How do we know at the operator-level that the tadpole $\langle\Omega|\phi(x)|\Omega\rangle=0$ vanishes in scalar $\phi^4$ theory?

Question

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!

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 $\langle \Omega|\hat{\phi}(x)|\Omega\rangle = 0$.

Edit:

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.