Considering that Poincaré algebra is given by the following relations

$$i[J^{\mu\nu},J^{\rho\sigma}]=\eta^{\nu\rho}J^{\mu\sigma}-\eta^{\mu\rho}J^{\nu\sigma}-\eta^{\sigma\mu}J^{\rho\nu}+\eta^{\sigma\nu}J^{\rho\mu}$$ $$i[P^{\mu},J^{\rho\sigma}]=\eta^{\mu\rho}P^{\sigma}-\eta^{\mu\sigma}P^{\rho}$$ $$[P^\mu, P^\rho]=0$$

and the fact that momentum is given by $P=(E,\vec{P})$, we can identify the zero component of the energy-momentum vector with the hamiltonian $P^0=H$, then I can say $$[H,P^i]=[H,J^{i}]=[H,P^0]=0$$ where $i$ is running on space indices. Hence we can say that energy, total momentum and angular momentum are conserved. But I don't know how to interpret this result. If I am studying a physical system that is symmetric under Poincaré transformations, these quantities are expected to be conserved, but the discussion on the algebra is purely mathematical, there isn't a physical system to which I can refer. The only meaning I can think about is that as Poincaré group is the group of isometries for Minkowski space then I can say that every system that doesn't break the symmetries of the space will have these quantities conserved but to me, it seems trivial and not rigorous. Moreover, if that is the interpretation I find strange that a similar situation isn't present in Galileo's group as their isometries are given by spatial translations and rotations so there isn't a hamiltonian which gives me the previous commutations rules. Thanks a lot for the help


1 Answer 1


It does sound a bit weird, at first, that using the Poincare algebra alone can give nontrivial conservation laws. After all, that's not how other algebras work. Many physical system carry a representation of the rotation group, but that doesn't imply that the system is rotationally symmetric (e.g. my shoe can rotate in space, but it's not a sphere), nor that angular momentum is conserved. By itself, saying there's a representation of the rotation group says little about how the system behaves; it only says that it makes sense to think about rotating it.

The Poincare group is different because it includes time translations, and hence you can't think about a representation about it without making a nontrivial statement about the dynamics. Yes, it's correct that assuming you have a representation of the Poincare group implies energy-momentum and angular momentum conservation. In cases where this is not true, you simply can't define a representation at all.

For example, consider a system in a fixed external field $\varphi(\mathbf{x})$. Then $P^0 = H$ contains explicit dependence on $\mathbf{x}$, so it doesn't commute with $P^i$. This could be as simple as, e.g. having two rooms in a lab, one of which contains an electric field. You can of course still define translation operators that move charges between the rooms, but they won't commute with time translation, because charges behave differently in an electric field. You could fix this by defining a formal $P^0$ that ignores the effect of the electric field, but then it no longer generates the actual time translation.

A slightly more subtle but analogous example is parity. When textbooks introduce the parity operator, they often phrase it in terms of obeying the expected commutation with the generators of the connected Poincare group. In other words, they assume there is a representation of the extended Poincare group and use this to explicitly construct $P$, and since that implies $[P, H] = 0$ it seems like they have inadvertently proven that parity is always conserved. This reasoning is incorrect, however, because if you look carefully at what the textbooks are doing, they're always defining parity on the free theory, for the "in" and "out" states. You can always do this (with the exception of theories with chiral spinors), and it doesn't preclude parity violation when interactions happen. In other words, for parity-violating theories, you cannot define a representation of the extended Poincare group on the full interacting theory.

You might have a residual unease here, which is: how can we ever fail to be able to define a representation? Isn't that something you can always do, independent of the physics?

Absolutely not! This attitude is just a consequence of thinking too much about examples like rotations of objects in empty space, where it seems trivial. Even for rotations, you can fail to have a representation. For example, consider sound waves inside a crystalline solid. Only some polarizations of the sound waves are allowed -- you simply cannot rotate the sound waves freely inside the crystal; if you did that naively, you wouldn't get a valid solution to the equations of motion. So here, there is no representation of the rotation group.

  • $\begingroup$ yes I think you undestood my problem, but i am still having trouble understanig truly the answer. It is not completely clear why the discussion is related to representations ( i have studied group theory, it is a notion that I know), In particular what I think is the central point of the answer: " In cases where this is not true, you simply can't define a representation at all". This is something unexpected, as i think that i can always act on physical sistem whith the transformations of the group and hence with its representations, as you said before in the case of rotations. $\endgroup$
    – Ratman
    Commented Apr 19, 2020 at 22:37
  • $\begingroup$ even though from a logic point of view it is clear that if I have a system with no translation simmetry this doesn't conserve total momentum and hence there is an incongruence with what I find in the Poincaré algebra (thanks a lot for the answer) $\endgroup$
    – Ratman
    Commented Apr 19, 2020 at 22:39
  • $\begingroup$ The fact that this could be related to a free theory is the interpretation i would have normally choose, it was what i meant with "doesn't break the simmetry of the space" (space and time homogenus and space isotropic). But this was just intuition, and i haven't find a formal explanation $\endgroup$
    – Ratman
    Commented Apr 19, 2020 at 22:47
  • $\begingroup$ @Frappa Well, suppose I have one room of a lab with an electric field, and one room without it. This breaks translational symmetry. It's still perfectly possible to define a translation operator that moves particles from one room to the other, and a time translation operator. But these operators will not commute, because charges will behave differently in each room. So assuming they do commute must be a nontrivial statement about the dynamics. $\endgroup$
    – knzhou
    Commented Apr 19, 2020 at 23:12
  • $\begingroup$ @Frappa I updated the answer with more clarification. $\endgroup$
    – knzhou
    Commented Apr 19, 2020 at 23:22

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