# Interpretation of $\vec{x}$ in QFT

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?

• The book is Zee, QFT in a nutshell? – innisfree Sep 22 '14 at 18:42
• @innisfree Yes. – user76568 Sep 22 '14 at 18:42

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.
• 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 $2$/$1$? – user76568 Sep 22 '14 at 19:06
• 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. – innisfree Sep 22 '14 at 19:09