11
$\begingroup$

The lorentz transform for spinors is not unitary, that is $S(\Lambda)^{\dagger}\neq S(\Lambda)^{-1}$. I understand that this is because it is impossible to choose a representation of the Clifford Algebra where all the $\gamma$ matrices are Hermitian.

However, does this not go against the conventional wisdom of Wigner's theorem which says that symmetry operations are needed to be either unitary of anti-unitary for the invariance of observable properties across frames? How can one reconcile this with the transformation above?

$\endgroup$
1
6
$\begingroup$

The most famous theorem by Wigner states that, in a complex Hilbert space $H$, every bijective map sending rays into rays (a ray is a unit vector up to a phase) and preserving the transition probabilities is represented (up to a phase) by a unitary or antiunitary (depending on the initial map if $\dim H>1$) map in $H$.

Dealing with spinors $\Psi \in \mathbb C^4$, $H= \mathbb C^4$ and there is no Hilbert space product (positive sesquilinear form) such that the transition probabilities are preserved under the action of $S(\Lambda)$, so Wigner theorem does not enter the game.

Furthermore $S$ deals with a finite dimensional Hilbert space $\mathbb C^4$ and it is possible to prove that in finite-dimensional Hilbert spaces no non-trivial unitary representation exists for a non-compact connected semisimple Lie group that does not include proper non-trivial closed normal subgroups. The orthochronous proper Lorentz group has this property. An easy argument extends the negative result to its universal covering $SL(2, \mathbb C)$.

Non-trivial unitary representations of $SL(2,\mathbb C)$ are necessarily infinite dimensional. One of the most elementary case is described by the Hilbert space $L^2(\mathbb R^3, dk)\otimes \mathbb C^4$ where the infinite-dimensional factor $L^2(\mathbb R^3, dk)$ shows up.

This representation is the building block for constructing other representations and in particular the Fock space of Dirac quantum field.

$\endgroup$
1
  • $\begingroup$ Nice post, Valter. Kurze Frage: do you know of a proof (book, article) for the this very dense statement: "non-trivial unitary representations do not exist for a non-compact connected Lie group that does not include proper non-trivial closed normal subgroups"? $\endgroup$
    – DanielC
    Sep 21 '17 at 8:23
7
$\begingroup$

This is a common misconception.

Lorentz group $SO(3,1)$ (or its double-cover $SL(2,\mathbb{C})$ if you want to follow Wigner's analysis of symmetries in QM) is non-compact. This means that it has no finite-dimensional unitary representations (the inability to choose hermitian Clifford basis elements is just a consequence of this).

When you're constructing classical Dirac spinors, you don't need unitarity. Indeed, there's no need for $S(\Lambda)$ to be a unitary representation. We are dealing with a classical field here, and unitarity is not required in classical physics.

In QFT we are dealing with a quantum Dirac field. The state space (fermionic Fock space) of the free QFT is infinite-dimensional and unitary. This doesn't contradict the original claim precisely because of infinite dimensionality.

Classification of the infinite-dimensional unitary irreps of the Poincare group (inhomogenous Lorentz group if you wish) was done by Wigner. It uses finite-dimensional nonunitary representation theory of $SL(2,\mathbb{C})$ (which is equivalent to the finite-dimensional representation theory of the complexified Lie algebra $\mathfrak{so}(4)\sim\mathfrak{D}_2$) heavily.

Summarizing: the difference is that Poincare generators of the QFT act unitary on the infinite-dimensional Fock space. The classical transformation of the spinor field is respected by this action, but does not have to be and is not unitary.

$\endgroup$
7
  • $\begingroup$ What does it mean for a state space to be unitary? Do you have some source where I can delve into all this? $\endgroup$ Sep 21 '17 at 4:42
  • 1
    $\begingroup$ @ArnabBarmanRay It means that there's a representation of the (10-dimensional) Poincare group on the fermionic Fock space of the free QFT, with all connected-to-identity elements of the group represented as unitary operators. Thus there's no conflict with the fundamental principle of QM which is: all symmetry transformations can be represented by either unitary or anti-unitary operators. The infinitesimal form of this claim is: all 10 generators of the Poincare group (including the Hamiltonian, the 3-momenta operator and 6 components of spacetime angular momentum) are hermitian operators. $\endgroup$ Sep 21 '17 at 5:09
  • 1
    $\begingroup$ It is false that a non-compact Lie group does not admit non-trivial finite-dimensional unitary representations. Take the additive non-compact Lie group $\mathbb R$ and any $n$-dimensional Hermitian matrix $A$. $ \mathbb R \ni r \mapsto e^{irA}$ is a non-trivial unitary continuous $n$-dimesional representation of the group. The example extends to direct products of $\mathbb R^n$ and compact lie groups immediately... $\endgroup$ Sep 21 '17 at 8:42
  • $\begingroup$ The impossibility depends on more precise requirements on the non-compact Lie group, and the representation has to be continuous. Sufficient conditions for non-existence of finite-dimensional unitary continuous representations of a connected non-compact Lie group are that the group does not contain a closed normal group different form the identity and the group itself. Notice that the normal group is not required to be connected. $\endgroup$ Sep 21 '17 at 8:46
  • 1
    $\begingroup$ @ AccidentalFourierTransform there are many procedures leading to the original Wigner classification. The idea was improved by Bargmann and, greately, by Mackey with the theory of induced representations as consequence of his imprimitivity theory. Both $SL(2,C)$ and $SO(3)$ ($SU(2)$) enter the procedure. The latter is the so-called little group of the former when classifying the orbits... $\endgroup$ Sep 21 '17 at 12:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.