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Question about the Noether charge algebra

+ 6 like - 0 dislike
483 views

I'm reading these notes - page 8 and 9 - and I'm a bit confused.

If we consider a field $\phi$ (which can be either bosonic or fermionic) transforming as: \begin{equation} \phi(x) \rightarrow \phi(x) + \delta \phi (x) \end{equation} with: \begin{equation} \delta \phi^a = t^a \phi(x) \end{equation} where $t^a$ is the generator of the transformation. The generators satisfy the Lie algebra: \begin{equation} [t^a,t^b] = if^{abc} t^c \tag{$*$} \end{equation} Let us suppose that the above transformation is a symmetry transformation such that the Noether charge corresponding to this symmetry is given by: \begin{equation} Q^a = \int \mathrm{d}^3 \mathbf{x} \; \pi \delta\phi^a = \int \mathrm{d}^3 \mathbf{x} \; \pi t^a \phi \end{equation} where $\pi$ is the canonical momentum density. It is then possible (but tedious) to show that the charges satisfy the so-called charge algebra: \begin{equation} [Q^a,Q^b] = i f^{abc} Q^c \tag{1} \end{equation} Until this point I understand it. But then the notes say on page 8:

[...] the charges generally have to satisfy the same algebras as the generators – in fact it is only because of this that the symmetry has any useful physical meaning. In particular it is the charges which are the physical observables that participate in interactions rather than gauge fields for example.

I don't really understand what is meant with the above statement. What does the quote have to do with the fact that Noether charges obey equation $(1)$?

Edit: I understand that the charges satisfy the same Lie algebra as the generators. But according to the quote above, if understand it correctly, we should also expect this based on logical/physical reasons. Apparently, according to the notes, "it is only because of this that the symmetry has any useful physical meaning." I don't understand why this is the case.

This post imported from StackExchange Physics at 2014-04-05 17:24 (UCT), posted by SE-user Hunter
asked Mar 20, 2014 in Theoretical Physics by Hunter (510 points) [ no revision ]
It is clear that equation $(1)$ and $(*\!)$ are of exactly the same form. Hence, the charges satisfy the same algebra as the generators. This is what the comment is about... If this wasn't what you were really asking, please clarify!

This post imported from StackExchange Physics at 2014-04-05 17:24 (UCT), posted by SE-user Danu
@Danu I can see that they both have the same form. I'm trying to basically understand the quote, please see my edit above.

This post imported from StackExchange Physics at 2014-04-05 17:24 (UCT), posted by SE-user Hunter

1 Answer

+ 2 like - 0 dislike

Well, this might not be exactly what OP is looking for, but the statement in Ref. 1 is in general not correct. That infinitesimal (global) symmetries (of an action) satisfy a Lie algebra does not imply that the corresponding Noether charges must also form a Lie algebra. There could be (classical) anomalies.

Example: One example is free Schrödinger theory, see e.g. Ref. 2. The symmetry transformations are a complex translation and a real phase rotation of the wave function field $\psi$. The Poisson algebra of the corresponding Noether charges develops a classical central charge.

References:

  1. Steven Abel, Anomalies, Lecture notes. The pdf file is available here.

  2. Tomas Brauner, Spontaneous Symmetry Breaking and Nambu-Goldstone Bosons in Quantum Many-Body Systems, Symmetry 2 (2010) 609; arXiv:1001.5212, page 6-7.

This post imported from StackExchange Physics at 2014-04-05 17:24 (UCT), posted by SE-user Qmechanic
answered Mar 20, 2014 by Qmechanic (2,790 points) [ no revision ]
Thanks for your reply. I'm just starting to learn about anomalies so I don't fully understand what this means. However, using $$Q^a = \int \mathrm{d}^3 \mathbf{x} \; \pi \delta\phi^a = \int \mathrm{d}^3 \mathbf{x} \; \pi t^a \phi$$ I have been able to show that the charge satisfy the Lie algebra both for fermionic and bosonic fields. Does that mean that charge is not always given by $Q^a = \int \mathrm{d}^3 \mathbf{x} \; \pi \delta\phi^a = \int \mathrm{d}^3 \mathbf{x} \; \pi t^a \phi$? Do you think I will better understand this problem if I just keep on reading the notes without worrying too

This post imported from StackExchange Physics at 2014-04-05 17:24 (UCT), posted by SE-user Hunter
much about this quote?

This post imported from StackExchange Physics at 2014-04-05 17:24 (UCT), posted by SE-user Hunter

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