Can symmetry generators be used for quantization?

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 $-m_0^2 s(s+1)$)?

• 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