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I am still at an early stage of studying Quantum Field Theory (I am reading QFT In A Nutshell by A. Zee).

In the book I'm reading, it starts from a discrete lattice of material "lumps" labeled by $a$, seperated from each other by a finite $l$, so that the position of lump $a$ is $q_a(t)$. The next step is to take $l \to 0$ and "promote":

$$ q \to \varphi $$ $$ a \to \vec{x} $$ $$ q_a(t) \to \varphi (t, \vec{x}) = \varphi (x) $$ $$ \sum_a \to \int d^Dx $$

If this was in the realm of (classical) continuum mechanics, $\vec{x}$ would simply label the simplified and theoretical (This aspect of physics benefits from this simplification at no cost) infinitesimal lump that was actually displaced by $\vec{x}$ from the origin at $t=0$.

In QFT though, by definition, the field is not meant to be a simplification of anything. It's meant to be quite the contrary. I am not yet confident enough to ask my concrete questions, because I realize that they carry with them implicit assumptions which are probably false. So I'll just ask:

If it's not the displacement $\vec{x}$, and It's not the classical label $\vec{x}$, then, what is it? What does it mean?

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  • $\begingroup$ The book is Zee, QFT in a nutshell? $\endgroup$ – innisfree Sep 22 '14 at 18:42
  • $\begingroup$ @innisfree Yes. $\endgroup$ – user76568 Sep 22 '14 at 18:42
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Initially the label $a$ represent the site on a lattice with sites separated by distance $l$. i.e., the discrete position for each of a collection of oscillators, $q_a$, with an oscillator at each lattice site.

When we take the continuum limit, $l\to0$, the label $a$ becomes continuous, i.e. it becomes the position variable $x$. We now have an infinite collection of oscillators, labelled by their position $x$ - an oscillator at each point in space.

The oscillators don't themselves move about, they're fixed at the lattice sites (and later points in space). But they are correlated. An oscillation in $q_1$ might be correlated with an oscillation in its next door neighbour, $q_2$. The propagation of oscillations in that way is basically particles propagating.

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  • $\begingroup$ It could have been an infinite collection in the discrete case as well, and replaced by uncountable in the continuum limit. But It's not essential :). $\endgroup$ – user76568 Sep 22 '14 at 19:02
  • $\begingroup$ @Dror, yes that's true. I hope I managed to answer your question, I couldn't really understand your confusion. QFT is a tough subject for self-study. If possible, I advise banding together with someone else and discussing the book. If that's not possible, keep going all the same! $\endgroup$ – innisfree Sep 22 '14 at 19:06
  • $\begingroup$ So, for example, if 2 particular oscillators ($1$ at $x_1$ and $2$ at $x_2$) switch their position at a later time, then $x_1$/$x_2$ becomes the label of $2$/$1$? $\endgroup$ – user76568 Sep 22 '14 at 19:06
  • $\begingroup$ I don't think that's the right picture. The oscillators don't themselves move about, they're fixed at the lattice sites (and later points in space). But they are correlated. An oscillation in $q_1$ might be correlated with an oscillation in its next door neighbour, $q_2$. The propagation of oscillations in that way is basically particles propagating. $\endgroup$ – innisfree Sep 22 '14 at 19:09
  • $\begingroup$ Given this question, how do you estimate my odds at grasping the entire book (By my self) "properly" enough? :) $\endgroup$ – user76568 Sep 22 '14 at 19:16

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