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2:43 AM
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Q: Why there's no tag of quantum chaos in physics.stackexchange.com?

MathematicalPhysicistalso in mathstackexchange there's no tag for quantum chaos, why is that? Thanks!

 
 
8 hours later…
10:59 AM
Supersymmetry algebras do not in general (or ever) contain the conformal algebra as a subalgebra do they?
Supersymmetry is definitely something different to CFT, not an extension in the same way CFT and SUSY are extensions of the Poincaré algebra
Oh there is something called a superconformal algebra
 
@Charlie sure, you can combine the two
note that except in 2d, this isn't anything new, since the conformal algebra in $(p,q)$ is just the Lorentz algebra in $(p+1,q+1)$
 
Is 2D SUSY special like 2D CFT?
Also just because maybe the wording so far has been unclear, the conformal algebra of $(p+q)D)$ spacetime is unique (just being the algebra of the group of transformations that locally rescale the metric), is the SUSY algebra of spacetime unique?
 
no, just because the CFT algebra in 2D is the Virasoro algebra, you get a "super-Virasoro" algebra for sCFT
 
ok, I'm not sure if my course touches scft
nvm it does not
 
@Charlie the SUSY extensions of Lorentz are never unique, you can have "copies" of the minimal supersymmetry, that's what the $\mathcal{N} = ...$ means you often see in SUSY contexts
 
11:12 AM
Ah, that explains the phrase "one can construct so called N-extended algebras"
 
in 2d, you get also a different kind of choice between Ramond and Neveu-Schwarz SUSY, corresponding to the similarly named boundary conditions of the superstring
 
I see
 
11:39 AM
welp, it's snowing again after we had at least 15°C last week
 
12:06 PM
There aren't that many SUSY extensions, IIRC
From what I remember there's only like 5 different SUSY algebras in all, although they come with parameters
 
 
1 hour later…
1:26 PM
whoo :-)
arnold is just 10 answers from his, so he should be getting it in no time
btw does anyone have any idea what's going on here:
@JohnRennie Silly question, but why doesn't the first curve pass through $(a,t)=(1,1)$? Isn't the standard normalization $a(t_0)=1$? — AccidentalFourierTransform Mar 16 at 18:00
 
1:55 PM
@AccidentalFourierTransform congrats
 
2:19 PM
AAAAAAAAAAAAAAAAAAAAA
sorry this last year i became a germanphobe
 
2:30 PM
hm?
 
2:41 PM
Oh, I get it. Awful pun, well done :P
 
3:17 PM
@AccidentalFourierTransform that last guy, lubos motl
I have heard his name in context of string theory..
it seems he was(is?) quite active on this site, damn
 
He used to be active, he's pretty renowned for being a big advocate of string theory as a complete theory of physics
 
I think these kind of people get a lot of hate?
 
@satan29 you can see for yourself here for example
 
well, Lubos is more (in)famous for his views not related to technical points of string theory. Google him and/or read his blog if you want to see why that is :P
 
In theoretical physics, the superconformal algebra is a graded Lie algebra or superalgebra that combines the conformal algebra and supersymmetry. In two dimensions, the superconformal algebra is infinite-dimensional. In higher dimensions, superconformal algebras are finite-dimensional and generate the superconformal group (in two Euclidean dimensions, the Lie superalgebra does not generate any Lie supergroup). == Superconformal algebra in dimension greater than 2 == The conformal group of the ( p + q ) {\displaystyle (p+q...
@Charlie it's what you get if instead of extending the Poincare algebra you extend the conformal algebra
 
3:33 PM
Ok that seems reasonable
Something that was/is surprising to me is that the supersymmetry generator $Q^\alpha$ has a spinor index? If I think of for instance the Poincaré algebra the generators have "4-vector" indices which I assumed was just a way of indexing them, it's strange to me that $Q$ has a different type of index despite just being a generator of the algebra
 
Coincidentally one of the nicest introductions to the superconformal group is in one of Kaku's (the guy in the other link driving people crazy) books
 
Because now you have in the supersymmetry algebra objects with 4-vector indices and objects with spinor indices, despite this all being at the Lie algebra level. Presumably this has something to do with a canonical representation or something
In fact no I am definitely confused about that, because $Q^\alpha$ should have in each entry a vector, so it's actually very confusing that we would have a spinor index there
 
One way to think about it is that the Poincare group is the Lorentz group and translations, spinors are representations of the Lorentz group, well what about spin analogues of translations? A vector in 4D decomposes into left and right spinors, so it's natural to consider an algebra that includes spinor analogues of translations
 
@Charlie the odd generators are anti-commuting (=fermionic), so why is it surprising they're spinors?
 
hmm
 
3:40 PM
it's exactly what you would expect in light of spin-statistics
 
But the Poincaré subalgebra generators are real vectors?
Or is that basically the point, that the supersymmetry generators have both
 
well, the translation generator $P^\mu$ is a vector. The generalized rotation/boost generator $M^{\mu\nu}$ is a 2-tensor. And now you also have the odd $Q^\alpha$, which are spinors
 
I'm a bit troubled now by the fact that I thought $M^{\mu\nu}$ was a sort of "vector valued matrix", since each component is a generator of the Lorentz/Poincaré algebra etc.
 
I think we're confusing values in representations and the abstract objects here
 
I definitely prefer separating my thinking about the abstract elements of the Lie algebra and anything in which we are implicitly thinking in a specific representation
 
3:47 PM
The abstract Poincaré algebra is generated by $M^{\mu\nu}, P^\mu$. If you let a Lorentz transformation act on these (the group can act on its adjoint after all), then they transform exactly as the indices indicate.
 
Hmm I've not really encountered the adjoint rep of the Lorentz group tbh
 
In any concrete representation, each of the $M^{\mu\nu}$ is a matrix itself, i.e. $\rho(M)$ is a "matrix of matrices"
 
ok yeah that I basically what I meant, i.e. $M^{01}$ is a vector in the Lie algebra
 
I don't know what "vector in the Lie algebra" means
 
I mean an element in the basis of the lie algebra maybe
 
3:50 PM
@Charlie it's just the representation on anti-symmetric 2-tensors, i.e. precisely the representation signified by the two indices on $M$
 
I'm not crazy right, each component of $M^{\mu\nu}$ before we've chosen a particular representation is just a basis vector in the Lorentz/poincare algebra
 
@Charlie Yes
And the two indices tell you that under some Lorentz transformation $\Lambda$, the result of the action of $\Lambda$ on some $M^{\mu\nu}$ will be $\Lambda^\mu_\sigma\Lambda^\nu_\rho M^{\sigma\rho}$
 
Actually I guess I would be happy if the reasoning was that the supersymmetry algebra is complex and the Poincaré algebra is a real Lie subalgebra, hence the supersymmetry generators $Q^\alpha$ are given spinor indices because they span a complex space
@ACuriousMind are we in the adjoint representation here?
 
Are you asking about why $Q^{\alpha}$ has a spinor index $\alpha$ when everything else seems to have $\mu,\nu$ indices
 
@Charlie Yes, this is the adjoint action of the Poincaré group on its algebra
 
3:53 PM
Yes that's exactly it
 
The $M$ carries two indices and the $P$ carries one index precisely because that's the correct index structure two show how the adjoint action of the group works on these objects
 
ok, my bad if I'm going in circles but ever since I first learned that in physics we usually implicitly talk in a representation it's converted me to being a purist wrt. separating the abstract stuff and the representation stuff out
 
exactly analogously, the $Q$ in the "super-case" carry spinor indices because that's the correct index structure two show how the adjoint action works on them
 
Ah it relates specifically to the adjoint rep, that is a satisfying answer I think, I'll have to go think it over for a bit
 
@Charlie the adjoint representation is special - it's the natural action of the group on its algebra - it's not a random choice
If you accept the $\mu$ on the $P^\mu$, then the $\alpha$ on the $Q^\alpha$ is really no different.
 
3:56 PM
Ok that is actually very generally useful information, because I was always slightly bothered that we use those indices on what I thought were entirely abstract things
Am I right to say that $M^{\mu\nu}$ isn't actually an element of the Poincaré algebra it's an element of the exterior algebra of the Poincaré algebra?
ok no that doesn't sound right actually
I guess I'm just whining about the fact that it's a matrix valued matrix
 
for each fixed $\mu\nu$, $M^{\mu\nu}$ is just an element of the algebra
there isn't really a space where the "full" $M$ would live
 
@ACuriousMind Ok, I was a bit thrown off by this which makes $M$ look like it lives in some higher tensor space
 
 
2 hours later…
5:39 PM
Is there version of the Wightman axiom that links the finite dimensional lorentz rep with the infinite dimensional unitary rep on the Fock space for other representations like conformal or super symmetry too?
Or are CF and supersymmetry only realised on the Fock space and not on the classical fields
 
 
3 hours later…
8:39 PM
@Charlie susy and cft are of course realized via fields if that's what you mean
 
8:54 PM
you can do CF and SUSY with classical fields, if you are mad
it's not too useful as far as I know but nothing is stopping you
Although you know
EM is conformally invariant in 4D
So there's that
 
9:06 PM
@Charlie Yes, the representation on the target space of the fields always has to be compatible with the unitary representation on the space of states
 
9:42 PM
He has a new book out, the title alone is going to drive some people nuts
 
I like string theory, and still claims like " Today, string theory forms the basis of much of the research being done in the world’s leading laboratories." admittedly drive me nuts :P
most of experimental research has nothing whatsoever to do with high energy physics, let alone string theory
that's just a garbage claim, and I think PR like this is ultimately more detrimental to string theory than just being honest
 
Anytime one cringes at something Kaku says they should just go look in one of his textbooks and they'll soon be humbled
 
9:57 PM
god that whole excerpt is just so pretentious
"Each time scientists have unraveled a new force, it has changed the course of civilization and altered the destiny of humanity." - please tell me more about how the strong force has changed civilization, Mr Kaku
he talks about fundamental forces and then just cites "quantum theory" as an example, this is not how any of this works
 
10:13 PM
@satan29 ah yes, lubos, the #NotLikeTheOtherGirls of white conservative men
"how the strong force has changed civilization" i'm sure the japanese might have something to say about this :-)
 
@AccidentalFourierTransform but we didn't even understand what the strong force was when we figured out how to build atom bombs!
 
if only we could build string bombs
 
fundamental understanding is not required for humans to wreak destruction upon one another with what they don't understand :P
 
fqq
@ACuriousMind I haven't read any of the books but the excerpts, tweets etc are generally awful
@bolbteppa so the textbooks are good? I never used those either
@Slereah SUSY can be useful for stochastic processes
 
To use a Kaku-ism, his textbooks are in a sense the holy grail, but they just throw too many un-derived formulas at you
 
11:04 PM
@ACuriousMind here is the same - it's over 20 degrees Celsius last Thursday.
 

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