(Disclaimer: Apologies for the long response.)
First, I agree with Upisoft. The premise is flawed because the theory of conduction in a solid that the circuit model is founded on, assumes the existence of a large number of 'free' electrons.
But you do raise an interesting question...
Deo said:
...the reason why the electron knows which path is the right one lies in quantum physics, meaning, it tries all paths at the same time and then chooses the easiest or right one... right?
I believe you are referring to the principle of least action, where the path actually chosen is one which satisfies the 'equations of motion' under the boundary conditions. From observation, we expect (classically) that the path the electron actually takes is the one from the battery to the short circuit and back. This can be confirmed by attaching an ammeter in series with one of the resistors.
Let's say that you do this experiment with some kind of a single electron source which you can 'fire at well' to get a single electron out. Let's also assume that the conducting wires are ideal.
The question is: would you make a statement about 'what path the electron takes' with just one experiment? The answer is no. Quantum mechanics will give you probability amplitudes for the electron to take various paths to return to the battery (including, believe it or not, one which doesn't even involve the conductor), but the computation of these amplitudes is founded on the assumption that you have a large number of identical copies of this system on which you can reproduce the experiment an arbitrarily large number of times.
The probabilities are to be interpreted as: "if I do the experiment N times -- one on each identical system, or on the same system after starting afresh -- (N is large), then the probability that the electron will cross the short is...p(N)" and of course you'd expect p(N) to be closer to 1 as N tends to infinity. Actually in this case, the relative smallness of quantum effects will make the difference between p(N) and 1 undetectable.
Note that the actual calculation for a circuit of this kind (or any kind!) will be complicated...I do not know how one would even incorporate the wires and their arrangement into a mathematical boundary condition. But this is where observation comes in. Quantum Mechanics will NOT tell you what will precisely happen. But it will give you a good enough idea. There will be temperatures, voltages, currents and materials (and combinations thereof) where the smallness of the Planck constant will not dwarf quantum effects.
Quantum mechanics
is at work and it
is why this circuit works the way it does. However, our interpretation that the electron takes all possible paths to reach its destination is one that theory tells us, and any act of measurement or observation will force it to 'collapse' to the classical path -- one that the experimenter will perceive when he measures the currents through the resistors to be zero (from which he would infer that the current passed through the short). This does not mean that the electron will pass through the resistors if you don't look (and not through the short at all)! We cannot make precise statements only probabilistic ones at best in quantum theory.
However, when we observe, measure or interact with the system, it's time evolution changes. In this case, the quantum effects are so small though, that the change in time evolution is imperceptible. How the electron 'knows' the right path isn't something we can really talk about. It just does
.
Have a look at Feynman's book "QED: The Strange New Theory of Light and Matter". Your friend will like it too. Its a wonderful exposition of the principle of least action, and also quantum physics.
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Attempt at a slightly more mathematical explanation...
I am taking the liberty of throwing in some mathematical arguments. If you are unfamiliar with some basic physics (Lagrangians, actions), then you can read it...otherwise, just ignore what follows and return to it later.
Suppose you were to write down the action (integral of the Lagrangian of quantum electrodynamics over time from t = 0 to t = T, the time it takes for the electron to make a round trip from the battery, through the circuit, and back home). Let's call it S. For any particular path, you would have a particular S, which you would compute by writing down the precise Lagrangian. The principle of least action says that the probability amplitude associated with this path would be proportional to exp(iS/\hbar). But if \hbar is very small, then this function will be highly oscillatory, and in the limit of \hbar going to 0, the nonzero contribution will only come from the path which extremizes the action (S), let's call it P_{cl}.
Now in the classical world, since \hbar is dwarfed out (as argued above), we will only get a nonzero contribution from 'path' P_{cl}. It can be argued and shown that the path will be the one which satisfies Maxwell's equations (ordinary electromagnetism, without observable quantum effects).
