Lorentz Generators: (J and M) vs. S

1. Dec 15, 2012

liorde

(if I am not even wrong, please let me know )
The generators of the (proper-orthochronous) Lorentz transformations are usually denoted by ${J_{\mu \nu }}$ or by ${M_{\mu \nu }}$. They consist of angular momentum generators and boost generators.
When discussing spinors, the notation changes to ${S_{\mu \nu }}$ (which is equal to the commutator of gamma matrices).

I am trying to understand if the change of notation reflects a difference between $S$ and $J$, or if it just a change of convention.

Is there any sense in which ${S_{\mu \nu }}$ is just spin while ${J_{\mu \nu }}$ is angular momentum?
Is there any sense in which ${S_{\mu \nu }}$ generates inner transformations while ${J_{\mu \nu }}$ does not?

Thanks

2. Dec 15, 2012

cosmic dust

When refernig generaly to Lorentz group, it's generators are not specific mathematical items, but they could be anything that satisfy Poicare algebra. When you want to find a specific representation of Lorentz group, then you are searching for matrices that satisfy Poincare aglebra. So, Sμν are a particular (up to similarity transormations) set of matrices, that satisfy Poincare algebra and they are used to describe spin-1/2 particles. These matrices can be constructed by the commutators of gamma matrices (you can check it by verifying that Sμν indeed satisfy Poincare algebra). Their difference with Jμν is that Sμν are the matrices of spinor representation while Jμν could be anything and that Sμν can be used only to rotate spin 1/2 systems while Jμν are used for every system.

3. Dec 15, 2012

andrien

NO.
what is inner transformation?

4. Dec 16, 2012

liorde

Maybe the word "internal" should have been used... I'm not sure what I mean, but I hear people referring to spin as generating an "internal" transformation.

In non relativistic quantum mechanics (of particles), when we want to discuss spin we add to the wave function a parameter $s$, which takes 2 values (for a spin 1/2 particle). That is in addition to the parameter $x$ which takes a continuum of values. The angular momentum operator is made out of the position and linear momentum operators. These operate on $x$. The spin operators are Pauli matrices which operate on $s$ (I know this was a lousy description, but you understand I'm sure....). Thus, you can say that spin is something "internal" since it does not depend on space coordinates, while orbital angular momentum is not "internal" in this sense. Also, it is known that the orbital angular momentum operators generate rotations in three-space, which makes them "not internal" in some sense.
(I know I am using the word "internal" in a obscured way... It's just that I think it is in the jargon and I'm hoping someone will explain it to me)

Now, in relativistic quantum field theory (and probably also in NRQM), we discuss space-time symmetries of the action, specifically Lorentz translations. These consist of rotations, and you can write down the corresponding generators. Do these generators correspond to orbital angular momentum or to spin? or both? How can I know how to decompose them to orbital and to spin? If orbital angular momentum themselves generate the rotations, what does spin generate? and where is it "hidden" in the Lorentz group? (I know that Lorentz can be generated by rotations and boosts, with no other generators needed, so what does spin do?)

5. Dec 16, 2012

samalkhaiat

6. Dec 16, 2012

strangerep

As always, your answers were excellently informative. But sometimes it's tricky to match the level of one's answers to the level of the questioner's educational background...

Liorde,
Perhaps you should give a quick summary of your background and what level of math/physics studies you've reached so far. That would help responders to tailor their answers appropriately. (Sam is right that, if you had trouble understanding his answers in thorough detail, then you should have given some hint of this.)

[Afterthought.... can you access a copy of Ballentine's textbook "QM -- A Modern Development". If so, are you comfortable with the level of math therein? If you can cope with it, you'll find good answers to lots of your questions...]

7. Dec 17, 2012

andrien

In correct theory of angular momentum as discovered by dirac,it is orbital angular momentum+spin which is a constant of motion and is a good quantum number.So spin is necessary.You can see for these related point in sakurai 'advanced quantum mechanics'

8. Dec 17, 2012

liorde

I'm sorry, but at the time (and for a long while) I didn't have any means of connecting to the internet, for technical reasons... And now I wanted to ask some more questions about this subject, but to approach it from a different angle, and that is why I opened a new thread (I also posted it in a more relevant forum - Quantum Physics as opposed to Relativity).

Anyway, I feel quite comfortable with the quantum field theory that I studied so far (and I am still studying), but I have this little yet substantial gap of understanding that drives me crazy. It has to do with the origin and justification of spin. I assume that my misunderstanding is due to the fact that this subject isn't presented to me with enough mathematical rigor (I am taking a course on QFT and I also use Peskin+Sredinski+Ramond+Greiner+.... this series does not converge...)

I guess that the problem is that I am not managing to state clearly my questions, as I haven't received satisfactory answers yet...

I'll try to rephrase my main question :

The generators of the Lorentz group (in 3+1 dimensions) are 3 boosts and 3 rotations. To the rotation elements of the group correspond three infinitesimal generators (which, of course, can be represented irreducibly in various vector spaces). Are these infinitesimal generators equal to orbital angular momentum (AM) operators? Or to total AM operators (=orbital+spin)? If it is the former, then where does spin come from - as it is not needed to generate Lorentz (since in this case we say that orbital AM generates the rotations)? If it is the latter, then why can we speak of two kinds of seperate things - orbital AM and spin - while the generator generates only one kind of thing, namely rotations, and what is the precise way to decompose AM to orbital and spin?

If you feel that you keep trying to explain something simple to me and that I just don't get it, let me know and I will move on and try to find the answer in books... I just thought that since this subject is not treated well in books, I could get a better answer here.

P.S.
Weinberg I and Ryder reason that spin originates from extra degrees of freedom when Lorentz transforming in the representations of the Poincare group. This is called "the little group method". If someone could explain this to me I would be grateful.

9. Dec 17, 2012

dextercioby

For the Lorentz group (more precisely its component connected to the identity), the 6 generators of the group are the 3 boost operators and the 3 orbital angular momentum operators, since by exponentiation they generate 3 independent rotations in the boosts space and 3 in the real 3-dimensional space respectively.

In addendum, spin doesn't come from the Lorentz group actually, but from the Poincaré group (the component connected to identity of the full Poincaré group), just like in non-relativistic QM spin is derived from the connected component of the Galilei group, not from the rotations subgroup.

Last edited: Dec 17, 2012
10. Dec 17, 2012

samalkhaiat

These texts explain to you that
$$SO(3) \subset SO(1,3) \approx SL(2, \mathbb{C}) \approx SU(2) \times SU^{*}(2)$$

The action of the generators of SO(3) subgroup on geometrical objects (fields) give you 3-dimentional orbital angular momentum plus INTEGER spin. This why we complexify the generators to get to SL(2,C).

If you study the above relations, you will understand the meaning of the transformation equation which I told you about in the other thread and its connection to Lorentz group representations, namely

$$\bar{\phi} (\bar{x})= U^{-1}\phi (\bar{x})U = D(\Lambda) \phi (x)$$

11. Dec 17, 2012

samalkhaiat

I would say, it has nothing to do with the Poincare group. Take a typical group element $(\Lambda , a)$ and set $\Lambda = 1$, Do you get spinors from $(1,a)$? No, you don't.
Spin is and Spinors are the natural objects of the abstract group $SL(2, \mathbb{c})$: the group of linear mapping in 2-dimentional, complex, symplectic space, namely the spinor space. This has direct relation to Lorentz group $SO(1,3) \approx SL(2, \mathbb{C})$. With or without translations, that relation still hold.

Sam

12. Dec 17, 2012

dextercioby

Actually, spin comes from the quantum mechanical symmetry group, which for a flat Minkowskian space time is the universal covering group of the connected component of the Poincaré group, by means of the second Casimir, first being linked to the mass of the (quanta of the) field.

13. Dec 18, 2012

samalkhaiat

In post #11, I said and meant SPIN ( $s = (n + m) /2$ ) and SPINORS ( $\psi_{\alpha}$, $\psi_{\dot{\alpha}}$) can be introduced NATURALLY in the Lorentz group $SO(3,1) \approx SL(2, \mathbb{C}) / \mathbb{Z}_{2}$. This relation means that the Lorentz algebra $so(1,3)$ is isomorphic to a direct sum of two mutually conjugate $sl(2, \mathbb{C})$ algebras and therefore has two Casimir operators
$$C_{1}\left( \frac{n}{2}, \frac{m}{2}\right) \equiv M^{\alpha \beta } M_{\alpha \beta}\left( \frac{n}{2}, \frac{m}{2}\right) = - 2 \frac{n}{2}( \frac{n}{2} + 1 ) \mathbb{E}$$
$$C_{2} \left( \frac{n}{2}, \frac{m}{2}\right) \equiv \bar{M}^{\dot{\alpha} \dot{ \beta}} \bar{M}_{ \dot{\alpha} \dot{\beta}}\left( \frac{n}{2} , \frac{m}{2}\right) = - 2 \frac{m}{2} ( \frac{m}{2} + 1 ) \mathbb{E}$$
where $(n/2 , m/2)$ is the $(n + 1)(m + 1)- \mbox{dimensional}$ irreducible representation (sequence) of the Lorentz algebra.
So, up to this point, SPIN and SPIONRS do not need the Poincare algebra. However, when we want to realize the SPIONRS in terms of FIELDS on Minkowski space-time we run into troubles because, in general, the $( n/2 , (2s – n)/2 )$ type spinor with $n \neq 0$ admits several realizations in terms of FIELDS. So, to select the spin-s fields, we need (the Momentum operator) to impose the following supplementary condition

$$P^{\alpha \dot{ \alpha}}\Psi_{\alpha \alpha_{1}…\alpha_{n -1}\dot{\alpha}\dot{\alpha}_{1}… \dot{\alpha}_{m - 1}}(x) = 0.$$

We also need our spin-s representation to satisfy the on-shell condition

$$( \partial^{2} - m^{2}) \Psi_{\alpha_{1}… \alpha_{n}\dot{\alpha}_{1}… \dot{\alpha}_{m}}(x) = 0.$$

So, you see that the Poincare generators

$$P^{\mu} = \frac{1}{2} ( \bar{ \sigma }^{ \mu } )_{ \dot{ \alpha } \alpha } P^{ \alpha \dot{ \alpha }},$$

and

$$M^{\mu \nu} = ( \sigma^{\mu \nu})_{\alpha \beta} M^{\alpha \beta} - ( \bar{\sigma}^{\mu \nu} )_{\dot{\alpha}\dot{\beta}}\bar{M}^{\dot{ \alpha} \dot{ \beta}}$$

come in to the picture when we need to talk about fields and their evolution on space-time. But, if you start from the Poincare group, then you get your spin by squaring the Pauli-Lubanski vector

$$W_{\alpha \dot{\alpha}} = -iM_{\alpha \beta}P^{\beta}{}_{\dot{\alpha}} + i \bar{M}_{\dot{\alpha}\dot{\beta}}P_{\alpha}{}^{ \dot{ \beta}}$$

Sam

14. Dec 18, 2012

Fredrik

Staff Emeritus
J (or M) consists only of angular momentum generators. It's an antisymmetric matrix (i.e. $J^T=-J$), so the diagonal elements are zero, and the ones below the diagonal are determined by the ones above it. So only 3 independent components.

15. Dec 19, 2012

vanhees71

Usually $J_{\mu \nu}$ are the generators of the special orthochronous Lorentz group (or its covering group, the SL(2,C)) in some of its representations. Thus indeed it generates both boosts ($J_{0\nu}$) and rotations ($J_{jk}$ with $j,k \in \{1,2,3\}$). It's antisymmetric under exchange of the two space-time indices, i.e., there are 6 independent generators as it must be for the six-dimensional Lie algebra sl(2,C).

You find a treatment of these matters in Appendix B of my QFT manuscript:
http://fias.uni-frankfurt.de/~hees/publ/lect.pdf

16. Dec 19, 2012

Fredrik

Staff Emeritus
Oh yeah, I had a little brain malfunction there. An antisymmetric 4×4 matrix has 6 independent components, not 3. But I'm pretty sure I've seen both J and M denote an antisymmetric 3×3 matrix whose independent components are the spin operators. (I don't have time to look at your notes right now).

17. Dec 19, 2012

samalkhaiat

As a vector space, the Lie algebra $sl(2, \mathbb{ C })$ is 3-dimensional, i.e. it has 3 generators. In general $sl(n , \mathbb{ C })$ is a $(n^{2} - 1)$-dimensional Lie algebra. The Lorentz algebra is 6-dimensional because of the relation
$$so(1,3) \sim sl(2 , \mathbb{ C }) \oplus sl(2 , \mathbb{ C })$$

Sam

18. Dec 19, 2012

dextercioby

Sam,

sl(2,C) is 3-dimensional only when viewed as a vector space over the field of complex numbers. When viewed as a vector space over the field of real numbers, it's 6 dimensional. The correct Lie algebra isomorphisms are

$$\mbox{so(1,3)} \simeq \mbox{sl}(2,\mathbb{C}) \simeq \mbox{su(2)} \oplus \mbox{su(2)}$$

$$\mbox{so(1,3)}_{C} \simeq \mbox{su(2)} \oplus \mbox{su(2)}$$

19. Dec 19, 2012

dextercioby

Actually, to assign the name <spin> to a particular weight (n+m)/2 of a finite-dimensional irreducible representation of SL(2,C) is quite confusing, since it automatically raises the question whether and how this number is linked to the eigenvalues of the squared Pauli-Lubanskii 4 vector (operator) when evaluated in the <particle's> own reference frame (if any).

If you can link these 2 numbers (by a proof), then you're guaranteed to be correct when using the word 'spin' when discussing the representation theory of SL(2,C). Else, you should call that number by the name it bears in mathematics, weight.

20. Dec 19, 2012

Zoot

Let me add my 2 cents worth, as this may help answer the original question in a simplistic manner (which is usually the way I think, unfortunately!). Let's say we have 3 types of fields: Scalar, Vector, and Spinor. We wish to look at these types of fields in another coordinate system (e.g. one that has been boosted and/or rotated). Now the scalar field is the simplest case, since it consists of just a single number at each point in space. So all we have to do is transform the coordinates of the field and we're done. The generators for such transformations are the orbital angular momentum operators Lαβ where the L0i generate boosts and the Lij generate rotations. For scalar fields there is no spin, so Jαβ = Lαβ. Now take a vector field and look at it from the same rotated/boosted coordinate frame. The vector field has 4 components at each point in spacetime. So just transforming the coordinates of where each vector is located is not enough. We also have to ROTATE/BOOST the vector which means mixing the COMPONENTS of the vector accordingly, as the components will be different in the new frame. So what operator mixes the components? The spin operators Sαβ. And the S0i mix the components for coordinate boosts, while the Sij mix the components for coordinate rotations. So the total angular momentum operator needed is equal to orbital angular momentum (to transform the coordinates) plus spin angular momentum (to mix the components): Jαβ = Lαβ + Sαβ. This means vector fields (particles) have spin, and it is not that difficult to show that for vector fields s = 1 by doing some work with the spin matrices. Now you are probably aware that Dirac spinors transform differently than vectors, thus the spin matrices for Dirac 4-spinor fields are different than for 4-vector fields, but again the operator needed is Jαβ = Lαβ + Sαβ, where again L transforms the coordinates (of where each spinor is located in the new frame) and S mixes the components (boosts/rotates the spinor itself). And these spin matrices for Dirac fields can be used to show that s = 1/2 for this field.

21. Dec 20, 2012

vanhees71

Another 2 cents. It's a bit dangerous to talk about "spin" and "orbital angular momentum" in relativistic quantum theory since this distinction usually is problematic, to say the least. That's why, for a full understanding of relativistic quantum theory (i.e., relativistic quantum field theory!), it is so important to understand the unitary representations of the Poincare group to a certain extent.

This analysis shows, expressed in a simplified way, that the single-particle states for (asymptotically) free particles can first of all be characterized by the particle's mass. Here, we have just to distinguish massive particles with $m>0$ and massless particles with $m=0$. In the following we'll use the momentum eigenstates as a basis of the single-particle space. These are also energy eigenstates with $E=\sqrt{m^2+\vec{p}^2}$ (using natural units with $\hbar=c=1$).

For massive particles, you can always transform into the restframe of the particle, where $\vec{p}=0$ and $E=m$. The corresponding energy eigenstate has to be further characterized by the transformation properties of this state under those proper orthochronous Lorentz transformations that keep this "standard momentum", $\vec{p}=0$, invariant (the "little group"), which of course is the subgroup of spatial rotations in this particular reference frame, the restframe of the particle. Now you can apply the usual representation theory of the rotation group, which is for quantum theoretical purposes the covering group of the usual rotationgroup, the SO(3), which is SU(2). The irresucible representations of this group are characterized by an integer or half-integer number $s \in \{0,1/2,1,\ldots \}$, and the basis states are given by the eigenstates of the corresponding threecomponent of the spin operator. Then the representation theory of the Poincare tells you how to extend this irrep. of the little group to a unitary representation of the full (proper orthochronous) Poincare group, the socalled Frobenius construction. Thus, the usual angular-momentum algebra for the spin of a particle is defined in the restframe of the particle. Since Lorentz boosts and rotation don't commute, there is no frame-independent separation into spin and orbital angular momentum. The spin-three vector in the restframe of the particle can be expressed in a Lorentz covariant way, which leads to the socalled Pauli-Ljubanski vector.

The massless case is more subtle. Since here the four momentum is lightlike, $E=|\vec{p}|$, you cannot transform into a reference frame, where $\vec{p}=0$. To find the irreps of the Poincare group, you rather have to define a standard momentum arbitrarily, which usually is taken to be [/itex]\vec{p}_0=E \vec{e}_z[/itex]. Now you do the same "Frobenius construction" as in the massless case, i.e., you first look for the irreps of the corresponding "little group", which is defined as the subgroup of the proper orthochronous Lorentz group that leaves the standard four-momentum $p_0=(E,0,0,E)$ invariant. Of course one part of this subgroup is the Abelian rotation SO(2) group, given by the rotations around the $z$ axis, but that's not the full little group. There are also two other independent one-parameter subgroups, the socalled "null rotations", which also leave $p_0$ invariant. All together you find that the little group is isomorphic to the ISO(2), i.e., the full symmetry group of the Euclidean two-dimensional plane. This is the same semidirect product of translations and rotations within this plane as in usual two-dimensional space, and you get its irreps. by the same construnction as in quantum theory, using the "Heisenberg algebra". The "translations" have only infinite-dimensional irreps., and the "momenta" have always a continuous spectrum, except the trivial one. This would mean that a particle described by such a general irrep. would have some strange continuous intrinsic quantum number, but such a thing nobody ever has found necessary to describe a real-world particle. Thus we must consider only those irreps. of the little group that represent the "translations" trivially, and for the real-world particles, we are left with the irreps. of the two-dimensional rotation group SO(2) or, more appropriate for the quantum theory, its covering group, which is the U(1).

The U(1) is abelian and has a lot of representations, but now, you must keep in mind that we are after the unitary irreps. of the full proper orthochronous Poincare group and not only the little group via the Frobenius construction. This of course always implies a irrep of the full (quantum) rotation group SU(2), which tells us that we have to use the representations $\exp(\mathrm{i} \lambda \varphi)$ for the rotations around the three axis in the subspace of fixed standard three momentum to induce the correct integer or half-integer representations of the full rotation group. Thus, one characterizes also the irreps. for the one-particle states of massless particles by a "spin" $s \in \{0,1/2,1,\ldots \}$, but there are always only two states $\lambda =\pm s$. Since this irrep. of the rotations around the three axis represents the rotations around the direction of the three momentum of the particle (in the reference frame, where the momenum points into $z$ direction), this is the helicity of the particle.

For massless particles with spin larger than 1/2 it turns out that, if you want to express them in terms of local quantum fields, one must take them as gauge fields in order to get rid of the non-physical degrees of freedom like for photons with spin 1, where you start with a four-vector field $A^{\mu}$ and have to get rid of the timelike and the longitudinal part, which both are unphysical, to end up with the two physical transverse polarization states. The helicity eigenstates are just left and right circular polarization states.

22. Dec 21, 2012

samalkhaiat

Yes, I am aware of that. Nearly all elements $A \in SL( 2 , \mathbb{ C } )$ can be parameterized by

$$A = \exp ( c_{ i } \sigma_{ i } )$$

where $c_{ i }$ is a complex 3-vector playing the role of local complex coordinates, and the Pauli matrices form a basis in the corresponding Lie algebra. We can also treat $SL( 2 , \mathbb{ C } )$ as a real six-dimensional Lie group using the following parameterization

$$A = \exp ( \frac{ 1 }{ 2 } \omega^{ ab } \sigma_{ ab } ),$$

where

$$( \omega^{ ab } )^{ * } = \omega^{ ab } = - \omega^{ ba }.$$

The two parameterizations are related by

$$2 c_{ 1 } = \omega^{ 01 } + i \omega^{ 23 }, \ 2 c_{ 2 } = \omega^{ 02 } + i \omega^{ 31 }, \ 2 c_{ 3 } = \omega^{ 03 } + i \omega^{ 12 }.$$

And [as a basis of the REAL Lie algebra of $SL( 2 , \mathbb{ C } )$]

$$\sigma_{ ab } = - \frac{ 1 }{ 4 } ( \sigma_{ a } \bar{ \sigma }_{ b } - \sigma_{ b } \bar{ \sigma }_{ a } ),$$

where

$$\sigma_{ a } \equiv ( I_{ 2 } , \sigma_{ i } ), \ \bar{ \sigma }_{ a } \equiv ( I_{ 2 } , - \sigma_{ i } )$$

These are related by

$$( \bar{ \sigma }_{ a } )^{ \dot{ \alpha } \alpha } = \epsilon^{ \dot{ \alpha } \dot{ \beta } } \epsilon^{ \alpha \beta } ( \sigma_{ a } )_{ \beta \dot{ \beta } }$$

Notice that Pauli matrices carry mixed indices whereas $( \sigma_{ ab } )_{ \alpha } {}^{ \beta }$ carry un-dotted indices. There are other generators carrying two dotted indices, I include them here for later use

$$( \bar{ \sigma } )^{ \dot{ \alpha }} {}_{ \dot{ \beta } } = - \frac{ 1 }{ 4 } ( \bar{ \sigma }_{ a } \sigma_{ b } - \bar{ \sigma }_{ b } \sigma_{ a } )^{ \dot{ \alpha } } {}_{ \dot{ \beta } }$$

We will also need the following identities

$$\mbox{ Tr } ( \sigma_{ a } \bar{ \sigma }_{ b } ) = - \eta_{ ab }$$
$$( \sigma^{ a } )_{ \alpha \dot{ \alpha } } ( \bar{ \sigma }_{ a } )^{ \dot{ \beta } \beta } = - 2 \delta^{ \beta }_{ \alpha } \delta^{ \dot{ \beta } }_{ \dot{ \alpha } }$$

Those who have not studied supersymmetry or the representations of $SL( 2 , \mathbb{ C } )$ and want to follow what I am about to do, they need work through the above and many other relations between the sigmas.

DEFINITION (to avoid pages of mathematical gibberish here, I will give a physicist’s definition):

A Lie algebra $\mathcal{ L }$ is the direct sum of two Lie algebras $\mathcal{ L }_{ 1 }$ and $\mathcal{ L }_{ 2 }$ if it is the vector sum and all the elements of $\mathcal{ L }_{ 1 }$ commute with all the elements of $\mathcal{ L }_{2}$. Symbolically, we represent this by:

$$\mathcal{ L } = \mathcal{ L }_{ 1 } \oplus \mathcal{ L }_{ 2 },$$

if

$$[ \mathcal{ L }_{ 1 } , \mathcal{ L }_{ 2 } ] \subset \mathcal{ L }_{ 1 } \cap \mathcal{ L }_{ 2 } = \varnothing$$

CLAIM 1:

$$\mathcal{ so }( 1 , 3 ) \cong \mathcal{ sl }( 2 , \mathbb{ C } ) \oplus \mathcal{ sl }( 2 , \mathbb{ C } ).$$

PROOF:

We start with the Lorentz algebra which we all know (I hope)

$$[ M_{ ab } , M_{ cd} ] = \eta_{ ad } M_{ bc } - \eta_{ ac } M_{ bd } + \eta_{ bc } M_{ ad } - \eta_{ bd } M_{ ac } .$$

Now, define the following three 2 by 2 matrices ( we met them in my previous post)

$$M_{ \alpha \beta } = \frac{ 1 }{ 2 } ( \sigma^{ ab } )_{ \alpha \beta } M_{ ab } = M_{ \beta \alpha } ,$$

and another 3 by

$$\bar{ M }_{ \dot{ \alpha } \dot{ \beta } } = - \frac{ 1 }{ 2 } ( \bar{ \sigma }^{ ab } )_{ \dot{ \alpha } \dot{ \beta } } M_{ ab }$$

Using these together with the properties of the sigmas, we can split the Lorentz algebra into two commuting algebras:

$$2 [ M_{ \alpha \beta } , M_{ \gamma \delta } ] = \epsilon_{ \alpha \gamma } M_{ \beta \delta } + \epsilon_{ \alpha \delta } M_{ \beta \gamma } + \epsilon_{ \beta \gamma } M_{ \alpha \delta } + \epsilon_{ \beta \delta } M_{ \alpha \gamma } \ \ (1)$$

similar one with the bared M and dotted indices $[ \bar{ M }_{ \dot{ \alpha } \dot{ \beta } } , \bar{ M }_{ \dot{ \gamma } \dot{ \delta } } ]$ and

$$[ M_{ \alpha \beta } , \bar{ M }_{ \dot{ \gamma } \dot{ \delta } } ] = 0$$

Ok, if we call $M_{ 11 } = E$ (for Elie Cartan), $M_{ 22 } = F$ (for Felix Klein) and $M_{ 12 } = M_{ 21 } = H/2$ (for Hermann Weyl), then eq(1) becomes

$$[ H , E ] = 2 E, \ \ [ E , F ] = H , \mbox{ and } \ [ F , H ] = 2 F ,$$

which (I hope) every body recognise as the Lie algebra $\mathcal{ sl } ( 2 , \mathbb{ C } )$. The bared M’s commutation relations lead to another $\mathcal{ sl } ( 2 , \mathbb{ C } )$ algebra. Thus as I claimed, the Lorentz algebra is isomorphic to a direct sum of two mutually conjugate $\mathcal{ sl } ( 2 , \mathbb{ C } )$ algebras.

CLAIM 2:

COMPLEX representation of the REAL Lie algebra $\mathcal{ su } ( 2 )$ is EQUIVALENT to representation of the COMPLEX Lie algebra

$$\mathcal{ su } ( 2 ) \otimes_{ \mathbb{ R } } \mathbb{ C } \left( \equiv \mathcal{ su } ( 2 ) \oplus i \ \mathcal{ su } ( 2 ) \right) = \mathcal{ sl } ( 2 , \mathbb{ C } )$$

PROOF:
$$\mathcal{ su } ( 2 ) = \left \{ H \in M_{ 2 } \mathbb{ C } : \mbox{ Tr } ( H ) = 0 , \ H^{ \dagger } = - H \right \}$$

$$i \ \mathcal{ su } ( 2 ) = \left \{ H \in M_{ 2 } \mathbb{ C } : \mbox{ Tr } ( H ) = 0 , \ H^{ \dagger } = H \right \}$$

Thus

$$\mathcal{ su } ( 2 ) \oplus i \ \mathcal{ su } ( 2 ) = \left \{ H \in M_{ 2 } \mathbb { C } : \mbox{ Tr } ( H ) = 0 \right \} \equiv \mathcal{ sl } ( 2 , \mathbb{ C } ) .$$

This is a special case of the general theorem which states that the complex representation of a real Lie algebra $\mathcal{ L }$ is equivalent to representation of the complex Lie algebra $\mathcal{ L } \oplus i \mathcal{ L } = \mathcal{ gl } ( n , \mathbb{ C } )$

Sam

Last edited: Dec 21, 2012
23. Dec 21, 2012

samalkhaiat

When you square the Pauli-Lubanski vector, you get
$$W^{ 2 } = - \frac{ m^{ 2 }}{ 2 } ( C_{ 1 } + C_{ 2 } ) + M \cdot P \cdot \bar{ M } \cdot P$$

where $C_{ 1 }$ and $C_{ 2 }$ are the two Casimir operators of the two $\mathcal{ sl } ( 2 , \mathbb{ C } )$ algebras appearing in the above mentioned direct sum. For massive field, the third term vanishes leaving only the two casimirs to determine the “SPIN”, whatever that means in the group theoretical context. In QM, students learn about the theory of spin through the representations of $\mathcal{ su } ( 2 )$ algebra, i.e. the eigen-values of the single Casimir $J ( J + 1 )$ and the WEIGHT vector $| j , m >$. The Lorentz group provides more elegant description of spin than that given by the poor single $SU(2)$ of QM.

Last edited: Dec 21, 2012