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Lorentz Generators: (J and M) vs. S 
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#1
Dec1512, 04:39 AM

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(if I am not even wrong, please let me know )
The generators of the (properorthochronous) Lorentz transformations are usually denoted by [itex]{J_{\mu \nu }}[/itex] or by [itex]{M_{\mu \nu }}[/itex]. They consist of angular momentum generators and boost generators. When discussing spinors, the notation changes to [itex]{S_{\mu \nu }}[/itex] (which is equal to the commutator of gamma matrices). I am trying to understand if the change of notation reflects a difference between [itex]S[/itex] and [itex]J[/itex], or if it just a change of convention. Is there any sense in which [itex]{S_{\mu \nu }}[/itex] is just spin while [itex]{J_{\mu \nu }}[/itex] is angular momentum? Is there any sense in which [itex]{S_{\mu \nu }}[/itex] generates inner transformations while [itex]{J_{\mu \nu }}[/itex] does not? Thanks 


#2
Dec1512, 06:13 AM

P: 123

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, S_{μν} are a particular (up to similarity transormations) set of matrices, that satisfy Poincare algebra and they are used to describe spin1/2 particles. These matrices can be constructed by the commutators of gamma matrices (you can check it by verifying that S_{μν} indeed satisfy Poincare algebra). Their difference with J_{μν} is that S_{μν} are the matrices of spinor representation while J_{μν} could be anything and that S_{μν} can be used only to rotate spin 1/2 systems while J_{μν} are used for every system.



#3
Dec1512, 07:21 AM

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#4
Dec1612, 03:27 PM

P: 15

Lorentz Generators: (J and M) vs. S
In non relativistic quantum mechanics (of particles), when we want to discuss spin we add to the wave function a parameter [itex]s[/itex], which takes 2 values (for a spin 1/2 particle). That is in addition to the parameter [itex]x[/itex] which takes a continuum of values. The angular momentum operator is made out of the position and linear momentum operators. These operate on [itex]x[/itex]. The spin operators are Pauli matrices which operate on [itex]s[/itex] (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 threespace, 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 spacetime 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
Dec1612, 04:11 PM

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www.physicsforums.com/showthread.php?t=650570 


#6
Dec1612, 05:17 PM

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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
Dec1712, 12:17 AM

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#8
Dec1712, 04:14 PM

P: 15

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
Dec1712, 04:50 PM

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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 3dimensional 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 nonrelativistic QM spin is derived from the connected component of the Galilei group, not from the rotations subgroup. 


#10
Dec1712, 05:18 PM

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[tex]SO(3) \subset SO(1,3) \approx SL(2, \mathbb{C}) \approx SU(2) \times SU^{*}(2)[/tex] The action of the generators of SO(3) subgroup on geometrical objects (fields) give you 3dimentional orbital angular momentum plus INTEGER spin. This why we complexify the generators to get to SL(2,C). [tex]\bar{\phi} (\bar{x})= U^{1}\phi (\bar{x})U = D(\Lambda) \phi (x)[/tex] 


#11
Dec1712, 05:43 PM

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Spin is and Spinors are the natural objects of the abstract group [itex]SL(2, \mathbb{c})[/itex]: the group of linear mapping in 2dimentional, complex, symplectic space, namely the spinor space. This has direct relation to Lorentz group [itex]SO(1,3) \approx SL(2, \mathbb{C})[/itex]. With or without translations, that relation still hold. Sam 


#12
Dec1712, 06:07 PM

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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.
Your remark is about spinorial / SL(2,C) indices, not about spin. 


#13
Dec1812, 10:43 PM

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In post #11, I said and meant SPIN ( [itex]s = (n + m) /2[/itex] ) and SPINORS ( [itex]\psi_{\alpha}[/itex], [itex]\psi_{\dot{\alpha}}[/itex]) can be introduced NATURALLY in the Lorentz group [itex]SO(3,1) \approx SL(2, \mathbb{C}) / \mathbb{Z}_{2}[/itex]. This relation means that the Lorentz algebra [itex]so(1,3)[/itex] is isomorphic to a direct sum of two mutually conjugate [itex]sl(2, \mathbb{C})[/itex] algebras and therefore has two Casimir operators [tex]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}[/tex] [tex]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}[/tex] where [itex](n/2 , m/2)[/itex] is the [itex](n + 1)(m + 1) \mbox{dimensional}[/itex] 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 spacetime we run into troubles because, in general, the [itex]( n/2 , (2s – n)/2 )[/itex] type spinor with [itex]n \neq 0[/itex] admits several realizations in terms of FIELDS. So, to select the spins fields, we need (the Momentum operator) to impose the following supplementary condition [tex] P^{\alpha \dot{ \alpha}}\Psi_{\alpha \alpha_{1}…\alpha_{n 1}\dot{\alpha}\dot{\alpha}_{1}… \dot{\alpha}_{m  1}}(x) = 0. [/tex] We also need our spins representation to satisfy the onshell condition [tex] ( \partial^{2}  m^{2}) \Psi_{\alpha_{1}… \alpha_{n}\dot{\alpha}_{1}… \dot{\alpha}_{m}}(x) = 0. [/tex] So, you see that the Poincare generators [tex] P^{\mu} = \frac{1}{2} ( \bar{ \sigma }^{ \mu } )_{ \dot{ \alpha } \alpha } P^{ \alpha \dot{ \alpha }}, [/tex] and [tex] 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}} [/tex] come in to the picture when we need to talk about fields and their evolution on spacetime. But, if you start from the Poincare group, then you get your spin by squaring the PauliLubanski vector [tex] W_{\alpha \dot{\alpha}} = iM_{\alpha \beta}P^{\beta}{}_{\dot{\alpha}} + i \bar{M}_{\dot{\alpha}\dot{\beta}}P_{\alpha}{}^{ \dot{ \beta}} [/tex] Sam 


#14
Dec1812, 10:54 PM

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#15
Dec1912, 02:15 AM

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Usually [itex]J_{\mu \nu}[/itex] 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 ([itex]J_{0\nu}[/itex]) and rotations ([itex]J_{jk}[/itex] with [itex]j,k \in \{1,2,3\}[/itex]). It's antisymmetric under exchange of the two spacetime indices, i.e., there are 6 independent generators as it must be for the sixdimensional Lie algebra sl(2,C).
You find a treatment of these matters in Appendix B of my QFT manuscript: http://fias.unifrankfurt.de/~hees/publ/lect.pdf 


#16
Dec1912, 04:06 AM

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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
Dec1912, 08:29 AM

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[tex]so(1,3) \sim sl(2 , \mathbb{ C }) \oplus sl(2 , \mathbb{ C })[/tex] Sam 


#18
Dec1912, 10:37 AM

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Sam,
sl(2,C) is 3dimensional 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 [tex] \mbox{so(1,3)} \simeq \mbox{sl}(2,\mathbb{C}) \simeq \mbox{su(2)} \oplus \mbox{su(2)} [/tex] [tex] \mbox{so(1,3)}_{C} \simeq \mbox{su(2)} \oplus \mbox{su(2)} [/tex] 


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