Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with hamiltonian-formalism
Search options not deleted
user 77590
The Hamiltonian formalism is a formalism in Classical Mechanics. Besides Lagrangian Mechanics, it is an effective way of reformulating classical mechanics in a simple way. Very useful in Quantum Mechanics, specifically the Heisenberg and Schrodinger formulations. Unlike Lagrangian Mechanics, this formalism relies on a "Hamiltonian" instead of a Lagrangian, which differs from the Lagrangian through a Legendre transformation.
0
votes
Non-relativistic QFT Lagrangian for fermions
Regarding question 2, in Weinberg's The Quantum Theory of Fields page 295, the author defines the functional derivative of an arbitrary bosonic functional $F[q,p]$:
$$\frac{\delta F[q,p]}{\delta q}=i …
- 31
3
votes
2
answers
1k
views
Non-relativistic QFT Lagrangian for fermions
Take the ordinary Hamiltonian from non-relativistic quantum mechanics expressed in terms of the fermi fields $\psi(\mathbf{x})$ and $\psi^\dagger(\mathbf{x})$ (as derived, for example, by A. L. Fetter …
- 31