Quarlep said:
Lets suppose I am in London I will go to Berlin but Before go there I must be somewhere between this two points.Lets call it Paris so The probability to go Berlin is London-Paris probability ( wave function squared) multiply by Paris Berlin probability. Is that true ?
Yes. That's right. The assumption is that you make a random choice of what city to go to from London, picking Paris with probability A and then you make an independent random choice of where to go from Paris, picking Berlin with probability B. Then the probability that you will go from London to Berlin via Paris is just A \cdot B
Let's make it a tiny bit more complicated. Suppose that you're flying out of London, and the way that you fly is to go to the airport, and pick a random destination. When you get to that airport, you pick another random destination. Etc. Now, if that's the way you're choosing your trip, then you can ask the probability that you will get to Berlin on your second flight.
So here's how you figure that out: You start in London. You pick your first destination at random. So you have a list of cities that are reachable from London on nonstop flight. To simplify, let's suppose that from London (L) you can get to Paris (P) or Amsterdam (A). From Paris, you can get to either London or Berlin (B). From Amsterdam, you can get to London or Berlin. Let Pr_1(X,Y) be the probability that you will go from X to Y in one hop.
Then there are two ways to get from London to Berlin in two hops:
- London to Paris to Berlin.
- London to Amsterdam to Berlin.
The probability of going the first route is Pr_1(L, P) Pr_1(P, B). The probability of going by the second route is Pr_1(L, A) Pr_1(A, B). To compute the probability of getting to Berlin on the second hop, you just add those probabilities together:
Pr_2(L, B) = Pr_1(L, P) Pr_1(P, B) + Pr_1(L, A) Pr_1(A, B)
If we generalize to starting at any city X and getting to any city Y, we conclude similarly:
Pr_2(X, Y) = \sum_Z Pr_1(X, Z) Pr_1(Z, Y)
Similarly, for N+1 hops, instead of just two:
Pr_{N+1}(X, Y) = \sum_{Z_1} \sum_{Z_2} ... \sum_{Z_N} Pr_1(X, Z_1) Pr_1(Z_1, Z_2) ... Pr_1(Z_N, Y)
So a trip itenerary is just a list of cities in the order that they are visited: X, Z_1, Z_2, ..., Z_N, Y. The probability of randomly choosing that itenerary is Pr_1(X, Z_1) Pr_1(Z_1, Z_2) ... Pr_1(Z_N, Y). The total probability of getting from X to Y in N+1 steps is Pr_{N+1}(X,Y) given above.
The idea behind a path integral is the same, except that instead of the fundamental quantity being probabilities, they are instead probability amplitudes. You have to square the total probability amplitude to get the probability.