To be specific, let us consider the double-well potential. If we study quantum mechanics, i.e. a particle in such a potential, then we know that there is no spontaneous symmetry breaking but a energy-level splitting. The ground state is presumably a superposition of the local minima. Now if we consider a quantum field in a box with finite volume $V$, still there is no SSB and the ground state is also a superposition of the local minima. If we now take $V\rightarrow \infty$, then we know that there must be SSB. I was wondering how the superposition state of the two minima reduces to one of them as we take the limit?

  • $\begingroup$ Your question is answered exactly in page 254-255 of Maggiore's QFT book: books.google.co.in/… $\endgroup$
    – SRS
    May 9, 2020 at 12:52
  • $\begingroup$ @SRS Thank you very much for bringing this book into my attention. I will read pages 254-255 $\endgroup$
    – Wein Eld
    May 9, 2020 at 14:18
  • $\begingroup$ @SRS I do not think pages 254-255 answer my question. The description there is something that I was aware of. What I want to know is the following. We first work with finite volume V, and obtain the symmetric vacuum. Now taking V to infinity, the symmetric vacuum is still there (though the energy splitting is going to be vanishing in that limit). Why SSB and how it happens? $\endgroup$
    – Wein Eld
    May 9, 2020 at 14:26
  • $\begingroup$ The infinite volume limit amounts to making the barrier between two wells infinitely high, formally! In that limit, if the system falls into one of the wells, it will remain permanently locked there because the transition amplitude between one ground state localized in the left well and the other ground state localized in the right well would be zero. @WeinEld $\endgroup$
    – SRS
    May 10, 2020 at 15:22

1 Answer 1


how the superposition state of the two minima reduces to one of them as we take the limit?

We can retain the symmetric vacuum state when taking the limit, and the result is mathematically well-defined (if the symmetry is discrete). The problem is that the result violates a physical principle, namely the cluster property.

Roughly, the cluster property says that the vacuum expectation value of a product of fields, like $\langle 0|\phi(x)\phi(y)|0\rangle$, should factorize into a product of vacuum expectation values, like $\langle 0|\phi(x)|0\rangle\,\langle 0|\phi(y)|0\rangle$, as the points $x,y$ become sufficiently separated from each other. The SSB vacuum states have this property, but the symmetric vacuum state does not. Both are well-defined (for discrete SSB), and both have the same minimum energy, but only the SSB vacua satisfy the cluster property.

To enforce this property when taking the infinite-volume limit, we can add a small explicit symmetry-breaking term to the action, then take the infinite-volume limit, and then remove the explicit symmetry-breaking term.

These references explain how the cluster property selects an SSB vacuum state:

  • In the context of spin systems (like the Ising model): Section 23.3, "Order Parameter and Cluster Properties", of Zinn-Justin's book Quantum Field Theory and Critical Phenomena.

  • In the context of QFT: Section 19.1 in Weinberg, The Quantum Theory of Fields, Volume II.

Weinberg's argument is reviewed below.

Why the symmetric vacuum violates the cluster property

Suppose that the symmetry in question is a $\mathbb{Z}_2$ symmetry (double-well potential).

Let $|{v}\rangle$ be a candidate for a vacuum state. In particular, $|{v}\rangle$ must have zero momentum, which implies $$ \newcommand{\la}{\langle} \newcommand{\ra}{\rangle} \la\psi|\phi(x)|{v}\ra = \la\psi|\phi(0)|{v}\ra. \tag{1} $$ Now consider the correlation function $\la{v}|\phi(x)\phi(y)|{v}\ra$. We can write this as $$ \la{v}|\phi(x)\phi(y)|{v}\ra = \la{v}|\phi(x)I\phi(y)|{v}\ra \tag{2} $$ where the identity operator $I$ can be written $$ I = \sum_{k=1,2}|{v}_k\ra\,\la{v}_k| +\sum_n \int dp\ |n,p\ra\,\la n,p| \tag{3} $$ where $|{v}_{1,2}\ra$ is any orthonormal basis for the set of lowest-energy states and where sum/integral term accounts for all other states orthogonal to these. The argument $p$ is momentum, and $n$ is all other degrees of freedom. Insert (3) into the right-hand side of (2) and use translation symmetry to get \begin{align} \la{v}|\phi(x)\phi(y)|{v}\ra &= \sum_{k=1,2}\la{v}|\phi(0)|{v}_k\ra \,\la{v}_k|\phi(0)|{v}\ra \\ &+\sum_n \int dp\ e^{ip(x-y)} \la{v}|\phi(0)|n,p\ra\,\la n,p|\phi(0)|{v}\ra. \tag{4} \end{align} Now assume that the quantities in the integrand are smooth enough so that the integral goes to zero as $|x-y|\to\infty$, leaving \begin{align} \lim_{|x-y|\to\infty}\la{v}|\phi(x)\phi(y)|{v}\ra &= \sum_{k=1,2}\la{v}|\phi(0)|{v}_k\ra \,\la{v}_k|\phi(0)|{v}\ra. \tag{5} \end{align} The $2\times 2$ matrix with components $$ M_{jk} := \la{v}_j|\phi(0)|{v}_k\ra \tag{6} $$ is not necessarily diagonal, but it is hermitian and therefore can be diagonalized by choosing a new basis $|{v}_{1,2}\ra$ if needed. Equation (5) shows that the cluster property is satisfied if and only if $|{v}\ra$ is one of the basis states in a basis that diagonalizes $M$. The "only if" part assumes that the eigenvalues of $M$ are distinct and non-zero, which requires that $\phi$ is not invariant under the symmetry in question (it must be an "order paraemter").

Altogether, this shows that the cluster property holds only for states that diagonalize the matrix (6), which in turn must be SSB states if the eigenvalues of $M$ are distinct and non-zero.

  • $\begingroup$ Then perhaps the problem is that why the symmetric vacuum violates the cluster decomposition principle. $\endgroup$
    – Wein Eld
    May 9, 2020 at 14:19
  • $\begingroup$ @WeinEld -- I added a calculation to address that. $\endgroup$ May 9, 2020 at 15:38
  • $\begingroup$ "We can retain the symmetric vacuum state when taking the limit" What do you mean by retaining the symmetric vacuum in the infinite volume limit? In this limit, the two vacua belong to two disjoint Hilbert spaces. @ChiralAnomaly $\endgroup$
    – SRS
    May 10, 2020 at 15:28
  • $\begingroup$ @SRS - I'm thinking of a euclidean path-integral formulation with unspecified initial/final states, so that the large-time limit automatically selects a lowest-energy state. If we don't add an explicit symmetry-breaking term before taking the limit, then the automatically-selected lowest-energy state will necessarily be symmetric in the large-time + large-volume limit, as long as the limit is well-defined. $\endgroup$ May 10, 2020 at 16:16
  • $\begingroup$ @SRS - We can also work in the operator-state formulation. The symmetry in question is discrete, so we can combine the two SSB Hilbert spaces into one Hilbert space (direct sum) and then form arbitrary superpositions of the two SSB vacua to get other states with the same lowest energy. This doesn't work for a continuous symmetry because we can't have a continuum of mutually orthogonal states in a separable Hilbert space, but it works fine for a discrete symmetry. $\endgroup$ May 10, 2020 at 16:19

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