Lorentz transformation of the vacuum state

In general, the Hamiltonian $H$ has non-zero vacuum expectation value (VEV): $$H \left.| \Omega \right> = E_0 \left.|\Omega \right>,$$ where $\left.|\Omega\right>$ is the vacuum state. The zero-point energy of the quantum oscillator is a good example to have non-vanishing VEV.

The problem strikes me is that how is this vacuum state transformed under Lorentz transformation?

On one hand, we can think of this vacuum state is the trivial representation of the Poincare group, which is a Lorentz invariant. ( In Wigner's classification, vacuum state is in the $(0,0,0,0)$ class, if I understand correctly.) So $$U(\Lambda) \left.| \Omega \right> = e^{-i \theta} \left.| \Omega \right>$$

On the other hand, the vacuum state has non-zero 4-momentum: $p = (E_0, \vec{0})$. So it seems it should transform like other irreps: $$U(\Lambda) \left.|p, \sigma\right> = \sum_{\sigma'}C_{\sigma'\sigma} \left.| \Lambda p, \sigma' \right>,$$ Now, the important thing is, $\Lambda p$ could have non-zero 3-momentum. But the vacuum state should have zero 3-momentum. Bang!

In textbooks, a common procedure to reconcile this contradiction is to shift the Hamiltonian $H \to H - E_0$; but, the Poincare algebra would be affected: $$[ K^i, P^j ] = \delta^{ij} H \to [ K^i, P^j ] = \delta^{ij} H + \delta^{ij} E_0$$

Thank you.

• This is way too many questions in one. Could you perhaps focus the question. You could also ask some of these as different SE questions. Commented Feb 20, 2014 at 19:55
• Sorry, I do not understand. In a relativistic QFT, the vacuum state is assumed to be the unique (so it is a ground state) state invariant under the whole Poincaré group and has, consequently, zero four momentum. Commented Feb 20, 2014 at 20:16
• @V.Moretti : But the vacuum is also a representation of the Poincare group, which should transform according to the rule of Poincare group. Therefore, it it has non-vanishing 4-momentum in one frame, it may acquire non-vanishing 3-momentum in another frame, which contradict the Lorentz invariance. Commented Feb 20, 2014 at 20:29
• A free quantum field is not a superposition of oscillators. That picture is only heuristic, it is a Fock representation of a suitable Weyl algebra. Commented Feb 20, 2014 at 20:37
• The vacuum in relativistic QFT is the trivial rep of Poincaré group, so it has no four momentum. Actually to prove it, it is sufficient that the state is invariant under four-translations (and it is assumed so). Commented Feb 20, 2014 at 20:40

As has been mentioned in the comments, it is an assumption that the QFT has a vacuum state which is annihilated by $P^\mu$. This is actually a very important point, since it is one of the crucial differences between flat space QFT and QFT in curved spacetime. This is explained in Wald's book QFT in Curved spacetime. Essentially in quantum theories, the "kinematic structure" is fixed by the canonical commutation relations between the canonical position and momenta (fields and conjugate momenta in QFT). In non-relativistic QM there is a finite number of degrees of freedom, and so the Stone von-Neumann theorem tells you there is a unique Hilbert space and choice of operators on the Hilbert space. The theorem does not hold for infinite number of d.o.f (QFT), and there are actually infinitely many inequivalent choices for the Hilbert space for quantum fields. In flat spacetime, the requirement that there exists a state such that $P^{\mu}|0\rangle=0$ picks out a unique Hilbert space. In curved spacetimes which in general have no killing vectors, it is unclear how to single out a unique Hilbert space, and this is a big difficulty in curved space QFT.

So the answer is that it is assumed $P^\mu|\Omega \rangle =0$ and this is a very important assumption.

Although the subtraction procedure you mention is used in many textbooks, I don't think it is really the correct solution. The argument is usually that the subtraction term is unobservable, since it only contributes a phase to the s-matrix and we only measure energy differences, but this really doesn't hold when gravity or Supersymmetry are in the picture. In fact, as you note it is not clear that the Poincare algebra closes under such a subtraction.

The operators in two theories with different Hamiltonians do not need to be the same, they just need to change in such a way that the Poincare Algebra still closes. If you had an original Hamiltonian $H_0$ and you perturbed it by $H_0\to H= H_0+V$, then sending $H \to H-E_0$ basically amounts to perturbing the Hamiltonian $H_0$ by $V-E_0$. For perturbations built out of local fields in a Lorentz invariant way, it is always possible to alter the algebra such that even in the interacting theory the algebra closes.

In fact, this is basically the answer to the question: What kinds of perturbations can I add to the Hamiltonian that result in a Lorentz invariant theory that obeys the cluster decomposition principle. I'll outline how the argument goes.

In the free theory we have states of the form $a_p^{\dagger}...|0\rangle$. This is the usual definition of our free Hilbert space. Consider $[K^0_i,H_0]=-iP_i$. We want the momentum label to mean the same thing after the perturbation (we only altered our energies) but we sent $H_0\to H_0+V$. So naively we pick up a term $[K_i^0,V]$. Requiring these two operators to commute is too strong a constraint. So instead we allow for $K_i^0 \to K_i^0+K_i^V$ such that $$[K^0,V]+[K^V,H_0]+[K^V,V]=0$$ Using the commutation relations $[a,a^{\dagger}]$, etc you may show that if V is of the form $$V=\int d^3x \mathcal{V}(\phi(\vec{x}))$$ where $\phi(\vec{x})$ is a local field built from the creation and annihilation operators, then a solution $$K^V_i=\int d^3x x_i\mathcal{V}(\phi)$$ closes the algebra (you really only need to check the commutator you mention). This is not to say that local QFT's are the only way to build Lorentz invariant quantum systems, it just says that local QFT's are some of the solutions.

The problem is that $E_0$ is not built out of local fields and can't really even be written in a local way in an infinite volume QFT. You would want something of the form $V=\int d^3x \frac{E_0}{V}$. For such a perturbation $K_V=0$ and we cannot satisfy $$[K^V_i,P_j]=\delta^{ij}E_0$$ This isn't that surprising since adding a constant to $P^0$ but not $P_i$ isn't really Lorentz invariant.

So in my opinion the subtraction is more of a cheat that happens to work since we only ever really calculate $|\langle out| S|in \rangle|^2$ for which it does not matter.

• two comments: 1. Your analysis is purely based on pQFT. Is there any possibility to generalize to a broader class of QFTs? ( I know it has been argued that pQFT perhaps is the only QFT that makes sense. ) 2. According to quantum mechanics (QM), unitary projective representation is allowed. The way to obtain a projective rep'n from a linear rep'n is to shift the algebra by a central charge extension (and imposing superselection rules if the topology of the Lie group is non-trivial). How does this operator affect the your argument? Commented Feb 21, 2014 at 22:38
• Projective representations may arise from central extensions or non-trivial topology of the group manifold. In the case of the Poincare group you may show that the projective representations are obtained by considering the universal cover, no central extension is permitted (this is done in Weinberg). If by pQFT you mean perturbative QFT then yes, my analysis is based purely on pQFT. I don't really know much non-perturbative QFT that doesn't involve SUSY, so maybe someone else can answer that.
– Dan
Commented Feb 22, 2014 at 6:37