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.

  • $\begingroup$ start with the simplest possible case - free particle, with a harmonic oscillator potential. If you can do that, then extend to fields. $\endgroup$ – levitopher Nov 3 '13 at 0:33
  • $\begingroup$ Related: physics.stackexchange.com/q/45808/2451 $\endgroup$ – Qmechanic Nov 22 '13 at 12:51

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

$$H(q,p,t)~=~H_{\rm free}+H_{\rm int},$$


$$ H_{\rm int}=-L_{\rm int}.$$

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


where $L_n$ is homogeneous in velocities $v$ with weight $n$, then

$$H~=~v^ip_i-L~=~\left(v^i\frac{\partial}{\partial v^i}-1\right) L =\sum_{n=0}^{2}(n-1)L_n = L_2 -L_0. $$

  • $\begingroup$ Thanks. But you used definition of Hamiltonian which breaks its interpretation as full energy of the system (or, in rel. case, zero component of energy-momentum tensor). Why? $\endgroup$ – John Taylor Nov 3 '13 at 10:22
  • $\begingroup$ The answer uses the standard definition of a Hamiltonian, which in many cases, such as, e.g. QED, is equal to the total energy. For discussions about Hamiltonian vs. total energy, see also e.g. this Phys.SE post and links therein. $\endgroup$ – Qmechanic Nov 3 '13 at 10:47

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 ${\cal L}_{int} \to - {\cal L}_{int}$.

  • $\begingroup$ But hamiltonian and lagrangian are really equal to each other (except the minus sign) for QED case (for Dirac spinor field, I forgot to write it). So do you know how to prove this? $\endgroup$ – John Taylor Nov 3 '13 at 10:25
  • $\begingroup$ QED is an example of a constrained system. In this case, time evolution is determined by the primary Hamiltonian, not the canonical Hamiltonian which are related by $H_p = H_c + u^m \varphi_m$ where $\varphi_m$ are primary constraints and $u^m$ are some multipliers that are determined by some consistency requirements on the constraints. arxiv.org/abs/hep-th/9312078 provides a very good discussion of such systems. $\endgroup$ – Prahar Nov 3 '13 at 14:47

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