I'm a mathematician slowly trying to teach myself quantum field theory. To test my understanding, I'm trying to tell myself the whole story from a Lagrangian to scattering amplitudes for scalar $\phi^4$ theory in a way that feels satisfying. There's one thing that's kind of bugging me.

One of the hypotheses of the LSZ formula is that the fields you're using have vanishing vacuum expectation value, so in the present case that $$\langle \Omega|\phi(x)|\Omega\rangle = 0.$$ For non-scalar fields, I buy that this follows from Lorentz-invariance, but I've also seen it claimed in a few books (e.g. in Section 10.4 of Ticciati's QFT for Mathematicians) that for $\phi^4$ theory in particular it follows from the fact that $$\phi\mapsto-\phi$$ is a symmetry of the Hamiltonian.

My question is: How exactly does this argument go?

The only idea I've been able to come up with is that maybe there's a unitary operator $U$ which flips the sign of $\phi$, that is, I guess, it has the property $$U^{-1}\phi(x)U=-\phi(x),$$ and then you'd argue that $U$ commutes with the Hamiltonian and therefore preserves its eigenstates, but I don't see why (or if) I should expect $U$ to exist, or if this is even the right idea at all.

Thanks, and of course please let me know if I made a mistake somewhere earlier in the process of arriving at this question!

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    $\begingroup$ I'm not so sure about relativistic QFT, but in the context of condensed matter, for example, I think the statement $\langle \Omega|\phi(x)|\Omega\rangle = 0$ is rather an assumption. More precisely, one assumes that the ground state of an interacting system preserves the same symmetries as the Lagrangian/action. This is not necessarily the case. It is is fact NOT the case when spontaneous symmetry breaking occurs, such as in the Ginzburg-Landau theory: en.wikipedia.org/wiki/Ginzburg%E2%80%93Landau_theory . $\endgroup$ Commented Mar 31 at 21:42
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    $\begingroup$ Yes, I definitely agree that we shouldn't expect this in the context of spontaneous symmetry breaking, and in fact my understanding is that $\phi^4$ theory with a negative mass term is a common toy example of that. If the story I sketched in this question makes sense, my hope is that the conclusion would be that the lowest-energy eigenspace is what's preserved by this $U$ operator, and so to know the vacuum is preserved you'd also need that this eigenspace is one-dimensional. $\endgroup$ Commented Mar 31 at 21:44
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    $\begingroup$ Generally speaking, we don't know this and must impose it (see ncatlab.org/nlab/show/tadpole+cancellation). In the $\phi^4$ case, there is a ${\mathbb Z}_2$ discrete symmetry which ensures that odd-point correlators vanish. $\endgroup$
    – Prahar
    Commented Mar 31 at 23:16
  • $\begingroup$ @Prahar I think we're on the same page about the tadpole cancellation. What I'm asking about here is exactly how the argument involving the $\mathbb{Z}_2$ discrete symmetry works. $\endgroup$ Commented Mar 31 at 23:48

1 Answer 1


I think you are basically there:

Let $\hat{U}$ be the unitary operation you defined, $\hat{U} \hat{\phi}(x) \hat{U}^\dagger \equiv -\hat{\phi}(x)$. This is a definition. We then assume (or check) that this operation is a symmetry of the Lagrangian.

We now assume that the ground state $|\Omega\rangle$ is also invariant under this operation, $\hat{U} |\Omega\rangle = |\Omega\rangle$ (there is in principle a phase factor, but we can take it to be $1$ by redefining $\hat{U}$).

With this assumption, \begin{align} \langle \Omega|\hat{\phi}(x)|\Omega\rangle &= \langle \Omega| \hat{U} \phi(x) \hat{U}^\dagger |\Omega\rangle \\ &= \langle \Omega| (-\hat{\phi}(x)) |\Omega\rangle \\ &= -\langle \Omega|\hat{\phi}(x)|\Omega\rangle, \end{align} which then implies $\langle \Omega|\hat{\phi}(x)|\Omega\rangle = 0$.


This is my attempt at showing that a $\hat{U}$ as above exists. I'm denoting operators in Fock space with a hat, $\hat{U}$, while unhatted operators act in the single-particle Hilbert space.

For a non-interacting theory, the field operator can be expanded as \begin{align} \hat{\phi}(\mathbf{x}) = \int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{\mathbf{p}}}}[\hat{a}_{\mathbf{p}}e^{i\mathbf{p}\cdot{\mathbf{x}}} + \hat{a}_{\mathbf{p}}^{\dagger}e^{-i\mathbf{p}\cdot{\mathbf{x}}}], \end{align} where $\hat{a}_{\mathbf{p}}$ denotes the annihilation operator for the mode with momentum $\mathbf{p}$.

A unitary transformation which preserves particle number acts on the annihilation operators as $\hat{U} \hat{a}_{\mathbf{p}} \hat{U}^\dagger = \int dp^\prime U(\mathbf{p};\mathbf{p}^\prime) \hat{a}_{\mathbf{p}^\prime}$, where $U(\mathbf{p};\mathbf{p}^\prime)$ are the matrix elements of the single-particle representation of the unitary operation. In particular, for $U(\mathbf{p};\mathbf{p}^\prime) \equiv -\delta(\mathbf{p} - \mathbf{p^\prime})$, we have $\hat{U} \hat{a}_{\mathbf{p}} \hat{U}^\dagger = - \hat{a}_{\mathbf{p}}$. A similar calculation holds for $\hat{a}^\dagger_{\mathbf{p}}$. Using this in the expression for $\hat{\phi}$ we obtain that $\hat{U} \hat{\phi}(x) \hat{U}^\dagger \equiv -\hat{\phi}(x)$, as required.

Then there is the question of how does this generalize to an interacting theory, which I don't know.

  • $\begingroup$ I may miss something obvious here, but how do you know that there exists such a unitary $U$? $\endgroup$ Commented Mar 31 at 22:02
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    $\begingroup$ Put differently: What you've shown is that if such a (unitary) $U$ exists, the statement follows. But for me it is not clear that a unitary $U$ exists which transforms the field as such (irrespective if it is a symmetry of the Hamiltonian or not and irrespective if the vacuum is invariant or not). As I said, I might miss something, perhaps obvious... $\endgroup$ Commented Mar 31 at 22:12
  • $\begingroup$ You're right. Perhaps there is no such $U$ then. I'm thinking about it. $\endgroup$ Commented Mar 31 at 22:14
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    $\begingroup$ One thought I had: this feels similar in many ways to the charge conjugation operator C. The whole discrete symmetries/CPT story is a part of this whole business I don't understand well, but it's possible that whatever makes that work would also make the thing I want work? $\endgroup$ Commented Mar 31 at 22:18
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    $\begingroup$ I don't understand your argument, but my (relativistic) QFT-knowledge is a bit rusty. Feel free to expand if you have time. $\endgroup$ Commented Mar 31 at 22:24

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