# Can symmetry generators be used for quantization?

+ 18 like - 0 dislike
165 views

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
retagged Apr 19, 2014
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)

+ 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:

http://ncatlab.org/nlab/show/spinning+particle

This post has been migrated from (A51.SE)
answered Sep 15, 2011 by (5,525 points)
+ 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 (340 points)

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysic$\varnothing$OverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.