# Lorentz Generators: (J and M) vs. S

by liorde
Tags: generators, lorentz, spin, spinor
HW Helper
P: 11,927
 Quote by 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
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.
 P: 9 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.
 Sci Advisor Thanks P: 2,497 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.
P: 910
 Quote by 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.
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.

 The correct Lie algebra isomorphisms are $$\mbox{so(1,3)} \simeq \mbox{sl}(2,\mathbb{C}) \simeq \mbox{su(2)} \oplus \mbox{su(2)}$$ $$\mbox{so(1,3)}_{C} \simeq \mbox{su(2)} \oplus \mbox{su(2)}$$

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
$$W^{ 2 } = - \frac{ m^{ 2 }}{ 2 } ( C_{ 1 } + C_{ 2 } ) + M \cdot P \cdot \bar{ M } \cdot P$$
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.