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In A. Zee's Quantum field theory in a Nutshell, he describes the QFT analogy of a matress, a 2D grid of points $q_a$ connected by springs (first page of first chapter, $q_a$ is the vertical displacement). Then (on page 20) he describes a source of excitation, for example by pressing down one point:

Obviously, pushing on the mass labeled by a in the mattress corresponds to adding a term such as $J_a(t) q_a$ to the potential V.

I don't understand the factor $q_a$ in that term. If the matress is completely at rest, then $q_a=0$ for all points. So this term would vanish, regardless how strong the source $J_(t)$ is.

What am I missing?

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    $\begingroup$ One adds $J q$ to the Lagrangian (or Hamiltonian). In deriving the equations of motion for $q$, one takes a partial derivative of the Lagrangian with respect to $q$. Thus adding $J q$ to the Lagrangian adds $J$ (without $q$) to the equation of motion. $\endgroup$
    – user2309840
    Commented Oct 17, 2015 at 14:55
  • $\begingroup$ Thanks, that makes sense. But you agree that for a source $J(\mathbf{x})\delta(t-t_0)$ that is active only at $t_0$, the Lagrangian (or Hamiltonian) at $t_0$ is not different with or without the source? $\endgroup$
    – Bass
    Commented Oct 17, 2015 at 16:11
  • $\begingroup$ We often make a distinction between the on-shell Lagrangian and the Lagrangian. The on-shell Lagrangian is the Lagrangian evaluated for a solution to the equations of motion. But the answer is no in either case. 1) The Lagrangian is different because there is now a $Jq$ term. 2) The on-shell Lagrangian is different because in the presence of $J$, the configuration $q = 0$ is (usually) no longer a solution of the equations of motion. $\endgroup$
    – user2309840
    Commented Oct 17, 2015 at 18:04
  • $\begingroup$ @user2309840 You should probably post your first comment as an answer. $\endgroup$
    – Javier
    Commented Oct 19, 2015 at 1:45

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One adds $Jq$ to the Lagrangian (or Hamiltonian). In deriving the equations of motion for $q$, one takes a partial derivative of the Lagrangian with respect to $q$. Thus adding $Jq$ to the Lagrangian adds $J$ (without $q$) to the equation of motion. The configuration $q=0$ is then (usually) no longer a solution of the equations of motion.

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