Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

Please welcome our new moderators!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

122 submissions , 103 unreviewed
3,497 questions , 1,172 unanswered
4,543 answers , 19,337 comments
1,470 users with positive rep
407 active unimported users
More ...

Can symmetry generators be used for quantization?

+ 18 like - 0 dislike
154 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 in Theoretical Physics by Tobias Kienzler (255 points) [ no revision ]
retagged Apr 19, 2014 by dimension10
@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)
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)

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:

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

This post has been migrated from (A51.SE)
answered Sep 15, 2011 by Urs Schreiber (5,085 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 ]

Your answer

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):
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$\varnothing$ysicsOverflow
Then 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.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...