The meaning of the addition of the spin generator to the Lorentz generatorsby liorde Tags: lorentz, representation, spacetime, spin, symmetry 

#1
Nov812, 10:48 AM

P: 15

When constructing the Lie algebra of the Lorentz Transformation, the references usually start with an infinitesimal properorthochronous transformation, and then find the infinitesimal generators. Let's call the set of these generators L. after finding L, the references usually compute the commutation relations of L, and then add to L the operators S, corresponding to the spin. The reason is that L+S obey the same commutation relations as L, and therefore L+S is more general then L.
My questions are: a) Is there any physical motivation to add the S (besides "it agrees with the phenomenon of spin")? I'm looking for a symmetricinspired motivation. b) Would the physics be any different if we didn't add S? c) I thought that the commutation relations of the algebra determine the group structure, so why aren't L and L+S equivalent? why do we need the extra S? d) could we add another term, let's call it T, to L+S such that L+S+T still obey's the algebra? e) If L represents a transformation which is originally defined as operating on spacetime coordinates, what do S and L+S represent in terms of spacetime coordinates? if the answer is that S doesn't care about spacetime but operates on a different space, can we say that T above (in question d) operates on a third space? f) I know that in quantum mechanics, angular momentum has integral eigenvalue while spin has half integral values. The reason is that angular momentum is defined via the position operator X and that leads to integral eigenvalue, while spin is defined to operate on "spin space" which is separate from "space space". What does that have to do with my previous discussion? Thanks 



#2
Nov812, 06:50 PM

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Since I don't understand the approach you're talking about, I'm wondering if what you're talking about is (somehow) that quantum physicists use the universal covering group of the symmetry group instead of the symmetry group itself? (For example, SU(2) instead of SO(3) and SL(2,ℂ) instead of SO(3,1)). If so, I might be able to partially answer some of your questions. (Not today though). 



#3
Nov912, 05:02 AM

P: 15

Ramond, Field Theory: A Modern Primer, Second Edition, Chapter 1, page 7.
Here is the link: http://books.google.co.il/books?id=C...page&q&f=false Look at the last paragraph. I've seen at least two more references using this method (although I might have exaggerated when I said "the references usually start with [...]"). I am not familiar with the concept of "universal covering group". I'll read about it to see if it helps me understand. Thanks 



#4
Nov912, 05:30 AM

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The meaning of the addition of the spin generator to the Lorentz generators
Thanks. I'm a bit busy right now, but I will take a look at that in a few hours. I should of course have posted a reference of my own. The approach that I'm somewhat familiar with is the one described in chapter 2, volume 1, of Steven Weinberg's "The quantum theory of fields". Unfortunately, it seems to be impossible to view the relevant pages at Google Books.




#5
Nov912, 08:07 PM

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Ramond isn't doing anything to find out what the Lie algebra of SO(3,1) is. He's just saying without proof that it's given by (1.2.27). Now it's easy to see that if two sets of operators ##\{L_{\mu\nu}\}## and ##\{S_{\mu\nu}\}## satisfy those commutation relations, and the two sets commute with each other, then the sums ##M_{\mu\nu}=L_{\mu\nu}+S_{\mu\nu}## satisfy those commutation relations too.
If ##\{M_{\mu\nu}\}## is an arbitrary set of operators that satisfy the commutation relations, we can define three new operators ##\{J_i\}## by (1.2.30). The operator ##J^2=J_iJ_i## commutes with each of the ##J_i##, so we can find simultaneous eigenvectors of e.g. ##J^2## and ##J_3##. If we write the eigenvalue equations as \begin{align} J^2jm\rangle &=j(j+1)jm\rangle\\ J_3jm\rangle &=mjm\rangle \end{align} we can prove that 2j must be a nonnegative integer and that j≤m≤j. (Nitpick: We don't actually prove that j is nonnegative. j can be ≤1. But j(j+1) is invariant under the change j→j1, so if j≤1 we can easily redefine j to be nonnegative). What's important to note here is that j can be a halfinteger. The possible values of j are 0, 1/2, 1, 3/2, 2,... If there's any doubt that there exist irreducible representations with halfinteger j, it can be removed by explicitly constructing a j=1/2 representation using the Pauli spin matrices. According to Weinberg's book, each particle species is associated with an irreducible representation of the Poincaré group. I think that this makes this j (which can be a halfinteger) one of the numbers that identify a particle species. We therefore take the particle's Hilbert space to be a direct sum of ##L^2(\mathbb R^3)## and ##\mathbb C^{2s+1}## where s is the particle's spin. Now we need a representation of the Lie algebra on each of these spaces. The L operators defined by (1.2.24) give us a representation on ##L^2(\mathbb R^3)##, so now we just need a representation on ##\mathbb C^{2j+1}##. This is what the spin operators are for. Note since the L+S operators are the Lorentz generators on the relevant Hilbert space, we're not adding the spin operators to the Lorentz generators. The spin operators are an essential part of the Lorentz generators, not something that we add to them. 



#6
Nov1012, 12:38 AM

P: 15

You say that
As an analogy, I feel that adding [itex] {{S_{\mu \nu }}}[/itex] to [itex] {{L_{\mu \nu }}}[/itex] is like multiplying a state vector (in quantum mechanics) by a phase. It doesn't change any matrix element just like in our case it doesn't change the commutation relations. I guess my mistake here is that I think that the commutation relations determine everything, but that is not true. Also, I'm used to thinking of a group or algebra as an abstract structure, which has realizations (i.e representations). Equation (1.2.24) is an example of a realization (of the group's infinitesimal generators) which operates on spacetime coordinates, while (1.2.27) is a more generalabstract relation that defines how the group behaves in any realizations. So the specific realization (1.2.24) operates on [itex]{L^2}({ℝ^3})[/itex], but that doesn't require that other representations will also operate on spacetime. They can just as well operate on spin space. I realize that there is some fundamental gap in my understanding of all of this. I hope my questions are understandable. Thanks 



#7
Nov1012, 04:59 AM

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To prove existence of a jm> with specific values of j and m, we need to define a specific representation. The definition of the L operators is the start of an existence proof for integer values of j. The definition of the Pauli spin matrices and the operators ##S_i=\frac 1 2 \sigma_i## is the start of the existence proof for j=1/2. For a more general existence proof, see e.g. theorem 4.9, page 102106 of "Lie grups, Lie algebras, and representations: an elementary introduction" by Brian Hall. Note that the jm> stuff is just what we do to represent the subalgebra su(2), but any representation of sl(2,ℂ) defines a representation of su(2) as well, so the jm> stuff will be part of what we consider when we define a representation of sl(2,ℂ). 



#8
Nov1012, 07:50 AM

P: 15

Thanks for your help!
I'm still a little bit confused... So you're saying that the commutation relations of the algebra (= the structure constants) do not include all the information about the group? Two algebras (e.g L and L+S) can have the exact same commutation relations but correspond to different groups? (in our example, L doesn't have a j=1/2 representation while L+S does, hence L and L+S are not isomorphic)? What additional structure do we need to define on the algebra so that it lets us tell apart two seemingly equivalent algebras? I mean, where is the information about the difference between the groups encoded when looking at the algebra? I thought that there is a one to one correspondence between the algebra and the group via exponentiation, but I guess this is not true since two equivalent algebras give rise to different groups. I'll put my question in another way: Suppose I define a new algebra thus: The basis operators are [itex]\left\{ {{Q_{\mu \nu }}} \right\}[/itex] where [itex]\mu ,\nu \in \left\{ {0\;,\;1\;,\;2\;,\;4} \right\}[/itex] and where the the set of operators [itex]\left\{ {{Q_{\mu \nu }}} \right\}[/itex] is antisymmetric in [itex]\mu ,\nu [/itex] (so there are only 6 different basis operators). The commutation relations are: [itex]\left[ {{Q_{\mu \nu }}\;,\;{Q_{\rho \sigma }}} \right] = i{\eta _{\nu \rho }}{Q_{\mu \sigma }}  i{\eta _{\mu \rho }}{Q_{\nu \sigma }}  i{\eta _{\nu \sigma }}{Q_{\mu \rho }} + i{\eta _{\mu \sigma }}{Q_{\nu \rho }}[/itex] where [itex]{\eta _{\nu \rho }} = {\rm{diag}}\left( {1\;,\;  1\;,\;  1\;,\;  1} \right)[/itex] and summation notation is used. Now define the group [itex]G[/itex] as the set of elements [itex]{e^{i{t_{\mu \nu }}{Q_{\mu \nu }}}}[/itex] for real parameters [itex]{t_{\mu \nu }}[/itex]. My set of questions is: Does the following question has a definite answer: "what are precisely the irreducible representations of the group [itex]G[/itex] ?" ? Can we know, without defining additional structure, if [itex]G[/itex] has irreducible representations in vector spaces of all finite dimensions? Or is there missing information and hence we can't determine which representations exist? Is [itex]G[/itex] isomorphic to the proper orthochronous Lorentz transformations? Since both [itex]{{L_{\mu \nu }}}[/itex] and [itex]{{L_{\mu \nu }} + {S_{\mu \nu }}}[/itex] obey the same commutation relations that [itex]{{Q_{\mu \nu }}}[/itex] does, but [itex]{{L_{\mu \nu }}}[/itex] doesn't have representations that [itex]{{L_{\mu \nu }} + {S_{\mu \nu }}}[/itex] does, I'm wondering where is the missing information. Thanks 



#9
Nov1012, 08:25 PM

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For example, if ##\big\{J_i\bigi\in\{1,2,3\}\big\}## is the basis of a 3dimensional Lie algebra, and ##[J_i,J_j]=i\varepsilon_{ijk}J_k##, then that Lie algebra is (isomorphic to) su(2). Suppose that R is a representation of su(2). This means that R is a Lie algebra homomorphism into a Lie algebra whose members are linear operators on a vector space, and whose Lie bracket is the commutator. So the definitions of "representation" and "su(2)" imply that ##[R(J_i),R(J_j)]=i\varepsilon_{ijk}R(J_k)##. Let's simplify the notation by defining ##K_i=R(J_i)##. The ##K_i## satisfy commutation relations that are identical to the Lie bracket relations of the ##J_i##. Since ##[K^2,K_3]=0##, these two operators may have simultaneous eigenvectors. The commutation relations for the ##K_i## imply that if there's a simultaneous eigenvector ##\mathbf{v}##, then there exist numbers (j,m) such that ##j\in\{0,1/2,1,3/2,\dots\}##, ##m\in\{j,j+1,\dots,j1,j\}##, and \begin{align} K^2\mathbf{v} &=j(j+1)\mathbf{v}\\ K_3\mathbf{v} &=m\mathbf{v}. \end{align} It's convenient to denote this ##\mathbf{v}## by ##jm\rangle##. The commutation relations do not imply that there exists such ##jm\rangle## for each (j,m) such that ##j\in\{0,1/2,1,3/2,\dots\}## and ##m\in\{j,j+1,\dots,j1,j\}##. Such things are determined by the function R, not by the commutation relations. That's the second important part of the answer to your question. For example, if \begin{align} \sigma_1 &=\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix},\qquad \sigma_2 =\begin{pmatrix}0 & i\\ i & 0\end{pmatrix},\qquad \sigma_3 =\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}\\ K_i &=R(J_i)=\frac{i}{2}\sigma_i, \end{align} then the ##K_i## satisfy the commutation relations, but every simultaneous eigenvector ##jm\rangle## has j=1/2 and m=±1/2. So there are only two of them. This was to be expected, because the vector space on which the linear operators (here represented by 2×2 matrices) are defined is 2dimensional. The third important part of the answer to your question is that even though the commutation relations define the Lie algebra, they do not determine the Lie group, because two Lie groups can have the same Lie algebra. For example SO(3) and SU(2) have the same Lie algebra, su(2). I'm not sure, but it may be the case that for each finitedimensional Lie algebra g there's exactly one simply connected Lie group with Lie algebra g. SO(3) isn't simply connected, but SU(2) is. SU(2) is homeomorphic to a 3sphere, while SO(3) is homeomorphic to a 3sphere with opposite points identified. In other words, SO(3) can be thought of as the collection of 1dimensional subspaces of ##\mathbb R^4##. 



#10
Nov1112, 12:45 PM

P: 15

Hi,
Thanks again for the answers! Your last paragraph raises some of my original questions again: Thanks 



#11
Nov1112, 09:06 PM

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[tex]\delta x^{\alpha} = (1/2) \omega^{\mu \nu} ( S_{\mu \nu})^{\alpha}{}_{\beta} x^{\beta}, \ \ (1)[/tex] where [itex]S_{\mu \nu}[/itex] are six 4 by 4 matrices with elements given by [tex]( S_{\mu \nu})^{\alpha}{}_{\beta} = \eta_{\mu \beta}\delta^{\alpha}_{\nu}  \eta_{\nu \beta} \delta^{\alpha}_{\nu}.[/tex] You can easily see that these matrices do form a representation of the Lorentz algebra. Indeed, they are the spin one matrices associated with Lorentz 4vector fields. They show up here (in that particular form) because the coordinates [itex]x^{\mu}[/itex] carry vector indices. In general, they act on abstract index space (we will come to that in a minute). We can also write the infinitesimal Lorentz transformation in terms of the generalized orbital angular momentum [tex]L_{\mu \nu} = x_{\mu}\partial_{\nu}  x_{\nu}\partial_{\mu},[/tex] as [tex]\delta x^{\alpha} = (1/2) \omega^{\mu \nu} L_{\mu \nu}x^{\alpha}. \ \ (2)[/tex] Using (1) and (2), we can write the infinitesimal Lorentz transformation as [tex] \delta x^{\alpha} = (1/4) \omega^{\mu \nu}\{ L_{\mu \nu} \delta^{\alpha}_{\beta} + ( S_{\mu \nu} )^{\alpha}{}_{\beta} \} x^{\beta}. \ \ (3)[/tex] Notice that L acts on the coordinates whereas S acts on the INDICES. So, to go over to abstract indices, only S will change. Now, at each spacetime point x, we define some generic FINITEDIMENTIONAL index space with elements [tex]\Phi^{A}(x) = \phi , V^{\mu}, \psi_{r}, F^{[\mu \nu ]}, G^{(\mu \nu)},\psi_{r}^{\mu}, \ … \ [/tex] (to be exact A = 1,4,4,6,10,16,…) The question about the transformation law of [itex]\Phi^{A}[/itex] can be answered in many different ways, but now we can just generalize eq(3) to [tex] \delta \Phi^{A}(x) = \frac{1}{2}\omega^{\mu \nu}\{ L_{\mu \nu} \delta^{A}_{B} + ( S_{\mu \nu} )^{A}{}_{B} \} \Phi^{B}(x). [/tex] This is exactly what you get when expanding the finite transformation [tex] \bar{\Phi}^{A}(x) = U^{1}(\Lambda ) \Phi^{A}(x) U(\Lambda ) = D^{A}{}_{B}( \Lambda ) \Phi^{B}( \Lambda^{1} x) [/tex] where [tex]D = \exp (\frac{1}{2} \omega \cdot S ), \ \ U = \exp ( \frac{1}{2} \omega \cdot J )[/tex] The [itex]D[/itex] forms a FINITEDIMENSIONAL NONUNITARY REPRESENTATION of the Lorentz group [itex]SO(1,3)[/itex] while the [itex]U[/itex] forms an INFINITEDIMENTIONAL UNIRARY REPRESENTATION. Sam 



#12
Nov1212, 04:32 AM

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$$\Lambda(\theta) =I+\theta^a\frac{\partial}{\partial\theta^a}\bigg_0\Lambda(\theta).$$ This firstorder term is what I'd denote by ##\omega## (suppressing ##\theta## from the notation). To first order in the parameters, ##x'=(I+\omega)x=x+\omega x##. In component form, $$x'^\mu =x^\mu+\omega^\mu{}_\nu x^\nu.$$ Is the second term on the right what you would denote by ##\delta x^\mu##? In that case, how do you get the S to appear, and why do we want it? 



#13
Nov1212, 05:45 AM

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[tex]\delta x^{\mu} = \omega^{\mu}{}_{\nu}x^{\nu}= \eta_{\nu \alpha}\delta^{\mu}_{\beta}\omega^{\beta \alpha}x^{\nu}[/tex] Now, interchange the dummy indices, [tex] \delta x^{\mu}= \{ \frac{1}{2} \eta_{\nu \alpha}\delta^{\mu}_{\beta}\omega^{\beta \alpha} + \frac{1}{2} \eta_{\nu \beta}\delta^{\mu}_{\alpha}\omega^{\alpha \beta}\}x^{\nu} [/tex] Now use [itex]\omega^{\alpha \beta} =  \omega^{\beta \alpha}[/itex], you find [tex] \delta x^{\mu} = \frac{1}{2} \omega^{\alpha \beta}( \eta_{\nu \beta}\delta^{\mu}_{\alpha}  \eta_{\nu \alpha} \delta^{\mu}_{\beta}) x^{\nu} [/tex] Sam 



#14
Nov1212, 06:33 AM

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Aaaah...I've been thinking that there should be a simple way to obtain the Lorentz algebra from the Lorentz group, but I didn't know how to do that until now. Thank you.




#15
Nov1212, 07:31 AM

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#16
Nov1212, 12:22 PM

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One needs a full discussion of specially relativistic invariance from the most general point of view, i.e. physics: special relativity (+ quantum mechanics); mathematics: differential geometry (+ harmonic analysis of locally compact groups on rigged Hilbert spaces).
There's no fully rigorous treatment, of course, each book has its own version, some more algebraic, other more analytic. I warmly reccomend the book by Fonda and Ghirardi, the Italian guys have a thorough treatment without using too much fancy mathematics. 


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