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Can symmetry generators be used for quantization?

+ 18 like - 0 dislike

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

This post has been migrated from (A51.SE)
asked Sep 15, 2011 in Theoretical Physics by Tobias Kienzler (255 points) [ no revision ]
retagged Apr 19, 2014 by dimension10
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.

This post has been migrated from (A51.SE)
@Joe I was thinking the other way around, _fixing_ the Eigenvalues to describe the particle kind (as in [Wigner's classification](http://dx.doi.org/10.2307%2F1968551), [wikipedia entry](http://en.wikipedia.org/wiki/Wigner%27s_classification)), and seeing if that yields the correct field equations.

This post has been migrated from (A51.SE)

2 Answers

+ 17 like - 0 dislike

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:


This post has been migrated from (A51.SE)
answered Sep 15, 2011 by Urs Schreiber (5,375 points) [ no revision ]
+ 6 like - 0 dislike

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.

This post has been migrated from (A51.SE)
answered Sep 15, 2011 by Nestoklon (340 points) [ no revision ]

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