# LOG#097. Group theory(XVII).

The case of Poincaré symmetry

There is a important symmetry group in (relativistic, quantum) Physics. This is the Poincaré group! What is the Poincaré group definition? There are some different equivalent definitions:

i) The Poincaré group is the isometry group leaving invariant the Minkovski space-time. It includes Lorentz boosts around the 3 planes (X,T) (Y,T) (Z,T) and the rotations around the 3 planes (X,Y) (Y,Z) and (Z,X), but it also includes traslations along any of the 4 coordinates (X,Y,Z,T). Moreover, the Poincaré group in 4D is a 10 dimensional group. In the case of a ND Poincaré group, it has $N(N-1)/2+N$ parameters/dimensions, i.e., the ND Poincaré group is $N(N+1)/2$ dimensional.

ii) The Poincaré group formed when you add traslations to the full Lorentz group. It is sometimes called the inhomogenous Lorentz group and it can be denoted by ISO(3,1). Generally speaking, we will generally have $ISO(d,1)$, a D-dimensional ($D=d+1$) Poincaré group.

The Poincaré group includes as subgroups, the proper Lorentz transformations such as parity symmetry and some other less common symmtries. Note that the time reversal is NOT a proper Lorentz transformation since the determinant is equal to minus one.

Then, the Poincaré group includes: rotations, traslations in space and time, proper Lorentz transformations (boosts). The combined group of rotations, traslations and proper Lorentz transformations of inertial reference frames (those moving with constant relative velocity) IS the Poincaré group. If you give up the traslations in space and time of this list, you get the (proper) Lorentz group.

The full Poincaré group is a NON-COMPACT Lie group with 10 “dimensions”/parameters in 4D spacetime and $N(N+1)/2$ in the ND case.  Note that the boost parameters are “imaginary angles” so some parameters are complex numbers, though. The traslation subgroup of the Poincaré group is an abelian group forming a normal subgroup of the Poincaré group while the Lorentz grou is only a mere subgroup (it is not a normal subgroup of the Poincaré group). The Poincaré group is said, due to these facts, to be a “semidirect” product of traslations in space and time with the group of Lorentz transformations.

The case of Galilean symmetry

We can go back in time to understand some stuff we have already studied with respect to groups. There is a well known example of group in Classical (non-relativistic) Physics.

The Galilean group is the set or family of non-relativistic continuous space-time (yes, there IS space-time in classical physics!) transformations in 3D with an absolute time. This group has some interesting subgroups: 3D rotations, spatial traslations, temporal traslations and proper Galilean transformations ( transformations leaving invariant inertial frames in 3D space with absolute time). Thereforem the number of parameters of the Galilean group is 3+3+1+3=10 parameters. So the Galileo group is 10 dimensional and every parameter is real (unlike Lorentz transformations where there are 3 imaginary rotation angles).

The general Galilean group can be written as follows:

$G\begin{cases} \mathbf{r}\longrightarrow \mathbf{r}'=R\mathbf{r}+\mathbf{x_0}+\mathbf{V}t\\ t\longrightarrow t'=t+t_0\end{cases}$

Any element of the Galileo group can be written as a family of transformations $G=G(R,\mathbf{x_0},\mathbf{v},t_0)$. The parameters are:

i) $R$, an orthogonal (real) matrix with size $3\times 3$. It satisfies $RR^T=R^TR=I$, a real version of the more general unitary matrix $UU^+=U^+U=I$.

ii) $\mathbf{x_0}$ is a 3 component vector, with real entries. It is a 3D traslation.

iii) $\mathbf{V}$ is a 3 component vector, with real entries. It gives a 3D non-relativistic (or galilean) boost for inertial observers.

iv) $t_0$ is a real constant associated to a traslation in time (temporal traslation).

Therefore, we have 10 continuous parameters in general: 3 angles (rotations) defining the matrix $R$, 3 real numbers (traslations $\mathbf{x_0}$), 3 real numbers (galilean boosts denoted by $\mathbf{V}$) and a real number (traslation in time). You can generalize the Galilean group to ND. You would get  $N(N-1)/2+N+N+1$ parameters, i.e, you would obtain a $N(N+3)/2+1$ dimensional group. Note that the total number of parameters of the Poincaré group and the Galilean group is different in general, the fact that in 3D the dimension of the Galilean group matches the dimension of the 4D Poincaré group is a mere “accident”.

The Galilean group is completely determined by its “composition rule” or “multiplication operation”. Suppose that:

$G_3(R_3,\mathbf{z_0},\mathbf{V}_3,t_z)=G_2\cdot G_1$

with

$G_1(R_1,\mathbf{x_0},\mathbf{V}_1,t_x)$

and

$G_2(R_2,\mathbf{y_0},\mathbf{V}_2,t_y)$

Then, $G_3$ gives the composition of two different Galilean transformations $G_1, G_2$ into a new one. The composition rule is provided by the following equations:

$R_3=R_2R_1$

$\mathbf{z_0}=\mathbf{y_0}+R_2\mathbf{x_0}+\mathbf{V}_2 t_x$

$\mathbf{V}_3=\mathbf{V}_2+R_2\mathbf{V}_1$

$t_z=t_x+t_y$

Why is all this important? According to the Wigner theorem, for every continuous space-time transformation $g\in G$ should exist a unitary operator $U(g)$ acting on the space of states and observables.

We have seen that every element in uniparametric groups can be expressed as the exponential of certain hermitian generator. The Galilean group or the Poincaré group depends on 10 parameters (sometimes called the dimension of the group but you should NOT confuse them with the space-time dimension where they are defined). Remarkably, one can see that the Galilean transformations also act on “spacetime” but where the time is “universal” (the same for every inertial observer). Then, we can define

$iK_\alpha=\dfrac{\partial G}{\partial \alpha}\bigg| _{\alpha=0}$

These generators, for every parameter $\alpha$, will be bound to dynamical observables such as: linear momentum, angular momentum, energy and many others. A general group transformation for a 10-parametric (sometimes said 10 dimensional) group can be written as follows:

$\displaystyle{G(\alpha_1,\ldots,\alpha_{10}=\prod_{k=1}^{10}e^{iK_{\alpha_k}\alpha_k}}$

We can apply the Baker-Campbell-Hausdorff (BCH) theorem or simply expand every exponential in order to get

$\displaystyle{G(\alpha_1,\ldots,\alpha_{10})=\prod_{k=1}^{10}e^{iK_{\alpha_k}\alpha_k}=\exp \sum_{k=1}^{10}\omega_k (\alpha_1,\ldots,\alpha_{10})K_{\alpha_k}}$

$\displaystyle{G(\alpha_1,\ldots,\alpha_{10})=I+i\sum_{k=1}^{10}\omega_k(\alpha_1,\ldots,\alpha_{10})K_{\alpha_k}+\ldots}$

The Lie algebra will be given by

$\displaystyle{\left[K_i,K_j\right]=i\sum_{k}c_{ij}^kK_k}$

and where the structure constants will encode the complete group multiplication rules. In the case of the Poincaré group Lie algebra, we can write the commutators as follows:

$\left[X_\mu,X_\nu\right]=\left[P_\mu,P_\nu\right]=0$

$\left[M_{\mu\nu},P_\alpha\right]=\eta_{\mu\alpha}P_\nu-\eta_{\nu\alpha}P_\mu$

$\left[M_{\mu\nu},M_{\alpha\beta}\right]=\eta_{\mu\alpha}M_{\nu\beta}-\eta_{\mu\beta}M_{\nu\alpha}-\eta_{\nu\alpha}M_{\mu\beta}+\eta_{\nu\beta}M_{\mu\alpha}$

Here, we have that:

i) $P$ are the generators of the traslation group in spacetime. Note that as they commute with theirselves, the traslation group is an abelian subgroup of the Lorentz group. The noncommutative geometry (Snyder was a pioneer in that idea) is based on the idea that $P$ and more generally even the coordinates $X$ are promoted to noncommutative operators/variables/numbers, so their own commutator would not vanish like the Poincaré case.

ii) $M$ are the generators of the Lorent group in spacetime.

If we study the Galilean group, there are some interesting commutation relationships fo the corresponding generators (rotations and traslations). There are 6 “interesting” operators:

$K_{i}\equiv \overrightarrow{J}$ if $i=1,2,3$

$K_{i}\equiv \overrightarrow{P}$ if $i=,4,5,6$

These equations provide

$\left[P_\alpha,P_\beta\right]=0$

$\left[J_\alpha,J_\beta\right]=i\varepsilon_{\alpha\beta}^\gamma J_\gamma$

$\left[J_\alpha,P_\beta\right]=i\varepsilon_{\alpha\beta}^\gamma P_\gamma$

$\forall\alpha,\beta=1,2,3$

The case of the traslation group

In Quantum Mechanics, traslations are defined in the space of states in the following sense:

$\vert\vec{r}\rangle\longrightarrow\vert\vec{r}'\rangle =\exp\left(-i\vec{x_0}\cdot \vec{p}\right)\vert \vec{r}\rangle=\vert\vec{r}+\vec{x_0}\rangle$

Let us define two linear operators, $R$ and $R'$ associated, respectively, to initial position and shifted position. Then the transformation defining the traslation over the states are defined by:

$R\longrightarrow R'=\exp\left(-i\vec{x_0}\cdot\vec{p}\right)R\exp \left(i\vec{x_0}\cdot \vec{p}\right)$

where

$R_i\vert\vec{r}'\rangle=\vec{r}_i\vert\vec{r}'\rangle$

Furthermore, we also have

$\left[\vec{x_0}\cdot \vec{p},\vec{y_0}\cdot R\right]=-i\vec{x_0}\cdot\vec{y_0}$

$\left[R_\alpha,p_\beta\right]=i\delta_{\alpha\beta}I$

The case of the rotation group

What about the rotation group? We must remember what a rotation means in the space $\mathbb{R}^n$. A rotation is a transformation group

$\displaystyle{X'=RX\longrightarrow \parallel X'\parallel^2=\parallel X\parallel^2 =\sum_{i=i}^n (x'_i)^2=\sum_{i=1}^n x_i^2}$

The matrix associated with this transformation belongs to the orthogonal group with unit determinant, i.e., it is an element of $SO(N)$. In the case of 3D space, it would be $SO(3)$. Moreover, the ND rotation matrix satisfy:

$\displaystyle{I=X^TX=XX^T\leftrightarrow \sum_{i=1}^N R_{ik}R_{ij}=R_{ik}R_{ij}=\delta_{kj}}$

The rotation matrices in 3D depends on 3 angles, and they are generally called the Euler angles in some texts. $R(\theta_1,\theta_2,\theta_3)=R(\theta)$. Therefore, the associated generators are defined by

$iM_j\equiv\dfrac{\partial R}{\partial \theta_j}\bigg|_{\theta_j=0}$

Any other rotation matric can be decomposed into a producto of 3 uniparametric rotations, rotation along certain 2d planes. Therefore,

$R(\theta_1,\theta_2,\theta_3)=R_1(\theta_1)R_2(\theta_2)R_3(\theta_3)$

where the elementary rotations are defined by

Rotation around the YZ plane:

$R_1(\theta_1)=\begin{pmatrix} 1 & 0 & 0\\ 0 & \cos\theta_1 & -\sin\theta_1\\ 0 & \sin\theta_1 & \cos\theta_1\end{pmatrix}$

Rotation around the XZ plane:

$R_2(\theta_2)=\begin{pmatrix} \cos\theta_2 & 0 & \sin\theta_2\\ 0 & 1 & 0\\ -\sin\theta_2 & 0 & \cos\theta_2\end{pmatrix}$

Rotation around the XY plane:

$R_3(\theta_3)=\begin{pmatrix} \cos\theta_3 & -\sin\theta_3 & 0\\ \sin\theta_3 & \cos\theta_3 & 0\\ 0 & 0 & 1\end{pmatrix}$

Using the above matrices, we can find an explicit representation for every group generator (3D rotation):

$M_1=-i\begin{pmatrix}0 & 0 & 0\\ 0 & 0 & 1\\ 0 & -1 & 0\end{pmatrix}$

$M_2=-i\begin{pmatrix}0 & 0 & -1\\ 0 & 0 & 0\\ 1 & 0 & 0\end{pmatrix}$

$M_3=-i\begin{pmatrix}0 & 1 & 0\\ -1 & 0 & 0\\ 0 & 0 & 0\end{pmatrix}$

and we also have

$\left[M_j,M_k\right]=i\varepsilon^{m}_{jk}M_{m}$

where the $\varepsilon^m_{jk}=\varepsilon_{mjk}$ is the completely antisymmetry Levi-Civita symbol/tensor with 3 indices. There is a “for all practical purposes” formula that represents a rotation with respect to some axis in certain direction $\vec{n}$. We can make an infinitesimal rotation with angle $d\theta$, due to the fact that rotation are continuous transformations, it commutes with itself and it is unitary, so that:

$R(d\theta)\vec{r}=\vec{r}+d\theta(\vec{n}\times\vec{r}+\mathcal{O}(d\theta^2)=\vec{r}-id\theta M_\alpha\vec{r}+\mathcal{O}(d\theta^2)$

In the space of physical states, with $\vec{k}=\theta\vec{n}$ some arbitrary vector

$\vec{r}'=R\vec{r}\longrightarrow\vert\vec{r}'\rangle=\vert R\vec{r}\rangle=U(R)\vert\vec{r}\rangle=e^{-i\vec{k}\cdot\vec{J}}\vert\vec{r}\rangle=e^{-i(k_xJ_x+k_yJ_y+k_zJ_z)}\vert \vec{r}\rangle$

Here, the operators $J=(J_x,J_y,J_z)$ are the infinitesimal generators in the space of physical states. The next goal is to relate these generators with position operators $Q$ through commutation rules. Let us begin with

$Q\longrightarrow Q'=e^{-i\vec{k}\cdot{J}}Qe^{i\vec{k}\cdot\vec{J}}$

$Q'\vert\vec{r}'\rangle =\vec{r}\vert\vec{r}'\rangle$

Using this last result, we can calculate for any 2 vectors $\vec{k},\vec{n}$:

$\left[\vec{k}\cdot\vec{J},\vec{n}\cdot\vec{Q}\right]=i(\vec{k}\times\vec{n})\cdot\vec{Q}$

or equivalent, in component form,

$\left[J_j,Q_k\right]=i\varepsilon_{jkm}Q_m$

These commutators complement the above commutation rules, and thus, we have in general

$\left[\vec{k}\cdot\vec{J},\vec{n}\cdot\vec{Q}\right]=i(\vec{k}\times\vec{n})\cdot\vec{Q}$

$\left[J_j,J_k\right]=i\varepsilon_{jkm}J_m$

$\left[J_j,Q_k\right]=i\varepsilon_{jkm}Q_m$

In summary: a triplet of rotation operators generates “a vector” somehow.

The case of spinning particles

In fact, these features provide two different cases in the case of a single particle:

i) Particles with no “internal structure” or “scalars”/spinless particles. A good example could it be the Higgs boson.

ii) Particles with “internal” degrees of freedom/structure/particles with spin.

In the case of a particle without spin in 3D we can define the angular momentum operator as we did in classical physics ($L=r\times p$), in such a way that

$J=Q\times P$

Note that the “cross product” or “vector product” in 3D is generally defined if $C=A\times B$ as

$C=A\times B=\begin{vmatrix}i & j & k\\ A_x & A_y & A_z\\ B_x & B_y & B_z\end{vmatrix}$

or by components, using the magical word XYZZY, we also have

$C_x=A_yB_z-A_zB_y$

$C_y=A_zB_x-A_xB_z$

$C_z=A_xB_y-A_yB_x$

Remember that the usual “dot” or “scalar” product is $A\cdot B=A_xB_x+A_yB_y+A_zB_z$

Therefore, the above operator $J$ defined in terms of the cross product satisfies the Lie algebra of $SO(3)$.

By the other hand, in the case of a spinning particle/particle with spin/internal structure/degrees of freedom, the internal degrees of freedom must be represented by some other operator, independently from $Q,P$. In particular, it must also commute with both operators. Then, by definition, for a particle with spin, the angular momentum will be a sum with two contributions: one contribution due to the “usual” angular momentum (orbital part) and an additional “internal” contribution (spin part). That is, mathematically speaking, we should have a decomposition

$J=Q\times P+S$

with $\left[Q,S\right]=\left[P,S\right]=0$

If $S$, the spin operator, satisfies the above commutation rules (in fact, the same relations than the usual angular momentum), we must impose

$\left[S_j,S_k\right]=i\varepsilon_{jkm}S_m$

The case of Parity P/Spatial inversions

This special transformation naturally arises in some applications. From the pure geometrical viewpoint, this transformation is very simple:

$\vec{r}'=-\vec{r}$

In coordinates and 3D, the spatial inversion or parity is represented by a simple matrix equals to minus the identity matrix

$P=\begin{pmatrix} -1 & 0 & 0\\ 0 & -1 & 0\\ 0& 0 & -1\end{pmatrix}$

This operator correspods, according to the theory we have been studying, to some operator P (please, don’t confuse P with momentum) that satisfies

$PqP^{-1}=-q$

$PpP^{-1}=-p$

and where $q, p$ are the usual position and momentum operators. Then, the operator

$L=q\times p$ is invariant by parity/spatial inversion P, and thus, this feature can be extended to any angular momentum operator like spin S or angular momentum J. That is,

$PJP^{-1}=J$ and $PSP^{-1}=S$

The Wigner’s theorem implies that corresponding to the operator P, a discrete transformation, must exist some unitary or antiunitary operator. In fact, it shows that P is indeed unitary

$P\left[Q_i,P_j\right]P^{-1}=\left[Q_i,P_j\right]=P(i\hbar\delta_{ij})P^{-1}$

If P were antiunitary we should get

$P\left[Q_i,P_j\right]P^{-1}=\left[Q_i,P_j\right]=P(i\hbar\delta_{ij})P^{-1}=-i\hbar\delta_{ij}$

Then, the parity operator P is unitary and $P^{-1}=P$. In fact, this can be easily proved from its own definition.

If we apply two succesive parity transformations we leave the state invariant, so $P^2=I$. We say that the parity operator is idempotent.  The check is quite straightforward

$\vert\Psi\rangle\longrightarrow\vert\Psi'\rangle=PP\vert\Psi\rangle\longrightarrow\vert\Psi\rangle=e^{i\omega}\vert\Psi\rangle$

Therefore, from this viewpoint, there are (in general) only 2 different ways to satisfy this as we have $PP=e^{i\omega}I$:

i) $e^{i\omega}=+1$. The phase is equal to $0$ modulus $2\pi$. We have hermitian operators

$P=P^{-1}=P^+$

Then, the effect on wavefunctions is that $\Psi (P^{-1}(\vec{r}))=\Psi (-\vec{r})$. That is the case of usual particles.

ii) The case $e^{i\omega}=-1$. The phase is equal to $\pi$ modulus $2\pi$. This is the case of an important class of particles. In fact, Steven Weinberg has showed that $P^2=(-1)^F$ where F is the fermion number operator in the SM. The fermionic number operator is defined to be the sum $F=L+B$ where L is now the leptonic number and B is the baryonic number. Moreover, for all particles in the Standard Model and since lepton number and baryon number are charges Q of continuous symmetries $e^{iQ}$  it is possible to redefine the parity operator so that $P^2=I$. However, if there exist Majorana neutrinos, which experimentalists today believe is quite possible or at least it is not forbidden by any experiment, their fermion number would be equal to one because they are neutrinos while their baryon and lepton numbers are zero because they are Majorana fermions, and so $(-1)^F$ would not be embedded in a continuous symmetry group. Thus Majorana neutrinos would have parity equal to $\pm i$. Beautiful and odd, isnt’t it? In fact, if some people are both worried or excited about having Majorana neutrinos is also due to the weird properties a Majorana neutrino would have under parity!

The strange case of time reversal T

In Quantum Mechanics, temporal inversions or more generally the time reversal is defined as the operator that inverts the “flow or direction” of time. We have

$T: t\longrightarrow t'=-t$

$\vec{r}'(-t)=\vec{r}(t)$

And it implies that $\vec{p}(-t)=-\vec{p}(t)$. Therefore, the time reversal operator $T$ satisfies

$TQT^{-1}=Q$

$TPT^{-1}=-P$

In summary: T is by definition the “inversion of time” so it also inverts the linear momentum while it leaves invariant the position operator.

Thus, we also have the following transformation of angular momentum under time reversal:

$TJT^{-1}=-J$

$TST^{-2}=-S$

Time reversal can not be a unitary operator, and it shows that the time reversal T is indeed an antiunitary operator. The check is quite easy:

$T\left[Q,P\right]T^{-1}=\left[TQT^{-1},TPT^{-1}\right]=-\left[Q,P\right]=Ti\hbar T^{-1}$

This equation matches the original definiton if and only if (IFF)

$TiT^{-1}=-i \leftrightarrow TT^{-1}=-1$

Time reversal is as consequence of this fact an antiunitary operator.

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