How do electrons choose their path?

                    Wire A
-----R-----
/           \
------I--->-[p]--             ---------I-->
\           /
-----R-----
Wire B


Wire A and B both have resistance $R$.
So current $\frac{I}{2}$ flows through each of the wires.

How come electrons know which wire path to choose?

What I am thinking is if we consider electrons at point P are in state $\frac{|A\rangle + |B\rangle}{\sqrt{2}}$ and at the junction some kind of measurement occurs which leaves half of the electrons in state $|A\rangle$ and other half in $|B\rangle$.

Please explain how exactly this works. Please point out why argument above is right or wrong (or at all relevant) in this situation.

EDIT: How can one derive basic laws for series and parallel ckts from more basic principles?

Thanks! :-)

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The formal theory of conductivity is probably of interest here: I'll just link to wikipedia's introductory article en.wikipedia.org/wiki/Classical_and_quantum_conductivity –  j.c. Nov 18 '10 at 18:38
This question is based on incorrect assumptions. Superposition only occurs mainly for free electrons (like the one in double-slit experiment). But in metals electrons are hitting the lattice all the time therefore essentially being in thermal equilibrium with the lattice. Considering that the wire is probably at room temperature, you can't really observe genuine quantum effects: they have all being smoothed out and it's all classical physics. –  Marek Nov 18 '10 at 19:12
Also, electricity is not really carried by free travelling electrons (just like water is not really carried by waves) en.wikipedia.org/wiki/Electric_current#Drift_speed –  Sklivvz Nov 18 '10 at 20:14
Can somebody show me some equations about this? Actual math behind this thing (wave nature, statistics etc whatever) but I want equations that are more basic than Ohm's laws. See edits. –  Pratik Deoghare Nov 18 '10 at 20:39

To speak of quantum state, the electrons should be coherent, have a non-negliglible de Broglie wavelength. On the other hand, a non-zero resistance imply a nonnegligible dissipation and is sure to break any coherence. Your pure quantum state is then quickly turned into a classical probabilistic mixture, and your electron behaves exactly like a water drop in a river which separates itself into to branch in front of an island: its trajectory depends on its position inside the cable, relative to impurity.

Edited :

To be more quantitative, the quantum effect can roughly be seen on a length of the scale Λ, where Λ is the thermal de Broglie wavelength, given by : $$\Lambda = \frac{h}{\sqrt{2\pi mkT}} \simeq 10^{-11}\mathrm{ m}$$ where the numerical application is for an electron at room temperature. The quantum effects are therefore already negligible at the nanometric scale.

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Think of it as a queue of people trying to enter a building with two entrances, both the same size and length. As people push each other, and as the people on the front move, queue proceeds, and statistically on equal amount in both entrances provided they are identical.

Same principle applies for different R's, R and 2R, where one entrance is smaller than the other one.

As for your argument. It is correct mathematically, in my opinion.

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I am getting the impression you do not follow the answers of the questions you asked =). –  Cem Nov 18 '10 at 19:38
Yalcin Wrong impression. Doing physics at workplace is bit problematic! (I don't work at swiss patent office ;) So you mean its statistics without quantum mechanics but QM could work right? –  Pratik Deoghare Nov 18 '10 at 19:41
Well by definition, QM should work everywhere except for cosmological scales where gravity dominates, however QM calculations for large objects are impossible to carry out. (sorry if my last comment offended you it didn't carry such a purpose) –  Cem Nov 18 '10 at 19:50
Wrong impression again! It didn't offend me at all. I took it very cheerfully. :-) –  Pratik Deoghare Nov 18 '10 at 20:36
Haha, I must say I'm pleased with being wrong this time. –  Cem Nov 18 '10 at 21:27
Quantum effects, as others have pointed out are negligible in most wires. If you want to derive from a particle like description of the wires, you'd be best off thinking in terms of gas/fluid dynamics (which is believed to in the continuous limit turn into the transport diffusion equations). You can simulate this on a computer by having an many electrons move ballistically through the material, and randomly scatter with some probability $p$ each time step. This model will give you Ohm's law.
Now granted, if you actually wanted to figure out what the probability $p$ is, and the effective resistance of the wire, things get really complicated and do indeed require quantum mechanics. If you want the math, you'd want to look into the theory of quantum mechanical scattering. The main idea is that an electron will be in a pure momentum p eigenstate, and get randomly placed into a new p' momentum eigenstate. Averaged over a bunch semi-classical particles, the I-V characteristics work out. The justification for this is in truth, somewhat sketchy (the above described simulation has not been formally proved to be the same as using the Boltzmann Transport and Diffusion Equations, we just think it does for appropriate limits).