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Take the Poincaré group for example. The conservation of rest-mass $m_0$ is generated by the invariance with respect to $p^2 = -\partial_\mu\partial^\mu$. Now if one simply claims

The state where the expectation value of a symmetry generator equals the conserved quantity must be stationary

one obtains

$$\begin{array}{rl} 0 &\stackrel!=\delta\langle\psi|p^2-m_0^2|\psi\rangle \\ \Rightarrow 0 &\stackrel!= (\square+m_0^2)\psi(x),\end{array}$$

that is, the Klein-Gordon equation. Now I wonder, is this generally a possible quantization? Does this e.g. yield the Dirac-equation for $s=\frac12$ when applied to the Pauli-Lubanski pseudo-vector $W_{\mu}=\frac{1}{2}\epsilon_{\mu \nu \rho \sigma} M^{\nu \rho} P^{\sigma}$ squared (which has the expectation value $-m_0^2 s(s+1)$)?

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A symmetry gives you a set of eigenstates, which is a step in the right direction, but you also need to be able to determine their corresponding eigenvalues, and a single generator doesn't do that. –  Joe Fitzsimons Sep 15 '11 at 9:13
@Joe I was thinking the other way around, fixing the Eigenvalues to describe the particle kind (as in Wigner's classification, wikipedia entry), and seeing if that yields the correct field equations. –  Tobias Kienzler Sep 15 '11 at 9:26

2 Answers 2

up vote 19 down vote accepted

What you observe is the general phenomenon that in relativistic theories time translation is replaced by "affine-parameter-translation" or "wordline translation symmetry" and hence the corresponding Hamiltonian becomes a constraint, the constraint that states must be invariant under this symmetry.

Yes, this works for the relativistic spinning particle and the Dirac equation, too. Here the translation symmetry on the worldline is refined to translation supersymmetry (for ordinary spinors even, this has nothing a priori to do with spacetime supersymmetry). The odd generator of the worldline supersymmetry turns out to be the Dirac operator. Again, states are required to be annihilated by it and this gives the Dirac eqation.

Plenty of pointers to details about how this works are here:


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Your example shows that you may use symmetry to get a Hamiltonian (which should be invariant) and for classification of its solutions: it is convenient to choose wavefunctions in a way that they form the basis of irreducible representations of the symmetry group.

To get the numbers you need to solve the equations, their symmetry is not enough. Symmetry may tell you only which states should have the same energy.

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