Lagrangian and hamiltonian of interaction How to prove that lagrangian of interaction is equal to hamiltonian of interaction with minus sign? For example, I can't prove it for special case - quantum electrodynamics. 
 A: Consider a general Lagrangian
$$L(q,v,t)~=~L_{\rm free}+L_{\rm int}.$$
It is implicitly understood that the free part $L_{\rm free}$ is at most quadratic in position and velocity variables. (In field theory the $q$ variables are fields, and the $v$ variables are time derivatives of the fields. They may be Grassmann-odd.)
Assume furthermore that the interaction term $L_{\rm int}=L_{\rm int}(q,t)$ does not depend on velocities $v$.
Then one may prove that the Hamiltonian
$$H(q,p,t)~=~H_{\rm free}+H_{\rm int},$$
satisfies
$$ H_{\rm int}=-L_{\rm int}.$$ 
This is most easily shown for regular Legendre transformations, but it also works quite generally for singular Legendre transformations, such as, e.g. QED. 
The main idea is that if 
$$L~=~\sum_{n=0}^{2}L_n,$$
where $L_n$ is homogeneous in velocities $v$ with weight $n$, then 
$$H~=~v^ip_i-L~=~\left(v^i\frac{\partial}{\partial v^i}-1\right) L =\sum_{n=0}^{2}(n-1)L_n = L_2 -L_0. $$
A: What you have said is true only when the interaction part of the Lagrangian has no dependence on the derivatives of fields AND when the free part of of the Lagrangian is precisely quadratic. In such cases,
$$
{\cal L}[\phi,\partial\phi] = \frac{1}{2} (\partial \phi)^2 - {\cal L}_{int}[\phi]
$$
The canonical Hamiltonian in this case is
\begin{equation}
\begin{split}
{\cal H} &= \partial \phi \frac{\delta {\cal L}}{\delta (\partial \phi)} - {\cal L}[\phi,\partial\phi] \\
&= \partial \phi \left( \partial \phi \right) - \left[ \frac{1}{2} (\partial \phi)^2 - {\cal L}_{int}[\phi] \right] \\
&= \frac{1}{2} (\partial \phi)^2 + {\cal L}_{int}[\phi]  
\end{split}
\end{equation}
In particular, QED does not have a Lagrangian of the type discussed above and therefore the canonical Hamiltonian cannot be obtained by replacing ${\cal L}_{int} \to - {\cal L}_{int}$. 
