39.4 Spontaneous symmetry breaking

39.4 Spontaneous symmetry breaking

Concept links · terms present in this machine draft; source roles are unverified: Lie algebra · Lie algebra representation · irreducible representation

In the standard Bargmann-Fock construction, there is a unique state |0⟩, and for the Hamiltonian of the free particle quantum field theory, this will be the lowest energy state. In interacting quantum field theories, one may have state spaces unitarily inequivalent to the standard Bargmann-Fock one. These can have their own annihilation and creation operators, and thus a notion of particle number and a particle number operator $\hat{N}$ , but the lowest energy |0⟩ may not have the properties

$$\hat{N}|0⟩=0,e^{-i\theta \hat{N}}|0⟩=|0⟩$$

Instead the state |0⟩ gets taken by $e^{-i\theta \hat{N}}$ to some other state, with

$$\hat{N}|0⟩\ne 0,e^{-i\theta \hat{N}}|0⟩\equiv |\theta ⟩\ne |0⟩(\mathrm{for}\theta \ne 0)$$

and the vacuum state not an eigenstate of $\hat{N}$ , so it does not have a well-defined particle number. If $[\hat{N},\hat{H}]=\bar{0}$ , the states |$\theta$⟩ will all have the same energy as $|0⟩$ and there will be a multiplicity of diferent vacuum states, labeled by $\theta$. In such a case the $U(1)$ symmetry is said to be “spontaneously broken”. This phenomenon occurs when non-relativistic quantum field theory is used to describe a superconductor. There the lowest energy state will be a state without a definite particle number, with electrons pairing up in a way that allows them to lower their energy, “condensing” in the lowest energy state.

When, as for the multi-component free particle (the Hamiltonian of equation $38.4)$ , the Hamiltonian is invariant under $U(n)$ transformations of the fields ${\psi }_j$ then we will have

$$[\hat{X},\hat{H}]=0$$

for $\hat{X}$ the operator giving the Lie algebra representation of $U(n)$ on the multiparticle state space (see section 38.2.2). In this case, if |0⟩ is invariant under the $U(n)$ symmetry, then energy eigenstates of the quantum field theory will break up into irreducible representations of $U(n)$ and can be labeled accordingly. As in the $U(1)$ case, the $U(n)$ symmetry may be spontaneously broken, with

$$\hat{X}|0⟩\ne 0$$

for some directions $X$ in $\mathfrak{u}(n)$ . When this happens, just as in the $U(1)$ case states did not have well-defined particle number, now they will not carry well-defined irreducible $U(n)$ representation labels.

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