the final result

   
   

   $$\lambda(\alpha)=-\frac{1}{\beta_{c}}\;\frac{\alpha}{3}.$$
   
   

   Obtain the same result in another way, using in equation (2.3.8) the asymptotic expansion of ${\bf D}_{\nu}$ [25], since for $\lambda\rightarrow 0$ the arguments of the ${\bf D}_{\nu}$ functions that appear in that equation go to infinity.

10. In order to determine the mean-field value of the critical exponent of $v_{R}$ in the $\lambda \varphi ^{4}$ model, write equation (2.3.5) in the form
    $$\begin{eqnarray*} \lefteqn{v_{R}\int_{0}^{\infty}{\rm d}\varphi\;\cosh(B) \;e^{-... ...^{-\left[(d+\alpha/2)\varphi^{2}+(\lambda/4)\varphi^{4}\right]}, \end{eqnarray*}$$
    
    
    
    

    where $B=2d v_{R}\varphi$. Then differentiate in terms of ${\rm d}\alpha$ and ${\rm d}\lambda$, keeping in mind that $v_{R}$ is a function of $\alpha$ and $\lambda$, in order to write the differential of $v_{R}$ in terms of its gradient as

    
    

    $${\rm d}v=\frac{C_{1}}{C_{0}}\;{\rm d}\alpha+\frac{C_{2}}{C_{0}}\;{\rm d}\lambda,$$
    
    

    where the coefficients are given by

    $$\begin{eqnarray*} C_{0} & = & \int_{0}^{\infty}{\rm d}\varphi\;e^{-\left[(d+\alp... ...ht]} & & \times\left[v\;\cosh(B) -\;\varphi\;\sinh(B)\right]. \end{eqnarray*}$$
    
    
    
    

    Next expand the hyperbolic functions in each one of these coefficients for small values of $B$, that is, in the vicinity of the critical curve. Observe that it is enough to expand $C_{1}$ e $C_{2}$ to first order but that $C_{0}$ must be expanded to the next non-vanishing order, because the equation of the critical curve implies that the terms of orders zero and one of its expansion cancel each other. In this way, obtain the differential of $v_{R}^{2}$ as

    
    

    $${\rm d}(v^{2})=C_{\alpha}\;{\rm d}\alpha+C_{\lambda}\;{\rm d}\lambda,$$
    
    

    where the coefficients, which define the gradient of $v_{R}^{2}$, are given by

    $$\begin{eqnarray*} C_{\alpha} & = & \frac{\displaystyle \frac{1}{3d}\int_{0}^{\in... ...-\left[(d+\alpha/2)\varphi^{2} +(\lambda/4)\varphi^{4}\right]}}. \end{eqnarray*}$$
    
    
    
    

    Stating from these expressions verify that $C_{\alpha}$ and $C_{\lambda}$ are finite and non-vanishing numbers in the vicinity of the critical curve. This suffices to show that the critical exponent of $v_{R}$ is once again $\gamma=1/2$ when we approach the critical curve in the parameter plane of the model, from any direction within the broken-symmetrical phase. Verify also that these two coefficients are negative, showing that the gradient of $v_{R}^{2}$ is oriented in the expected direction.