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I actually have two different sub-questions, both based on the understanding of the quadratic term of the lagrangian, so the answer is probably linked. I will use the example of the linear-$\sigma$ model because it's the one I'm more comfortable with, so

$$\mathcal{L}=\frac{1}{2}\sum\partial^\mu\phi_i\partial_\mu\phi_i-\frac{1}{2}\mu^2\sum\phi_i^2-\frac{\lambda}{4}(\sum(\phi_i)^2)^2.$$

Does the symmetry break, or is it either broken or unbroken?

With this I mean that I know that if $\mu^2>0$ the ground state is nondegenerate (so, no SSB) and if $\mu^2<0$ the ground state is degenerate (so, SSB). But does the value of $\mu^2$ change? Does my theory goes from being broken to be restored and vice-versa, or is every theory either broken OR unbroken, and then it stays that way? From what I read, it seems that SSB is a proerty of the theory, so it either is or isn't broken. But if this is the case, what does it mean for the electroweak theory to restore the symmetry at high temperatures? Is $\mu^2$ a function of $T$?

How does the mass come in?

From what I understood, the mass is defined as the quantity $m$ in the equation of motion: for the KG field $(\square-m^2)\phi=0$. Therefore, for the linear-$\sigma$ model, the mass should be the second derivative of $V(\phi_i^2)$ at the minimum, so if $\mu^2>0$ then $m=\mu$. If $\mu^2<0$, the value of the second derivative of $V$ at the minimum is different (because the minimum is different), so the mass in the broken case is different than the one in the restored case. Is my understanding correct? I'm doubtful because I also read that in the broken case, being $\mu^2<0$, the mass is not physical because $\mu$ isn't real.

These are the main questions that I have. There may be some deeper conceptual misunderstanding regarding the whole SSB idea: if you see some in the questions, please show it to me.

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    $\begingroup$ I think you might need to make the distinction between explicit and spontaneous symmetry breaking. $\endgroup$ – probably_someone Jun 29 at 16:43
  • $\begingroup$ @probably_someone why is this needed? Do the answers to my questions depend on whether the breaking is spontaneous or not? I'm interested in the symmetry breaking that is considered in the Goldstone theorem, in Higgs mechanism and in electroweak symmetry. Is this spontaneous or explicit? $\endgroup$ – Mauro Giliberti Jun 29 at 16:46
  • $\begingroup$ Never mind, I just realized you specify that you're talking about spontaneous symmetry breaking (SSB). $\endgroup$ – probably_someone Jun 29 at 17:24
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    $\begingroup$ Linked, or else. $\mu^2$ is definitely a function of T. It should be covered in your text. $\endgroup$ – Cosmas Zachos Jun 29 at 19:42
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  1. You're supposed to think of $\mu^2$ as a parameter, and there's no need to consider if it $>0$ or $<0$. You proceed by minimising the potential and then seeing that the nature of vacuum/vacua is different for $\mu^2>0$ and $\mu^2<0$. As @CosmasZachos mentioned in the comments, it is certainly a function of $T$. The exact function can be determined from a more microscopic description. At any rate, these parameters are often thought of as observables in a UV complete theory-at the given energy scale we need to use them as parameters, but a more complete theory will predict them.

  2. Roughly, it is defined in terms of an expansion about some critical temperature $\mu^2(T)\sim (T-T_c)$ and whether or not SSB occurs is determined by $T>T_c$ or $T<T_c$. This is the sign of $\mu^2$. If you wish, you could replace the mass with this $(T-T_c)$ and repeat the procedure in 1) above. This relies on the assumption that the full microscopic potential can be obtained via taylor expanding in $(T-T_c)$.

  3. I'm not sure what you mean by $\mu^2$ 'changing', but I'm assuming you're referring to the pole of the propagator shifting because of the interaction term. That's a different issue-the parameter in the Lagrangian is the bare mass at this stage(i.e. you haven't split it into physical mass and counterterms etc). You are minimising a CLASSICAL object, then later at tree level equating it to the vev of some quantum field. In general, you have to minimise not $V(\phi)$ but the the effective potential $V_{eff}(\phi)$. At one loop, it is given by the Coleman Mandula correction-$$V_{eff}(\phi)=V(\phi)-\frac{i\hbar}{2}\int \frac{d^4k}{(2\pi)^4}\log\bigg(\frac{k^2-V''(\phi)}{\hbar^2}\bigg)+...$$

  4. $\mu^2$ is not the mass. It is a parameter in the Lagrangian. The mass is the pole of the propagator. As it turns out, if $\mu^2>0$, then for a free field theory, it indeed is the pole of the propagator, therefore the mass. But in general it changes under interactions(unless there is some principle protecting it eg. the Ward identities for photons). Your argument about the second derivative of the potential doesn't really work because you need to consider loop corrections to find the actual mass. At the tree level, of course, you cannot interpret it as a mass-if you do then you have to allow for tachyonic excitations. Just play around with the dispersion relations to convince yourself of this. In this case, you cannot hope to call it a mass-and it fortunately isn't. That's the whole point of the SSB mechanicsm-the physical mass comes out with the right sign.

  5. Addendum-Why can't we blindly interpret $\mu^2$ as the mass? Because this interpretation revolves around the assumption that we have a perturbative QFT about the true vaccuum. If this term has the wrong sign, then $\phi=0$ is no longer the minima, and so when you quantize it, it will not correspond to the vacuum state over which $\phi$ acts. You must then, in some sense, 'subtract off' this deviation from the vacuum to find the actual perturbative field, which turns out to have a physical mass indeed.

Hope this helped.

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  • $\begingroup$ Thank you for your answer. I have a few comments: (1) do you mean $mu$ or $mu^2$? (3) no, I didn't mean that. I'm aware of those effects, although I didn't study them in the context of SSB. I mean precisely what you said in points 1 and 2, so that $mu^2$ is a parameter and not a constant, and if I consider the same system with different temperatures, the value of $mu^2$ will change. (4) so if I understand correctly, it's not like $mu^2<0$ forbids $mu$ to be a mass, but instead $mu^2<0$ creates a system whose pole of the propagator isn't $mu$ and is (luckily) a real quantity. Right? $\endgroup$ – Mauro Giliberti Jun 30 at 8:10
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    $\begingroup$ (1) My bad, I meant $\mu^2$. I'll make edits. (3) Yes, it will change. (4) It forbids us to interpret $\mu$ as a mass because (a) well, it is imaginary now (b) what saves us is that while $\phi$ no longer has a valid particle interpretation(because imaginary mass), it is not the actual field('perturbation on the vacuum') anyway. You get the physical field by expanding around the minimum i.e. vacuum, $\phi=\phi_0+\tilde{\phi}$ and $\tilde{\phi}$ has a physically correct mass, and this is the actual field(to be quantized). $\endgroup$ – GRrocks Jun 30 at 8:17
  • $\begingroup$ The pole of the propagator for the wrong sign of $\mu^2$ will not correspond to a physical mass because the dispersion relation will not quite work out. You will have $p^2=m^2<0$, and this is a tachyonic excitation. $\endgroup$ – GRrocks Jun 30 at 8:21

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