[Weinberg QFT] 2.4 The Poincare Algebra

  This article is one of the posts in the Textbook Commentary Project.

As we saw in Section 2.2, much of the information about any Lie symmetry group is contained in properties of the gorup elements near the identity. For the inhomogeneous Lorentz group, the identity is the transformation $\Lambda^\mu_\nu = \delta^\mu_\nu$, $a^\mu=0$, so we want to study those transformations with $$\begin{align}\Lambda^\mu_\nu = \delta^\mu_\nu + \omega^\mu_\nu,\ a^\mu = \epsilon^\mu,\end{align}$$ both $\omega^\mu_\nu$ and $\epsilon^\mu$ being taken infinitesimal. The Lorentz condition (5) reads here $$\eta_{\rho\sigma} = \eta_{\mu\nu}(\delta^\mu_\rho + \omega^\mu_\rho)(\delta^\nu_\sigma + \omega^\nu_\sigma) = \eta_{\sigma\rho} + \omega_{\sigma\rho} + \omega_{\rho\sigma} + O(\omega^2).$$

We are here using the convention, to be used throughout this book, that indices may be lowered or raised by contraction with $\eta_{\mu\nu}$ or $\eta^{\mu\nu}$ $$\begin{align*}\omega_{\sigma\rho} & \equiv \eta_{\mu\sigma} \omega^\mu_\rho \\  \omega^\mu_\rho &\equiv \eta^{\mu\sigma} \omega_{\sigma\rho}.\end{align*}$$ 

Keeping only the terms of first order in $\omega$ in the Lorentz condition (2.3.5), we see that this condition now reduces to the ansitymmetry of $\omega_{\mu\nu}$ $$\begin{align}\omega_{\mu\nu} = -\omega_{\nu\mu}.\end{align}$$

An antisymmetric second-rank  tensor in four dimensions has $(4\times 3)/2=6$ idependent components, to uncluding the four components of $\epsilon^\mu$, an inhomogeneous Lorentz transformation is described by $6+4=10$ parameters.

Since $U(1,0)$ carries any ray into itself, it must be proportional to the unit operator, and by a choice of phase may be made euqal to it. For an infinitesimal Lorentz transformation (1), $U(1+\omega,\epsilon)$ must then equal $1$ plus terms linear in $\omega_{\rho\sigma}$ and $\epsilon_\rho$. We write this as $$\begin{align}U(1 + \omega, \epsilon) = 1 + \frac{1}{2} i \omega_{\rho\sigma} J^{\rho\sigma} - i\epsilon_\rho P^\rho + \cdots.\end{align}$$

Here $J^{\rho\sigma}$ and $P^\rho$ are $\omega$-and $\epsilon$-independent operators, and the dots denote terms of higher order in $\omega$ and $/$ pr $\epsilon$. In order for $U(1+\omega,\epsilon)$ to be unitary, the operators $J^{\rho\sigma}$ and $P^\rho$ must be Hermitian $$\begin{align}J^{\rho\sigma\dagger} = J^{\rho\sigma},\ P^{\rho\dagger} = P^{\rho}.\end{align}$$ 

Since $\omega_{\rho\sigma}$ is antisymmetric, we can take its coefficient $J^{\rho\sigma}$ to be antisymmetric also $$\begin{align}J^{\rho\sigma} = -J^{\sigma\rho}.\end{align}$$

As we shall see, $P^1,P^2$, and $P^3$ are the components of the momentum operators, $J^{23},J^{31},$and $J^{12}$ are the components of the angular momentum vector, and $P^0$ is the energy operator, or Hamiltonian.

We now examine the Lorentz transformation properties of $J^{rho\sigma}$ and $P^\rho$. We consider the product $$U(\Lambda, a) U(1+\omega, \epsilon) U^{-1}(\Lambda, a),$$ where $\Lambda^\mu_\nu$ and $a^\mu$ are here the parameters of a new transformation, unrelated to $\omega$ and $\epsilon$. According ti Eq. (2.3.11), the product $U(\Lambda^{-1}, -\Lambda^{-1} a)U(\Lambda,a)$ equals $U(1,0)$, so $U(\Lambda^{-1}, -\Lambda^{-1} a)$ is the inverse of $U(\Lambda,a)$. If follows then from (2.3.11) that $$\begin{align}U(\Lambda, a) U(1+\omega, \epsilon) U^{-1} (\Lambda, a) = U\left( \Lambda ( 1+\epsilon)\Lambda^{-1}, \Lambda \epsilon - \Lambda \omega \Lambda^{-1} a\right) .\end{align}$$

To first order in $\omega$ nad $\epsilon$, we have then $$\begin{align}U(\Lambda, a)\left[\frac{1}{2}\omega_{\rho\sigma} J^{\rho\sigma} - \epsilon_\rho P^\rho\right] U^{-1}(\Lambda, a) = \frac{1}{2} \left(\Lambda \omega \Lambda^{-1}\right)_{\mu\nu} J^{\mu\nu} - \left(\Lambda \epsilon - \Lambda \omega \Lambda^{-1} a \right)_\mu P^\mu.\end{align}$$

Equating coefficients of $\omega_{rho\sigma}$ and $\epsilon_\rho$ on both sides of this equation (and using (2.3.10)), we find $$\begin{align}&U(\Lambda, a) J^{\rho\sigma} U^{-1} (\Lambda, a) = \Lambda_\mu^\rho \Lambda_\nu^\sigma (J^{\mu\nu} - a^\mu P^\nu + a^\nu P^\mu ),\\ &U(\Lambda, a) P^\rho U^{-1}(\Lambda, a) = \Lambda_\mu^\rho P^\mu.\end{align}$$

For homogeneous Lorentz transformations (with $a^\mu=0$), these transformation rules simply say that $J^{\mu\nu}$ is a tensor and $P^\mu$ is a vector. For pure translations (with $\Lambda^\mu_\nu=\delta^\mu_\nu$), they tell us that $P^\rho$ is translation-invariant, but $J^{\rho\sigma}$ is not. In particular, the change of the space-space components of $J^{\rho\sigma}$ under a spatial translation is just the usual change of the angular momentum under a change of the origin relative to which the anglar momentum is calculated.

Next, let's apply rules (8), (9) to a transformation that is itself infinitesimal, i.e., $\Lambda^\mu_\nu=\delta^\mu_\nu+\omega^\mu_\nu$ and $a^\mu=\epsilon^\mu$, with infinitesimal $\omega^\mu_\nu$ and $\epsilon^\mu$ unrelated to the previous $\omega$ and $\epsilon$. Using Eq. (3), and keeping only terms of first order in $\omega^\mu_\nu$ and $\epsilon^\mu$, Eqs. (8) and (9) now become $$\begin{align}i\left[ \frac{1}{2} \omega_{\mu\nu} J^{\mu\nu} -\epsilon_\mu P^\mu, J^{\rho\sigma}\right] &= \omega_\mu^\rho J^{\mu\sigma} + \omega_\nu^\sigma J^{\rho\nu} - \epsilon^\rho P^\sigma + \epsilon^\sigma P^\rho ,\\i\left[\frac{1}{2} \omega_{\mu\nu} J^{\mu\nu} - \epsilon_\mu P^\mu, P^\rho\right] &= \omega_\mu^\rho P^\mu.\end{align}$$

Equating coefficients of $\omega_{\mu\nu}$ and $\epsilon_\mu$ on both sides of these equations, we find the commutation rules $$\begin{align}i[J^{\mu\nu}, J^{\rho\sigma}] &= \eta^{\nu\rho} J^{\mu\sigma} - \eta^{\mu\rho}J^{\nu\sigma} - \eta^{\sigma\mu}J^{\rho\nu} + \eta^{\sigma\nu} J^{\rho\mu},\\ i[P^\mu, J^{\rho\sigma}] &= \eta^{\mu\rho}P^\sigma - \eta^{\mu \sigma} P^\rho,\\ [P^\mu,P^\rho] &= 0.\end{align}$$

This is the Lie algebra of the Poincare group.

In quantum mechanics a special role is played by thos operators that are conserved, i.e., that commute with the energy opeerator $H=P^0$. Inspection of Eqs. (13) and (14) shows that these are the momentum three-vector $$\begin{align}\mathbf{P} = \left\{ P^1, P^2, P^3 \right\} \end{align}$$ and the angular-momentum three-vector $$\begin{align}\mathbf{J} = \left\{ J^{23}, J^{31}, J^{12} \right\} \end{align}$$ and, of course, the energy $P^0$ itself. The remaining generators form what is called the 'boost' three-vector $$\begin{align}\mathbf{K} = \left\{ J^{10}, J^{20}, J^{30} \right\} .\end{align}$$

These are not conserved, which is why we do not use the eigenvalues of $\mathbf{K}$ to label physical states. In a three-dimensional notation, the commutation relations (12), (13), (14) may be written $$\begin{align}[J_i, J_j] = i \epsilon_{ijk} J_k,\\ [J_i, K_j] = i \epsilon_{ijk} K_k,\\ [K_i, K_j] = -i \epsilon_{ijk} J_k,\\ [J_i, P_j] = i \epsilon_{ijk} P_k,\\  [K_i, P_j] = i H \delta_{ij},\\ [J_i, H] = [P_i, H] = [H,H] = 0,\\ [K_i, H] = iP_i,\end{align}$$ where $i,j,k$ etc. run over the values $1,2,$ and $3,$ and $\epsilon_{ijk}$ is the totally antisymmetric quantity with $\epsilon_{123}=+1$. The commutation relation (18) will be recognized as that of the angular-momentum operator.

The pure translations $T(1,a)$ form a subgroup of the inhomogeneous Lorentz group with a group multiplication rule given by (2.3.7) as $$\begin{align}T(1, \bar{a}) T(1, a) = T(1, \bar{a} + a).\end{align}$$

This is additive in the same sense as (2.2.24), so by using (3) and repeating the same arguments that led to (2.2.26), we find that finite translations are represented on the physical Hilbert space by $$\begin{align}U(1, a) = \exp (-iP^\mu a_\mu).\end{align}$$

In exactly the same way, we can show that a rotation $R_\theta$ by an angle $\left| \theta\right|$ around the direction of $\theta$ is represented on the physical Hilbert space by $$\begin{align}U(R_\theta, 0) = \exp (i\mathbf{J}\cdot \mathbf{\theta}).\end{align}$$

It is interseting to compare the Poincare algebra with the Lie algebra of the symmetry group of Newtonian mechanics, the Galiliean group. We could derive this algebra by starting with transformation rules of the Galiliean group and then following the same procedure that was used here to derive the Poincare algebra. However, since we already have Eqs. (18)-(24), it is easier to obtain the Galilean algebra as a low-velocity limit of the Poincare algebra, by what is known as an Inonu-Wigner contraction. For a system of particles of typical mass $m$ and typlical velocity $v$, the momemtum and the angular-momentum operators are expected to be of order $J\sim 1$, $\mathbf{P}\sim mv$. On the other hand, the energy operator is $H=M+W$ with a total mass $M$ and non-mass energy $W$ (kinetic plus potential) of order $M\sim m$, $W\sim mv^2$. Inspection of Eqs. (18)-(24) shows that these commutation relations have a limit for $v<<1$ of the form $$ [J_i,J_j]=i \epsilon_{ijk} J_k,\ [J_i,K_j] = i\epsilon_{ijk} K_k,\ [K_i,K_j]=0,\\ [J_i,P_j]=i \epsilon_{ijk} P_k,\ [K_i,P_j]=i M \delta_{ij},\\ [J_i,W] = [P_i, W] = 0,\ [K_i, W]=i P_i,\\ [J_i,M]=[P_i,M]=[K_i,M]=[W,M]=0$$ with $K$ of order $1/v$. Note that the product of a translation $\mathbf{x}\rightarrow \mathbf{x}+\mathbf{a}$ and a'boost' $\mathbf{x}\rightarrow \mathbf{x}+\mathbf{v}t$ should be the transformation $\mathbf{x}\rightarrow \mathbf{x}+\mathbf{v}t+\mathbf{a}$, but this is not true for the action of these operators on Hilbert space: $$\exp (i\mathbf{K}\cdot \mathbf{v}) \exp (-i\mathbf{P}\cdot \mathbf{a})= \exp (iM\mathbf{a}\cdot \mathbf{v}/2) \exp (i(\mathbf{K}\cdot \mathbf{v} - \mathbf{P}\cdot \mathbf{a})).$$

The appearance of the phase factor $\exp(iM\mathbf{a}\cdot \mathbf{v}/2)$ shows that this is a projective representation, with a superselection rule forbidding the super-position of states of different mass. In this respect, the mathematics of the Poincare group is simpler than that of the Galiliean group. However, there is nothing to prevent us from formally enlarging the Galilean group, by adding one more generator to its Lie algebra, which commutes with all the other generators, and whose eigenvalues are the masses of the various states. In theis case physical states provide an ordinary rather than a projective representation of the expanded symmetry group. The difference appears to be a mrer matter of notation, except that with this reinterpretation of the Galilean group there is no need for a mass superselection rule.