Why in the field theory, particle's motion is described by 0+1 dimensional field theory? I started reading the lecture notes on Path integral formulation by Ashoke Das. At the very first page of the introduction chapter, he says that - "a theory describing the motion of a particle can be regarded as a special case, namely, we can think of such a theory as a $(0+1)$ dimensional field theory".
I don't understand why this is so. Maybe it is a very trivial question but I couldn't find a straightforward answer. I am puzzled by thinking why the time is enough to describe the motion? A particle surely has spatial dimension and can move in a 3-dimensional space as well. In absence of any external potential, a particle moves in a straight-line with constant velocity. This certainly requires a $1+1$ description. Doesn't it?
 A: Field theory in $(0+1)$ dimensions is formally equivalent to particle mechanics.
To see why let's first consider a scalar field $\phi$ in $(d+1)$ dimensions: it specifies a field value $\phi(t, x_1,...,x_d)$ for each point $(t, x_1,...,x_d)$ in spacetime. Therefore, the field is a map from spacetime to real numbers:
$$(t, x_1,...,x_d) \rightarrow \phi(t, x_1,...,x_d).$$
On the other hand, the trajectory of a particle in $(d + 1)$ spacetime dimensions gives you a point in space for each time.
The trajectory can be viewed as a map
$$t \rightarrow (\phi_1(t), . . . , \phi_d(t)).$$
I used the symbol $\phi_i(t)$  instead of the more common $x_i(t)$ to make more evident the analogy with the field case: the "output" of the "trajectory map" is $d$ real numbers but the "input" space is just the real line, formally a $(0+1)$ spacetime (i.e. only time!).
In short: a collection of $d$ scalar fields $\phi_i$ over $(0+1)$  spacetime is equivalent to the trajectory (or world line) of a particle in $(d+1)$ spacetime.
Note: it is also possible to consider $0+0$ "field theory", but this is another story, see this question.
