# Dirac Gamma Matrices

Hi everyone,

From the condition:

$$\gamma_{\mu}\gamma_{\nu}+\gamma_{\nu}\gamma_{\mu} = 2g_{\mu\nu}$$​

how does one formally proceed to show that the objects $$\gamma_{\mu}$$ must be 4x4 matrices? I unfortunately know very little about Clifford algebras, and for this special relativity project of mine I'd much rather not need brute force!

Cheerio!

Kane

PS: I'm using the signature (+---) for the metric tensor, although this should only change the content of the matrices, not the proof itself, I suspect.

PPS: I quite realize that it probably cannot be shown that the gamma matrices *must* be 4x4 matrices, what I want to know is if the anticommutator conditions are precisely the defining relations for a R(1,3) Clifford algebra or something like that, and how we eliminate the possibility of lower-dimensional 'isomorphisms' (don't know the correct algebra mapping term) existing.

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Kane O'Donnell said:
Hi everyone,

From the condition:

$$\gamma_{\mu}\gamma_{\nu}+\gamma_{\nu}\gamma_{\mu} = 2g_{\mu\nu}$$​

how does one formally proceed to show that the objects $$\gamma_{\mu}$$ must be 4x4 matrices? I unfortunately know very little about Clifford algebras, and for this special relativity project of mine I'd much rather not need brute force!

Cheerio!

Kane

PS: I'm using the signature (+---) for the metric tensor, although this should only change the content of the matrices, not the proof itself, I suspect.

PPS: I quite realize that it probably cannot be shown that the gamma matrices *must* be 4x4 matrices, what I want to know is if the anticommutator conditions are precisely the defining relations for a R(1,3) Clifford algebra or something like that, and how we eliminate the possibility of lower-dimensional 'isomorphisms' (don't know the correct algebra mapping term) existing.

I don't know any sophisticated demonstration with a lot of jargon, but I know the simple, dumb approach.

From $\gamma_0^2 =1$ one sees that the eigenvalues are $\pm 1$ (and $\pm i$ for the other gamma matrices). Also it's easy to show from the anticommutation relations that the matrices must be traceless. From those two conditions, a representation must be even dimensional. Since we need 4 linearly independent matrices, 2 dimensions is not enough (there's only the 3 Pauli matrices available). So the next possibility is 4 dimensions.

Pat

pat_connell
I could be wrong here, forgive me if I am. I haven't really gotten around to calculating specifically Dirac's gamma matrices. however, note that the metric tensor g is symmetric under interchange of indices, maybe proceed from there.

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No it's fine , I am really just trying to understand how to justify the selections step by step other than just guessing the things.

Thanks,

Kane

I get all of it now except why the matrices must be of even dimension. I've read on the net about a hundred times that it is because the eigenvalues are so and so and the matrices are traceless, but I can't see it - why can't you have a traceless odd-dimension square matrix with eigenvalues of say plus or minus 1?

Cheers,

Kane

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You can find a good proof in the case D=4 in the article written by Wolfgang Pauli:"Contributions mathématiques à la théorie des matrices de Dirac"Ann.Inst.Henri Poincaré 6,109-136(1936).You can find this article in the book :"Wolfgang Pauli Collected Scientific Papers" edited by R.Kronig & V.F.Weisskopf,Interscience Publishers,a division of John Wiley & Sons,Inc.,1964,volume 2,page 753.

He uses $$\displaystyle{x_{4} = x^4 = ict}$$ so be careful with the transcription to coordinates with $$\displaystyle{\eta_{\mu\nu} = diag (+1\-1\-1\-1)}$$.

Good luck in all!

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Kane O'Donnell said:
I get all of it now except why the matrices must be of even dimension. I've read on the net about a hundred times that it is because the eigenvalues are so and so and the matrices are traceless, but I can't see it - why can't you have a traceless odd-dimension square matrix with eigenvalues of say plus or minus 1?

Cheers,

Kane

Umm, so the eigenvalues are all $$\pm 1 and \pm i$$. We assume the matrices are of full rank (to check the case dim=n) and that they are diagonal in the eigenvalue basis. So the trace is the sum of the elements on the diagonal, i.e. the eigenvalues. Now if there are an even number of elements, you can arrange the 1's and i's to cancel out, but in an odd number of dimensions, you can't, there will always be an unmatched term, and the trace in an odd dimension cannot therefore be 0.

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Kane O'Donnell said:
I get all of it now except why the matrices must be of even dimension. I've read on the net about a hundred times that it is because the eigenvalues are so and so and the matrices are traceless, but I can't see it - why can't you have a traceless odd-dimension square matrix with eigenvalues of say plus or minus 1?

Cheers,

Kane

Because (as SelfAdjoint already mentioned) if you go to a basis where the gamma are diagonal, the trace is the sum of the eigenvalues. If the eigenvalues are $\pm 1$ and the sum of the eigenvalues is 0, therefore...

Pat

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nrqed said:
Because (as SelfAdjoint already mentioned) if you go to a basis where the gamma are diagonal, the trace is the sum of the eigenvalues. If the eigenvalues are $\pm 1$ and the sum of the eigenvalues is 0, therefore...

That's the simplest explanation,but it's got so much mathematics in it.In order to go to a basis in which those $$\gamma$$ matrices are diagonal,you've got to use a theorem which enables passing from one to another irreductible representation of the Clifford algebra.This theorem (proved in many QFT books for the case D=4;e.g.Jauch,Rohrlich,Appendix A2,but also the article by Wolfgang Pauli (see above))
states that if $$\gamma_{\mu}$$ and $$\gamma_{\mu}\prime$$ are 2 irreductible representations of the Clifford algebra (and hence satisfy the anticommutation relations),then there is a NONSINGULAR MATRIX "S" such that
$$\gamma_{\mu}\prime = S\gamma_{\mu}S^-1$$,and that this matrix is unique,except for an arbitrary multiplicative factor.
To quote Jauch,Rohrlich,Appendix A2:
"The proof of the main theorem is greatly facilitated by the powerful lemma of Schur (I.Schur,<<Neue Begruendung der Theorie der Gruppencharaktere",Sitzungsber.Preuss.Akad.,1905,p.406) which,for our purpose,may be formulated as follows:Let $$\gamma_{r}$$ and $$\gamma_{r}\prime$$ two irreductible representations of degree n,n' ($$n\leq n\prime$$) and let S be a matrix with n' rows and n columns which connects the two representations by
$$\gamma_{r}\prime S = S\gamma_{r}$$.Then S is either the null matrix (the matrix which consists only of zeros) or it is nonsingular.In the latter case,n=n'"
And then a demonstration of Schur's lemma is given.
It's also the Schur's lemma that enebles us to prove Burnside's theorem:"The matrices of an irreductible n-dimensional representation of any group contain $$n^2$$ LINIARLY INDEPENDENT MATRICES".I quoted from Francis D.Murnaghan's book:" The theory of group representations",The John Hopkins Press,Baltimore,1938,p51 (for a reference to Schur's lemma,v p.47).

To conclude:irreductible matrix representations of the Clifford algebra constructed as the liniar space of complex n*n matrices together with the anticommutation relation cannot have odd number of lines and columns.It follows that n can be only even.For n=2 you find the Pauli matrices+unit matrix.For n=4,you have the Dirac matrices,etc.
The anticommutation relation enebles us to find EXACTLY 16 liniarly independent elements of the Clifford algebra,and hence,using Burnsides theorem to find that for Clifford algebra given by $$\gamma_{(\mu} \gamma_{(\nu} = 2g_{\mu\nu} I_n$$ "n" MUST be 4.

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sorry

dextercioby said:
... by $$\gamma_{(\mu} \gamma_{(\nu} = 2g_{\mu\nu} I_n$$ "n" MUST be 4.
that should have been of course:
$$\gamma_{(\mu} \gamma_{\nu)} = 2g_{\mu\nu} I_n$$

I'm still a novice in editing TEX :tongue2:

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The proof of the main theorem is greatly facilitated by the powerful lemma of Schur (I.Schur,<<Neue Begruendung der Theorie der Gruppencharaktere",Sitzungsber.Preuss.Akad.,1905,p.406)

Sure, Shur. Shur's lemma is ordinarily taught as a part of modern undergraduate courses in linear algebra. Not always proved, but stated and used.

That's the simplest explanation,but it's got so much mathematics in it.In order to go to a basis in which those matrices are diagonal,you've got to use a theorem which enables passing from one to another irreductible representation of the Clifford algebra.

Thankyou to everyone that replied! This quote above is precisely what I was after, as I realized that if we could pass to a diagonal matrix we wouldn't have a problem - the problem of course is that since the gamma matrices are not self-adjoint they don't have a basis of eigenvectors and hence there was (to my knowledge) no diagonal representation of each of the gammas as linear transformations. If there *is* a way to pass to a diagonal form, then my issue is resolved

Thanks again,

Kane

Ok. I have been looking at this problem quite a bit in order to find a good balance between simplicity and rigour in making this argument. So, my question is - is the following correct?

1. The operators $$\gamma_{\mu}$$ together with their anticommuting property have the structure of a Clifford algebra. We assume the Clifford algebra is finite dimensional and over some complex space (can't justify it, don't know enough about Clifford algebras - help?). Hence it is isomorphic to a matrix algebra over R, C or H, and we have a matrix representation. For each gamma operator it's matrix must be traceless (sum of eigenvalues is zero).

2. From the four operators you can (by multiplication) construct exactly 16 operators. Using the traceless property one can show the 16 operators are linearly independent. As such the dimension of the Clifford algebra is 16. This implies the underlying vector space has dimension 4 (16 = 2^4). Therefore our Clifford algebra is over a 4D vector space and (since n is even) is isomorphic to the 4x4 complex matrix algebra.

The reason I've taken this approach is that I still can't justify to myself that the gammas are diagonalisable directly, hence I can't use the simple trace/eigenvalue pair argument.

Kane

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Kane O'Donnell said:
2. From the four operators you can (by multiplication) construct exactly 16 operators. Using the traceless property one can show the 16 operators are linearly independent. As such the dimension of the Clifford algebra is 16. This implies the underlying vector space has dimension 4 (16 = 2^4). Kane

Actually Burnside's theorem implies:$$4=\sqrt{16}$$.Of course,u don't need Bunside's theorem to tell u that $$4=\sqrt{16}$$,it's just that it makes the connection between the dimension of an irreductible representation and the number of (linearly independent) generators of the Clifford algebra.

Kane O'Donnell said:
Therefore our Clifford algebra is over a 4D vector space and (since n is even) is isomorphic to the 4x4 complex matrix algebra.

The reason I've taken this approach is that I still can't justify to myself that the gammas are diagonalisable directly, hence I can't use the simple trace/eigenvalue pair argument.

Kane