1
$\begingroup$

I currently take a lecture called Symmetries in Particle Physics. And when talking about the spontaneous breaking of a global continuous symmetry the lecturer said that the field is expanded around the minimum of the potential $V$. And it is not clear to me why it is considered an expansion of the field.

To exemplify what I mean lets consider a theory given by the Lagrangian $$ \mathcal L = \frac 1 2 (\partial_\mu \sigma)^\dagger (\partial^\mu \sigma) - \lambda (\sigma^2 - \mu^2)^2, $$ where $\sigma = (\sigma_1, \sigma_2, \sigma_3)^T$ is a real scalar triplet. So it is clear to see that the Lagrangian is invariant under the global transformation $$ \sigma \to R(\vec\theta) \sigma, $$ where $R(\vec\theta)$ is the 3 dimensional matrix representation of $SO(3)$ and $\mu^2,\lambda > 0$ are real parameters. The degenerate minimum of this potential is given by the condition $$ ||\sigma_\text{min.}|| = \mu $$ We go on by imposing a shift (or as the lecturer said expand around the potential minimum) $$\sigma = \sigma' + \begin{pmatrix} 0 \\ 0 \\ \mu \end{pmatrix}. $$ Here is the thing I do not understand. My professor said we expand the field for small perturbations around the ground state configuration of the theory. But as far as I understand this is just a coordinate transformation in configuration space, which is fine to do anyway as far as I know. Am I missing something here? For me an expansion would imply a loss of generality, hence only small deviations from the ground state are allowed in the resulting theory. But as I said I do not see why we have to "expand" at all.

EDIT:

My confusion is probably based on the fact that when discussing spontaneous symmetry breaking one mixes things from the classical perspective with the quantum perspective.

To add to the example from my lecture above I add some citations from textbooks I use:

We can now define a set of shifted fields by writing $$ \phi^i (x) = (\psi_k, v + \sigma(x)),\quad k=1,\dots,N-1 $$

from Chapter 11 in "An Introduction to QFT"-Peskin and Schroeder

If we expand $\phi$ around one of the minima, say $\phi = \sqrt{\frac{6m^2}{\lambda}} + \tilde\phi$

from Chapter 28.1 in "QFT and the SM"-Schwartz

In QFTs it is possible to use the fieldconfiguration $\phi_0$ (which minimizes the potential) as an center of expansion for functional integrals. The field is therefore expanded in the constant $\phi_0$ and a fluctuation $$ \phi(x) = \phi_0 + \frac{1}{\sqrt{2}} (\rho(x) + i \varphi (x)) $$

from chapter 10.2.1 in "Von der Quantenfeldtheorie zum SM" by Münster (I translated the cited part from German)

To me shifted field is a more suitable terminology, since it is just a reparametrization of the Lagrangian. But often authors use the term expansion and I cannot see why.

$\endgroup$
3
  • $\begingroup$ I quite honestly am not sure what your question is, as much as I might like to help. I'm sorry. $\endgroup$ – kaylimekay Jan 5 at 16:53
  • $\begingroup$ @kaylimekay I added some more detail to my question. It might be semantics... $\endgroup$ – AlmostClueless Jan 5 at 21:57
  • 1
    $\begingroup$ Recall what your quantum field does: it adds an excitation on your ground state (vacuum) or else destroys one on it. You may rewrite it in terms of anything you wish, as long as you can compute the same quantity easily rather than with difficulty. For instance, the additional energy introduced by an excitation. $\endgroup$ – Cosmas Zachos Jan 5 at 22:08
2
$\begingroup$

Let me share the way I currently understand this (which may be flawed since I'm also studying it right now - corrections are welcome!). IMHO this is all about how we quantize a field and relate it to particles. Recall that if $\phi(x)$ is a free scalar field we quantize it by expanding it into creation and annihilation operators $$\phi(x)=\int\dfrac{d^3p}{(2\pi)^3}\dfrac{1}{2\omega_p}\bigg(a(\mathbf{p})e^{ipx}+a^\dagger(\mathbf{p})e^{-ipx}\bigg)\tag{1}.$$

If the field is not free this still holds asymptotically with in/out fields and this reveals the spectrum of particles in the in/out states of scattering experiments.

Now take this and evaluate the vacuum expectation value. As you will immediately recognize because $a(\mathbf{p})$ annihilates the vacuum we have: $$\langle 0|\phi(x)|0\rangle=0\tag{2}.$$

Now suppose $\phi(x)$ is a scalar multiplet, so that we have a column of $\phi_i(x)$ transforming into one another by the global symmetry. If instead of (2) have $$\langle 0|\phi(x)|0\rangle=\alpha\tag{3}$$

where $\alpha$ is some non-zero column vector then the only way (1) and (3) can hold at the same time is if $a_i(\mathbf{p})|0\rangle\neq 0$, because this being zero already implies (2).

The Physics behind this is that as you can see from (1) in a sense we are viewing the field as a collection of harmonic oscilators. We are quantizing excitations about a ground state.

Then what we do is to realize that if $\phi(x)$ is our dynamical variable writing it as $$\phi(x)=\alpha+\chi(x)\tag{4}$$

we loose no information at all. We are just shifting to a more convenient variable. In fact $\langle 0|\chi(x)|0\rangle=0$ by definition and $\chi(x)$ can be quantized as we often do, according to (1), without problem.

As Goldstone, Salam and Weinberg put in their 1962 paper ("Broken Symmetries"): "It is inconvenient to work with fields with nonzero vacuum expectation value".

Because of this we perform that shift working with an equivalent, but more convenient variable, whose quantization can be understood as "quantizing excitations about a ground state". In particular observe that once this shift is done and the Lagrangian is expressed in terms of the more convenient fields with zero vacuum expectation value one readily identifies the implications of SSB upon the particle content of the theory: the Goldstone bosons.

$\endgroup$
1
$\begingroup$

When studying a quantum field theory without spontaneous symmetry breaking, we are typically only able to access the states which perturb around the vacuum (practically speaking). These include particle states, which only make sense as an excitation from the vacuum.

The same is true in the spontaneously broken case, the physical particle states only make sense as an excitation from the vacuum. The tricky case in this situation is that the true vacuum is not at the origin, it is instead at some nonzero value of the fields. To access these particle states it makes sense to expand your fields from those values, since particle states are defined as a perturbation from that point.

$\endgroup$
2
  • $\begingroup$ So classically speaking there is no need for expansion, but when quantizing the theory the terminology of an expansion makes sense, since the vacuum is not well defined whithin QFTs? $\endgroup$ – AlmostClueless Jan 5 at 19:17
  • $\begingroup$ Well, in some sense you should also expand in the classical picture, but that is more of a computational convenience. Where as in the quantum picture it has to do with what the physical states are. $\endgroup$ – fewfew4 Jan 5 at 22:58

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.