They state $$\langle\Omega|\phi(x)|\lambda_{\bf p}\rangle=\langle\Omega|e^{iP\cdot x}\phi(0)e^{-iP\cdot x}|\lambda_{\bf p}\rangle \tag{7.4}$$ where $|\lambda_{\bf p}\rangle$ is a state of momentum ${\bf p}$. They then rewrite this as $$\langle\Omega|\phi(0)|\lambda_{\bf p}\rangle e^{-ip\cdot x}\tag{7.4}$$ with $p^0=E_p$.

So they've basically said $$\langle\Omega|e^{iP\cdot x}=\langle\Omega|.$$ This would be fine if the interacting vacuum had zero energy and momentum. It does have zero momentum but on page 86 they've defined $E_0=\langle\Omega| H|\Omega\rangle$ so really I'd expect $$\langle\Omega| e^{iP\cdot x}=\langle\Omega| e^{iE_0t}.$$

I've tried seeing what happens if I just subtract the term $E_0$ from the Hamiltonian to give the interacting vacuum zero energy. Subtracting it from the interaction term basically gave me an infinite S-matrix, while subtracting it from the free part gives asymptotic states at infinity negative energy and seems to mess up their proof of the expression of the two point correlator.

What should I be doing?

  • 1
    $\begingroup$ Have they perhaps forgotten to tell you they set $E_0 = 0$? $\endgroup$
    – ACuriousMind
    Commented Aug 22, 2015 at 23:25
  • $\begingroup$ @ACuriousMind Yes that's the most obvious explanation but why is that justified? Why can we just set it to zero? $\endgroup$
    – Okazaki
    Commented Aug 23, 2015 at 8:25
  • 2
    $\begingroup$ That was the kind of thing that made me abandon Perskin for good $\endgroup$
    – Hydro Guy
    Commented Sep 5, 2015 at 19:58

2 Answers 2


The short answer: Peskin & Schroeder assume that the vacuum is invariant under translations and Lorentz transformations (which implies $P^\mu|\Omega\rangle=0$). Alternatively, translation invariance implies $\vec P|\Omega\rangle=0$ in all reference frames, and combining this with invariance under boosts gives $P^0|\Omega\rangle=0$.

The less short answer: Since P&S is perturbative up to chapter 7, one thing you could ask is this: what interactions, when added perturbatively to a given free theory, will automatically lead to the interacting ground state $|\Omega\rangle$ being Lorentz invariant. Perturbation theory assumes that the interacting ground state $|\Omega\rangle$ overlaps with the free vacuum $|0\rangle$, i.e. $\langle\Omega|0\rangle\neq 0$, so let's set $|\Omega\rangle=a|0\rangle+b|\Delta\rangle$ where $\langle \Delta|0\rangle=0$ and $a\neq 0$. By definition of $|0\rangle$ and the 4-momentum for the free theory $P_0^\mu$, we have $P_0^\mu|0\rangle=0$. The 3-momentum operator for the interacting theory is usually the same as that for the free theory (certainly in the absence of derivative couplings), so let's first suppose $\vec P|0\rangle=0$. In order to have $\vec P|\Omega\rangle = \vec p|\Omega\rangle$ with $\vec p\neq 0$ we would need $\vec P|\Delta\rangle \propto |0\rangle$. This is impossible, however, because the orthogonal complement of $|0\rangle$ is preserved by $P_0^\mu$ ($\vec P$ doesn't change particle number). Together with Lorentz invariance, this implies that $P^\mu|\Omega\rangle=0$ when $\vec P=\vec P_0$ perturbatively in the coupling.

One example where $\vec P \neq \vec P_0$ is in an effective field theory for the dynamics of electrons that ignores photons. Here, there is an ambiguity in the ground state from the choice of background photons that could originate from a distant source. If the background field of photons is sufficiently strong, then the 'lowest energy state'* could be a thermalized 'gas' of electron-positron pairs. The 'free' theory for electrons (Dirac fermions) is Lorentz-invariant, but the interacting theory needs to include terms that account for things like finite propagation times (if the gas is sufficiently dilute). Essentially, there is momentum stored in photons that isn't accounted for in the free Fock space of electrons.

In general, the ground state can differ from the (interacting) vacuum state, spontaneously breaking Lorentz symmetry and leading to a preferred frame. Spontaneous symmetry breaking of this type is necessary if $P^\mu|\Omega\rangle\neq 0$, and the preferred frame is the one in which 3-momentum vanishes. Lorentz invariance can be broken in other ways too, e.g. through a vacuum with nonzero electromagnetic charge. Additional examples can be found here.

  • 1
    $\begingroup$ Thanks a lot. Indeed combining translation invariance and boost invariance gave me the result. The extended answer was also very illuminating! $\endgroup$
    – Okazaki
    Commented Sep 5, 2015 at 11:07

Perhaps I am overlooking the real problem here, but isn't it clear that by doing $$ H\rightarrow H+E_0 $$ both the vacuum and the one particle state energies get shifted by the same amount $$ H|\Omega\rangle = E_0\,,\qquad H|p\rangle =E_0+E_p $$ so that $E_0$ Actually cancels out $$ \langle\Omega|\phi(x)|p\rangle=e^{iE_0 t}\langle\Omega|\phi(0)|p\rangle e^{-i px}=\langle\Omega|\phi(0)|p\rangle e^{-i E_p t+i\vec{p}\vec{x}} $$ that is you are free to set $E_0=0$ without affecting the result.

Moreover, subtracting a constant term from the Hamiltonian can't possibly give you a divergent S-matrix because it simply rescales all fields by the same identical phase, and in QM the states are defined defined up to a phase.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.