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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}$?

Qmechanic
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Yotam Vaknin
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    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
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    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. –  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
  • The field is a superposition of an infinity of oscillators. Write it as such. Observe how both propagation terms amount to the same thing! Stick to free fields. – Cosmas Zachos Dec 15 '18 at 20:44
  • @YotamVaknin, I was not referring to the specific form of that operator but the formalism of field theory. There is no analog in classical particle mechanics either but the formalism of QM brings it forth. There are classical field Poisson brackets and all the other items that are elevated to QM by application of i*h_bar. My point is that if you see how the field is described in CFT the QM prescription follows as it does for particles. –  Dec 15 '18 at 20:46
  • @CosmasZachos Thanks :) I see what you are saying, that the field solves both equations, but is this always true? Since we use the "schrodinger" approach for the perturbation theory, does it mean that the Hamiltonian gives the "correct time evolution?" – Yotam Vaknin Dec 15 '18 at 20:49
  • @ggcg Thanks for your response! I can see that they both follow the same prescription. But when we try to solve using perturbation methods for a field in QFT, we assume, at different points in the derivation, both types of time evolution. I am asking if they are somehow equivalent? which is the "true" time evolution for a state? – Yotam Vaknin Dec 15 '18 at 20:52
  • Define a state. They are equivalent, and related by a field transformation. That said it doesn't hurt to use the representation that is easiest to "prove" a theorem or derive a result. –  Dec 15 '18 at 20:56
  • @ggcg Thanks for that! A state would be a vector in the Fock space of the system I guess, if that changes anything. Anyway I would very much appreciate a link to a source about this! thx – Yotam Vaknin Dec 15 '18 at 20:59
  • Interaction picture. – Cosmas Zachos Dec 15 '18 at 21:02
  • @CosmasZachos Hey! Thanks for the link. I understand that if for a free field E.L. == $e^{iHt}$ then everything follows (using the Interaction picture). My beef is with this equality. Where does it come from? How do you show this for fermions? – Yotam Vaknin Dec 15 '18 at 21:06
  • Yes Fock space. One view is that a state is a classical "shape" of the field at all points in space at some instant in time. Then the momentum operator is i*h_bar times the functional derivative with respect to the field. –  Dec 15 '18 at 21:29

2 Answers2

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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.

tparker
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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.

Yotam Vaknin
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