Why in the field theory, particle's motion is described by 0+1 dimensional field theory?

Question

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?

Quillo · Accepted Answer

Field theory in $(0+1)$ dimensions is formally equivalent to particle mechanics.
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, it can be viewed as a mapping
of spacetime into the real numbers: $(t, x_1,...,x_d) rightarrow phi(t, x_1,...,x_d)$.
The trajectory of a particle in $(d + 1)$ dimensions specifies a point in space for each moment of time. It can be viewed as a mapping of the real line (i.e. $t in mathbb{R} $) into space, namely: $t rightarrow (phi_1(t), . . . , phi_d(t))$. I used the symbol $phi$ to indicate the coordinated of the particle just to make more evident the analogy with the initial scalar field: now the "outcome" of the mapping are $d$ numbers but the initial ambient space is just a $(0+1)$ spacetime (i.e. only time). Loosely speaking: a collection of $d$ scalar fields in $(0+1)$  spacetime are equivalent to the motion of a particle in $(d+1)$ spacetime.

Why in the field theory, particle's motion is described by 0+1 dimensional field theory?

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