Widdekind Posted September 26, 2013 Share 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 Link to comment Share on other sites More sharing options...

ajb Posted September 26, 2013 Share Posted September 26, 2013 You can treat them as independent variables, and you should be thinking in terms of fermionic fields here not wavefunctions. Link to comment Share on other sites More sharing options...

## 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 account## Sign in

Already have an account? Sign in here.

Sign In Now