Deriving Dirac Hamiltonian with (+,---) Metric Signature

[pleasehelpmeno]

Hi can anyone explain how to derive an expression for the Dirac Hamiltonian, I thought the procedure was to use $\mathcal{H}= i\psi^{\dagger}\Pi -\mathcal{L}$, but in these papers the have derived two different forms of the Dirac equation $H=\int d^{3}x \psi^{\dagger}i\partial_{0}\psi$http://arxiv.org/abs/hep-ph/9905242 and $H=\int d^{3}x -\psi^{\dagger}i\partial_{0}\psi$http://arxiv.org/abs/hep-ph/0003045v3 yet both use the (+,---) metric signature.

a) does anyone know how to derive a Hamiltonian that only contains the $\partial_0$ operator?
b) is it possible to have this - sign in place using the same metric tensor convention?

 

[dextercioby]

The derivation is simple, if you know the theory of Dirac (!) for constrained systems, as the Dirac field is an example of dynamical system with class II constraints.

Check out the textbook by Henneaux and Teitelboim. Even though it's particularly addressed to gauge systems, I suspect it has a general overview of all constrained systems.

 

[Avodyne]

Neither of these expressions is valid, since the hamiltonian must be expressed in terms of the coordinates and momenta, and not their time derivatives.

 

[jfy4]

The Lagrangian is
$$\mathcal{L} = i\bar{\psi}\gamma^{\mu}\partial_{\mu}\psi - m\bar{\psi}\psi$$
$$= i\bar{\psi}(-\gamma^0 \partial_0 + \gamma^i \partial_i )\psi - m\bar{\psi}\psi$$
$$\mathcal{H} = \Pi \dot{\psi} - \mathcal{L}$$
so
$$\Pi = \frac{\partial \mathcal{L}}{\partial \dot{\psi}} = -i\psi^{\dagger}$$
then
$$\mathcal{H} =- i\psi^{\dagger}\dot{\psi} - i\bar{\psi}(-\gamma^0 \partial_0 + \gamma^i \partial_i )\psi + m\bar{\psi}\psi = -i\bar{\psi}\gamma^i \partial_i \psi + m\bar{\psi}\psi$$
with metric $(-+++)$

 

[pleasehelpmeno]

thats what i though, which is why i am confused that the above papers don't have hamiltonians in that form.

Can you think of any reason why there hamiltonains have time derivatives?