Widdekind Posted September 26, 2013 Posted September 26, 2013 http://en.wikipedia.org/wiki/Quantum_electrodynamics#Equations_of_motion When you take the variational derivative, of the QED Lagrangian, with respect to the wave function [math]\Psi[/math]... why doesn't the derivative include terms, due to the conjugate transpose of the wave function [math]\bar{\Psi}[/math] ? In Classical analogy, for a Lagrangian with the KE term [math]\left( \frac{1}{2m} \vec{p}^T \circ \vec{p} \right)[/math], derivatives with respect to momentum would include (one) terms, from the transpose of momentum, which is essentially the same mathematical object
ajb Posted September 26, 2013 Posted September 26, 2013 You can treat them as independent variables, and you should be thinking in terms of fermionic fields here not wavefunctions.
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now