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In nonrelativistic quantum mechanics, a symmetry is a transformation on states in the Hilbert space which keeps the Hamiltonian invariant and this implies that the generator of the transformation must commute with the Hamiltonian. However, we know that a relativistic quantum field theory has Lorentz invariance built into it. However, all the Lorentz generators (boosts) do not commute with the Hamiltonian. Then how does one implement and what does it mean to have Lorentz symmetry at the quantum level. At the classical level there is no problem to implement Lorentz symmetry. But at the quantum level, comparing with non-relativistic quantum mechanics, in quantum field theory we do NOT have, $[H,J^{\mu\nu}]=0$ for all $\mu,\nu$. Then how is Poincare symmetry is a symmetry of quantum field theory?

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  • $\begingroup$ The transformation that implements the required symmetry on the Hilbert space can mix whichever complete system of state vectors you have, and in fact it generally does so. For example, $U(1)$ symmetry transformations mix particle states with antiparticle ones, and they're different eigenstates of the four-momentum. Having a symmetry in a relativistic theory (in the sense in which I guess you are understanding it) only requires the commutativity with the hamiltonian, not with every generator of the Poincaré group. $\endgroup$ – Giorgio Comitini Jan 1 '16 at 18:17
  • $\begingroup$ I should have added, I guessed that you are talking about symmetries which have as a consequence the conservation in time of some quantity. $\endgroup$ – Giorgio Comitini Jan 1 '16 at 18:21
  • $\begingroup$ Have you read Weinberg Vol. 1? There is a grand explanation for precisely this question. $\endgroup$ – Prahar Jan 5 '16 at 15:40
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A symmetry in QM is a map from the Hilbert space of states $\mathcal{H}$ into itself such that the $ray$ $product$ $|(\psi,\phi)|^2$ remains unchanged. This is the most natural definition of what a symmetry should be: you have a state $\psi$ and you transform it into $\psi'$, either "actively" or "passively", for example by looking at the same system from another reference frame. You have a symmetry because you have the same probability of observing $\phi$ given $\psi$ and of observing $\phi'$ given $\psi'$.

Now, given the above definition, Wigner's theorem states that such a transformation should be a linear unitary or anti-linear anti-unitary map.

If you want to implement a group of symmetries $\mathcal{G}$, the most natural choice to do it is via a representation $U : \mathcal{G} \to M(\mathcal{H})$, where $M(A)$ is the set of maps on A, so that the symmetry $g \circ h$ is implemented by $U(g) \circ U(h)$. Wigner theorem states that this will be a (anti)linear (anti)unitary representation.

In this sense you implement the Lorentz symmetry in a QFT, not in the sense you mentioned in your question.

Actually the story goes further, and you have to define the S matrix (that is what is ultimately produced as observable quantity by a QFT) and require it to be Lorentz symmetric in a certain sense, and THIS is what is usually meant for a relativistic QFT.

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You're in fact wrong in the claim that the equality $[J^{\mu\nu}, H] = 0$ holds for all $\mu ,\nu$. The commutator is exactly zero for $\mu ,\nu \in 1,2,3$, i.e., for spacelike indices. But it is not zero for the case when $\mu = 0, \nu \neq 0$, i.e., for the Lorentz boost generator; precisely, for $K_{i}\equiv J_{0i}$ and $H\equiv P_{0}$ we obtain $$ \tag 1 [K_{i}, H] = -iP_{i} $$ These statements come from general commutation relations for generators of the Poincare group, and thus must hold in each Poincare invariant theory. But how relation $(1)$ may be true in Poincare invariant theory? The answer is simple. Existence of symmetry in Heisenberg picture means nothing but that corresponding generators are integrals of motion: $$ \frac{dA}{dt} = [A,H] + \frac{\partial A}{\partial t} = 0 $$ For $K_{i}$, we have $$ \tag 2 \frac{dK_{i}}{dt}= -iP_{i} + \frac{\partial K_{i}}{\partial t} = 0, $$ from which follows that in each Poincare invariant theory $K_{i}$ must be time dependent, and the law of time dependence is given by $(2)$.

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There is indeed a subtlety with boosts. Suppose your field is at some state at a given time. After you boost, you get a different spacelike slice to define a state on. So you are actually trying to relate different Hilbert spaces, unlike making an automorphism as in nonrelativistic QM.

The resolution is to work with local operators. For instance, you can say that a theory is Lorentz invariant if all the vevs of local operators transform properly.

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