# Physics Notes by Jakob Schwichtenberg

### Sidebar

the_standard_model:poincare_group

# The Poincare Group

also known as inhomogeneous Lorentz group.

The Poincare group is not semisimple, because it is a semidirect product of the translation group $\mathbb{R}(3,1)$ and the Lorentz group $O(3,1)$:

$$\text{Poincare} = O(3,1) \ltimes \mathbb{R}(3,1)$$

In addition, the Poincare group is non-compact, because we can boost to arbitrarily high values of the momentum and hence the parameter space is infinite.

Take note that in practice, we use its universal covering group $$SL(2,\mathbb{C}) \ltimes \mathbb{R}(3,1),$$ where $SL(2,\mathbb{C})$ is the group of complex $2 \times 2$ matrices with unit determinant. The reason for this is that representations of $O(3,1)$ can't describe particles with spin $1/2$, i.e. fermions.

consider the Lorentz group, SO(3,1). Again it's not really this group but its double cover that matters in physics; its double cover is SL(2,C).http://math.ucr.edu/home/baez/twf_ascii/week109

## Why is it interesting?

The double cover of the Poincare group is the fundamental spacetime symmetry of modern physics. For example, it lies at the heart of the standard model of particle physics.

The Poincare group is the set of all transformations that leave the speed of light invariant. Thus, the Poincare group yields all possible transformations between allowed frames of reference. This is incredibly useful, when we want to write down fundamental laws of nature. The fundamental laws should be valid in all allowed frames of reference, otherwise they would be quite useless.

In practice, we can use our knowledge of all transformations inside the Poincare group to write down equations that are invariant under all these transformations. These equations then hold in all allowed frames of reference. This is such a strong restriction on the possible equations that is is almost enough to derive the most important equations of fundamental physics: the Dirac equation, the Klein-Gordon equation and the Maxwell-Equations.

The Hilbert space of one-particle states is always an irreducible representation space of the Poincare group. […] The construction of the unitary irreducible representations of the Poincare group is probably the most successful part of special relativity (in particle physics, not in gravitation theory, for which it is a disaster). It permits us to classify all kinds of particles and implies the main conservation laws (energy-momentum and angular momentum). […] The translation generators are responsible for the energy-momentum conservation laws, the rotation generators of the conservation of angular momentum, and the boost generators of the conservation of initial position. ” from Reflections on the Evolution of Physical Theories by Henri Bacry
The enlargement of the Lorentz group to the Poincare group was proposed [ 13] as a way of describing the quantum states of relativistic particles without using the wave equations. The states of a free particle are then given by the unitary irreducible representations of the Poincare group.” from Deformed Poincare containing the exact Lorentz algebra by Alexandros A. Kehagias et. al.

## Representation Theory

### Representations of the Lorentz Group

At the heart of the representation theory of the Poincare group is the representation theory of the proper orthochronous Lorentz group $SO(1,3)^{\uparrow}$. We can concentrate on this subset of the Lorentz group, because the Lorentz group can be decomposed as follows:

$$\label{eq:completedef03} O(1,3) = \{ SO(1,3)^{\uparrow}, \Lambda_P SO(1,3)^{\uparrow} , \Lambda_T SO(1,3)^{\uparrow} , \Lambda_P \Lambda_T SO(1,3)^{\uparrow} \} \, .$$ These four disconnected components are connected through parity transformations $\Lambda_P$ and time-reversal transformations $\Lambda_T$. This means in practice that once we have derived the representations of $SO(1,3)^{\uparrow}$, we get the representations of the other components (and hence of the full Lorentz group) simply by acting with the corresponding representations of the parity and time-reversal transformations on our representations.

The lowest-dimensional irreducible representation of the Lorentz group are

Name Dimension $(j_L,j_R)$
scalar 1 $(0,0)$
left-chiral spinor 2 $(\frac{1}{2},0)$
right-chiral spinor 2 $(0,\frac{1}{2})$
vector 4 $(\frac{1}{2},\frac{1}{2})$
self dual antisymmetric 3 $(1,0)$
anti-self dual antisymmetric 3 $(0,1)$
traceless symmetric 9 $(1,1)$

Moreover, in physical application the reducible Dirac representation is important: $$(\frac{1}{2},0) \oplus (0,\frac{1}{2}) .$$ This means, the famous Dirac spinors transform according to this reducible representation. Further, we have $$(\frac{1}{2},0) \otimes (0,\frac{1}{2}) = (\frac{1}{2},\frac{1}{2})$$ $$(\frac{1}{2},0) \otimes (\frac{1}{2},0) = (1,0) \oplus (0,0).$$

The adjoint representation of the Lorentz group is $6$-dimensional and given by $$(1,0) \oplus (0,1) .$$ (Source: page 42 in http://solenodonus.com/file/electromagnetic-duality-for-children.pdf )

#### Derivation of the Representations of the Lorentz Group

To derive the representations of $SO(3,1)$, we employ Weyl's unitary trick. This trick allows us to derive irreducible non-unitary representations of the Lorentz group, by starting with the known unitary representations of $SO(4)$. However this

works only for the finite or spin representations and for the Lorentz group these by no means exhaust all representations.” from INVARIANCE IN ELEMENTARY PARTICLE PHYSICS BY N . KEMMER et. al.

Take note that this application of Weyl's unitary trick is so important that is has its own name: Wick rotation. This means the Weyl unitary trick that deformes $SO(3,1) \rightarrow SO(4)$ is called a Wick rotation. (c.f. p. 28 in http://math.ucr.edu/home/baez/thesis_wise.pdf)

We can use that the Lie algebra of $SO(4)$, denoted by $\mathfrak{so}(4)$, is isomorphic to to the Lie algebra $\mathfrak{su}(2)\times \mathfrak{su}(2)$. (In fact, $SU(2) \times SU(2)$ is the double cover of $SO(4)$ and therefore we have the isomorphism $SO(4) \simeq SU(2) \times SU(2) / \mathbb{Z}_2$).

The understand this explicitly, we write down the commutation relations of the Lorentz group:

$$\label{eq:JJcommutator} [J_i,J_j]=i \epsilon_{ijk} J_k$$ $$\label{eq:JKcommutator} [J_i,K_j]=i \epsilon_{ijk} K_k$$ $$\label{eq:KKcommutator} [K_i,K_j]=- i \epsilon_{ijk} J_k \, ,$$ where $J_i$ are the generators of rotations and $K_i$ are the generators of boosts. This means, a general Lorentz transformation can be written as $$\Lambda = {\mathrm{e }}^{i \vec J \cdot \vec \theta + i \vec K \cdot \vec \Phi} .$$

Now, we employ Weyl's unitary trick and multiply the boost generators with $i$: $K_i \rightarrow i K_i$. This is a deformation of the Lie algebra. The deformed Lie algebra is simply the Lie algebra $\mathfrak{so}(4)$. Then, we can introduce the following new generators from the old ones $$\label{eq:defN} N^{+}_i \equiv \frac{1}{2}(J_i +i K_i)$$ $$N^{-}_i \equiv \frac{1}{2}(J_i -i K_i) .$$ (Another point of view is that we here complexify the Lorentz algebra: $\mathfrak{so}(3,1) \rightarrow \mathfrak{so}(3,1)_\mathbb{C}$, because we now allow complex linear combinations of the generators).

The commutation relations for these new set of generators are $$[N^{+}_i,N^{+}_j]=i \epsilon_{ijk} N^{+}_k$$ $$[N^{-}_i,N^{-}_j]=i \epsilon_{ijk} N^{-}_k$$ $$[N^{+}_i,N^{-}_j]= 0 \, .$$ Thus, we can see that we have here simply two times the Lie algebra $\mathfrak{su}(2)$. We conclude that through the introduction of the new set of generators $N^{+}_i$ and $N^{-}_i$, we have found an isomorphism between $\mathfrak{so}(4)$ and $\mathfrak{su}(2) \times \mathfrak{su}(2)$.

To summarize: <diagram>

 $\mathfrak{so}(3,1)$ - - Weyl Trick - [email protected] $\mathfrak{so}(4)$ [email protected] $\mathfrak{su}(2) \times \mathfrak{su}(2)$

</diagram>

Take note that we have changed something fundamental through the Lie algebra deformation (the Weyl trick) $K_i \rightarrow i K_i$. The resulting complexified Lorentz algebra $\mathfrak{so}(3,1)_\mathbb{C}$ is isomorphic to $\mathfrak{sl}(2,\mathbb{C})_\mathbb{C}$. Thus by deriving the representations of this Lie algebra we get the representations of the group $SL(2,\mathbb{C})$. $SL(2,\mathbb{C})$ is the group of complex $2 \times 2$ matrices with unit determinant. The difference between $SL(2,\mathbb{C}$ and $SU(2)$ is that the former group contains transformations that aren't necesarrily unitary). This group is the universal cover of the Lorentz group. ( $SL(2,\mathbb{C})$ double covers $SO(3,1)^\uparrow$. This means there is a map that maps $SL(2,\mathbb{C})$ elements to $SO(3,1)^\uparrow$ elements and this map is not one-to-one, but two-to-one. Two elements of $SL(2,\mathbb{C})$ get mapped to the same element of $SO(3,1)^\uparrow$.) Therefore by studying the irreducible representations of $\mathfrak{so}(3,1)_\mathbb{C}$, we do not only get representations of the Lorentz group, but additional representations. One additional representations is, for example, the famous spin $1/2$ representation that we use to describe fermions. This means the representations of the Lorentz group aren't enough to describe all fundamental systems. We actually need to representations of the double cover of the Lorentz group to describe fermions.

the Lorentz group is not compact because it contains boosts (hence all unitary representations are infinite-dimensional); and it is not simply connected because it contains rotations (so we need to study the representations of its universal covering group SL(2, C)).http://www.staff.uni-giessen.de/~gd1632/2014-hadron-physics/hadron-app-2.pdf
physical experiments show that a connected double cover of the Poincare is more appropriate in creating the symmetry group actual spacetime.https://apps.carleton.edu/curricular/math/assets/symmetrycompsproject.pdf

The usual way how the representation theory of the Lorentz algebra is explained makes use of the complexification of the Lie algebra and the fact that the complexified Lorentz algebra looks like two copies of the $SU(2)$ Lie algebra $\mathfrak{su}(2)$: $$\mathfrak{so}(3,1)\otimes \mathbf C = \mathfrak{sl}(2,\mathbf C) + \mathfrak{sl}(2,\mathbf C)$$ However, the complexification of the Lie algebra of an orthogonal group is the same thing no matter what signature you start with $$\mathfrak{so}(3,1)\otimes \mathbf C =\mathfrak{so}(4)\otimes \mathbf C =\mathfrak{so}(2,2)\otimes \mathbf C = \mathfrak{sl}(2,\mathbf C) + \mathfrak{sl}(2,\mathbf C)$$

The Lie algebras we are interested in in physics are real Lie algebras. By restricting to real Lie algebras, we can still consider different real forms of a given Lie algebra: \begin{align} \mathfrak{so}(3,1)&=\mathfrak{sl}(2,\mathbf C) \\ \mathfrak{so}(4)&=\mathfrak {su}(2) + \mathfrak {su}(2) \\ \mathfrak{so}(2,2)&=\mathfrak{sl}(2,R) +\mathfrak{sl}(2,R) \end{align}

### Representations of the Poincare Group

To make sense of the probabilistic interpretation of quantum theory, we need to use unitary representations (see e.g. page 69 in Schwartz's book).

However, the Poincare group is non-compact and there is a theorem that “unitary representations of non-compact groups are infinite-dimensional” (Ref).

The infinite-dimensional representations are considered unphysical because we never see particle states in nature labelled by extra continuous parameters.http://homepages.uc.edu/~argyrepc/cu661-gr-SUSY/susy2001.pdf

The Poincare group is non-compact, because we can boost to arbitrarily high values of the momentum and hence the parameter space is infinite. (“The Lorentz group has both finite-dimensional and infinite-dimensional representations. However, it is non-compact, therefore its finite-dimensional representations are not unitary (the generators are not Hermitian). The generators of the infinite-dimensional representations can be chosen to be Hermitian.Source).

Thus it seems highly problematic how we can make sense of the representation of the Poincare group in quantum theory.

However, by looking at this problem a bit closer a solution can be found: The infinite dimensionality of the unitary representations of the Poincare group stems from the fact that we label particle states with the continuous parameter $p_\mu$, called four-momenta.

Therefore, we can organize the representations of the Poincare group, with the help of so called little groups. These little groups are the subgroups of the Poincare group that are left over after we have fixed the non-compact transformations (boosts and translations).

This means that we organize the representations of the Poincare group, while holding the non-compact transformations fixed. (The non-compact part of the Poincare group are the boosts and translations. )

We label representations usually with the help of Casimir operators. One Casimir operator of the Poincare group is $p_\mu p^\mu$, i.e. the momentum squares. The Casimir operator yields simply a number for each representation. The number that corresponds to $p_\mu p^\mu$ is what we call the mass in physics.

With this understanding, we are able to classify the representations. For the Poincare group the physically relevant classes of representation are the positive mass representation and the massless representation. (The other representations are either trivial or describe, for example, tachyons.)

Formulated differently, we fix a representation of the translation group which is labelled by some number $M$ (= the value of the quadratic Casimir operator). Then we investigate the effect of the Lorentz transformations on such a fixed representation. The subgroup of the Lorentz group that leaves the fixed representation of the translation group invariant, is called the little group of this translation representation.

For example, we fix the representation of the translation generator $$p_\mu = (m,0,0,0)$$ and then investigate the representations of the subgroup that leaves this representation invariant. The little group is in this case $SO(3)$, the group that rotates the spatial part of the four-vector. This simply means that we label massive particles through their spin $j$, i.e. the quantum number that labels $SO(3)$ representations and through the fixed mass $m$.

Similarly, we can consider the case

$$p_\mu = (E,0,0,E)$$

for which $p_\mu p^\mu =m^2 = 0$ holds. We can choose the four-vector for this massless case without any loss of generality, because it is always possible to boost a massless particle into such a state.

The little group in this case is $SO(2)$, i.e. rotations around the $z$-axis. Therefore, we label massless particles by just one number, called helicity $\lambda$.

A good discussion for the Poincare group can be found here and chapter 4 “The Poincaré transformations” in Symmetries and Group Theory in Particle Physics by Costa and Fogly. To quote from it:

“The unitary representation realized by the operators U (a, Λ) is, of course, infinite-dimensional, since the Hilbert space of the state vectors |Φ> is infinite- dimensional, but it is, in general, reducible. To obtain the unitary irreducible representations, one has to find the invariant subspaces, i.e. invariant under the transformations (a, Λ).|Φ >”

### Possible Points of Confusion

For the massless case: “Actually, the little group preserving p µ is isomorphic to the non-compact group of Euclidean motions on the plane—SO(2) plus two “translations” generated by the linear combinations K1 + J2 and K2 − J1 of boosts and rotations. However, being a non-compact group itself, this little group’s unitary representations are infinite-dimensional, except when the eigenvalues of the “translations” are zero, in which case it effectively reduces to SO(2). The infinite-dimensional representations are considered unphysical because we never see particle states in nature labelled by extra continuous parameters.http://homepages.uc.edu/~argyrepc/cu661-gr-SUSY/susy2001.pdf

1. The lengths of the momentum four-vector: $P_\mu P^\mu$, where $P^\mu$ are the generators of the translation group. The physical interpretation of the characteristic numbers $P_\mu P^\mu \equiv m^2$ is that it's the square of the mass.
2. The Pauli-Lubanski four-vector $$W^2=W_0^2-\vec{W}\cdot \vec{W} ,$$ where \begin{eqnarray} W_0&=&\vec{J}\cdot \vec{P}, \nonumber \\ W_i&=& P_0J_i+\epsilon_{ijk}P_jK_k , \end{eqnarray} .

## Discussion of the Double Cover of the Poincare Group

As mentioned above, in physics we actually use the double cover of the Poincare group $SL(2,\mathbb{C}) \ltimes \mathbb{R}(3,1),$ because the original Poincare group $O(3,1) \ltimes \mathbb{R}(3,1)$ is not able to describe fermions.

The Lorentz group $O(3,1)$ is the set of transformations that leave the speed of light invariant. This can be translated mathematically into the condition that $O(3,1)$ contains all transformations $M$ that leave the Minkowksi metric $\eta = diag(1,-1,-1,-1)$ invariant: $$M^T \eta M \stackrel{!}{=} \eta .$$ This in turn means that $O(3,1)$ must leave the spacetime interval $$s^2= t^2-x^2-y^2-z^2 = V^T \eta V$$ invariant, where $V = (t,x,y,z)$ is a usual four-vector. For an element $O \in O(3,1)$, we have under the transformation $V \rightarrow OV$ $$s^2 = V^T \eta V \rightarrow s'^2 = (O V)^T \eta (OV)= V^T O^T \eta O V.$$ Therefore, if $O^T \eta O = \eta$, we have, $s^2=s'^2$.

How is this condition encoded in the double cover group $SL(2,\mathbb{C})$?

The vector representation ( = the $(\frac{1}{2},\frac{1}{2})$ representation) of $SL(2,\mathbb{C})$ acts on the space of hermitian $(2\times 2)$ matrices. (The Minkowski space is isomorphic to the space of hermitian $(2\times 2)$ matrices). A basis for this space is given by the usual Pauli matrices $\sigma_j$, together with the $(2\times 2)$ unit matrix. Concretely this means that we can represent an event in spacetime as $$V = t\,\mathrm{id} + x\,\sigma_x+y\,\sigma_y+z\,\sigma_z.$$ Take note that this is a matrix, in contrast to the usual four-vectors that are used in special relativity. Therefore, an $\zeta$ element of $SL(2,\mathbb{C})$ acts on such an event $V$ as $$V\mapsto \zeta^\dagger\,V\,\zeta .$$ With this interpretation, we can now check if the Minkowski metric is invariant under $SL(2,\mathbb{C})$ representations. First, we note that $$\det{V}=t^2-x^2-y^2-z^2.$$ Thus, we check $$\det(\zeta^\dagger\,V\,\zeta) = |\det\zeta|^2 \det V = \det V \quad \blacksquare$$ Take note that $\det\zeta$ holds by definition, because $\zeta$ is an element of the group $SL(2,\mathbb{C})$. ($SL(2,\mathbb{C})$ is the group of $(2 \times 2)$ matrices with unit determinant).

## The Poincare Algebra

\begin{eqnarray} {}[J_i,J_j]&=&i \epsilon_{ijk}J_k, \nonumber \\ {} [J_i,K_j]&=& i\epsilon_{ijk} K_k, \nonumber \\ {} [K_i,K_j]&=& -i \epsilon_{ijk}J_k ,\nonumber \\ {} [J_i,P_0]&=&0, \nonumber \\ {} [J_i,P_j]&=& i\epsilon_{ijk} P_k, \nonumber \\ {} [P_0,P_i]&=&0, \nonumber \\ {} [K_i,P_0]&=&i P_i, \nonumber \\ {}[K_i,P_j]&=& i\ P_0 \delta_{ij} , \label{c} \end{eqnarray}