# 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.