Let's say I have some Lorentz-violating theory. For concreteness we could imagine a scalar field \begin{equation} S=\int {\rm d}^4 x \left(-\frac{1}{2} (\partial \phi)^2 + V(x)^\mu \partial_\mu \phi\right) \end{equation} where $V(x)^\mu$ is some space-time dependent vector field (ie a non-trivial external current sourcing $\partial_\mu \phi$).

Classically, I can define a Lorentz transformation that acts in the normal way, $x^\mu \rightarrow \Lambda^{\mu}_{\ \ \nu}x^\nu$, $\phi(x) \rightarrow \phi(\Lambda^{-1}x)$. Of course, this transformation won't be a symmetry because the action will not be invariant.

What I want to understand is what the analog of this logic is in quantum field theory. Ie: I would expect that I could define a Lorentz transformation, but I should find there is some condition defining a symmetry that is violated by this transformation.

Now let's say I quantize the theory, and work in the canonical formalism. Surely I can define the action of Lorentz transformations $\Lambda$ on the Hilbert space. After all, given a state $|\Psi\rangle$, I should be able to ask what the state looks like under a boost. Furthermore, I would expect the set of transformations to form a representation on Hilbert space, $R(\Lambda)$, since the Lorentz transformations themselves form a group.

However, it's not clear to me if the transformations can be taken to be unitary. I basically see two possibilities... (1) it is possible to construct a unitary representation of the Lorentz group on the Hilbert space of the theory, but it won't have any physical relevance because the operators controlling the dynamics (4-momentum operator, angular momentum operator) won't transform as tensors under these transformations, or (2) it is not possible to construct a unitary representation.

In favor of possibility (1): in non-relativistic quantum mechanics, I could always define angular momentum operators $L_i = \epsilon_{ijk}x_j p_k$ which obey the algebra for $SU(2)$, even if the Hamiltonian is not Lorentz invariant, so it seems I could always construct a unitary representation of rotation operators by exponentiating these.

My questions are:

  • Is constructing $R(\Lambda)$ possible in a Lorentz-violating theory? (If not, what stops you from constructing it?)
  • Can we choose a unitary representation for $R(\Lambda)$ (possibility (1)), or not (possibility (2))?
  • $\begingroup$ Exactly! This is what I want to know. I think this is my option 1 (unless I am missing something!) $\endgroup$
    – Andrew
    Jul 26, 2020 at 3:40
  • 2
    $\begingroup$ The notion of quantum symmetry is various. One may have a symmetry in Wigner's sense (a unitary representation of a Lie group with s desired effect on observables) and here there is a twofold nonexclusive possibility. This symmetry leaves invariant a physically relevant pure state (the "vacuum" of the theory), the symmetry commutes with time evolution (dynamical symmetry). Breaking of a symmetry may occur at each of the three steps of the previous discussion. $\endgroup$ Jul 26, 2020 at 9:17
  • $\begingroup$ Note that the presence of the space-time dependent vector field $V(x)^\mu$ violates translation invariance as well as Lorentz invariance, so this might not be the cleanest setup in which to talk about Lorentz violation. $\endgroup$ Jul 26, 2020 at 13:55
  • $\begingroup$ @Andrew, I do not understand why you say the action shown in your post is Lorentz violating. A scalar field $\phi (x)$ is invariant by definition, the composition of a vector $J^\mu$ (for a current usually $J$ is used) with the partial derivative $\partial_\mu$ is a scalar as well. Please comment. $\endgroup$ Jul 27, 2020 at 15:00
  • $\begingroup$ @ValterMoretti Thanks -- these are fine distinctions and my confusion may simply boil down to mixing these up. I will think about it. $\endgroup$
    – Andrew
    Jul 27, 2020 at 16:10

2 Answers 2


If I understand the question correctly, then your question is more general than relativity. For example, you can ask the same question about rotations in a non-relativistic theory. In the spirit of trying to address the problem in the simplest situation possible, allow me to even downgrade to regular particle quantum mechanics, not a QFT. The Hamiltonian is of course, $$H = \frac{\textbf{p}^2}{2m} + V(\textbf{x}).$$ Let's look at the rotation $R_{\hat{n}}(\theta)$, which is the rotation around the unit vector $\hat{n}$ by an angle $\theta$. The action of rotations on the Hilbert space is defined regardless of what is the Hamiltonian. For example we might simply define on the transformation operation on the Hilbert space as, $$U(R_{\hat{n}}(\theta))|\textbf{x}\rangle = |R_{\hat{n}}(\theta) \ \textbf{x}\rangle.$$ This is true whether the rotation is a symmetry of the Hamiltonian or not.

Whenever you meet a new operation, you need to figure out how it acts on the Hilbert space. Even though all symmetries of the Hamiltonian are unitary, not all unitary operations are symmetry of the Hamiltonian/Lagrangian. I hope this helps.

  • $\begingroup$ I want to mull over it some more, but I think this is the answer. The question is definitely not specific to field theory, it arose as a pedantic point when answering this question: physics.stackexchange.com/questions/566963/…. I think the net result is that the existence of a unitary representation of a given symmetry group on Hilbert space is a necessary, but not sufficient, condition for the quantum theory to possess this symmetry (just as you have said). Thanks! $\endgroup$
    – Andrew
    Jul 28, 2020 at 23:51
  • $\begingroup$ Another simple example along the lines you suggested would be that the translation operator should be unitary in single particle quantum mechanics, even if there is a non-zero potential. $\endgroup$
    – Andrew
    Jul 29, 2020 at 0:21
  • $\begingroup$ Yes that's also another good example. All of these examples, rotations, boots, or translations, are actual symmetries of spacetime. That is, our models that break any of these symmetries only does so because they have "external" terms. So maybe one more example that is different in that regard is the parity operator. Its action is unitary, and it's not an actual symmetry of physics. $\endgroup$
    – A. Jahin
    Jul 29, 2020 at 12:40

If we want to describe the real world, QFT (quantum field theory) should respect the apparent symmetries of our universe. One symmetry is the translation invariance, another symmetry is the Lorentz invariance, which compound the Poincaré group ISO(1,3), the isometry group of Minkowski space. Formally $\vert \psi \rangle \to P \vert \psi \rangle$ where $P$ is a Poincaré transformation. In addition unitarity is requested to guarantee that the matrix elements $M = \langle \psi_1 \vert \psi_2 \rangle$ are Poincaré invariant. Under a Poincaré transformation we have $M = \langle \psi_1 \vert P^\dagger P \vert \psi_2 \rangle$ asking for $P^\dagger P = 1$.

The unitary representations of the Poincaré group are only a small subset of all of the representations of the Poincaré group. The unitary irreducible representations of the Poincaré group were classified by E. Wigner in 1939 and they are all infinite dimensional.

As for your question, I do not think it is a realistic description of nature to build a unitary representation out of a Lagrangian which is not Lorentz invariant. For instance in non-relativistic quantum mechanics you can have a particle spin state which is a linear combination of spin up and spin down and the norm is invariant under rotation, however in non-relativistic quantum mechanics you can not describe particles creation and annihilation.

  • $\begingroup$ Thanks for the comment! I agree with the first two paragraphs. I also agree my question is not directly relevant for a realistic description of nature. It's more of a "what if" question... We can apply a Poincaire transformation to a non-Poincaire invariant theory; we will find that the Lagrangian looks different in different frames. Then my question is, in this quantum theory, what are the properties of operators that implement Poincaire transformations? I think the operators should form a representation of the Poincaire group, but are they or can we choose them to be unitary operators? $\endgroup$
    – Andrew
    Jul 28, 2020 at 17:20
  • $\begingroup$ @Andrew, For instance the boost matrix (a representation of the Poincaré group) is not unitary, that is $\Lambda^\dagger_{boost} \neq \Lambda^{-1}_{boost}$. $\endgroup$ Jul 29, 2020 at 11:51

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