The h Bar

Manas Dogra

02:08

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

02:23

@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

02:27

@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

02:33

@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

02:42

@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

03:00

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

03:16

@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

03:32

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

03:43

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

Slereah

09:35

"Given a Lagarngian density L"

What's a lagarngian

More Anonymous

11:38

My head hurts ...

Slereah

that's what you get when you start thinking about the physical world

I'm surprised I don't get a bloody nose from just thinking about it

More Anonymous

I was actually thinking about the social constructs of the world

J.K. Rowling | ContraPoints

►

Is there any trans intellect? (not a youtube intellect)

ACuriousMind

12:25

@ManasDogra Focusing on the characteristic of Pauli or $\gamma$ matrices as bases for matrices misses kind of the point

the reason we care about the $\gamma$-matrices is because they define the Dirac spinor representation as their unique irrep

that in matrix form they form a basis for matrices is more of an accident

bolbteppa

It's not an accident, that's the point I was trying to make above

ACuriousMind

well, it works in d=2,4,8,16 etc, but not in others

when you look for "generalizations of $\gamma$-matrices to arbitrary dimensions" the characteristic that generalizes is not that they form a basis of matrices

bolbteppa

It works in the dimensions where your matrix is a direct product of two Weyl spinors at least

cOnnectOrTR 12

The no. Of field lines crossing a unit area placed normal to the field at a point is a measure of strength of electric field at that point. This means if we place a small planar element of area delta s normal to E at a point, the no of field lines crossing it is proportional to Edelta s.
How Edelta s. It should be only delta s.

bolbteppa

Yeah it would miss the point to think it generalizes to every dimension I think

ACuriousMind

12:33

abstractly, the Clifford algebra has some dimensions where it is isomorphic to the real matrices of that dimension, and some where it isn't

bolbteppa

One issue is that '$\gamma_5$' is not traceless in odd dimensions right

ACuriousMind

From O'Farrill's excellent primer:

it's the $\mathrm{Mat}_N(\mathbb{R})$ and $\mathrm{Mat}_N(\mathbb{C})$ case where you get them as the "basis of matrices"

cOnnectOrTR 12

@ACuriousMind can u answer?

ACuriousMind

@cOnnectOrTR12 no, I'm not interested in trying to discuss vague notions of "field lines"

the proper way to do EM is to do actual vector calculus

cOnnectOrTR 12

You mean faraday was duffer.

ACuriousMind

12:41

No, I just mean we have better tools nowadays

cOnnectOrTR 12

Ok! But in my course book it’s only defined this way and I am trying to understand it. Highschool level or in India 10+2 level

bolbteppa

13:01

$\mathbb{H}$ is not even a field, to say anything about the irrep you would have to invoke module theory :z

cOnnectOrTR 12

@ACuriousMind but to do the calculus you need to establish flux formula?

Maybe later!

bolbteppa

In the real case I'm not sure that the $s-t = 0,6$ mod $8$ thing affects representing an arbitrary (real) $2^{d/2} \times 2^{d/2}$ matrix for $d$ even as a sum of all the traceless $\gamma_A$'s in the Clifford algebra (e.g. $3 - 1 = 2$ in $d = 4$ Minkowski?), the signature doesn't seem to matter

ACuriousMind

@bolbteppa the products of the $\gamma$-matrices change depending on the signature (their anticommutator depends on it, after all)

note that in the $\mathbb{H}$ case, you can't really start with "matrices" unless you want quaternionic entries

bolbteppa

Yeah but I don't think the signs in the metric matter for this point, the anti-symmetric combinations still form an orthogonal basis, they are traceless (in even $d$) and orthogonal w.r.t. the trace, that's all you need for the argument I think

I can vaguely think of a non-absurd example where you'd want to work with a Clifford algebra over the quaternion's, but they can be avoided I think

Semiclassical

14:42

D***, John Baez knows everything: link.springer.com/article/10.1007%2FBF00955201

And he knew it before I was even born!

Ishwaran

14:54

Hi !, Been a month joining SE. we can clarify questions here in the chat room too ?. or chat room is meant for something else ?

ACuriousMind

15:20

@Ishwaran the chat room is to chat about whatever you want

You can ask questions here, but if they could actually be main site questions it's better to post them on the main site! Chat is better for more open-ended or opinion-based questions

if you want to discuss exercises in particular, there's a dedicated problem solving strategies room for that

Ishwaran

15:50

@ACuriousMind Thank you for your guidance !

bolbteppa

Is $S' = \int d^2 \sigma' \eta'^{\alpha \beta} \partial_{\alpha'} X^{\mu} \partial_{\beta'} X_{\mu} = \int (d^2 \sigma J) (\frac{\partial \sigma^{\gamma}}{\partial \sigma'^{\alpha}} \Lambda \eta^{\alpha \beta} \frac{\partial \sigma^{\delta}}{\partial \sigma'^{\beta}}) \partial_{\gamma} X^{\mu} \partial_{\delta} X_{\mu} = \int d^2 \sigma \eta^{\gamma \delta} \partial_{\gamma} X^{\mu} \partial_{\delta} X_{\mu}$ how the residual reparameterization and Weyl symmetry go 'globally'

Then if $\frac{\partial \sigma^{\gamma}}{\partial \sigma'^{\alpha}} \eta^{\alpha \beta} \frac{\partial \sigma^{\delta}}{\partial \sigma'^{\beta}} = \rho \eta^{\gamma \delta}$ we have $\Lambda = (\rho J)^{-1}$ so you get conformal transformations, but the primes might be off or something

Slereah

16:32

Remember to put fancy mathjax if you're gonna put in a big equation like that

bolbteppa

Is that the double $

Slereah

double $ is better than one, but for something that big I will even recommend eqnarray

\begin{eqnarray}
S' &=& \int d^2 \sigma' \eta'^{\alpha \beta} \partial_{\alpha'} X^{\mu} \partial_{\beta'} X_{\mu} \\
&=& \int (d^2 \sigma J) (\frac{\partial \sigma^{\gamma}}{\partial \sigma'^{\alpha}} \Lambda \eta^{\alpha \beta} \frac{\partial \sigma^{\delta}}{\partial \sigma'^{\beta}}) \partial_{\gamma} X^{\mu} \partial_{\delta} X_{\mu}\\
&=& \int d^2 \sigma \eta^{\gamma \delta} \partial_{\gamma} X^{\mu} \partial_{\delta} X_{\mu}
\end{eqnarray}

PM 2Ring

18:29

Would this question be acceptable if it asked how to calculate what would happen to the iron sphere? In particular, how fast would it collapse, and how much of the gravitational PE would be converted to heat. I have no idea how to calculate the rate of collapse. And even if I had that info, I don't know how to calculate the heat because I don't know how much energy would be stored as stress in the resulting non-uniform density sphere.

FWIW, a uniform density sphere of iron with the volume of the Sun would have a mass around 5.6 solar masses, maybe add ~1% if the sphere is at CMB temperature.

Manas Dogra

19:51

@ACuriousMind Hmm...I like @bolbteppa's approach although which is not exactly what I was looking for but quite general

I am wondering how these great theoretical physicists almost always gets it right..
Beta decay spin problem? Introduce a particle...and yes indeed there is a particle not something like "spin is not conserved here".

Then there's two different kind of kaons---so introduce some more particles??? NO..."parity is violated here"--- just like that..and yeah wu comes along to say "yes" to yang-lee's theories.

They could have said the other way round---say spin is not conserved and maybe then introduce a particle for parity violation instead but they did JUST the right thing as if they somehow knew the experiments before...and got nobel prizes...i mean--how everytime? :)

Slereah

20:32

That's because you've been fooled

People don't remember the thousands of theories that were wrong

Manas Dogra

@Slereah sed lyf :(

Slereah

if you go back far enough on any topic you can see the myriad of papers on weird theories that didn't go anywhere

I mean you know, usually they're right on some level, because otherwise they wouldn't publish it

but they do not describe physical reality

Manas Dogra

Yeah I heard that there were a lot of other theories of gravity after Newton and before einstein came along

So there MAY exist some percentage of sheer luck and accident even in these kind of discoveries..

Slereah

20:52

There's a ton

Gravity as pressure from a fluid

Gravity with different dropoff exponent

Gravity with extra terms

Gravity from imbalance of electromagnetism

etc etc

There's a bunch of weird experiments that were done to rule out some of them

Like "gravity as a fluid" was tested by trying to see if you could shield gravitational effects

also a lot of principle of equivalence testing because that one is really weird

Manas Dogra

Is there a one place where all of these are assimilated?
Or we gotta do some diggin'

Slereah

There's a few books on the topic

History of science books

Avantgarde

Gravity as an entropic force

Slereah

that one is more recent I think

Avantgarde

yeah I know

Nothing can beat GR as of yet. It's incredibly hard.

Slereah

21:06

Reading actual "breakthrough" papers is kind of a chore

all the papers just when a phenomenon gets discovered and nobody knows what that's about

Avantgarde

you mean like Einstein's papers?

Slereah

there's a lot of random ideas going on

Partly, yeah

Avantgarde

yeah I can imagine

Slereah

Although special relativity stuff goes back to like 1896

with the Michelson-Morley experiment

Tons of papers trying to explain the discrepency

Avantgarde

SR is nice too

idk where hep-th is going. Seems dormant as far as breakthroughs are concerned

Slereah

21:10

Not much going on rn unless some marginally unlikely experiment goes off

Avantgarde

The only thing that I think is quite interesting is color-kinematics duality

Slereah

there's always plenty of experiments that rely on very unlikely events

Just monitoring cosmic rays and stuff like that

sometimes you get weird results

Avantgarde

the next decades belong to GW astronomy. Take this up, future grad students. Don't waste your hopes on quantum gravity

Slereah

also nice, but unlikely to reveal anything too wild I think

I mean maybe something wild, but more in the realm of cosmology or astrophysics

not so much GR itself

Avantgarde

I have a similar feeling. Everything is consistent with GR. Not sure when we'll see the first deviation in any expt

Slereah

21:13

Universe found to be 13.657 billion years old instead of 13.656

Avantgarde

lol

Slereah

I mean it's still a million years

Pretty long

Avantgarde

for humans

Slereah

I'm a human

Avantgarde

21:25

We lost some giants in the last few years

Dyson, Gell-Mann, Weinberg, Hawking

Slereah

23:59

Gell Mann famously said "If I have seen further than others, it is because I am surrounded by dwarfs"

A true giant

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