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Symmetry and Physics | Not Even Incorrect

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Someone is wrong on the internet

It’s getting late, however I can’t assist myself. Studying too many mistaken issues about symmetry and physics on Twitter has compelled me to do that. And, John Baez says I don’t clarify issues. So, right here’s what the connection between symmetry and physics actually is.

Within the language of mathematicians, speaking about “symmetries” means you’re speaking about teams (typically Lie teams, or their infinitesimal variations, Lie algebras) and representations. The relation to physics is:

Classical mechanics (Hamiltonian type)

In classical mechanics the state of a system with $n$ levels of freedom is given by a degree in section house $P=mathbf R^{2n}$ with $n$ place coordinates $q_j$ and $n$ momentum coordinates $p_j$. Capabilities on this house are a Lie algebra, with Lie bracket the Poisson bracket
${f,g}$. Dynamics is given by selecting a distinguished operate, the Hamiltonian $h$. Then the worth of any operate on $P$ evolves in time in line with
$$frac {df}{dt}={f,h}$$
The Hamiltonian $h$ generates the motion of time translations. Making use of the identical formulation, different features generate the motion of different teams (spatial translations, rotations, and so forth.). In case your operate satisfies ${f,h}=0$, it generates a “symmetry”, and doesn’t change with time (is a conserved amount).

Quantum mechanics

Quantization of a classical system is one thing mathematically apparent: go from the above Lie algebra to a unitary illustration of the Lie algebra. This takes components of the Lie algebra (features on $P$) to skew-adjoint operators on a Hilbert house, the house of quantum states. There’s a theorem (Stone-von Neumann) that claims that (modulo technicalities) there’s just one method to do that, and it provides an irreducible unitary illustration that works for polynomials as much as diploma two. For greater diploma polynomials there’ll at all times be “operator ordering ambiguities”. The illustration is given by
$$1rightarrow -imathbf 1, q_jrightarrow -iQ_j, p_jrightarrow -iP_j$$
It is a illustration as a result of
$${q_j,p_k}=delta_{jk}rightarrow [-iQ_j,-iP_k]=-idelta_{jk}mathbf 1$$
The correct-hand aspect is the Heisenberg commutation relations for $hbar=1$.

For extra particulars, I wrote an entire ebook about this.

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