2
$\begingroup$

Generally speaking, phase transitions divide into two types: First order and second order. To me, Higgs field's SSB sounds like a second-order one though I don't know the dependency of Higgs field's VEV on temperature(Energy or whatever, etc.) to say if it's continuous or not[considering the VEV as the only order parameter](another point is that the VEV doesn't receive any loop correction by renormalization as we impose it as a renormalization condition like what's done in Peskin, and it can be interpreted as independence of VEV as an order parameter on temperature! so one might say it's first-order phase transition.)

I want to know with some reliable reasoning that if Higgs' SSB is a first or second-order phase transition?

*Regarding that the Higgs' lagrangian wrong-signed mass term is zero and the phase transition happens as a result of quantum fluctuations or more explicitly the loop corrections in the effective action(Coleman-Weinberg mechanism).


Edit: For simplicity let's consider the charge less lamda phi^4 theory with parity symmetry, the mass term being zero and the mass term being generated by loop corrections(Coleman-Weinberg mechanism) after spontaneous symmetry breaking. In this case the phase transition is not thermal but quantum mechanical ane the control parameter is coupling constant(I guess this can be the control parameter if it's not the mass term!)

$\endgroup$
  • 1
    $\begingroup$ This depends on certain important details like dimensionality. Are you referring to scalar QED in $(3+1)$ dimensions, a la Coleman-Weinberg? In this case the transition is first order, which you can show in the process of solving the "Final Project" after Chapter 13 in Peskin & Schroeder. $\endgroup$ – Seth Whitsitt Dec 14 '19 at 19:04
  • $\begingroup$ What if there's no kind of gauge boson and interactions are only due to the self-interactions and the VEV effective interactions with the scalar field. And I'll be glad to be given a general answer in (d+1) dimension. $\endgroup$ – Bastam Tajik Dec 14 '19 at 22:13
  • 1
    $\begingroup$ Are you familiar with the effective action, as described here: en.wikipedia.org/wiki/Effective_action ? If so I can write a detailed answer describing how this object can be used to determine whether a transition is first-order or continuous, and then discuss its predictions for $\phi^4$ theory and the abelian Higgs model. If not, then I would recommend studying it in your favorite QFT textbook, as it is a very important object in determining the properties of a QFT. $\endgroup$ – Seth Whitsitt Dec 15 '19 at 22:23
  • $\begingroup$ Yes. I've studied effective action on Peskin and Schwartz. And I do understand the page as well. I'll be grateful if you write the general answer. Thanks 🙏 $\endgroup$ – Bastam Tajik Dec 16 '19 at 9:56
2
$\begingroup$

First of all, I should say that everything here is given in Peskin & Schroeder, but it is spread through several chapters. I've focused on the particular question and added some of my own perspectives on the logical progression of the arguments.

You can determine whether a phase transition will be first- versus second-order by studying the effective action, $\Gamma[\phi_{\mathrm{cl}}]$. Recall that this object is a functional of the fields $\phi_{\mathrm{cl}}(x)$ which satisfies $$ \frac{\delta}{\delta \phi_{\mathrm{cl}}(x)} \Gamma[\phi_{\mathrm{cl}}] = 0 \tag{1}\label{eq1} $$ in the absence of external sources. Here, the field $\phi_{\mathrm{cl}}$ is given by $$ \phi_{\mathrm{cl}}(x) = \langle \phi(x) \rangle, $$ where the expectation value is taken with respect to the exact ground state of the QFT. Therefore, assuming one knows the functional $\Gamma[\phi_{\mathrm{cl}}]$ exactly, one can determine the vacuum expectation value of your fields by solving Eq. (\ref{eq1}). One is usually interested in using $\phi_{\mathrm{cl}}$ as a so-called "order parameter," where the solution $\phi_{\mathrm{cl}} = 0$ is the "disordered phase" while $\phi_{\mathrm{cl}} \neq 0$ is the "ordered phase." The phase transitions between the two phases is called first order or discontinuous if the value of $\phi_{\mathrm{cl}}$ changes discontinuously from zero at the phase transition, while it is called continuous if it changes continuously from zero to its non-zero values.

Of course, it is totally impossible to calculate $\Gamma[\phi_{\mathrm{cl}}]$ exactly in a generic interacting QFT, but we can use perturbation theory to obtain it. Let's consider (Euclidean) O($N$)-symmetric $\phi^4$ theory in four dimensions (the properties of the phase transition are not altered by Wick rotation): $$ \int d^4 x \left[ \frac{1}{2} \left( \partial_{\mu} \phi_{\alpha} \right)^2 + \frac{s_0}{2} \phi^2 + \frac{\lambda_0}{4} \phi^4 \right]. $$ Here, we will take $\lambda_0$ to be small so we can use perturbation theory, and $s$ is a tuning parameter. The index $\alpha = 1,2,..,N$, and $\phi^2 = \sum_i \phi_{\alpha} \phi_{\alpha}$. The effective action is calculated to leading order in Peskin and Schroeder, and in terms of renormalized parameters it is $$ \frac{1}{\mathrm{volume}} \Gamma[\phi_{\mathrm{cl}}] = \frac{s}{2} \phi_{\mathrm{cl}}^2 + \frac{\lambda}{4} \phi_{\mathrm{cl}}^4 + \frac{1}{64 \pi^2} \left\{ (N-1) (\lambda \phi_{\mathrm{cl}}^2 + s)^2\left( \log\left[ (\lambda \phi_{\mathrm{cl}}^2 + s)/M^2 \right] - \frac{3}{2} \right) + (3 \lambda \phi_{\mathrm{cl}}^2 + s)^2\left( \log\left[ (3 \lambda \phi_{\mathrm{cl}}^2 + s)/M^2 \right] - \frac{3}{2} \right) \right\}. $$ Here, $M$ is an arbitrary mass scale introduced to renormalize the theory.

Without interactions, there is clearly a continuous transition at $s=0$, where $\phi_{\mathrm{cl}} = -\sqrt{-s/\lambda}$ for $s<0$ which smoothly goes to $\phi_{\mathrm{cl}} = 0$ for $s>0$. But now including our extra term, if we look at the $s \rightarrow 0$ limit, we find $$ \frac{1}{\mathrm{volume}} \Gamma[\phi_{\mathrm{cl}}] \approx \lambda\frac{\phi_{\mathrm{cl}}}{4} \left\{ 1 + \frac{\lambda}{16 \pi^2} \left[ (N+8) \left(\log(\lambda \phi_{\mathrm{cl}}^2/M^2) - \frac{3}{2}\right) + 9 \log 3\right] \right\}. \tag{2}\label{eq2} $$ Here, we notice a major problem - one which generally haunts massless QFTs. Although we have obtained the second term in the brackets as a perturbation of the first, we see that it is proportional to $\lambda \log \phi_{\mathrm{cl}}/M$. Then no matter how small $\lambda$ is, there is still always a small field value of $\phi_{\mathrm{cl}} \sim M e^{-1/\lambda}$ where the second term will be comparable to the first, invalidating perturbation theory. This just gets worse at higher orders, where we expect terms of the form $\lambda^n \log^n \phi_{\mathrm{cl}}/M$. Since we are precisely interested in the small $\phi_{\mathrm{cl}}$ limit, our expression cannot be trusted to determine the phase transition.

But this seems kind of stupid, since the mass scale $M$ was completely arbitrary from the start. If we began with one arbitrary $M$ and find perturbation theory breaks down at $\phi_{\mathrm{cl}} \sim M e^{-1/\lambda}$, what's to stop us from now defining a much smaller scale $M'$ where now perturbation theory works all the way down to $\phi_{\mathrm{cl}} \sim M' e^{-1/\lambda}$? And indeed, interating this process, it seems clear that we should be able to understand our problem all the way down to $\phi_{\mathrm{cl}} = 0$, provided perturbation theory remains valid.

The way to systematically deal with this is using the renormalization group (RG). The idea is to take into account how your observables depend on $M$. This answer isn't the best place to give a thorough introduction to RG, so I will simply use some major results. The effective action turns out to satisfy the following "Callan-Symanzik equation": $$ \left[ M \frac{\partial}{\partial M} + \left( M \frac{d \lambda}{dM} \right) \frac{\partial}{\partial \lambda} \right] \Gamma[\phi_{\mathrm{cl}}] = 0 $$ (I should mention that this is especially simple because $\phi^4$ theory does not have wave function renormalization at one-loop). The general solution to this differential equation is $$ \Gamma[\phi_{\mathrm{cl}}] = F(\bar{\lambda}(M,\phi_{\mathrm{cl}})) \phi_{\mathrm{cl}}^4, $$ where we now have a "running" coupling constant $$ \bar{\lambda} = \frac{\lambda}{1 - (\lambda/8 \pi^2) (N+8) \log(\phi_{\mathrm{cl}}/M)}. $$ Comparing this general functional form with Eq. \ref{eq2}, we see that we can write $$ \frac{1}{\mathrm{volume}} \Gamma[\phi_{\mathrm{cl}}] \approx \bar{\lambda}\frac{\phi_{\mathrm{cl}}}{4} \left\{ 1 + \frac{\bar{\lambda}}{16 \pi^2} \left[ (N+8) \left(\log(\bar{\lambda} \phi_{\mathrm{cl}}^2/M^2) - \frac{3}{2}\right) + 9 \log 3\right] \right\}. $$ Using the definition of $\bar{\lambda}$ and expanding to leading order in $\lambda$ gives Eq. \ref{eq2}, but now we have used our knowledge of how the theory continuously changes with the scale $M$ to improve perturbation theory.

If we now consider the $\phi_{\mathrm{cl}} \rightarrow 0$ limit, we see that $\bar{\lambda}$ smoothly goes to zero, and the analysis proceeds without issue. The minimum remains at $\phi_{\mathrm{cl}}$ at the transition, and therefore the phase transition is continuous.

Of course, our result is particular to our model. In the four-dimensional abelian Higgs model (scalar QED), where a single scalar field is coupled to a gauge field, it turns out that the system undergoes a first-order phase transition (this is worked out in detail in the "Final Project" in P&S after Chapter 13). Finally, I have not discussed theories in less than four dimensions. This is because massless super-renormalizable theories are plagued by incurable IR divergences in perturbation theory, because these phase transitions are truly strongly coupled and non-perturbative. They are usually studied either by non-perturbative methods, or by a formal expansion in dimensionality close to $d=4$.

$\endgroup$
  • $\begingroup$ Thanks for answering. So the only controling parameter is the renormalized mass of the scalar particle? $\endgroup$ – Bastam Tajik Dec 22 '19 at 23:35
  • 1
    $\begingroup$ Yes, since the coupling $\lambda$ is irrelevant, only by tuning the mass can you access the two phases and their transition. $\endgroup$ – Seth Whitsitt Dec 24 '19 at 5:08
2
$\begingroup$

The answer may very much depend on what concrete field-theoretic model and situation you have in mind. I will assume that we talk about the actual Higgs mechanism, that is, generation of mass for gauge fields by interaction with a scalar (Higgs) field. I will also assume that by phase transition you mean a thermal transition, that is one driven by changing temperature while keeping all other parameters of the theory fixed.

In this case, there need not be any phase transition associated with the Higgs VEV at all, as long as the VEV does not break any exact global symmetry of the theory. This is the case of the Standard Model. For a quick reference, check e.g. Figure 1 of this paper. You can see that the electroweak phase transition would be first order if the physical Higgs mass were below ca 70 GeV. Above this value, the transition is a smooth crossover. The two regimes are separated by a critical endpoint where the transition is of second order.

Should you rather have in mind a model without gauge fields, e.g. a purely scalar model with self-interaction of the Higgs field, then the Higgs VEV will break some genuine global symmetry. In this case, there must be some phase transition, separating two phases where the symmetry of the ground state is different. Whether this phase transition is of first or second order is a dynamical question that generally requires a detailed analysis of the model.

$\endgroup$
  • $\begingroup$ Temperature dependence of the VEV facilitates the problem very much but from Coleman-Weinberg mechanism we know that the phase transition is not thermal and it's a quantum phase transition. In other words the control parameter can't be the temperature from this point of view. You can even consider the simplest case of scalar field theory with self interaction without any mass term. $\endgroup$ – Bastam Tajik Dec 18 '19 at 15:55
  • 1
    $\begingroup$ I think you need to formulate your question more precisely and concretely if you want to get a concrete answer. What is the specific model you are considering? Ideally, write down its Lagrangian. What is the control parameter that drives the phase transition? $\endgroup$ – Tomáš Brauner Dec 18 '19 at 16:05
  • $\begingroup$ I edited the question, while I think the controling parameter can be the coupling constant or the mass term. $\endgroup$ – Bastam Tajik Dec 18 '19 at 17:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.