# Lorentz transformation implemented by a non-unitary operator.

One often come across in QFT sentences like the following, for instance:

...under a Lorentz transformation $\Lambda$ implemented by the unitary operator $U(\Lambda)$, a Dirac field transforms as

$$U(\Lambda)^{-1}\Psi(x)U(\Lambda) = D(\Lambda) \Psi(\Lambda^{-1}x),$$ where the matrix $D$...

What would be the other case, i.e. implemented by a non-unitary operator? Is $N(x)$ an example of such an operator?

$N(x):=\mathrm{i}x+x^2$.

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Poincaré invariance is a fundamental requirement of relativistic (quantum) physics. In particular if $U_g : {\cal H} \to {\cal H}$ represents the (non-necessarily linear) action of a Poincaré transformation $g$ on (normalized) vectors $\psi$ of the Hilbert space $\cal H$ of the considered system, transition probabilities have to be preserved: $$|\langle U_g(\psi) | U_g(\phi) \rangle|^2 = |\langle \psi | \phi \rangle|^2 \quad \forall \psi, \phi \in {\cal H}\:, ||\psi||=||\phi||=1\:. \quad (1)$$ A celebrated theorem due to Wigner establishes that a (bijective) map $U : {\cal H} \to {\cal H}$ verifying (1) must necessarily be linear and unitary or anti linear and anti unitary, depending on the physical nature of the transformation.

Concerning representations of Poincaré group $\cal P$, by definition they have to satisfy, in addition to (1), ${\cal P} \ni g \mapsto U_g$ with $U_g U_h = U_{g\cdot h}$ ($^*$) and $U_e = id$, where $\cdot$ is the group product in $\cal P$, just in view of the definition of group representation. In principle each $U_g$ has to be unitary or anti unitary.

If $g\in \cal P$ belongs to the proper orthochronous subgroup ${\cal P}_+^\uparrow$, it can always be decomposed as $g= h\cdot h$ where $h$ still belongs to the same subgroup. Therefore $U_g= U_{h} U_{h}$, thus $U_g$ must be unitary (even if $U_h$ is anti unitary,as the composition of a pair of anti unitary operator is always unitary).

We conclude that the orthochronous Poincaré group ${\cal P}_+^\uparrow$ (and consequently the orthochronous Lorentz group $SO(1,3)^\uparrow$) can only be represented by unitary operators in quantum theories, when the action of the group is on states.

Non unitary representations arise dropping the last requirement. For instance dealing with Dirac or Weyl spinors.

($^*$) Actually a phase could take place, since states are represented by normalized vectors up to phases: $U_gU_h = e^{i\alpha(g,h)}U_{g\cdot h}$, however it does not change the result of the subsequent reasoning. As a matter of fact, it is possible to prove that continuous (projective) unitary representations of ${\cal P}_+^\uparrow$ are not affected by such phases, differently form representations of Galileo's group where those phases play a crucial role.

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Wonderfully explained (I like the bold to emphasize). Do you have any idea where I can find the Wigner theorem with a kind of easy proof? – Faq Jan 28 '14 at 11:40
Wigner theorem has a very technical proof with many cases unfortunately, I do not know easy proofs. You could try this paper, for a relatively recent attempt to simplify original Wigner's proof arXiv:0808.0779 – Valter Moretti Jan 28 '14 at 12:52
@LoveLearning: you can also make use of the fundamental theorem of projective geometry, but if you want to start from scratch, that just shifts the burden around – Christoph Jan 28 '14 at 13:09
Sure, that is a nice way, but it is not elementary: One should know the fundamental theorem of p.g. (and its proof). Here is a review on that theorem and Wigner theorem is proved as an application. Claude-Alain Faure: "An Elementary Proof of the Fundamental Theorem of Projective Geometry" Geometriae Dedicata, Vol. 90, Issue 1, pp 145-151, March 2002 – Valter Moretti Jan 28 '14 at 13:16
– Christoph Jan 28 '14 at 13:19