It is generally said that Ohm's law isn't a real law since it is only a fine model found through experiments to explain the nature of circuits. And resistance is defined as the ratio of the voltage difference between two poles and the current flow through the circuit. Often resistance is explained as electrons having collision with the atoms of the conductor and is some kind of drag (?) through which electrons loose energy. This explanation doesn't seem to be exact case.

What is actually going on if we zoom into the quantum level? How are the concepts of resistance and voltage drop interpreted then?


To me, the most interesting characteristics of electrical resistance is the orbital positions, which feeds common quantum calculations. That is the most electrical conductivity, thereby the lowest electrical resistance, comes in the middle of the periodic chart. It is the column, 29-Cu Copper, 47-Ag Silver, and 79-Au Gold.

The challenge for thinking of these using the quantum model is that the electrons are in the toward the max of the current d-shell. These are all d-shell (Xd9) which you get the quantum number from:

What is the right quantum numbers of Copper?

That reference is important because it states that the last electron quantum number for Copper is the same as the last electron quantum number of Zinc, which has very different - low - electrical conductivity -- the opposite of Copper. As such, one would not expect the expression with the same last electron to yield such opposite if the quantum number is the same. One would expect that the last electron is the critical one for electrical conductivity - being the most removed (lowest energy to release).

I believe the answer lies in other research about 'transition metals' which exhibit a number of contradictory. Specifically, at:


It states that d-Shell transition provides a) magnetic, b) when in solution color, and c) oxidation, but not electrical resistance.

So, to me the linkage of electrostatic conductivity to current quantum theory remains inadequate. Quantum theory is good at only certain properties.


When I started learning QM in the first lecture we tried a "semi classical" way to imagine these things. Imagine a metallic cube. Let's assume that some current $I$ is flowing through it. The current through an $A$ cross section will be : $$I=qnvA.$$ where $q$ is the unit charge, $v$ is the average speed of the electrons and $n$ is the volumetric concentrations. We can assume that the speed is proportional to the potential energy $E$ $$v=\mu E$$ Where $\mu$ represents the ability of the electrons to move. Then $$I=qn\mu EA.$$ And the current density $j$ is $$j=qn\mu E.$$ Hence the conductivity $\gamma$ is $\gamma=\frac{j}{E}$ then it is $$\gamma=qn\mu$$ And the resistance is $$\rho=\frac{1}{\gamma}=\frac{1}{qn\mu}$$ And where quantum mechanics comes in? It is neccessary when we want knowledge about $\mu$.At first assume that we are working with a metal which has a perfect crystalline structure, then the electrons won't scatter much on the atoms so a perfect crystal would not have any resistance. The resistance is from the defects of the crystal where electrons scatter. So the resistance is due to scattering, not collision, is an electron hits a nucleus it is called electron capture which is another interesting topic. Since $\mu$ is not a number I should better call it $\mu(x,y,z)$ and $n$ is really $n(x,y,z)$ and they are both dependent on the shape of the metal piece and the position too. so it won't be a simple equation to use(And we did not even calculated with the electrons effects on each other!). The voltage drop sould be easy to calculate for you then.I hope my answer was good enough :)

  • $\begingroup$ Thanks for your answer. But what did you mean exactly by electron scattering? And how does electron scattering arise resistance? It will be helpful if you elaborate a little bit. $\endgroup$ – user102426 Jan 24 '18 at 6:26
  • $\begingroup$ When an electron gets close to an atom it's path can be changed by the electromagnetic force of the nucleus or electromagnetic force of it's electrons. When this happens, the projection of the velocity in the original direction will be smaller and if you substitute a smaller velocity to my first equation you will get a smaller current, which means, voltage drop and resistance are occuring. $\endgroup$ – L.Gyula Jan 24 '18 at 12:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.