The QH QM QA thread.

Or is it the other way round? id est assume the existence of these particles, and then find a piece of mathematics to describe them?
Click to expand...

I'd say historically this was the case.

The reason I mentioned lattices was that, historically, Gell-Mann constructed something called the ``eight-fold way'', an allusion to his buddhist beliefs. This ``eight-fold way'' essentially corresponded to the ways to make mesons out of quarks.

Sorry if I seem to be drifting...but there is a method to my madness!
 
BenTheMan said:
...but there is a method to my madness!
Click to expand...
Ha! Well, this promised much, but has so far delivered nothing. Ah well.

But look guys, I seem to have painted myself into a corner here; we agreed, did we not, that by chopping my orinigal bases for su(2), this implied that the generators of SU(2) have t = t + 4 pi symmetry, which in turn implies 1/2-integer spin. At least, I thought we had so agreed.

But I also took on faith that the 3 bases for su(2) in some sense "are" the mediators of the electro-weak force, the B, W and Z bosons.

But as these are bosons, they must have integer spin! So what gives? I am thoroughly confused (but what's new?)
 
Ben, if I didn't know you better, I'd say you were teasing me. What do you mean? Do you mean that if I choose to apply a factor of 1/2 to the canonical basis for su(2), as I did earlier, I will be describing a half-integer-spin particle, but if I don't so choose, I am describing an integer-spin particle?

I tell you straight, I find that a repellent thought. Thankfully for you, I don't assume for a moment that this is what you mean. Maybe you mean it's possible to find an alternative basis? I can't find it!
 
Sad sod that I am, I have nothing better to do this Saturday night.

Poor me!

So, I remind you that Lie theory states, quite categorically, that for each Lie group I may associate at least one, and at most one algebra. It seems the converse may not be true.

Recall I found a possible algebra for SU(2) as

$$[X,Y]= -2Z$$
$$[Z,X] = -2Y$$
$$[Z,Y] = 2X$$

I remind you also that Lie algebras are vector spaces, so, by the axioms of this theory, any spaces with the same dimension (= same number of bases ) are isomorphic, and recalling that the constraints on the matrices that constitute the representation of the group SO(n) implies that the algebra has $$\frac{n^2 -n}{2}$$ basis vectors.

Obviously, for SO(3), this implies a 3-basis, so it looks like so(3) and su(2) are the same, up to isomorphism. In fact, we can do better. I copy from above

$$[X,Y]= -2Z$$
$$[Z,X] = -2Y$$
$$[Z,Y] = 2X$$

I then worked the algebra for SO(3) and found, calling my basis vectors A ("=" X), B ("=" Y) and C ("=" Z)

$$[A,B]= -C$$
$$[C,A] = -B$$
$$[C,B] = A$$

So SU(2) and SO(3) have the same algebra, up to isomorphism!!

Maybe this is what Ben meant by "different su(2)'s"?
 
Hmmm. One su(2) is a symmetry OF space-time, and one su(2) is a symmetry ON space-time. when we say ``spin'' we are refering to the particle's transformation properties under the Lorentz group. I have failed miserably at explaining most of these things to you, but that still won't stop me!

So, for example, let's pretend to be naughty physicists for a moment and ignore the fact that so(3,1) is non-compact, and treat is as so(4) (bear with me). We can decompose so(4) as su(2)xsu(2). Actually, there is an isomorphism, as people who are smarter than me tell me, and we write $$so(4)\sim su(2)_L \otimes su(2)_R$$ where the `L' and `R' stand for `left' and `right' respectively.

Then ``particles'' in our universe must be representations of this group, which basically dictates how they transform under the isometries of space-time. We can label the particles that we see as representations of su(2)xsu(2). The smallest representations we have are the 2 (fundamental) and 3 (adjoint). ``Scalar'' particles, which I talked about before are left alone by Lorentz transformations, and so under su(2)xsu(2) they are written as (1,1). Fermions live in (2,0) or (0,2) representations, which is the genisis for the name ``left-handed'' and ``right-handed''.
 
What I have to convince you of is that the (2,0) representation that we need is still a representation of SO(3,1).

I will rely on AlphaNumeric for this, if he knows. I think it is in Ryder's QFT textbook, so if I have time I will try to work it out tonight. But you can procede by noting that $$SO(3,1)\sim SL(2,\mathbb{C})$$, where SL(2,C) can be thought of as the group of 2x2 matrices with real entries, of determinant 1.
 
So SU(2) and SO(3) have the same algebra, up to isomorphism!!

Maybe this is what Ben meant by "different su(2)'s"?
Click to expand...

Actually, you've just found one of my favorite ways to show crackpots that they are wrong.

There is one class of crackpot called ``time isn't a dimension'' crackpot. Trust me, they are out there. Then very generally this implies that the Lorentz symmetry should be SO(3)---the group of rotations you can do on a three vector and leave it's length unchanged. BUT SO(3), which you have pointed out, is the same as SU(2). THIS then implies, if you want your particles as reps of the spatial rotation group (you do), that positrons don't exist!

The way to see this is in the above---we need su(2)xsu(2) reps to get electrons (2,0) and positrons (0,2). Electrons and positrons are in different reps under the Lorentz group, which is not possible if ``time isn't a dimension'' crackpot is correct.
 
BenTheMan said:
Hmmm. One su(2) is a symmetry OF space-time, and one su(2) is a symmetry ON space-time. when we say ``spin'' we are refering to the particle's transformation properties under the Lorentz group. I have failed
miserably at explaining most of these things to you, but that still won't stop me!
Click to expand...
Good! Part of the problem here is we are using rather different notations, but this time I'll resist the temptation to be overly pedantic.

First, as far as I am aware, symmetry is described by the group, so what you call su(2) I would prefer to call SU(2).

Second, the distinction you make between symmetries OF and ON, I think I would call the difference between a realization and a representation, respectively. This is because the former tells us what the group IS, the latter tells us what it DOES (to a vector space), loosely speaking
We can decompose so(4) as su(2)xsu(2). Actually, there is an isomorphism, as people who are smarter than me tell me, and we write $$so(4)\sim su(2)_L \otimes su(2)_R$$ where the `L' and `R' stand for `left' and `right' respectively.
Click to expand...
Once again, the "otimes" is an operation on groups (the group direct product), and I think that chirality is a group property, since it is a form of (anti)symmetry. All this is to clarify, not criticize!

Let's accept my interpretation for now and ask what might be the algebra of $$SU(2) \otimes SU(2)$$. I'm guessing that it will be the direct sum of their two algebras $$su_2 \oplus su_2$$.

I think we can do better than this by noting that there is an isomorphism $$su_2 \otimes \mathbb{C} \sim sl_2(\mathbb{C})$$. I'd like to make use of this, because I was once shown how to work out the roots of $$sl_2(\mathbb{C}) \oplus sl_2(\mathbb{C})$$ which, when expressed as row vectors on the dual basis turned out to be precisely (2,0) and (0,2)!
Fermions live in (2,0) or (0,2) representations, which is the genisis for the name ``left-handed'' and ``right-handed''.
Click to expand...
I have no real reason (other that this possible coincidence) to think this what you are getting at, although I was once told that physical observables are the eigenvalues for the action of some operator on a vector space.

Since a root is a linear functional that returns the eigenvalue for each vector in my space, it is certainly possible this is what you mean. I have no intention of showing the full derivation (it runs to 5 pages in portable document format), though I could, I think, if anyone were sufficiently interested, succinctly outline the steps (that is,assuming I'm not talking out my ear again)
 
Last edited:
There's a group isomorphism $$\frac{SL(2,\mathbb{C})}{\mathbb{Z}_{2}} \to SO(3,1)$$. You can do all your Lorentz algebra using the the extended Pauli matrices $$\sigma^{\mu} = (\mathbb{I}_{2},\sigma^{i})$$ and noting that you have nice things such as $$X = x_{\mu}\sigma^{\mu}$$ transforms under SL(2,C) but that det(X) = $$x_{\mu}x_{\nu}\eta^{\mu\nu}$$. In the Pauli construction the det(X) is invariant under SL(2,C), but we know that a space-time interval is invariant under SO(3,1).

All you want to know about Lorentz and SU(2) and SL(2,C) transformations and plenty you didn't want to know (Chapter 5 is the relevent part now, but other chapters have been discussed here).

And some more
 
AlphaNumeric said:
There's a group isomorphism $$\frac{SL(2,\mathbb{C})}{\mathbb{Z}_{2}} \to SO(3,1)$$.
Click to expand...
Either you are taking the piss here, or you flatter me. Please don't do either.

What you seem to be doing here is quotienting a complex Lie group by a real integer group which is itself a quotient group. I am lost; how does this work?

I'm going for a beer, then I will get back to my knitting.
 
SL(2,C) is the universal cover of SO(3,1), with a 2-1 relationship. For each X in SO(3,1) there's a Y and also -Y, because they lead to equivalent SO(3,1) structures. Have a gander at the 5th chapter of my 1st link.

What's wrong with quotienting SL(2,C) by Z_2? It just means that you're taking all the elements of SL(2,C) in pairs due to two elements in SL(2,C) being equivalent to the same element in SO(3,1).

Just as SU(2) is a double covering of SO(3).
 
Yes, OK, but I'm trouble seeing how the partition works. Is it that the equivalence classes are $$x,\;-x \in [x]\; \forall x, \;-x \in SL(2,\mathcal{C})$$? It looks like that's what it should be, but it's not obvious to me how this is induced by the quotient.

Of course, I always have the option of swallowing it whole, but usually I don't like to do that (though I concede it is a fine way to make my life more difficult than it needs to be!)
 
While technicaly $$\mathbb{Z}_{2}$$ is something like {0,1}, since it's $$\frac{\mathbb{Z}}{2\mathbb{Z}}$$, the group isomorphism expression is physicist shorthand for "There's 2 matrices in SL(2,C) for every one in SO(3,1)". In terms of the physics thing comes from the fact that the Lorentz group is defined to be the part of SO(3,1) which has positive determinant and it's 0,0 entry greater than 1, ie it's connected to the identity. However, since SL(2,C) is complex it doesn't have this partitioning, it's simply connected so you can move smoothly from N to -N, it's all part of the group SL(2,C). Due to the quadratic nature of the transforms (ie $$\tilde{x} \to N\hat{x}N^{-1}$$, N and -N are equivalent, so there's a double counting compared to SO(3,1).

It's important to remember that 'The Lorentz group' isn't strictly SO(3,1). SO(3,1) partitions into 4 sections and the continuous nature of physical transforms and needing to connect to the identity means we only take one of these 4. Hence there's no -N in the physicists SO(3,1) for a given N, even though -N is actually in the full SO(3,1) due to the quadratic nature of what defines SO(3,1).
 
AlphaNumeric said:
the Lorentz group is defined to be the part of SO(3,1) which has positive determinant
Click to expand...
Not getting you here. Surely the monicker "special" defines SO(3,1) to have positive determinant? So what gives with the "part of" qualifier?
It's important to remember that 'The Lorentz group' isn't strictly SO(3,1). SO(3,1) partitions into 4 sections and the continuous nature of physical transforms and needing to connect to the identity means we only take one of these 4.
Click to expand...
Well, I don't see this at all. Are you suggesting that SO(3,1), as a topological space, is the disjoint union of 4 connected components? I can see there may be 2, but four? And if there are 4 connected components (I don't doubt your authority, by the way), how would you set about choosing which one to take?

Anyhow, is this relevant to my quest, or is it a digression in the Lie interpretation of relativity? Or are they intimately related?
 
Sorry, I should have worded things better.

On a naive level, the Lorentz group just needs to satisfy $$\Lambda^{T} \eta \Lambda = \eta$$, ie the metric is preserved.

You thus find that the total space of matrices/transforms which satisfy this are O(3,1). This has 4 seperate connected components, depending on the decomposition explained in the first 3 pages of chapter 5 of the first set of notes I linked to. So the Lorentz group is actually the entirity of O(3,1). We then want to take the connected part which forms a group, so it's the 'proper orthochronous subgroup' of O(3,1), $$SO(3,1)^{\uparrow}$$. So the 'physicists Lorentz group' is actually 1/4 of the transformations which preserve the metric.

I turns out that the other 3 parts can be obtained by factoring in the effects of the discrete symmetry operators P, T and PT (since they swap the spacial, time and all of the entries in the Lorentz group)
 
O(4) has 2 components: SO(4) that leaves a sphere invariant, and the compositions of SO(4) maps with P. I think in intuitive level, O(3,1) has 4 components because the 'sphere' in this indefinite metric has 2 components.
 
Gosh, you guys leave me breathless! WTF is P (parity??) What is T? Please , please for the sake of we poor outsiders, define your terms!

Sorry to be a bore about this, but I am trying hard to keep up. And assuming I knew what the operators P and T are (I don't), what is the difference between P,T and PT?
 
If I understood correctly, P is the inversion of all the coordinates, T is the inversion of time coordinate. In Lorentzian space I think P is usually defined as the inversion of only spacial coordinates, but this does not change the essence; by adopting this definition PT will be P and vice versa.
 
You must log in or register to reply here.
Back
Top