Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  Different representations of the Lorentz algebra

+ 4 like - 0 dislike
1956 views

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:

$$M_{\mu\nu}=i(x_\mu\partial_\nu-x_\nu\partial_\mu)+S_{\mu\nu}$$

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?

This post imported from StackExchange Physics at 2014-06-21 21:36 (UCT), posted by SE-user jinawee
asked May 3, 2013 in Theoretical Physics by jinawee (120 points) [ no revision ]

1 Answer

+ 7 like - 0 dislike

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 $$ \Phi_a(x) \to \sum_b \left[ \exp \left( \frac{i}{2} \omega^{\mu\nu} \left( S_{\mu\nu} + M_{\mu\nu} \right) \right) \right]_{ab} \Phi_b(x) $$

This post imported from StackExchange Physics at 2014-06-21 21:36 (UCT), posted by SE-user Prahar
answered May 3, 2013 by prahar21 (545 points) [ no revision ]
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).

This post imported from StackExchange Physics at 2014-06-21 21:36 (UCT), posted by SE-user Neuneck
@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

This post imported from StackExchange Physics at 2014-06-21 21:36 (UCT), posted by SE-user ramanujan_dirac
@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.

This post imported from StackExchange Physics at 2014-06-21 21:36 (UCT), posted by SE-user ramanujan_dirac
@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.

This post imported from StackExchange Physics at 2014-06-21 21:36 (UCT), posted by SE-user Prahar

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOver$\varnothing$low
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...