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 , 22,786 comments
1,470 users with positive rep
820 active unimported users
More ...

  How does one prove Fierz identities?

+ 6 like - 0 dislike
1665 views

Fierz identities are discussed in the wikipedia article:
http://en.wikipedia.org/wiki/Fierz_identity
but the article doesn't give any derivation. The article implies that they arise from the blade structure of a Clifford algebra.


Some notes to be deleted after we have a solution:

They seem to arise in QFT calculations and (perhaps as a result) they can always be put into "pure density matrix" form. That is, the spinors can be arranged to appear in an order that makes them natural for a trace operation.

For example, with $U,V,W$ as unspecified operators and $\theta,\psi$ as spinors, one might have:
$(\bar{\theta}U\theta)(\bar{\psi} W\theta)(\bar{\theta} V\psi)$
where the unbarred and barred spinors do not always match as they do in "$...\theta)(\bar{\theta}...$". But this can be rearranged to give:
$(\bar{\theta}U\theta)(\bar{\theta} V\psi)(\bar{\psi} W\theta)$
One can then turn the $\psi\bar{\psi}$ type elements into (almost) pure density matrices and treat the problem as one of computing traces:
$\textrm{tr} [ (\theta\bar{\theta}U\theta\bar{\theta} V\psi\bar{\psi} W ]$
And then one can use various properties such as the fact that with pure density matrices such as $|a\rangle\langle a|^2 = |a\rangle\langle a|$ one has that anything that begins and ends with such a thing has to be a complex multiple of the pure density matrix.

This would give the Fierz identities without explicitly looking at the blade structure of the Clifford algebra and that is very suspicious to me. I doubt it works.

On a previous question: Some Majorana fermion identities I intuitively convinced myself that Fierz identities could be established this way (and some very smart people seemed to agree) but when I tried to solve that problem this way I quickly was buried in unsuccessful calculations. Now it could be that that particular problem couldn't easily be so attacked, or my method was misapplied, etc.; I'm not sure.

This post imported from StackExchange Physics at 2014-05-20 08:37 (UCT), posted by SE-user Carl Brannen
asked Mar 5, 2011 in Theoretical Physics by Carl Brannen (240 points) [ no revision ]
Hi Carl. My answer below shows how it is the blade/matrix completion relations that are important. As it's a sunday arvo, I didn't spend too much time thinking about it. I'm sure that there's a more pure Clifford algebra approach. Is this the type of result you were expecting? It would be worthwhile reproducing the standard Fierz identities found in standard texts and the paper I linked to. I've also seen some SU(N) relations referred to as Fierz identities. Maybe we (or preferably someone that's not me) can add more details to the Wikipedia page?

This post imported from StackExchange Physics at 2014-05-20 08:37 (UCT), posted by SE-user Simon

1 Answer

+ 6 like - 0 dislike

I always thought of Fierz identities as a kind of completeness relation (*) for products of spinors. To use the bra-ket notation: $$|a\rangle\langle b| = \sum k_i \langle b|M_i|a\rangle M_i$$ for some convenient trace orthogonal basis.

To find the $k_i$ for the specific basis and space that you're working in, you multiply by some $M_j$ and take the trace: $$ tr(M_j |a\rangle\langle b|) = \langle b|M_j|a\rangle = \sum_i k_i tr( M_i M_j ) \langle b|M_i|a\rangle $$ since the basis is orthogonal with respect to the trace, we see that $k_i^{-1} = tr( M_i M_i )$.

This is then used to prove the Fierz identities as $$ \langle a|U|b\rangle \langle c|V|d\rangle = \sum_i\langle a|U (k_i \langle c|M_i|b\rangle M_i) V|d\rangle = \sum_i k_i \langle c|M_i|b\rangle\langle a|U M_i V|d\rangle $$

Thus you get your Fierz rearrangements. Note that depending on your definitions, if you're deriving the Fierz identities for anticommuting spinors, then you might have a sign factor in the first trace formula and definition of $k_i$ that I gave.

The standard Fierz identities are for 4-dimensional spinors, with the basis of gamma matrices $$ 1\,,\quad \gamma_\mu\,,\quad \Sigma_{\mu\nu}\;(\mu<\nu)\,,\quad \gamma_5\gamma_\mu\,,\quad \gamma_5 $$ which has $1 + 4 + 6 + 4 + 1 = 16$ elements, which is what you'd expect for 4*4 matrices. The basis elements are normally on both the left and right hand side of the Fierz identity. The traces for the above basis can be calculated from the Wikipedia Gamma matrix page.

In the previous question, Some Majorana fermion identities, I used 2-component notation to check the identities. This is convenient since the completeness relation is really simple. Using the conventions of Buchbinder and Kuzenko the spinor completeness relation is simply the decomposition into an antisymmetric and symmetric part: $$ \psi_\alpha \chi_\beta = \frac12\varepsilon_{\alpha\beta}\psi\chi - \frac12(\sigma^{ab})_{\alpha\beta}\psi\sigma_{ab}\chi \,, $$ and similarly for the dotted-spinors (complex conjugate representation). This means that you don't need a table of coefficients to do Fierz rearrangements, all the steps fit easily in your memory.

I'm not sure what the best reference for Fierz identities is. Maybe have a look at Generalized Fierz identities and references within.


Footnote (*)

You can also think of the completeness relation purely in terms of the matrices. This is an alternate approach to derive the Fierz identities and is the one used in Generalized Fierz identities.

This post imported from StackExchange Physics at 2014-05-20 08:37 (UCT), posted by SE-user Simon
answered Mar 6, 2011 by Simon (325 points) [ no revision ]

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$\hbar$ysi$\varnothing$sOverflow
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
...