I've found many definitions of Lorentz generators that satisfy the Lorentz algebra: $$[L_{\mu\nu},L_{\rho\sigma}]=i(\eta_{\mu\sigma}L_{\nu\rho}-\eta_{\mu\rho}L_{\nu\sigma}-\eta_{\nu\sigma}L_{\mu\rho}+\eta_{\nu\rho}L_{\mu\sigma}),$$ but I don't know the difference between them.

Firstly, there is the straightforward deduction evaluating the derivate of the Lorentz transformation at zero and multiplying it by $-i$. It's a very physical approach.

Another possibility is to define:

$$\left(J_{\mu\nu}\right)_{ab}=-i(\eta_{\mu a}\eta_{\nu b}-\eta_{\nu a}\eta_{\mu b})$$

This will hold for any dimension. I find it a bit confusing because we mix matrix indices with component indices.

We could also define:


Where $S_{\mu\nu}$ is Hermitian, conmutes with $M_{\mu\nu}$ and satisfies the Lorentz algebra. I think this way is more geometrical because we can see a Lorentz transformation as a rotation mixing space and time.

The two last options look quite similar to me.

Lastly, we could start with the gamma matrices $\gamma^\mu$, that obey the Clifford algebra: $$\{\gamma^{\mu},\gamma^{\nu}\}=2\eta^{\mu\nu}\mathbb{I}$$ (this is easy to prove in QFT using Dirac's and KG's equations). And define: $$S^{\mu\nu}=\frac{i}{4}[\gamma^{\mu},\gamma^{\nu}]$$

It seems that this is the most abstract definition. By the way, how are Clifford algebras used in QFT, besides gamma matrices (I know they are related to quaternions and octonions, but I never saw these applied to Physics)?

Are there any more possible definitions?

Which are the advantages and disadvantages of each?

Are some of them more fundamental and general than the others?


1 Answer 1


UPDATE - Answer edited to be consistent with the latest version of the question.

The different definitions you mentioned are NOT definitions. In fact, what you are describing are different representations of the Lorentz Algebra. Representation theory plays a very important role in physics.

As far as the Lie algebra are concerned, the generators $L_{\mu\nu}$ are simply some operators with some defined commutation properties.

The choices $L_{\mu\nu} = J_{\mu\nu}, S_{\mu\nu}$ and $M_{\mu\nu}$ are different realizations or representations of the same algebra. Here, I am defining \begin{align} \left( J_{\mu\nu} \right)_{ab} &= - i \left( \eta_{\mu a} \eta_{\nu b} - \eta_{\mu b} \eta_{\nu a} \right) \\ \left( S_{\mu\nu}\right)_{ab} &= \frac{i}{4} [ \gamma_\mu , \gamma_\nu ]_{ab} \\ M_{\mu\nu} &= i \left( x_\mu \partial_\nu + x_\nu \partial_\mu \right) \end{align} Another possible representation is the trivial one, where $L_{\mu\nu}=0$.

Why is it important to have these different representations?

In physics, one has several different fields (denoting particles). We know that these fields must transform in some way under the Lorentz group (among other things). The question then is, How do fields transform under the Lorentz group? The answer is simple. We pick different representations of the Lorentz algebra, and then define the fields to transform under that representation! For example

  1. Objects transforming under the trivial representation are called scalars.
  2. Objects transforming under $S_{\mu\nu}$ are called spinors.
  3. Objects transforming under $J_{\mu\nu}$ are called vectors.

One can come up with other representations as well, but these ones are the most common.

What about $M_{\mu\nu}$ you ask? The objects I described above are actually how NON-fields transform (for lack of a better term. I am simply referring to objects with no space-time dependence). On the other hand, in physics, we care about FIELDS. In order to describe these guys, one needs to define not only the transformation of their components but also the space time dependences. This is done by including the $M_{\mu\nu}$ representation to all the definitions described above. We then have

  1. Fields transforming under the trivial representation $L_{\mu\nu}= 0 + M_{\mu\nu}$ are called scalar fields.
  2. Fields transforming under $S_{\mu\nu} + M_{\mu\nu} $ are called spinor fields.
  3. Fields transforming under $J_{\mu\nu} + M_{\mu\nu}$ are called vector fields.

Mathematically, nothing makes these representations any more fundamental than the others. However, most of the particles in nature can be grouped into scalars (Higgs, pion), spinors (quarks, leptons) and vectors (photon, W-boson, Z-boson). Thus, the above representations are often all that one talks about.

As far as I know, Clifford Algebras are used only in constructing spinor representations of the Lorentz algebra. There maybe some obscure context in some other part of physics where this pops up, but I haven't seen it. Of course, I am no expert in all of physics, so don't take my word for it. Others might have a different perspective of this.

Finally, just to be explicit about how fields transform (as requested) I mention it here. A general field $\Phi_a(x)$ transforms under a Lorentz transformation as $$ \Phi_a(x) \to \sum_b \left[ \exp \left( \frac{i}{2} \omega^{\mu\nu} L_{\mu\nu} \right) \right]_{ab} \Phi_b(x) $$ where $L_{\mu\nu}$ is the representation corresponding to the type of field $\Phi_a(x)$ and $\omega^{\mu\nu}$ is the parameter of the Lorentz transformation. For example, if $\Phi_a(x)$ is a spinor, then $$ \Phi_a(x) \to \sum_b \left[ \exp \left( \frac{i}{2} \omega^{\mu\nu} \left( S_{\mu\nu} + M_{\mu\nu} \right) \right) \right]_{ab} \Phi_b(x) $$

  • $\begingroup$ Very nice explanation! Clifford algebras and Lie algebras of orthogonal groups (the Lorentz group is SO(3,1)) have a close relation. This is because Spin(n) is the double cover of SO(n). $\endgroup$
    – Neuneck
    May 6, 2013 at 7:14
  • $\begingroup$ @Prahar: Please could you also add the explicit formula for how fields transform? Is it just $\exp(i \omega_{\mu \nu}L^{\mu \nu})$ (What is the convention used for the constants multiplying the parameters in the exponent generally?) where $L_{\mu \nu }$ are the appropriate generators? Excellent answer BTW, it helps very much when stuff is summarized as simply as possible, and connected with more elementary stuff. +1 $\endgroup$
    – user7757
    Jun 20, 2014 at 11:59
  • $\begingroup$ @Prahar: Also how do you define the $S_{\mu \nu}$ in $M_{\mu \nu}$? I know the partial derivative part comes by taylor expanding the field. Also isn't there a typo in the question the $S^{\mu \nu}$ mentioned after the clifford algebra should be $\Sigma^{\mu \nu}$ according to your answer. $\endgroup$
    – user7757
    Jun 20, 2014 at 12:10
  • $\begingroup$ @ramanujan_dirac - I think the question was edited after I wrote the answer so some of the notation might be inconsistent. I'll edit it now. $\endgroup$
    – Prahar
    Jun 20, 2014 at 15:13

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