Effects of a non-Lorentz-invariant vacuum state

Question

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:

https://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.

Answer 1

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.

Answer 2

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.

Answer 3

You can make everything work consistently, if you go back to fundamentals. In the middle of the 20th century, it was (belatedly) discovered that the kinematic symmetry group of non-relativistic physics is not the homogeneous Galilei group, which has only 10 dimensions, but its central extension to the Bargmann group, which has an 11th dimension.

This upgrade was not reflected by a similar upgrade on the Relativity side of the (Relativistic ⇐ Non-Relativistic) correspondence limit arrow, so as a result: the correspondence limit arrow was broken, and Correspondence Principle was violated.

To repair this requires making a similar extension on the relativistic side of the arrow, thereby lifting the symmetry group for Relativity - Poincaré - from 10 dimensions to a group, that one might call, "Relativistic Bargmann", that has 11. The place where this change makes its presence felt is precisely with the issue of vacuum energy and - for other representations - internal energy.

Let's see how this works...

In Relativity, the role of energy is played by a quantity $E$ that is meant to fully embody mass-energy unification. For an ordinary slower-than-light body, it includes its kinetic + internal energy, which will be denoted $H$, and the energy $mc^2$ associated with its mass $m$, resulting in $E = mc^2 + H$. Both $H$ and $m$ have non-relativistic limits, but $m$ is not actually related to any generator per se. Rather it arises as the magnitude of the energy-momentum vector $(E,𝐏)$: $\left(mc^2\right)^2 = E^2 - |𝐏|^2c^2$. In contrast, what does arise from a generator is the relativistic mass $M = E/c^2$. This also has the mass as its non-relativistic limit.

As is, $E$, $m$, $M$ and $H$ are not, and can't be made, independent of one another, but are all tied directly to $(E,𝐏)$. The Poincaré group is too constrained. In contrast, in the Bargmann group, $(H,𝐏,m)$ gives rise to a 5-vector in which each component is independent. There is no constraint. Instead, in place of the mass-shell invariant $\left(mc^2\right)^2 = E^2 - |𝐏|^2 c^2$, one has two invariants: $|𝐏|^2 - 2mH$ and $m$.

To create the 11 dimensions of the Relativistic version of the Bargmann group out of the 10 dimensions of the Poincaré group requires making $(H,M)$ independent and this, in turn, requires disassociating the difference $(E-H)/c^2$ from the rest mass $m$ and generalizing it into an invariant $μ$, whose non-relativistic limit will be the mass $m$, too. Then, writing $E = H + μc^2$, and defining the internal energy $U$ as the extra contribution to $μ$ required to give you the rest mass: $U = (m - μ)c^2$, the mass shell invariant becomes: $$\begin{align} |𝐏|^2 &= \frac{E^2}{c^2} - m^2 c^2\\ &= \left(\frac{H^2}{c^2} + 2μH + μ^2c^2\right) - \left(μ^2c^2 + 2μU + \frac{U^2}{c^2}\right)\\ &= \frac{H^2}{c^2} + 2μH - 2μU - \frac{U^2}{c^2}, \end{align}$$ or just: $$|𝐏|^2 - 2μH - \frac{H^2}{c^2} = -2μU - \frac{U^2}{c^2}.$$

In terms of the relativistic mass $M$, the two resulting invariants can be written as: $$|𝐏|^2 - 2MH + \frac{H^2}{c^2} = -2μU - \frac{U^2}{c^2}, \quad M - \frac{H}{c^2} = μ.$$ And now: the correspondence limit to non-relativistic world is repaired. These invariants are a deformation of the invariants for the Bargmann group, parametrized by $α$ as: $$|𝐏|^2 - 2MH + αH^2, \quad M - αH.$$ The correspondence arrow is given by $$α = \frac{1}{c^2}\quad⇐\quad α = 0.$$

The "relativistic Bargmann" group, itself, may then be presented as a one-parameter deformation of the Bargmann group, itself. The Lie brackets that are affected are: $$\begin{align} \left\{K_i,P_j\right\} = δ_{ij} \frac{E}{c^2} = δ_{ij} M = δ_{ij} \left(μ + \frac{H}{c^2}\right)&\quad⇐\quad \left\{K_i,P_j\right\} = δ_{ij} μ,\\ \left\{K_i,K_j\right\} = -ε^k_{ij} \frac{J_k}{c^2}&\quad⇐\quad \left\{K_i,K_j\right\} = 0, \end{align}$$ all other brackets remaining the same; where $𝐊 = \left(K_1,K_2,K_3\right)$ and $𝐉 = \left(J_1,J_2,J_3\right)$ respectively go with boosts and rotations. I am writing the brackets as Poisson brackets (as they actually are), reserving the rectangular brackets for the quantized version $[\_,\_] = iħ\left\{\_,\_\right\}$.

In all, the parametrized family of Lie algebras has the following set of brackets: $$ \left\{J_i,J_j\right\} = ε^k_{ij} J_k,\quad \left\{J_i,K_j\right\} = ε^k_{ij} K_k,\quad \left\{J_i,P_j\right\} = ε^k_{ij} P_k,\\ \left\{K_i,K_j\right\} = -αε^k_{ij} J_k,\quad \left\{K_i,P_j\right\} = δ_{ij} \left(μ + αH\right) = δ_{ij} M,\quad \left\{P_i,P_j\right\} = 0,\\ \left\{J_i,H\right\} = 0,\quad \left\{K_i,H\right\} = P_i,\quad \left\{P_i,H\right\} = 0,\\ \left\{J_i,μ\right\} = 0,\quad \left\{K_i,μ\right\} = 0,\quad \left\{P_i,μ\right\} = 0,\quad \left\{H,μ\right\} = 0. $$ From this also follows the brackets for $M = μ + αH$: $$ \left\{J_i,M\right\} = 0,\quad \left\{K_i,M\right\} = αP_i,\quad \left\{P_i,M\right\} = 0,\quad \left\{H,M\right\} = 0,\quad \left\{H,M\right\} = 0. $$ These could be used in place of the brackets for $μ$.

The representations for this group are the same as for the Poincaré group - except that there is an additional energy term corresponding to $U$. Energy becomes additively-relative, once again, as it is in the non-relativistic case. That's a consequence of re-establishing the correspondence limit arrow with the lifted version of Galilei to a lifted version of Poincaré.

For the ground state $|0⟩⟨0|$, the relativistic condition $⟨0|E|0⟩ = 0$ - which is imposed by the requirements $𝐊|0⟩ = 𝟬$ and $𝐏|0⟩ = 𝟬$ and the Lie bracket $\left[K_i,P_j\right] = iħ α δ_{ij} E$ now becomes $⟨0|M|0⟩ = 0$. Now, you can also see why the constraint on energy disappears in the non-relativistic case $α = 0$. Instead, it becomes a condition on the mass $\left[K_i,P_j\right] = iħ δ_{ij} M$ - which is actually what it was, in the Relativistic case, in the first place!

Instead of $⟨0|E|0⟩ = 0$ telling you that the vacuum energy is zero, now it's just telling you $⟨0|M|0⟩ = 0$ that the relativistic mass is zero. The train collision has been diverted, because the rails have been disconnected and separated and the trains zoomed right past each other on parallel tracks. The time translation generator is not connected with $E$ anymore, but with $H$, and there is no constraint on $H$! It doesn't appear on the right-hand side of any Lie brackets. Instead, you get $⟨0|H|0⟩ = ⟨0|U|0⟩$.

For the vacuum state, the "internal energy" $U$ plays the role of vacuum energy. All of the Poincaré representations have an extra energy $U$ under the extension of Poincaré group to the "relativistic Bargmann" group.

For ordinary slower-than-light bodies, it provides an additive contribution to the rest mass $m = μ + αU$ - relative to the invariant mass $μ$, which compensates for the non-additivity of $m$ in Relativity. As you'll recall, when you take the sum of two bodies with respective rest-masses $m_0$ and $m_1$, then the total body's rest mass $m > m_0 + m_1$ will be greater than the sum of the component rest masses, since part of what went into $m$ was the kinetic energy of the component bodies by virtue of their motion around the center of mass of the combined body. So, even though the invariant masses are additive $μ = μ_0 + μ_1$, the rest masses need not be, and their non-additivity is what $U$ compensates for, even with $μ = μ_0 + μ_1$.

So, another feature that arises from the extension of the Poincaré group is that a distinction is drawn between an invariant "bare" mass $μ$ and the rest mass $m$. For the reason just explained, it is a necessary feature for composite bodies; but there's no reason to assume that the same can't also be applied to elementary bodies - to particles. So, it may actually provide a means to encapsulate the distinction between "bare" versus "dressed" mass in quantum field theory.

Pay careful attention to how the "central charge" $μ$ appears in the Lie brackets in relation to the deformation $α$. At $α = 0$, it is a (non-trivial) central charge that provides the extra 11th dimension that turns Galilei into Bargmann. However, at $α = 1/c^2$, it does not provide the extra dimension for the lifting of Poincaré but stays intact, when dropping back down from 11 dimensions to 10! Instead, it's $U$ that's filtered out. So, even though $μ$ is a "trivial" central charge to the extension of Poincaré, it is not trivial. The reduction to Poincaré does not consist of zero'ing out $μ$, but constraining $(H,M)$, removing their independence by tying them both to $(E,𝐏)$ ... in such a way as to force $U = 0$.

The reduction from 11 dimensions to 10 hits the extension by the central charge diagonally, not head-on.

The absence of vacuum energy in the Poincaré representations is therefore directly tied to and expressed by the reduction condition $U = 0$, itself; and this shows in stark relief precisely what the deficit is in Poincaré that forces the condition. So, you need the 11th dimension to get vacuum energy.