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?


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.

  • $\begingroup$ Okay. So in particle mechanics, the objective is to find the spatial co-ordinates as a function of time. Thus the field for particle mechanics is actually the spatial co-ordinates. Or in other words, the field value $\phi$ is the value of the co-ordinates $(x,y,z)$ at some particular time $t$. Is this what you are trying to say? $\endgroup$ – Samapan Bhadury Oct 2 '20 at 13:22
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    $\begingroup$ Yes, the coordinates of a particles are its "internal degrees of freedom", similarly to the $\phi$ value of the field, which is its internal degree of freedom. $\endgroup$ – Quillo Oct 2 '20 at 13:24
  • $\begingroup$ Okay. Thanks for clarifying the issue. $\endgroup$ – Samapan Bhadury Oct 2 '20 at 13:26

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