# Euler-Lagrangian equation of motion of quantum fields in QFT

A canonical way of doing quantum field theory is by starting with some Lagrangian, for example, that of free scalar field $$L=\frac{1}{2}\partial_{\mu}\phi \partial^{\mu}\phi-\frac{1}{2}m\phi^2$$ Then by employing the Euler-Lagrangian equation, i.e. $$\delta L=0$$, which would produce Klein-Gordon equation for the field $$(\square+m^2)\phi=0$$ Then we proceed to various quantization procedure that lead us to express $$\phi$$ in terms of creation/annihilation operator.

However, when I am reading QFT text, it is often said that the Euler-Lagrangian equation does not hold exactly in QFT, and there are various quantum fluctuation that is characteristic of QFT.

I don't understand this statement, didn't we start doing QFT by employing the Euler-Lagrangian equation, and in this case, just the KG equation? Didn't we do quantization on the basis of this equation? Why is it said that the quantization would make the original E-L equation be violated by quantum fluctuation? Can anyone give an explicit example of quantum fluctuation violating the E-L equation?

• In the path integral formalism, EL equations only hold in the saddle point approximation (which is a 'good' approximation at first sight). All other paths contributing to the transition amplitudes are thought of as deviations from the EL equations. Commented Mar 1, 2022 at 10:36
• I am copying my comment to the answer below:Hi, I understand that in the Path Integral Approach, EL is probably not needed. But if you have Srednicki's QFT book, you can have a look at section 3, where the EL equation (3.7) leads to the field solution (3.14). Then he employed the quantization in (3.28). I think in Schwartz's QFT book, he follows essentially the same way. This is the point where I don't get Commented Mar 1, 2022 at 10:53
• @JeanbaptisteRoux This is not completely correct. The equations of motion hold in all correlators up to contact terms. Commented Mar 1, 2022 at 21:18

No, we don't start by assuming EL equations for the quantum fields. We start by assuming the action of the quantum fields. The job that the EL equations do in classical field theory, i.e., predicting the time-evolution of the field configuration, is done by evaluating the propagator $$\langle \phi(x)\phi(y)\rangle$$ via performing the path-integral in the action formulation of QFT. Only when you take the large action limit (i.e., $$S\gg1$$) does the path integral tell you that the only surviving contribution to the propagation amplitudes of fields will come from those classical paths that obey the EL equations. The Schwinger-Dyson equations give you the QFT analog of classical EL equations. You can explicitly see the corrections to the EL equations, for example, when you do the perturbative expansion of the SD equations for a weakly coupled QFT.

• Hi, I understand that in the Path Integral Approach, EL is probably not needed. But if you have Srednicki's QFT book, you can have a look at section 3, where the EL equation (3.7) leads to the field solution (3.14). Then he employed the quantization in (3.28). I think in Schwartz's QFT book, he follows essentially the same way. This is the point where I don't get. Commented Mar 1, 2022 at 10:53
• @TanTixuan I don't have Schwartz with me at the moment but I had a look at Section $3$ of Srednicki. In a free field theory, the classical Green's function and the quantum Green's function remain the same -- simply because there are no loops in a free field theory. Srednicki is dealing with a free field theory in Section $3$. In particular, he's simply using the classical EL equations to motivate the machinery that is needed to solve the eigenvalue problem of the KG Hamiltonian. However, in general, you just start with $3.28$ and the quantum version of $3.25$ as the definition of your QFT.
– user87745
Commented Mar 1, 2022 at 11:10
• Hi, The free field Schwinger Dyson equation, in my understanding, reads something like (copied from Schwartz) $(\square_{x_1}+m^2)<\phi(x_1)\phi(x_2)\phi(x_4)\phi(x_3)>=-i\delta(x_1-x_2)<\phi(x_3)\phi(x_4)>...$. If EOM is satisfied exactly, then I assume there shouldn't be terms on the RHS, i.e. the delta terms. Commented Mar 1, 2022 at 11:18
1. It is true that the operators in the operator formalism (say in the Heisenberg picture) satisfy the Heisenberg EOMs exactly (which in Klein-Gordon theory is the Klein-Gordon equation).

2. However, when going from the operator formalism to the path integral formalism (by inserting infinitely many completeness relations$$^1$$), the path integral becomes an infinite-dimensional integral over (both on-shell and) off-shell/virtual field configurations, where the EOMs are not necessarily satisfied.

In particular, the EOMs inside correlator functions in the path integral formalism are only satisfied in a certain quantum average sense, cf. the Schwinger-Dyson (SD) equations $$\left< \Omega \left| T_{\rm cov}\left\{ F[\phi]\frac{\delta S[\phi;J]}{\delta \phi(x)}\right\}\right| \Omega \right>_J~=~i\hbar\left< \Omega \left| T_{\rm cov}\left\{\frac{\delta F[\phi]}{\delta \phi(x)} \right\}\right| \Omega \right>_J ~.$$ Further subtleties arise because of the time-ordering symbol $$T_{\rm cov}$$ in the correlator functions, cf. e.g. this related Phys.SE post.

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$$^1$$ For the derivation, see any good textbook on QM/QFT.

The Heisenberg field operators $$\hat\phi(\mathbf{x},t)$$ do in fact obey the E-L equations, however expectation values of these operators don't and require corrections. For example, an expectation value containing two field operators obeys $$(\partial_x^2-m^2)\langle0|T\phi(x)\phi(y)|\rangle = -i\delta^3(x-y)$$ the term on the RHS is called a contact term.