# Effects of a non-Lorentz-invariant vacuum state

I'm here asking about real or though experiments (i.e., physical effects) where, at least in principle, one can see some consequence of a non-Lorentz-invariant vacuum state in an otherwise Poincare invariant theory.

Let me develop the question. Assume a theory in which the Hamiltonian (that closes Poincare algebra with the rest of generators) $H$ acts non-trivially on the vacuum state $|0>$, where 'non-trivially' simply means:

$$H|0> = E \, |0>$$

with $E$ a positive constant. Since the Hamiltonian transforms as the temporal component of a 4-vector, the vacuum state is not Lorentz invariant. Therefore, the theory is not Lorentz invariant (it is usually claimed that this is an additional condition besides the Poincare algebra). However, I'm not able to see any consequence of this fact. I think that this does not affect any cross-section or decay rate.

I think that one can redefine the Hamiltonian so that $H'=H-E$ and $H'|0>=0 \,$ (this does not affect the Poincare algebra if one also redefines the boost generators properly). I know this seems obvious (this redefinition is usually done in canonical quantization when one doesn't adopt normal ordering), but I've just read this in this forum:

http://physics.stackexchange.com/a/8360/10522

However, in special relativity, energy is the time component of a 4-vector and it matters a great deal whether it is zero or nonzero. In particular, the energy of the empty Minkowski space has to be exactly zero because if it were nonzero, the state wouldn't be Lorentz-invariant: Lorentz transformations would transform the nonzero energy (time component of a vector) to a nonzero momentum (spatial components).

One has a family of vacuum states related by Poincare transformations which are unitary, I don't think this is a problem... what do you think?

Added: 1) I'm not thinking about a Poincare invariant Lagrangian with a potential of the form $(A^2(x)-v)^2$, where $A_{\mu}(x)$ is a vectorial field that acquires a vacuum expectation value $v$. Assume that every field has zero vev.

2) I'm looking for Lorentz violating effects instead of vacuum energy effects unless you argue that these vacuum energy effects (Lamb shift, spontaneous emission, etc.) break Lorentz symmetry.

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– Qmechanic Jul 20 '12 at 22:06
I think it is not exactly the same question. The Hamiltonian of my theory is the special-relativistic one with one constant added. I'm going to edit the question to be more clear. – Diego Mazón Jul 20 '12 at 22:16
If you redefine the hamiltonian and the momentum you not need to mess with the Boost. In QFT you have $E_0=\sum \frac{\omega _{k}}{2}$ and $\vec{P}_0=\sum \frac{\vec{k}}{2}$...so individually, each momentum redefinition matchs the energy redefinition ;) – Nogueira Dec 12 '15 at 20:43
Thanks, @Nogueira . I'm not able to understand what you mean, can you elaborate on your point? – Diego Mazón Dec 20 '15 at 17:28
@Drake If you take the normal ordering of the Hamiltonian, you still have a generator of translation, and the commutation relations don't change. – Nogueira Dec 20 '15 at 17:32

The formulation of this question assumes that it is impossible to have a vacuum state where $\langle0|H|0\rangle > 0$ without violating Lorentz invariance. This is not true. Generally, when there is energy density in the vacuum, you have the appropriate pressure to keep Lorentz invariance, because the stress tensor ends up proportional to $g_{\mu\nu}$ is Lorentz invariant.

In infinite space, the energy would be infinite, since it's a finite energy density. It you cut off the theory in a big box to regulate the energy, you do get a Lorentz breaking total energy in the vacum, but the breaking of Lorentz invariance is only coming from the fact that there are walls or identifications which pick out a Lorentz frame. If you boost the box, you have a momentum in the box, but that's just because the pressures on the edges of the boosted box are not balanced anymore, so that there is a net momentum coming in from the box walls, or if it is periodic, you have a net momentum from the moving periodic boundaries.

So it is not true that vacuum energy breaks Lorentz invariance. This is a little counterintuitive, because we are used to energies being localized in a particle, so that pressure can't be constantly moving momentum around. In the vacuum, the energy is everywhere, and you can have pressure stresses which keep the whole formulation Lorentz invariant.

This is why the vacuum energy infinity in field theories is said lead to cosmological constant renormalization, not Lorentz breaking.

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Nice answer and good point the one of the proportionality between energy-momentum tensor and the Minkowski metric (upvote for you), however I'm not sure your answer is correct (maybe I'm missing something). I know that a box would break Lorentz, so let's assume that the volume is infinite and the energy momentum tensor is proportional to Minkowski metric and their components are finite. – Diego Mazón Jul 23 '12 at 17:42
Continuation: To have a Lorentz invariant theory one needs an invariant vacuum state so $B_i |0>=0$, where $B_i$ is the boost generator. And thus one should have $<0|H|0>=0$ since Poincare algebra contains $[B_i, P_j] \sim H \delta _{ij}$. In others words, if vacuum state carries energy-momentum vector $(E,0)$ (with $E$ different from zero), the vacuum cannot be Lorentz invariant. This argument suggests that (renormalized?) vacuum energy must be zero in a Lorentz invariant theory. – Diego Mazón Jul 23 '12 at 17:43
One more point: the fact that the energy-momentum tensor (=stress tensor) be proportional to the Minkowski metric is a consequence of having a Poincare invariant action functional, but it doesn't say anything about the symmetries of the vacuum state, right? – Diego Mazón Jul 23 '12 at 17:58
@drake: One could also have $\langle 0 | H 0 \rangle = \pm\infty$ consistently--- that's the constant energy density and a cosmological constant. Of course you can't have a finite energy--- the space is infinite, you have a finite density. You are right that if it is any finite value, then it must be zero, but it's not deep at all. – Ron Maimon Jul 23 '12 at 18:46
Dear Ron, $B_i^{\dagger}=B_i$, $P_i|0>$ finite, $B_i|0>=0$, $[B_i, P_j]=i\,H \delta_{ij}$ $\implies$ $<0|H|0>=0$. Just take the vacuum expectation value of the commutator. – Diego Mazón Jul 23 '12 at 19:01

In axiomatic quantum field theory, it is assumed that there is a unique Poincare invariant (projective) state. An arbitrary normalized representative $|0\rangle$ of this state is called the vacuum state. Poincare invariance implies that for all $x$, the state $e^{x\cdot P/\hbar}|0\rangle$ is a multiple of $|0\rangle$. This implies that $P|0\rangle=p|0\rangle$ for some $p$. Redefining $P$ as $P-p$ gives another representation of the Poincare group in which the vacuum state has zero 4-momentum, as usually assumed.

On the other hand, Rn Maimon considers a situation not covered by axiomatic QFT, where a state $|0\rangle$ he calles (in my view mistakenly) the vacuum state is not in the domain of the translation group. Thus $e^{x\cdot P/\hbar}|0\rangle$ is (fro nonzero $x$) not a state vector in the physical Hilbert space (but a proper distribution), the vacuum expectation value of the momentum vector does not exist, and the usual covariance arguments break down. As I understand him, he takes this to be the situation in an infinite universe with finite local momentum density but infinite total energy. However, in such a situation, Poincare invariance is no longer relevant. Indeed, in quantum gravity, there doesn't seem to be an observer-independent notion of a vacuum state, due to the Unruh effect. Instead opne has a family of Hadamard states that, taken together, replace the vacuum state of flat QFT. This family as a whole is indeed Poincare invariant, even invariant under the group of volume-preserving diffeomorphisms.

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