I was reading about how reflection and refraction work at a quantum scale. What I understood was that the photon "can" take multiple paths to reach the end point and when we consider all these possibilities, most of them cancel out as they interfere destructively, while there's constructive interference around a certain direction. That is what we observe is the the sum of all the probabilities of the possibile paths the photon could take.

Please correct me if I'm wrong.

My question is how do we calculate that probability of each path? Or do we only calculate the sum of the probabilities directly?

  • $\begingroup$ Destructive interference is a as much a mathematical concept as a physics one. In a lake full of waves you can see many "nulls" where waves have cancelled but in fact all the waves will reemerge .... there is only true destruction when they hit the beach. In all mediums the energy is stored (water is elastic) and never lost .... until absorption. Same for the EM medium. $\endgroup$ Aug 11, 2023 at 21:28
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    $\begingroup$ See this classic link to Feynman's online book QED: The strange Theory of Light and Matter. It's intended audience is for those without extensive mathematical background: "It takes seven years -- four undergraduate and three graduate -- to train our physics students to do that in a tricky, efficient way. That's where we are going to skip seven years of education in physics: By explaining quantum electrodynamics to you in terms of what we are really doing, I hope you will be able to understand it better than some of the students!" $\endgroup$ Aug 14, 2023 at 12:55

2 Answers 2


The probability to go from point $x_a$ at time $t_a$ to point $x_b$ at time $t_b$ is $$ P(b,a)=|K(b,a)|^2 $$ where $K$ is a propagator.

To get closer to a specific path, pick an intermediate point $c$. The probability to go from $a$ to $c$ to $b$ is $$ P(b,c,a)=|K(b,c)K(c,a)|^2 $$

So it is possible in principle to get closer and closer to a specific path by choosing intermediate points and multiplying the propagators. However, the probability $P(b,\ldots,a)$ almost certainly gets smaller and smaller because $|K|\le1$ for all $K$.


I think there is a vivid demonstration from the viewpoint of Feynman's path integral. The path integral scheme states that if a particle is at the initial position $x_0$ when $t=t_0$, then the probability that the particle is observed in $x_1$ when $t=t_1$ is given by path integral $$P(x_0,t_0;x_1,t_1)=\int\mathcal{D}x(t)e^{iS[x(t)]}\,.$$ where $S$ is the action of theory you consider.

On the other hand, this formula has a physical interpretation which could be seen in the "proof" of it e.g. this lecture, that this transition probability is a sum over all probabilities w.r.t possible paths $x(t)$ this particle could travel $$P(x_0,t_0;x_1,t_1)=\sum_{\mathrm{possible}\ x(t)}P[x(t);x(t_0)=x_0,x(t_1)=x_1]\, .$$ And thus $$\sum_{|\overline{x}(t)-x(t)|<\epsilon}P[\overline{x}(t);\overline{x}(t_0)=x_0,\overline{x}(t_1)=x_1]=\mathcal{D}x(t)e^{iS[x(t)]}\rightarrow 0\, .$$ It is just like the similar calculation in a normal integral $$\int_{x_0}^{x_0+\epsilon}\mathrm{d}x f(x)=\epsilon f(x_0)\rightarrow 0$$

Now there is finally an answer to all your questions

  1. We can calculate this probability by propagator or path integral method. And we find for a given path, this quantity must run to 0;
  2. According to the previous arguments, yes.

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