Poisson brackets for Hamiltonian descriptions

In summary: I hope this summary answered your question about the natural assumption made in the Hamiltonian formulation of gauge theories regarding the Poisson algebra of charges and the Lie algebra of symmetries. Thank you for your question and for allowing me to share my knowledge with you.
  • #1
haushofer
Science Advisor
Insights Author
2,952
1,497
Hi, I have a (maybe rather technical) question about the Hamiltonian formulation of gauge theories, which I don't get.

With an infinitesimal symmetry on your space-time M one can look at the corresponding transformation of the canonical variables in phase-space PS. This can be done by a phase space function H[]:M --> PS. The generators of these symmetries give you global charges.

If I generate my symmetry with a vector field [itex]\xi[/itex] or [itex]\alpha [/itex], the statement which is often made is that

[tex]
\{H[\xi],H[\alpha] \} = H [[\xi,\alpha]]
[/tex]

In words: the Poisson algebra {} of the charges (LHS) is isomorphic (=) to the Lie algebra of the corresponding symmetries on your space-time (RHS).

My question is: why is this so natural to assume? Henneaux and Brown wrote an article about the details and adjustments of this assumption (central charges in the canonical realization...), but I don't see why this should be "natural" in the first place. I'm missing something here.

Thanks in forward :)
 
Physics news on Phys.org
  • #2
haushofer said:
Hi, I have a (maybe rather technical) question about the Hamiltonian formulation of gauge theories, which I don't get.

With an infinitesimal symmetry on your space-time M one can look at the corresponding transformation of the canonical variables in phase-space PS. This can be done by a phase space function H[]:M --> PS. The generators of these symmetries give you global charges.

If I generate my symmetry with a vector field [itex]\xi[/itex] or [itex]\alpha [/itex], the statement which is often made is that

[tex]
\{H[\xi],H[\alpha] \} = H [[\xi,\alpha]]
[/tex]

In words: the Poisson algebra {} of the charges (LHS) is isomorphic (=) to the Lie algebra of the corresponding symmetries on your space-time (RHS).

My question is: why is this so natural to assume? Henneaux and Brown wrote an article about the details and adjustments of this assumption (central charges in the canonical realization...), but I don't see why this should be "natural" in the first place. I'm missing something here.

Thanks in forward :)

I hope, I have uderstood your question correctly!

In the real life, we use Noether procedure to write [itex]H_{\alpha}[/itex] in terms of the dynamical variables on PS:

[tex]H_{\alpha} = \int dx \pi \mathcal{L}_{\alpha} \phi \ \ \ (R)[/tex]

where [itex]\mathcal{L}_{\alpha}[/itex] is Lie derivative along the vector field [itex]\alpha[/itex].
It follows immediately from eq(R) that

1) canonical transformations generated by the charge/constraint correspond precisely to the diffeomorphisms generated by the vector field [itex]\alpha[/itex] ,i.e.,

[tex]
\{ H_{\alpha} , \phi \} = \mathcal{L}_{\alpha} \phi, \ \ \mbox{etc.} \ (1)
[/tex]

2) charges/constraints satisfy the algebra

[tex]\{ H_{\alpha}, H_{\xi} \} = H_{[\alpha , \xi ]} \ \ \ (2)[/tex]

To confirm the consistency between eq{1) and eq(2), we use the Jacobi identity

[tex]\{ \{ H_{\alpha} , H_{\xi} \} , \phi \} = \{\{H_{\alpha}, \phi\} , H_{\xi} \} - \{\{H_{\xi} , \phi \} , H_{\alpha} \}[/tex]

This gives

[tex]\{\{H_{\alpha} , H_{\xi} \} , \phi \} = \mathcal{L}_{\alpha} \mathcal{L}_{\xi} \phi - \mathcal{L}_{\xi}\mathcal{L}_{\alpha} \phi = \mathcal{L}_{[\alpha , \xi ]} \phi[/tex]

However, the algebra of constraints is not, in general, a proper Lie algebra: the RHS of eq(2) may contain structure functions rather than structure constants. This means, in the BRST terminology, that we may have an open algebra. Indeed, it has been shown [Bargmann and Komar] that the algebra of constraints is not isomorphic to the Lie algebra of the "gauge group" of tetrad-gravity.

*****

Ok, here is a math question for you:

Given two function [itex] f,g : PS \rightarrow \mathbb{R}[/itex], their Poisson bracket is defined by

[tex]\{ f , g \} = \mathcal{L}_{X_{f}}g = - \mathcal{L}_{X_{g}}f[/tex]

this means that PB turns the vector space of functions on PS into Lie algebra. Use this Lie-bracket to show that the linear map [itex] f \rightarrow X_{f}[/itex] (which associates to f its (Hamiltonian) vector field) takes Poisson brackets of functions into commutators of vector fields:

[tex][X_{f} , X_{g}] = X_{\{f,g\}} \ \ (3)[/tex]

compair this with eq(2).

regards

sam
 
  • #3
A, nice! I will look at this tomorrow! Sam to the rescue, again :D
 
  • #4
haushofer said:
A, nice! I will look at this tomorrow! Sam to the rescue, again :D

It is a pleasure, very time.
 

1. What are Poisson brackets and how are they used in Hamiltonian descriptions?

Poisson brackets are mathematical operators used in classical mechanics to describe the dynamics of a system. In Hamiltonian descriptions, they are used to calculate the time evolution of a system's variables and determine the equations of motion.

2. How do Poisson brackets relate to the Hamiltonian of a system?

Poisson brackets are directly related to the Hamiltonian of a system. The Hamiltonian is a function that describes the total energy of a system, and Poisson brackets are used to calculate the change in the Hamiltonian over time.

3. Can Poisson brackets be used for any type of system or are they limited to classical mechanics?

Poisson brackets are primarily used in classical mechanics, but they can also be applied to other systems such as quantum mechanics and field theory. However, their interpretation and use may differ in these other contexts.

4. Are there any important properties of Poisson brackets that I should be aware of?

Yes, there are several important properties of Poisson brackets that are essential for understanding their use in Hamiltonian descriptions. These include the Jacobi identity, linearity, and the Leibniz rule.

5. How are Poisson brackets calculated and what do the results represent?

Poisson brackets are calculated using a specific formula that involves taking the partial derivatives of the Hamiltonian with respect to the variables of the system. The resulting value represents the rate of change of one variable with respect to another, and is used to determine the system's equations of motion.

Similar threads

  • Special and General Relativity
Replies
4
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
691
Replies
7
Views
789
Replies
1
Views
2K
Replies
27
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Beyond the Standard Models
2
Replies
61
Views
6K
  • Advanced Physics Homework Help
Replies
4
Views
3K
Replies
16
Views
1K
Back
Top