# Interaction Hamiltonian of Scalar QED

1. Aug 22, 2013

### AuraCrystal

1. The problem statement, all variables and given/known data

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

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

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

3. 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.

Last edited: Aug 22, 2013
2. Aug 22, 2013

### 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.

3. Aug 22, 2013

### AuraCrystal

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.

Last edited: Aug 22, 2013
4. Aug 22, 2013

### Vic Sandler

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.

5. Aug 22, 2013

### 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.

6. Aug 22, 2013

### Vic Sandler

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

7. Aug 22, 2013

### AuraCrystal

Against what? Time? Space? Both? :)

8. Aug 22, 2013

### 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.

9. Aug 22, 2013

### AuraCrystal

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

10. Aug 23, 2013

### 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 $A_{\mu}$.