If I have two probability distributions $P$ at $t$ and $P’$ at $t’$ separated by some time interval. Then, can I describe the transform between the two distributions as $$P’(x) = \int P(a) D(a, x-a, t’, t) da$$ over all space. Where $D(a, b, t, t’)$ is some function that returns the probability of a particle at position $a$ at time $t$, ending up at position $b$ at time $t’$?

From my layman understanding, this seems analogous to saying that the probability of a particle being at some point $x’$ in $P’$ is equal to the sum of the probability that for all points in space a particle was at some position $x$ in $P$ times the probability that a particle at $x$ ended up at $x’$ after the time interval.

This seems to be similar to the definition of the Feynman Propagator, but I lack enough knowledge in field theory to say if these really are similar or if I am misunderstanding the concept of the propagator.


1 Answer 1


Then, can I describe the transform between the two distributions as $$P’(x) = \int P(a) D(a, x-a, t’, t)da $$ over all space.

No. Not in general, if you want to include quantum mechanics.

It is well known in quantum mechanics that the wavefunction $\psi(a,t')$ propagates according to the kernel $K(x,t, a, t')$ as $$ \psi(x,t) = \int da K(x,t; a,t')\psi(a,t')\;. $$

Further, it is well known that $$ P(x,t) = |\psi(x,t)|^2 $$ so $$ P(x,t) = \int da K(x,t; a,t')\psi(a,t')\int da' K^*(x,t; a',t')\psi^*(a',t') $$

  • $\begingroup$ What does the * denote here? I seem to remember it having some thing do to with the complex values. $\endgroup$ Commented Apr 13 at 17:37
  • $\begingroup$ The $*$ symbol here denotes complex conjugation. For a complex number $z=x+iy$, where $x$ and $y$ are real, we have $z^* = x - iy$. $\endgroup$
    – hft
    Commented Apr 13 at 17:38
  • $\begingroup$ Sorry to keep asking questions, but what does it mean here that we are integrating the complex conjugation over da’ as opposed to da? Is that just to specify that these are separate integrals? $\endgroup$ Commented Apr 13 at 17:42
  • $\begingroup$ Yes, they are separate integrals. Both $a$ and $a'$ are "dummy variables" (meaning they are integrated over, so there is no dependence on either). The final expression can also be written as $|\int da K(x,t;a,t')\psi(a,t')|^2$ $\endgroup$
    – hft
    Commented Apr 13 at 17:49
  • $\begingroup$ Can we put bounds on K? Like can we assume that the probability amplitude of K is zero when |x-a|/|t’-t| >= c ? Can we also make the assumption that K is smooth and continuous over space and time? $\endgroup$ Commented Apr 13 at 17:55

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