# The proof of Goldstone's theorem

On page 352 of Peskin and Shroeder.
The athors show the proof of Goldstone's theorem.

A general continuous symmetry transformation has the form $$\phi^a \to \phi^a + \alpha\Delta^a (\phi) ,\tag{11.12}$$ where $\alpha$ is an infinitesimal parameter and $\Delta^a$ is some function of all the $\phi$'s. Specialize to constant fields; then the derivative terms in $\cal{L}$ vanish and the potential alone must be invariant under (11.12). This condition can be written $$V(\phi^a)=V(\phi^a+ \alpha\Delta^a (\phi)) \quad \text{or} \quad \Delta^a (\phi)\frac{\partial}{\partial \phi^a}V(\phi)=0.$$ Now differentiate with respect to $\phi^b$, and set $\phi=\phi_0$: $$0=\left( \frac{\partial\Delta^a}{\partial \phi^b} \right)_{\phi_0} \left( \frac{\partial V}{\partial \phi^a} \right)_{\phi_0} + \Delta^a (\phi_0) \left( \frac{\partial^2}{\partial \phi^a \partial \phi^b}V \right)_{\phi_0} .\tag{11 .13}$$ The first term vanishes since $\phi_0$ is a minimum of $V$, so the second term must also vanish. If the transformation leaves $\phi_0$ unchanged (i.e., if the symmetry is respected by the ground state), then $\Delta^a (\phi_0)=0$ and this relation is trivial. A spontaneously broken symmetry is precisely one for which $\Delta^a (\phi_0)\ne 0$; in this case $\Delta^a (\phi_0)$ is our desired vector with eigenvalue zero, so Goldstone's theorem is proved.

Why in spontaneously broken symmetry $\Delta^a (\phi_0)\ne 0$?
And on the other hand when the symmetry is not spontaneously broken, $\Delta^a (\phi_0)= 0$?
Thanks.

• I don't know what sort of answer you're looking for: The definition 11.12 of $\Delta^a(\phi_0)$ means that $\Delta^a(\phi_0) = 0$ if and only if $\phi_0$ is invariant under the symmetry. The definition of SSB says that SSB occurs when $\phi_0$ is not invariant under the symmetry. There are no further steps, it's literally just comparing the two definitions. – ACuriousMind Aug 13 '17 at 8:50
• Is OP perhaps asking what an invariance of $\phi_0$ has to do with an invariance of $|\Omega\rangle$, the ground state vector? SSB is [perhaps sometimes confusingly] stated in terms of invariance of the minimum of the potential, and invariance of the vacuum vector. Might this be the source of confusion here? – AccidentalFourierTransform Aug 13 '17 at 8:57
• @AccidentalFourierTransform : In the same section of this book, they discuss the linear sigma model which involves a set of $N$ real scalar field $\phi^i(x)$: $$\cal{L}=\frac{1}{2}(\partial_{\mu}\phi^i)^2 + \frac{1}{2}{\mu}^2(\phi^i)^2 - \frac{\lambda}{4}[(\phi^i)^2]^2 , \tag{11.5}$$ This Lagrangian has an $O(N)$ symmetry. When this symmetry is not spontaneously broken, $\phi_0^i$(a constant field which has lowest-energy) moves along the surface of the sphere $(\phi_0^i)^2=\frac{\mu^2}{\lambda}$ where the potential is minimized. – GotchaP Aug 16 '17 at 0:55
• When this symmetry is spontaneously broken, $\phi_0^i$ is fixed at one point of the surface(Actually they choose coordinates so that $\phi_0^i$ points in the $N$th direction.). So at first I thougt that when the symmetry is not spontaneously broken, $\Delta^a (\phi_0)\ne 0$ and when the symmetry is spontaneously broken, $\Delta^a (\phi_0)= 0$. – GotchaP Aug 16 '17 at 0:55

Let's pretend the charge $Q^a$ in question is well defined (which isn't, by Fabri-Picasso's theorem, but let's not fuss, as something of the sort effects field rotations anyway.)

P&S first look at the constant pieces of the fields which will perforce add to the energy of the vacuum, so they minimize the potential w.r.t. these constants. Then, having shifted fields suitably by the potential minimum constants, you have $\langle \phi \rangle=0$.

Now inspect $\langle \Delta^a \phi \rangle\sim \langle 0|[Q^a, \phi ]|0\rangle$. If the vacuum is symmetric, $\exp (i\theta Q^a) |0\rangle=|0\rangle$, so $Q^a| 0\rangle=0$ and therefore $\langle \Delta^a \phi \rangle$ vanishes.

The contrapositive of this obtains if $Q^a| 0\rangle\neq 0$, the definition of SSB, so a symmetry transformation alters the vacuum--shifts it to another state (a different vacuum). So for an asymmetric vacuum, $\langle \Delta^a \phi \rangle \neq 0$, anymore, in contrast to the above symmetric vacuum $\langle \Delta^a \phi \rangle=0$.

In this $\langle \Delta^a \phi \rangle \neq 0$ case, this v.e.v. must constitute the null vector of the mass matrix, and $\langle \Delta^a \phi \rangle \neq 0$ identifies the Goldstone modes n. That is, $\langle 0|\phi|0\rangle=0$, $\langle n|\phi|0\rangle\neq 0$, and these combinations of $\phi$s drop off the collection of mass terms. So, e.g. in the standard model, it is not the Higgs component getting the v.e.v. that is the goldstone; but, instead, all the three modes that rotate to it by the 3 broken generators, so their transforms get a v.e.v.

The currents of these broken charges basically start with linear expressions of the goldstons. In effect, the broken generators $Q^a$ pump goldstons in and out of the vacuum to turn it into another state, degenerate with it (vacuum again), since the $Q^a$s commute with the Hamiltonian so the new states have the same energy.

Why in spontaneously broken symmetry $\Delta^a (\phi_0)\ne 0$?

And on the other hand when the symmetry is not spontaneously broken, $\Delta^a (\phi_0)= 0$?

The reason is perhaps easier to understand from a geometrical point of view. If there is a symmetry described by a compact group transformation, then either:

• the vacuum must lie at the origin, which represents a unique point, invariant to transformations. This would imply that $\Delta^a (\phi_0)= 0$, or

• the vacuum must lie along a connect set of points at equal distance away from the origin, (think of a valley like a sombrero hat) all of which share the same energy (they are degenerate). In this case the vacuum is not invariant to transformations. In fact, the transformation would move the point along the valley. So the symmetry is not explicitly broken, but $\Delta^a (\phi_0)\ne 0$, which leads to a spontaneously broken symmetry.