Interaction Hamiltonian of Scalar QED

[AuraCrystal]

Homework Statement

Problem 7.15 from Aitchison and Hey, Volume I, 3rd Edition. Verify the forum (7.139) of the interaction Hamiltonian \mathcal{H_{S}^{'}}, in charged spin-0 electrodynamics.

Equation 7.139 is

\mathcal{H_{S}^{'}}= - \mathcal{L}_{int} - q^2 (A^0)^2 \phi^{\dagger} \phi

Homework Equations

The interaction Lagrangian:
\mathcal{L}_{int}= -iq(\phi^\dagger \partial^\mu \phi - (\partial^\mu \phi^\dagger)\phi) + q^2 A^\mu A_\mu \phi^\dagger \phi

The Attempt at a Solution

There are no time derivatives of A in the interaction Lagrangian, so
\mathcal{H_S^{'}} = \pi_{\phi}^{'} \dot{\phi}+\pi_{\phi^\dagger}^{'} \dot{\phi^\dagger}-\mathcal{L}_int
We have that
\pi_{\phi}^{'} = \frac { \partial \mathcal{L}_{int}}{ \partial \dot{\phi}}=-iq \phi^\dagger A_0
and similarly,
\pi_{\phi^{\dagger}}^{'} = \frac { \partial \mathcal{L}_{int}}{ \partial \dot{\phi^\dagger}}=iq \phi A_0
Thus
\mathcal{H_S^{'}}=- \mathcal{L} +iqA_0(\phi \dot{\phi^{\dagger}} - \phi^{\dagger} \dot{\phi}).

Which obviously does not match 7.139 above. This book uses a (+,-,-,-) sign convention for the metric.

 

[Vic Sandler]

I don't have a copy of the book. Are you sure about eqn (7.139)? It has the hamiltonian on the lhs, but the lagrangian density on the rhs. Once that's cleared up, I think an integration by parts might do the trick. Of course, there is a missing factor of A_{\mu} in the first term of the lagrangian in the relevant equations section.

 

[AuraCrystal]

I don't have a copy of the book. Are you sure about eqn (7.139)? It has the hamiltonian on the lhs, but the lagrangian density on the rhs. Once that's cleared up, I think an integration by parts might do the trick. Of course, there is a missing factor of A_{\mu} in the first term of the lagrangian in the relevant equations section.

Well, the idea is that the Hamiltonian is given by \mathcal{H}=\Sigma_i \pi_i \dot{\phi} - \mathcal{L}, and when we include interactions between \phi and A_\mu we get that it is the sum between the free K.G. Lagrangian, the free Maxwell Lagrangian, and the interaction Lagrangian I gave above. Now, since the interaction Lagrangian depends on the time derivatives of \phi and \phi^\dagger, their canonical momenta will change, and so we have to account for that in determining the perturbation in the Hamiltonian from the free case.

Also, the problem is talking about Lagrangian and Hamiltonian densities, so integration by parts isn't really helpful.

 

[Vic Sandler]

Well, the idea is that the Hamiltonian is given by \mathcal{H}=\Sigma_i \pi_i \dot{\phi} - L,.

Now you've got the hamiltonian density on the lhs and the lagrangian on the rhs. I would expect hamiltonian = F(lagrangian), or hamiltonian density = G(lagrangian density). I don't think you can mix them like you (and the book?) are doing.

 

[AuraCrystal]

Yeah sorry. I keep forgetting the \mathcal. And yeah, all of the H's and L's refer to the Lagrangian and Hamiltonian densities.

 

[Vic Sandler]

In that case, you should be able to apply integration by parts. First, integrate both sides of the equation.

 

[AuraCrystal]

Against what? Time? Space? Both? :)

 

[Vic Sandler]

Space. The integral of the hamiltonian/lagrangian density over all of space is the hamiltonian/lagrangian. There might be some statement in the book to the effect that \phi goes to zero rapidly as the space coordinates go to infinity.

 

[AuraCrystal]

Yeah, but there's no spatial derivatives in \mathcal{H}_S^{'}

 

[Vic Sandler]

Yes there are. I see that you have edited your posts to properly represent densities. But you haven't yet multiplied the first term in the lagrangian by A_{\mu}.