# A proof that Heisenberg's and Euler Lagrange's equations are equivalent in QFT [closed]

I asked this before (link, link) but I think people didn't understand what I was asking, so I am going to try again . Thanks for everyone that helped so far.

In QFT, Heisenberg's equation is equivalent to Euler Lagrange's equation for any specific Lagrangian. They have the exact same form, since Representations of the Lorentz group limit the possible equations we can have in the first place.

In classical QM, this would mean: $$\dot{X} = P \\ \dot{P} = -\frac{dV}{dX}$$

which is true in the Heisenberg picture for operators, and it is also true classically. This is also true in QFT, and it is an important feature of the theory.

I am looking for a proof of this for a general QFT Lagrangian, for both Bosonic and Fermionic (anti-)commutation relations.

A few important notes:

1. Notice that the Poisson brackets have very different form in field theory, so please don't write things that don't take this into account. link
2. If you just set Fermionic anti-commutation relations, some classical Hamiltonians will have different Hamilton and Heisenberg equations. For example $$H=P, H=P_1P_2$$.
3. For fermions, the answers I got point somehow to Grassman-valued-numbers. This won't always work (see 2), but more importantly I don't understand what the actual argument is. This is why I am asking for a proof.

Thanks! (and thanks specifically to @Qmechanic)

## closed as too broad by Qmechanic♦Feb 9 at 21:52

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• It seems that OP is asking (v2) for a general proof of the equivalence between Lagrangian and Hamiltonian theories at the quantum level. That seems too broad. At the classical level for non-degenerate bosonic theories, see e.g. this Phys.SE post. – Qmechanic Feb 6 at 12:58
• Not exactly, because in the quantum level Heisenberg and Hamiltonian formulation are not necessarily equivalent. – Yotam Vaknin Feb 7 at 15:45
• According to your definitions how do you distinguish the Heisenberg vs. the Hamiltonian formulation at the quantum level? – Qmechanic Feb 7 at 16:05
• The Hamilton formulation is $\dot{A} = \{ A,H \}_{PB}$, while Heisenberg is $\dot{A} = -i [A,H]$. These are different formal operations, and I don't see why they should always give the same result. (and in some edge cases they actually don't...) – Yotam Vaknin Feb 7 at 17:05
• That's just the classical Hamiltonian formulation. There are also quantum mechanical Hamiltonian formulations. – Qmechanic Feb 7 at 17:14