I'm trying to understand the Poincaré invariance of the vacuum state in Minkowski spacetime, how it implies the uniqueness of the vacuum state, and why there's no unique vacuum state in general spacetimes. For Poincaré invariance I found the following demonstration:

Under a Lorentz transformation $\Lambda$ a generic field transforms as: $$U^{\dagger}(\Lambda) \phi(x) U(\Lambda) = S(\Lambda)\phi(\Lambda x),$$ where $U(\Lambda)$ belongs to the representation of the Lorentz group acting on the physical states while $S(\Lambda)$ belongs to the representation acting on the operators.

The vacuum is clearly Lorentz invariant. If you want for your VEV to be Lorentz invariant it must be: $$\langle 0 |\phi(x)|0\rangle = \langle 0|\phi(\Lambda x)|0\rangle,$$ however, because of the invariance of the vacuum: $$\langle 0 |\phi(x)|0\rangle = \langle 0 |U^{\dagger}(\Lambda)\phi(x)U(\Lambda)|0\rangle = S(\Lambda)\langle 0 |\phi(\Lambda x)|0\rangle ,$$ and so it must $S(\Lambda)=1$ be which is true for a scalar field.

Reference: https://www.physicsforums.com/threads/poincare-invariance-of-vacuum.752840/

I've got several questions:

  1. In $U^{\dagger}(\Lambda) \phi(x) U(\Lambda) = S(\Lambda)\phi(\Lambda x)$ I understand that we transform $\phi(x)$ as $U^{\dagger}(\Lambda) \phi(x) U(\Lambda)$, but I don't understand the right hand side. Why is it $\phi(\Lambda x)$ on one side and $\phi(x)$ on the other? Why are both sides equal?
  2. Accepting this as a demonstration of Poincaré invariance of the vacuum state, how does this imply uniqueness of the vacuum state?
  • 5
    $\begingroup$ It does NOT imply uniqueness. That is an extra assumption (which we now know to be false in any gauge or gravitational theory). Also, Lorentz transformations map $x \to \Lambda x$, so the fields must have different arguments. $\endgroup$
    – Prahar
    Feb 12 at 14:38

1 Answer 1


The equation $$ U^\dagger(\Lambda) \phi(x) U(\Lambda) = S(\Lambda) \phi(\Lambda x)$$ is essentially the consistency requirement that applying a Lorentz transformation commutes with quantization. The l.h.s. is the quantum viewpoint of how the operator $\phi(x)$ transforms under the unitary action of the Lorentz group on the space of states, the r.h.s. is the classical viewpoint of how fields transform under the ordinary action of the Lorentz group on their domain.

Requiring these to be equal means that you can either quantize and then go to another frame or go to another frame and then quantize and both procedures will yield the same result. If this was not the case, we could not call this quantum theory "relativistic" with a straight face since it would not be properly covariant under the relativistic Lorentz transformations.

This has nothing to do with uniqueness of the vacuum; that the vacuum is often unique is an additional assumption and in fact there can be situations where there is more than one "vacuum", e.g. with the instantonic $\theta$-vacua of QCD.


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