Working out a Second Class Constraint in Gory detail?

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SUMMARY

The discussion centers on the concept of second class constraints in the context of quantum field theory. The user, Daniel, presents a specific constraint defined as C^i = \pi^i - p^i = 0, highlighting the non-commutativity of these constraints with [C^i, C^j] ≠ 0. Daniel expresses a need for foundational resources on the topic, as he lacks traditional training in quantum field theory. Ultimately, he indicates progress in understanding the Lagrangian density related to his equations.

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  • Understanding of second class constraints in Hamiltonian mechanics
  • Familiarity with conjugate momentum in classical mechanics
  • Basic knowledge of quantum field theory principles
  • Ability to interpret Lagrangian and Hamiltonian formulations
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  • Study the mathematical formulation of second class constraints in Hamiltonian systems
  • Explore the implications of non-commuting constraints in quantum mechanics
  • Learn about the role of conjugate momentum in quantum field theory
  • Review Lagrangian density derivations in various physical contexts
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Students and researchers in theoretical physics, particularly those focusing on quantum field theory and Hamiltonian mechanics, will benefit from this discussion.

pqnelson
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Hello,

I've been fooling around with some equations, and I've managed to land a second class constraint (I don't know whether to laugh or cry). Well, I was wondering is there any good introduction to the subject?

The problem I have is that I made the conjugate momentum a function of position, so I have the constraint:
C^i = \pi^i - p^i = 0
and thus
[C^i, C^j] \neq 0
and I would like to know what exactly [C^i, C^j] is equal to; however, I've never really learned quantum field theory "traditionally" and so I'm completely stuck.

Any help would be greatly appreciated!
 
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If i don't know who \pi^{i} and p^{i} are, i can't really help you.

Daniel.

P.S. It's enough to post the lagrangian (density) you started with.
 
Well, I think I've got it figured out now, thanks anyways.
 

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