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 \begin{equation} [K_i, P_j] = -iM\delta_{ij} \end{equation} 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): \begin{equation} 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})} \end{equation} 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?


2 Answers 2


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.


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.

  • $\begingroup$ 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? $\endgroup$ Jul 5, 2011 at 17:49
  • $\begingroup$ 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)? $\endgroup$ Jul 5, 2011 at 17:54
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    $\begingroup$ Typo(?) in the first formula of the update (v2): Planck's constant $\hbar$ should appear in the numerator rather than the denominator. $\endgroup$
    – Qmechanic
    Jul 6, 2011 at 17:04
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    $\begingroup$ @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 $SU(2)$. $\endgroup$ Jul 7, 2011 at 14:01
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    $\begingroup$ 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. $\endgroup$ Jul 7, 2011 at 14:08

The central charge $M$ corresponds to the translation operator for a fifth coordinate. The Galilei group has a five-dimensional representation that is best seen by considering the effect of an infinitesimal boost $𝛖$ on the kinetic energy $H$, momentum $𝐏$ and mass $M$: $$ΔH = -𝛖·𝐏, \hspace 1em Δ𝐏 = -𝛖M, \hspace 1em ΔM = 0.$$ Together, it transforms as a 5-vector $(H,𝐏,M)$. For the coordinates $(𝐫,t)$, however, we have only the transform $$Δ𝐫 = -𝛖t, \hspace 1em Δt = 0,$$ (with a similar transform for the coordinate differentials $(d𝐫,dt)$), and are unable to form an invariant 1-form with the coordinate differentials and mass-energy-momentum vector. What we get is: $$Δ(𝐏·d𝐫 - H dt) = (-𝛖M)·d𝐫 + 𝐏·(-𝛖dt) - (-𝛖·𝐏)dt - H(0) = -M𝛖·d𝐫.$$ Thus suggests adding in an extra coordinate $u$ (and its differential $du$) with the transform law $$Δu = 𝛖·𝐫, \hspace 1em Δ(du) = 𝛖·d𝐫$$ which would allow us to complete the invariant $$Δ(𝐏·d𝐫 - H dt + M du) = 0.$$

This can be seen from the standpoint of Relativity in the following way. First, we have the proper time, denoted here as $s$ and a quadratic invariant which expresses it in terms of the differentials of the coordinate invariants: $$ds^2 = dt^2 - \frac{1}{c^2} |d𝐫|^2.$$ Now consider the time dilation $s - t$. In the non-relativistic limit as $(1/c)^2 → 0$, it also approaches 0. However, when scaled up by $c^2$, the corresponding quantity $u = c^2 (s - t)$ does have a non-trivial non-relativistic limit! Upon substitution of $s$ by $u$, the quadratic invariant in Relativity becomes the following two invariants: $$ds = dt + \frac{1}{c^2} du, \hspace 1em |d𝐫|^2 + 2 dt du + \frac{1}{c^2} du^2 = 0.$$ In the non-relativistic limit, this reduces to: $$ds = dt, \hspace 1em |d𝐫|^2 + 2 dt du = 0.$$ The 4-dimensional geometry of non-relativistic theory embeds as a light cone in a 5-dimensional space whose extra coordinate $u$ bears the same relation to $M$ as $𝐫$ does to $𝐏$. This geometry is the Bargmann geometry and it goes with the central extension of the Galilei group which, itself, is called the Bargmann group.

In Relativity, instead of kinetic energy $H$ and mass $M$, one has the "total energy" $E$, which has the same transform law as $H$. However, the coordinates in relativity transform - under boost - differently: $$Δ𝐫 = -𝛖t, \hspace 1em Δt = -\frac{1}{c^2} 𝛖·𝐫.$$ So, one can form an invariant one-form with them: $$\begin{align} Δ(𝐏·d𝐫 - E dt) &= (-𝛖M)·d𝐫 + 𝐏·(-𝛖t) - (-𝛖·𝐏) dt - E \left(-\frac{𝛖·d𝐫}{c^2} \right) \\ &= \left(\frac{E}{c^2} - M\right) 𝛖·d𝐫 \\ &= 0, \end{align}$$ provided one equates $E = M c^2$, with $M$, this time, corresponding to what used to be called the "moving mass", which is no longer invariant, but transforms in Relativity as $ΔM = -𝛖·𝐏/c^2$.

You can connect this back to the non-relativistic case by treating the kinetic energy $H$ as $E$, itself with an invariant "rest value" $μ c^2$ of the energy subtracted out: $H = E - μ c^2$, replacing $E$ by $M$, and adding in the invariant $μ c^2 ds$ to write: $$𝐏·d𝐫 - E dt + μ c^2 ds = 𝐏·d𝐫 - (H + μ c^2) dt + μ c^2 \left(dt + \frac{du}{c^2}\right) = 𝐏·d𝐫 - H dt + μ du.$$ In effect, that's the Relativistic version of the Bargmann geometry: a 5 dimensional representation of the Poincaré group with the invariants $$μ = M - \frac{H}{c^2}, \hspace 1 em P^2 - 2 M H + \frac{H^2}{c^2},$$ which recovers Relativity, itself, as the special case where the linear invariant is set to a "rest mass" $μ = m$, and the quadratic invariant is set to 0: $$0 = P^2 - 2 M H + \frac{H^2}{c^2} = P^2 - \frac{E^2}{c^2} + m^2 c^2.$$ More generally, it corresponds to the Poincaré group, trivially centrally extended with the inclusion of the central charge $μ$, which - in this generalization - may not be identified as any rest mass at all.

(Edit: I just looked at Weinberg. For notation, his "$H = M + W$" corresponds to what I'm writing as "$E = μ c^2 + H$". I always use $E$ to total energy - kinetic plus mass-energy - for Relativity, and $H$ for just the kinetic part, and non-relativistically for energy. He's using $H$ and $W$, respectively, for these in the discussion in section 2.4. So, for dimensional correctness, with my notation, his (2.4.22) would be $$[K_i, P_j] = iħ \frac{E}{c^2} δ_{ij} = iħ \left(μ + \frac{H}{c^2}\right) δ_{ij} = iħ M δ_{ij},$$ and the same would apply in the non-relativistic case, except $M = μ$.)

As to your other question, in a Lie algebra, a central charge is any Lie vector that has zero Lie brackets with everyone else. Effectively, that serves just as well as a definition (it's not the actual definition, but it's a necessary condition), since you can always treat any linear invariant of a Lie algebra as the central charge of the central extension of a reduced Lie algebra; namely the Lie algebra obtained by setting the invariant to 0. For example, for the Bargmann group, the linear invariant is just $μ = M$, itself, and the Galilei group is obtained from it by setting $M = 0$, while the Bargmann group is recovered back as the central extension of the Galilei group with $μ$ as the central charge.

Now ... about boosts and translations; consider the effect of an infinitesimal translation by $𝛆$: $$Δ'𝐫 = 𝛆, \hspace 1em Δ't = 0, \hspace 1em Δ'u = 0.$$ This time, we look at the coordinates themselves $(𝐫, t, u)$, not the coordinate differentials $(d𝐫, dt, du)$, since coordinate differences and differentials are translation-invariant. Does the boost $Δ$ and translation $Δ'$ commute? $$ΔΔ'𝐫 = Δ𝛆 = 𝟎, \hspace 1em ΔΔ't = Δ0 = 0, \hspace 1em ΔΔ'u = Δ0 = 0,$$ but $$Δ'Δ𝐫 = Δ'(-𝛖t) = -𝛖Δ't = 𝟎, \hspace 1em Δ'Δt = Δ'0 = 0, \hspace 1em Δ'Δu = Δ'(-𝛖·𝐫) = -𝛖·Δ'𝐫 = -𝛖·𝛆.$$ Thus $$[Δ,Δ']𝐫 = 𝟎, \hspace 1em [Δ,Δ']t = 0, \hspace 1em [Δ,Δ']u = 𝛖·𝛆.$$ The result is a translation on the $u$ coordinate - i.e. the effect of the central charge $μ = M$ - as I originally characterized it.

How does this show up in the non-relativistic Schrödinger equation? Just consider the free non-interacting case, here. The equation is mis-characterized a part quadratic and part linear in the differential operators, and written like this $$-\frac{iħ}{2m}∇^2ψ = iħ\frac{∂}{∂t}ψ.$$ In actuality, it is an expression of the linear and quadratic invariants $$μ = M = m, \hspace 1em P^2 - 2MH = 0,$$ with the central charge $μ$ now identified as the mass $m$, itself.

So, it should actually be written as: $$\left(|-iħ∇|^2 - 2 m \left(iħ\frac{∂}{∂t}\right)\right) ψ = 0.$$ The invariant 1-form $$𝐏·d𝐫 - H dt + μ du$$ is directly connected to the correspondences $$𝐏 = -iħ∇, \hspace 1em H = iħ\frac{∂}{∂t}, \hspace 1em μ = -iħ\frac{∂}{∂u},$$ which informs us what the operator form of $μ$ should be. So, when this is back-substituted with the choice of value $μ = m$, the result are the equations $$-iħ\frac{∂}{∂u}ψ = mψ, \hspace 1em \left(|-iħ∇|^2 + 2 \left(iħ\frac{∂}{∂t}\right)\left(iħ\frac{∂}{∂u}\right)\right) ψ = 0.$$

The solution, as a function of $u$ is $$ψ(u) = ψ(0) e^{imu/ħ}.$$ There's the extra phase factor. The coordinate $u$ shows up, essentially, as that extra phase you were wondering about.

Likewise, in Relativity, if you identity $μ = m$ as the rest mass, then the quadratic invariant previously derived is the mass shell invariant: $$P^2 - \frac{E^2}{c^2} + m^2 c^2 = 0.$$ If you apply the same operator correspondence, as before, the result will contain the solutions for the Klein-Gordon equation.

As for "super-selection rules": a quantity that commutes with everything else gives rise to a super-selection rule. Every state is an eigen-state of a quantity that commutes with everything else; so the state space is block-diagonal, with one block for each value of the quantity.

So, the quantum state space is actually a hybrid between purely quantum and classical. In the extreme case of classical physics, everything commutes with everything, so there's a superselection on everything, and the state space is totally diagonal. In the opposite extreme case of purely quantum systems, there are no universally-commuting quantities, and the state space has only one diagonal block.

Recall that - in the Heisenberg Picture - the time differential of anything in connected to its commutator. So, if a quantity commutes with everybody and is not explicitly time-dependent, then it has zero time derivative. Correspondingly, there are no state transitions in which it changes. That's actually one of the main obstacles, by the way, of trying to hybridize classical and quantum dynamics: there's no clear way to get a back-reaction to the classical quantities from the quantum dynamics part of a hybrid system.

Weinberg is effectively treating the linear invariant $μ$ as something commutes not just with the other generators of the Bargmann group, but with everything else. Another way of stating this, in light of the operator correspondence just described, is that observables all be independent of $u$: a gauge invariance with respect to $u$.


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