# Does the non-relativistic conservation law of particles have an underlying (approximate) symmetry?

In momentum and energy is low enough, we end up with the same number of neutrons, protons and electrons after a collision as before it. This can be considered an approximate conservation law. Shouldn't there be a corresponding (approximate) symmetry corresponding to each of these particle groups?

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You mean like isospin or lepton number? –  pfnuesel Aug 9 '13 at 13:35
No just the number of particles - you start out with say five protons, and end up with five protons. Same for neutrons and electrons. –  Sebastian Henckel Aug 9 '13 at 13:36
I don't know how relevant this is, but in low-energy nuclear physics we apply the Bogoliubov transformation to describe pairing, and because the system has a finite number of particles, the resulting wavefunctions show fluctuations in particle number. There is a phase transition between the paired and unpaired states. The relevant symmetry is discussed on p. 6 of this paper: arxiv.org/abs/nucl-th/0303010 . Dunno if this sheds any light on the relativistic/nonrelativistic situation. –  Ben Crowell Aug 9 '13 at 13:59

As Sebastian Henckel said, there are many such conservation laws in non-relativistic physics which prohibits particle creation and annihilation. The number of particles of each allowed type is conserved separately.

To see the symmetry associated with these numbers, e.g. the total number $N$ of elementary particles, one has to realize that the symmetry variation of an observable $L$ is equal to $$\delta L = \epsilon\{N,L\}$$ where $N$ is the symmetry generator and the bracket is a Poisson bracket. Clearly, $N$ depends neither on positions $x$ nor on the momenta $p$ which is why the bracket vanishes and the symmetry acts trivially – nothing transforms under it. So you may say that there is a symmetry and it's formally isomorphic to $U(1)$ but Nature only allows objects that are invariant under it so you will never see any "change" linked to the symmetry.

Quantum mechanically, the bracket is replaced by $1/i\hbar$ times the commutator which is nonzero. The ket vectors simply transform as $$|\psi\rangle \to \exp(i\lambda N)|\psi\rangle$$ under the symmetry you're looking for. This is just the overall change of the phase which doesn't change the physical (measurable) properties of the ket vector. Because the number of particles is conserved so perfectly, any complete enough measurement of the system will find it in an eigenstate of $N$. The conservation law implies that the final state will be an eigenstate, too. So after one complete enough measurement, the subspaces of the Hilbert space with different eigenvalues of $N$ are pretty much separated from each other – we say that they live in different superselection sectors.

Let me comment on a word in the title, "approximate". Because the conservation law is exact, the symmetry has to be exact as well, of course. It's uninteresting not because it's violated or approximate but because it acts trivially.

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