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The quantum fields are operator valued distributions. In some sloppy books like Peskin and Schroeder the Euler-Lagrange equation are used to get the equations of motion.

What does it mean to take a derivative with respect to some operator?

Qmechanic
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  • Related: https://physics.stackexchange.com/q/48030/2451 – Qmechanic Jun 28 '16 at 17:25
  • The Euler Lagrange equations, variations and the like are done on classical fields. The equations of motion therefore are for real/vector etc valued functions. Then the theory is quantised, and it's enforced that the operators satisfy the same eom. – snulty Jul 02 '16 at 17:41

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For a (sufficiently nice) expression $f(X)$, where $X$ ranges over a vector space), the directional derivative $Ydf(X)$ with respect to $Y$ (in the same vector space) is the coefficient of $\epsilon$ in an expansion $f(X+\epsilon Y)-f(X)$ where $\epsilon$ is a formal variable with $\epsilon^2=0$.

The functional derivative $\delta f(x)/\delta(X)$ is the linear mapping that maps $Y$ to $Ydf(X)$.

Arnold Neumaier
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