# Representation of the Galileo Group and Central Charges

I've arrived at this question because I've been reading Weinberg's Quantum Theory of Fields Volume I, and I'm in the second chapter about relativistic quantum mechanics. Weinberg discusses the representations of the Poincare group, and after deriving the structure constants he gets the structure constants for the Galilean group as well. I have a few questions about this:

He introduces these operators $M$ and $W$--where do they come from? (This question may make sense only to those reading Weinberg, I'm not sure, so it's less important).

How can we conclude that in the Galilean Algebra $$[K_i, P_j] = -iM\delta_{ij}$$ Shouldn't boosts and translations commute? Why is this equal to $M$ instead of total energy? Is it because $M$ is a "central charge"? What exactly is a central charge?

Does it have to do with the fact that $M$ is in the Lie Algebra, it is a generator of a symmetry transformation, and because it is in the center and commutes, its charge is conserved. But how do we decide what its conserved charge is? What is to prevent us from adding more of these central charges?

Also, look at the product of a translation and a boost (K is the generator of boosts): $$e^{-i\vec{K}\cdot \vec{v}}e^{-i\vec{P}\cdot \vec{a}}=e^{iM\vec{a}\cdot \vec{v}/2}e^{-i(\vec{K}\cdot \vec{v} + \vec{P}\cdot \vec{a})}$$ Weinberg says that this phase factor (with the M) shows we have a projective representation, "with a superselection rule forbidding the superposition of states of different mass". Where did this phase factor come from? What are these superselection rules about?

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## 1 Answer

The boosts and translations do not commute in neither relativistic or nor relativistic systems, please see for example the case of the Poincare group.

Since $K_i = M_{0i}$ , $(i=1,2,3)$, we get for the Poincare group: $[K_i, P_j] =i (\eta_{0j}P_i-\eta_{ij}P_0) = -i \delta_{ij}P_0$

Now, The Galilean group can be obtained from the Poincare group by means of Wigner-Inonu contraction, see for example this link.

In the non relativistic limit the rest momentum is dominated by the mass term thus under the contraction the above relation becomes:

$[K_i, P_j] =-i \delta_{ij}M$

The above reasoning shows that the parameter $M$ is the mass.

Considered as an Lie algebra element M commutes with all other generators and cannot be removed by a smooth redefinition of the generators, this is the reason that it is called a central extension, see for example chapter 3 of Ballentine's book, where several representations of the Galilean group are constructed.

The consequence is that in each irreducible representation of The Galilean group $M$ must be represented by a scalar (the particle's mass), and two representations with different masses are unitarily inequivalent.

The formula of the product of a translation and a boost can be directly obtained the cation of the Baker–Campbell–Hausdorff formula.

This formula means that the wavefunctions acquires a phase factor in a transformed inertial frame (in other words, the wavefunction representation is projective). In quantum mechanics a global phase is not measurable, thus no physical consequence is caused

However, Representations corresponding to different masses fall into different superselection sectors, this means that linear superpositions of wavefunctions of particles with different masses are unphysical, because in this case each component would acquire a different phase which contradicts experiment because differential phases are observable in quantum mechanics.

It is worthwhile to mention that this is exactly the example which was originally analyzed by V. Bargmann in his paper "On unitary ray representations of continuous groups", Ann. of Math.,59 (1954), 1–46.

Update

This update contains the answers to the questions appearing in Aruns comments:

The noncommutativity of the momentum and boost operators is quantum mechanical. The commutators are written in $\hbar= 1$ units, working in general units, the boost-momentum commutator has the form:

$[K_i, P_j] =i \hbar \delta_{ij}M$

The noncommutativity is observable only in the action of the operators on the wave functions. Let me elaborate this point for the case a free particle considered in Ballentine's book.

First, the phase space (i.e., the manifold of initial data) is $\mathbb{R}^6$ covered with the coordinates $\{ \mathbf{q}, \mathbf{v} \}$, which upon quantization satisfy the commutation relations:

$[ q_i, v_j] = \frac{i}{M} \delta_{ij}$

Consider the operators:

The finite translation operator $\mathcal{D}_{q_0} =\exp(i \frac{ M\mathbf{v}. \mathbf{q_0}}{\hbar})$.

The finite boost operator $\mathcal{B}_{v_0} = \exp(i \frac{\mathbf{K}. \mathbf{v_0}}{\hbar}) = \exp(i \frac{M \mathbf{q}. \mathbf{v_0}}{\hbar})$

Consider a wave function on the phase space $\psi(\mathbf{q})$. It is not difficult to see that:

$\mathcal{B}_{v_0} \mathcal{D}_{q_0} \psi(\mathbf{q})= \exp(\frac{i M \mathbf{v_0}.\mathbf{q_0}}{\hbar}) \mathcal{D}_{q_0} \mathcal{B}_{v_0} \psi(\mathbf{q})$

The fourth relativistic momentum components is given by

$P_0 = \sqrt{M^2 c^4 + p^2 c^2}$

In the non-relativistic limit $c \rightarrow \infty$, the first term dominates and we get $P_0 \approx M c^2$. Now, you absorb the factor $c^2$ into the definition of the generators, (or work in units c=1). This is essentially the Wigner-Inonu contraction of the Poincare group to the Galilean group.

I'll give you here the standard argument of the uniatry inequivallence between the quantization of two free particles with different masses $M_1 \ne M_2$. Let us use the subscript 1 and 2 for the individual particle operators (Each acting on a different Hilbert space ). If the two representations are unitarily equivalent means that there is an isometry $U$ such that

$\mathbf{K_2} = U \mathbf{K_1}U^{*}$ and $\mathbf{P_2} = U \mathbf{P_1}U^{*}$, thus we obtain $[\mathbf{K_2}, \mathbf{P_2}] = U [\mathbf{K_1}, \mathbf{P_1}]U^{*}$ , which means $M_1 = M_2$, a contradiction.

Unitary inequivalence is similar to representations of $SU(2)$ having different $J$, consquently having different dimensions. In the finite dimensional case, it is obvious that we cannot find an isometry between different dimensional Hilbert spaces. One might think that all infinite dimensional representations of the Galilean are unitarily equivalent which is not the case.

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I thought non-relativistic boosts and translations commutted--boost plus translations sends $(x,t) \rightarrow (x+vt,t) \rightarrow (x+vt+x_0,t)$ and translation plus boost should be $(x,t) \rightarrow (x+x_0,t) \rightarrow (x+x_0+vt,t)$. I'm getting that the end result is the same either way. Also I'm still not sure how you get $P_0$ becomes $M$ in the nonrelativistic limit, and why $M$ is the mass. Do let $P_0 = H = \frac{1}{2}mv^2 + U$, and neglect a term or something? –  Arun Nanduri Jul 5 '11 at 17:49
Lastly, could you clarify what unitarily equivalent means? Are the different values of $M$ you can have analogous to the different values of $J_z$ you can have in a representation of SO(3)? –  Arun Nanduri Jul 5 '11 at 17:54
Typo(?) in the first formula of the update (v2): Planck's constant $\hbar$ should appear in the numerator rather than the denominator. –  Qmechanic Jul 6 '11 at 17:04
@Arun, correct, the free particle representation is a projective representation of the Galilean group without central extension. The same representation is a true unitary representation of the centrally extended Galilean group. This is analogeous to the spin $\frac{1}{2}$ representation of $SU(2)$, which is a projective representation when restricted to $SO(3)$ but a true representation of $SU(2)$. –  David Bar Moshe Jul 7 '11 at 14:01
The extra phase factors in the action on wavefunctions do not necessarily imply that the representation is projective, one needs to prove that the composition law has the form $U(g_1)U(g_2) = c(g_1, g_2) U(g_1 g_2)$, where the phase factor $c(g_1, g_2)$ is a nontrivial cocycle. –  David Bar Moshe Jul 7 '11 at 14:08