# Can't the Higgs vacuum be a superposition?

Supose you have a complex scalar field $\phi$ statisfying the typical Higgs lagrangian \begin{equation} L = \partial_{\mu}\phi\partial^{\mu}\phi + \mu^2\phi^*\phi-\frac{\lambda^2}{2}(\phi^*\phi)^2 \end{equation}

It has an infinitely degenerated vacuum state parametrized by an angle $\theta$ like $\phi_0=e^{i\theta}v$ where $\phi_0$ is a posible vacuum state for each posible theta.

Now, any election of a vacuum would indeed break the symmetry of the system and, classically (I'm thinking in magnets now) this seems kind of inevitable. But since this is a quantum theory.. why can't we just make a vacuum that is a superposition of every posible vacuum? Something like:

\begin{equation} |\phi_0'\rangle=\frac{1}{2\pi}\int_0^{2\pi}|{\phi_0(\theta)}\rangle d\theta \end{equation}

would this vacuum state break symmetry? is it even a posible state? is it more stable that just a regular symmetry-breaking one?

Any comments on the idea of a superposition of symmetry-breaking vacuum states will be welcome!

P.S.: I'm not saying that $\phi_0'$ is the correct superposition, I just wrote it down to give an example of what I was thinking.

• I recently asked a very similar question: physics.stackexchange.com/questions/321867/… – jak May 7 '17 at 13:48
• Maybe it is helpful to have a look at how the Higgs mechanism can be formulated in a completely gauge invariant way: there is a paper by Higgs himself about it, or, for example, arxiv.org/abs/1001.1176. In this formulation there is no parameter $\theta$, because there is no $SU(2)_L$ gauge freedom in the theory to begin with. – jak May 7 '17 at 14:01
• But.. but... @JakobH , the OP is not talking about the Higgs mechanism... there are no gauge fields mentioned. He is misusing "Higgs lagrangian" for the Goldstone model. He is talking about plain SSB ... the 1961 variety, no? – Cosmas Zachos May 7 '17 at 22:39
• Aren't these vacua disjoint? – SRS Jun 6 '17 at 11:04

## 1 Answer

Goldstone's theorem is unforgiving. It somehow feels your idiosyncratic notation for the vacuum is expressly designed to moot it. Work in the polar parameterization, $$\phi(x)\equiv r(x) e^{i\Theta(x)},$$ so $\langle r(x)\rangle =v$ for any and all vacua at the bottom of the Goldstone sombrero.

I imagine, hoping I am right, that your vacua are defined/labelled by $$\langle \theta| \Theta(x)|\theta\rangle=\theta.$$ Recall a small rotation by angle ε will always shift the Goldstone field and thus its v.e.v. by ε--no choice there.

Any vacuum $|\Omega\rangle$, in any case, will have $$\langle \Omega| \phi(x)/v|\Omega\rangle=e^{i\theta'},$$ for some θ'.

For your $|\theta\rangle$, θ'=θ. For your exemplar Ansatz $|\phi_0'\rangle$, (unnormalizable, multi-superselectioned),the integral of θ would net you θ'=0, etc.

The key point: it really does not matter what your θ' is. You simply redefine your goldston to $\Theta'\equiv \Theta -\theta'$ so $$\langle \Theta' \rangle =0,$$ but $\langle \delta_\epsilon \Theta' \rangle =\epsilon\neq 0$, which is all that matters: a shift from an arbitrary origin. SSB appears inescapable, in the absence of other fields.