The h Bar

2:08 AM

The Pauli matrices and the identity matrix forms a basis for the 2x2 matrices..
The gamma matrices and certain combinations of them forms one for the 4x4 matrices..(the same one used to construct fermion bilinears)..

Do these nice sets of baes have a name? and how to construct them from scratch for any arbitrary dimension? I mean is there something like Gram-Schmidt orthogonalization for them?

and is there one for 3x3 ones(i seem not to find them)

Semiclassical

@ManasDogra the question you should probably be asking, if you're thinking Gram-Schmidt, is what inner product to use for said basis @ManasDogra

Manas Dogra

Haven't checked or thought..but won't ordinary matrix multiplication do?
And btw I am not wanting a GS process to be specific..anything..any way to produce those special looking n^2 basis matrices would do

Semiclassical

Gram-Schmidt requires an inner product

otherwise how are you going to compute components?

(and matrix multiplication by itself is not an inner product. inner products give numbers, not matrices)

the 3-by-3 case may be suggestive, though: en.wikipedia.org/wiki/Gell-Mann_matrices

(need to add in the identity matrix as well, or you'll only get traceless 3-by-3 hermitian matrices)

Manas Dogra

2:23 AM

@Semiclassical Then maybe defining some operation which takes two matrices and gives out a number would do..If I knew how to define the inner product and all I wouldn't be asking :(

But as it seems now, GS can't be used...Any other way?

Semiclassical

well, think about Pauli matrices for a bit. what do you know about products of them?

say, sigmaX * sigmaY

Manas Dogra

@Semiclassical commutator of two pauli matrices gives another pauli matrix(upto some constants infront)

@Semiclassical is the * matrix multiplication or direct product?

Semiclassical

the former

right. in fact, you don't even need the commutator to make that claim (b/c of pauli matrices anticommuting)

Manas Dogra

@Semiclassical Yes

Semiclassical

so products of Pauli matrices are still Pauli matrices, and in particular when multiplied by themselves they're the identity matrix

Now, what's something different about the identity matrix as compared to the three Pauli matrices?

they're all hermitian and 2-by-2

Manas Dogra

2:27 AM

@Semiclassical It's not traceless?

Semiclassical

right. and that's something we can take advantage of:

Manas Dogra

@Semiclassical I think I see a pattern---generators of SU(n) and identity matrix forms a basis for n*n matrices?

Semiclassical

yeah, or at least the Hermitian matrices

(which shouldn't really be a surprise: you get unitary matrices by exponentiating tracless Hermitian matrices)

Manas Dogra

OH YES!

Semiclassical

but to finish up here: if you compute trace(sigma(i) sigma(j)), then this equals 2 if i=j and 0 otherwise

so $\text{tr}(\sigma_i \sigma_j)=2\delta_{ij}$. which looks a lot like orthogonality relations

and, indeed, they are. hermitian matrices have the Frobenius inner product tr(AB)

and the Pauli matrices + identity constitute an orthogonal basis w/r/t to this inner product

Manas Dogra

2:33 AM

@Semiclassical Well this Frobenius product is something completely new...The rest was something I had somewhere inside my mind but couldn't apply well to understand the problem I was having...Thanks a lot for clearing it up.

Semiclassical

now, at this point I have to point out I was leading you on a little diversion. the above was to show that Hermitian matrices do indeed have an orthogonal basis and Gram-Schmidt does make sense. But in terms of identifying a basis for larger matrices, it's better to just straightaway imitate the Pauli matrices

and there's more than one basis, for better or worse

for this I'll use Dirac notation instead, denoting the basis vectors as $|1\rangle,|2\rangle,\ldots,|n\rangle$

analogous to the x-Pauli matrix, we have the matrices $|j\rangle \langle k|+|k\rangle\langle j|$ for $j< k$

Manas Dogra

@Semiclassical Yes I was going to that...the matrices used to make the basis for 4*4 matrices, the ones I mentioned--1,4 gamma matrices, 6 of their commutators and other combinations involving $\gamma_5$...these aren't the ones which one would get from generators of SU(4)..

Semiclassical

analogous to the y-Pauli matrix, we toss in $i$ on the first term and $-j$ on the second.

Manas Dogra

Hmm I understand now..but what about the gamma matrices basis I talked about in the earlier text..how to get that?

Semiclassical

and then there's a few ways to imitate the z-Pauli matrix. one is to do $|j\rangle\langle j+1|-|j+1\rangle \langle j|$ for $j=1,2,\ldots,n-1$

i'm less sure of how this works for the n=4 case tbh, mostly b/c the gamma matrices are somewhat special

they do not, for instance, constitute a basis by themselves of 4-by-4 complex matrices

Manas Dogra

2:42 AM

@Semiclassical Yes that's why we need the others

Semiclassical

right, c.f. physics.stackexchange.com/a/102078/55641

Manas Dogra

Yes I was talking of that group only--the one used to construct fermion bilinears

@Semiclassical And about this...where can i learn about this generalization of GS process? Will replacing the usual inner product with the Frobenius one simply do?

Semiclassical

in essence, yes, though of course you need to pick non-orthogonal 'vectors' to orthogonalize first

GS is a process for converting a non-orthogonal set of vectors into a basis, after all

if you're just looking for a basis for n-by-n hermitian matrices, here's a picture of one (slightly different than the one I said above):

i don't know how one picks out the gamma-matrix basis for complex 4-by-4 matrices, by contrast.

(the main difference with the picture's version vs. mine is that the picture doesn't bother trying to imitate the z-Pauli matrices, and instead just picks all the projectors onto basis vectors. this means that one doesn't need to include the identity matrix.)

incidentally, that basis goes back at least as far as von Neumann in the 1920's :)

Manas Dogra

@Semiclassical Yeah that's another problem.. Because choosing different set of non-orthogonal 'vectors' would give bases.

I could even start from the matrices in which one element is 1 and all other elements are 0.

Semiclassical

3:00 AM

the problem with that, for the Hermitian case, is that such matrices are in general not Hermitian

they're fine for complex matrices, but in that case you run into another issue

take all the matrices of the kind you describe, along with those with an $i$ instead (b/c complex)

then the natural inner product for n-by-n complex matrices is tr(A*B), i.e., Hermitian conjugate on the first one

and the bad news is that 2n^2 matrices I just described, are already an orthonormal basis for complex n-by-n matrices

so GS is done before it begins

so GS doesn't really help you towards gamma matrices

i don't think that's entirely surprising. gamma matrices have a bunch of spinor stuff going on, after all, and none of what i've said takes that into account

so gamma matrices are intrinsically fancier than Pauli matrices

@ManasDogra there are ways to generalize the gamma matrices, mind. see for instance en.wikipedia.org/wiki/Higher-dimensional_gamma_matrices

see also en.wikipedia.org/wiki/Weyl%E2%80%93Brauer_matrices

Manas Dogra

@Semiclassical Nice..I get it now..

@Semiclassical and now there's lot of stuff to read for now...thank you so much

I will drop out now--my college classes start (online) in ~10 minutes. Thanks again.

Semiclassical

/wave

bolbteppa

3:16 AM

@ManasDogra here's the generalization

In two dimensions, the two matrices $\{\sigma_x,\sigma_y \}$ generate a $\sum_{k=0}^{n=2} {n \choose k} = (1 + 1)^2 2^2 = 4$ dimensional Clifford algebra $\sigma_{\mu} = \{I,\sigma_x,\sigma_y, \sigma_x \sigma_y = i \sigma_z \}$. Treated as vectors, these 'vectors' (matrices) are orthogonal under the inner product $(A,B) = \mathrm{tr}(A B)$. Thus any $2 \times 2$ matrix with $2^2 = 4$ entries can be expressed as $A = a^{\mu} \sigma_{\mu}$ since $2 a^{\mu} = \mathrm{tr}(A \sigma^{\mu})$.

In four dimensions, the four matrices $\{\gamma_{\mu} \}$ generate a $\sum_{k=0}^{n=4} {n \choose k} = (1 + 1)^4 = 2^4 = 16$ dimensional Clifford algebra $\gamma_A \in \{I \gamma_{\mu}, \gamma_{\mu \nu} , \gamma_{\mu \nu \rho}, \gamma_{\mu \nu \rho \sigma} = \varepsilon_{\mu \nu \rho \sigma} \gamma_5 \}$, where there are ${4 \choose 0}$ $I$ elements, ${4 \choose 1} = 4$ $\gamma_{\mu}$ elements, ${4 \choose 2} = 6$ elements $\gamma_{\mu \nu} = \frac{1}{2}(\gamma_{\mu} \gamma_{\nu} - \gamma_{\nu} \gamma_{\mu})$ , ${4 \choose 3} = 4$ elements $\gamma_{\mu \nu \rho} = \frac{1}{3!}(\gamma_{\mu}…

It's sometimes called the Fierz decomposition of a product of two spinors, where the product of two spinors (thought of as 'column vectors') generate a matrix which can then be written as a linear combination of Clifford elements

Semiclassical

those three-index gamma matrices are interesting

i mean, i can see they're the same as the 'usual' others (the bit about gamma_5)

bolbteppa

I think there's an even deeper reason a general matrix breaks up into a sum of matrices we think of as 'scalars, vectors, anti-symmetric rank two tensors, anti-symmetric rank three tensors, etc...'

Semiclassical

that is suggestive

bolbteppa

The dimension needs to be $2^{n/2} \times 2^{n/2}$ first off I think, which is a restriction, so the count is $(1 + 1) = 2^n = \sum_{k=0}^n {n \choose k}$ which shows the count as a sum of quantities whose number of components are ${n \choose k}$ (anti-symmetric tensors) so that makes it make more sense, but there's still something deeper in here I think

Something like that

(Should have used $- i \sigma_x \sigma_y = \sigma_z$, and it's $(1+1)^2=2^2=4$ in the first line of the first post above)

More in this style in chapter 3 of this

Semiclassical

3:32 AM

what i find amusing, with that picture I linked earlier

is that that's a basis for matrices, and von Neumann wrote it down

but (at least in that paper) he didn't use the fact that it's an orthonormal basis

didn't even use the inner product at all

bolbteppa

Looks like those matrices come from decomposing a matrix $A_{ab}$ into the sum of a traceless symmetric matrix, an anti-symmetric matrix, and it's trace, and writing each of these in terms of it's natural basis (given by those three types of matrices)

Semiclassical

yup

it's the obvious basis for Hermitian matrices

analogous to Pauli X/Y along with the two Z-projectors

bolbteppa

$M_{ab} = [M_{(ab)} - \frac{1}{n} \mathrm{tr}(M) \delta_{ab}] + M_{[ab]} + \frac{1}{n} \mathrm{tr}(M) \delta_{ab} = (b_{\mu \nu} B_{\mu \nu} + c_{\mu \nu} C _{\mu \nu} + a_{\mu} A_{\mu})_{ab}$

Semiclassical

right

bolbteppa

Think that's right, yeah that's a good point about the similarity to Pauli

Semiclassical

3:43 AM

Main difference with something like, say, Gell-Mann matrices or standard Pauli matrices, is that the identity matrix isn't one of the basis vectors

so you can't drop that to get the traceless matrices

hence not so useful for SU(2)

bolbteppa

Maybe you can decompose those terms into the anti-symmetric Clifford matrices only in $2^{n/2}$ dimensions using e.g. the Levi-Civita tensor to dualize or something

Semiclassical

von Neumann's basis is more useful for stuff like "how do you figure out a density matrix based on expectation values"

indeed, that's what he used the basis to establish in the paper I'm thinking of (though in a way that seems unnecessarily coordinate-dependent)

nowadays I think you'd just appeal to the inner product on matrices and then invoke Riesz representation theorem

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