# E.L. Equations in QFT

In QFT, we use the Lagrangian to construct the Hamiltonian, and in the Interaction Picture (with regards to the Free Field Hamiltonian) use the full Hamiltonian to calculate the changes in the field (or the wave function) over time. By that I mean, for a field $$\Psi$$:

$$\Psi (t) = e^{iH t} \Psi e^{-iH t}$$

(This is usually more complex, using the time ordered exponential etc, but bare with me)

On the other hand (!), to solve for the free field, we use the E.L. equations. For example in the K.G. case:

$$(\partial _ \mu \partial^\mu + m^2) \Psi = 0$$

Which one is it? Are they equivalent in some sense?

To put it clearly: Which equation describes the time evolution of operators and states in QFT? E.L. or ~schrodinger equation~ in the sense of $$e^{-i H t}$$?

• Well, most field theories have both a Lagrangian and a Hamiltonian formulation. The corresponding EOMs are the Lagrangian and Hamiltonian equations, respectively. – Qmechanic Dec 15 '18 at 20:28
• Write down $\Psi$ for the free field, and demonstrate the consistency of the two equations. – Cosmas Zachos Dec 15 '18 at 20:29
• They are equivalent and related by a transformation. We use solutions to the E.L. equations to describe "free field states" then perturb the system to describe interactions and scattering using the interaction Hamiltonian. I would recommend reviewing the formalism from the point of view of classical mechanics and classical field theory, then apply to QFT. – ggcg Dec 15 '18 at 20:33
• @ggcg Can you please elaborate on that? because I cannot think of any classical analog of $e ^ {iHt}$. – Yotam Vaknin Dec 15 '18 at 20:35
• @Qmechanic But the Hamilton equations and exponentiation by the Hamiltonian are two different equation, aren't they? – Yotam Vaknin Dec 15 '18 at 20:36

In general, the Schrodinger equation. The EL equation describes the time evolution of a classical field without quantum fluctuations.

For a free field theory, it turns out that the solutions to the EL and the Schrodinger equations are very closely related: once you've solved the classical EL equation, converting the resulting classical Hamiltonian (expressed in terms of the plane-wave solutions to the EL equation) into a quantum Hamiltonian is basically a single step - namely, imposing the canonical commutation relations. After you'd done this, you're actually describing a different problem (the time-evolution of the free quantum rather than classical field theory). The solutions look very similar when written out, but their interpretation is different.

Once you go to an interacting theory, the EL equation becomes much less useful. Since it can't be solved analytically, you can't even figure out the exact dynamics of the interacting classical field, let alone the quantum one. So the solutions to the exact (as opposed to perturbative) EL equation are rarely used.

People wrote very good answers, but as a novice to the field I honestly couldn't understand them. I finally have a simple derivation that gives this result, so I'll share it here. The end result is - they are equivalent, and it is particularly hard to prove. Here is a BAD proof, that can set some minds at ease. It has many different problems, the most important one being that it doesn't work for Fermions.

First of all, notice that for a function $$F(A,B)$$ of two operators $$A,B$$ we always have:

$$[A,F] = \frac{\partial F}{\partial B} [A,B]$$

Assuming [A,B] is a scalar. This can be proven by expanding $$F$$ to a series.

In classical mechanics, if we have a Field $$\phi$$ with a Lagrangian density $$L$$, we can define $$H$$ with the property of:

$$\dot{\phi} = \{ \phi, H \}_{PB} = \frac{\partial H}{\partial \Pi} \{ \phi, \Pi \}_{PB}$$

Which is basically Hamilton's equation. If we define $$\phi$$ in such a way that:

$$[\phi, \Pi] = i \{\phi, \Pi\}_{PB}$$

as we do in QFT, we will finally get the defining equation of QM:

$$\dot{\phi} = \frac{\partial H}{\partial \Pi} \{ \phi, \Pi \}_{PB} = i \frac{\partial H}{\partial \Pi} [ \phi, \Pi ] = i [\phi, H]$$

So all 3 formulations (E.L., Hamilton and Heisenberg) are equivelent.