$\bullet$ Consider the following Hamiltonian (density) $$\mathscr{H}=\frac{1}{2}(\partial^\mu\phi)(\partial_\mu\phi)+\mathcal{V}(\phi) \hspace{0.4cm}\text{where}\hspace{0.4cm}\mathcal{V}(\phi)=\frac{\lambda}{4!}(\phi^2-v^2)^2\tag{1}$$ is a double-well potential. Expanding the potential $\mathcal{V}(\phi)=\frac{\lambda}{4!}(\phi^2-v^2)^2$ about one of the minima (say, $\phi=+v$) and retaining up to quadratic term in the potential, one obtains a perturbative vacuum $|0+\rangle$. This is not the exact vacuum of the full Hamiltonian because higher order terms in the potential have been neglected. Similarly, expanding $V(\phi)$ about $\phi=-v$, and again retaining up to quadratic term one obtains another perturbative vacuum $|0-\rangle$. This too, by the same reason, is not the exact vacuum of the full Hamiltonian.

$\bullet$ Similar procedure in a quantum mechanical problem with the Hamiltonian $$H=\frac{p^2}{2m}+V(x)=\frac{p^2}{2m}+k(x^2-a^2)^2,$$ will lead to two perturbative ground states $|0+\rangle$ and $|0-\rangle$. They are not the exact ground states of $H$. The exact ground state is either given by $|S\rangle$ or $|A\rangle$ which are the symmetric and antisymmetric linear combinations of the perturbative ground states $|0+\rangle$ and $|0-\rangle$.

In this case, one cannot talk about the energy of the states $|0+\rangle$ or $|0-\rangle$ because they are not the eigenstates of H. One can only talk about energies of $|S\rangle$ and $|A\rangle$.

Question 1: If this is true, how is it that in QFT, we talk about the states $|0+\rangle$ and $|0-\rangle$ being degenerate (in energy)? They are just the perturbative vacua and not the exact vacua. Hence, not the eigenstates of $\mathscr{H}$. For a reference, see A Modern Introduction to Quantum Field Theory by Michele Maggiore, page 254-255.

Question 2: What is/are the exact ground states of $\mathscr{H}$ and what is their relation to $|0+\rangle$ and $|0-\rangle$?

  • $\begingroup$ A closely related question of mine: physics.stackexchange.com/questions/485861/…. I independently came across essentially the same explanation that is given by Dan in their answer to your post. However, I am not quite sure I understand the claims being made. In particular, I am confused about exactly the same points that you raise in your comment on Dan's answer. Have you found an answer to these issues? Thanks :) $\endgroup$
    – user87745
    Jun 28, 2021 at 10:17

1 Answer 1


The two cases are different. Your second example does not exhibit spontaneous symmetry breaking because there can be tunneling between the two semiclassical vacua. The real vacuum is a superposition of the two.

In QFT in infinite volume, the tunneling amplitude between the two classical minima vanishes (essentially it would take infinite energy to change the field everywhere in space) so the system gets stuck in one and exhibits SSB.

  • 1
    $\begingroup$ If I understand you correctly then you're saying that in the QFT, the perturbative vacua $|0_+\rangle$ and $|0_-\rangle$ become the true vacua of the full Hamiltonian $H$. Well, that sounds correct but is there a way one can show/motivate that? If there is no tunneling amplitude one can only say that $\langle 0_+|H|0_-\rangle=\langle 0_-|H|0_+\rangle=0$. But can one claim $H|0_+\rangle=|H|0_-\rangle$? Do you argue by saying that the Hamiltonian is diagonal in the subspace $|0_{\pm}\rangle$? But that doesn't prove that the diagonal entries are equal. @Dan $\endgroup$
    – SRS
    Jul 20, 2017 at 11:10
  • $\begingroup$ @SRS I'm not sure I understand the question. For this question to make sense in field theory, you need to have already calculated the effective action $\Gamma[\phi]$, integrating out all loop effects. The true vacuum is then defined by minimizing the effective potential for $\phi$. It is by definition the lowest energy state (vacuum). If there is more than one solution to the minimization and they are degenerate, there will be SSB. $\endgroup$
    – Dan
    Aug 4, 2017 at 18:06

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