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I Quantum-Mechanical treatment of circuits

  1. Mar 26, 2017 #1
    Hi everyone. I have to say I never get satisfied with introductory explanations of how electric phenomena occurs. Specifically, I would like to know how Quantum Mechanics explain the raise of a electric current and especially what makes a resistor being (maybe) the most dissipative element in a circuit. I guess we would need to know about how the atoms are interacting with each other as with the free (less bounded in QM language) electrons as well in a specific material to answer these questions. But that would then make it very hard to explain these things using Quantum Mechanics.
     
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  3. Mar 26, 2017 #2

    Henryk

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    Gold Member

    Davidge,

    There is no short answer to your question. What you want to know is essentially a large chunk of Solid State Physics.
    First thing is a general solution to Schrodinger equation given periodic potential due to regular arrangement of atoms in a solid. The solution is in form of energy bands, that have certain width (kinetic energy) and are separated by energy gaps.
    Next thing, you fill the bands with available (valence) electrons starting from the lowest energy state and going up. If you fill some bands completely and all other bands are empty, you have an insulator. If you ran out of available electrons in a middle of a band, you have a metal. A semiconductor is an insulator with a small energy gap (for example, silicon has an energy gap of 1.11 eV, gallium arsenide gap is 1.45 eV, 'true' insulators would have an energy gap of 3 eV or more).
    Next thing is the detailed band structure, that is relationship between the energy and (quasi) momentum of an electron. In vacuum, the energy is given by ##E = \frac {p^2}{2m} ## where m is the electron mass.
    In a solid, the relationship is not necessary quadratic and the 'effective mass' is, generally, different from the free electron mass and can even be negative (that give you 'holes' in semiconductors).
    Next thing is electron scattering. The band structure is obtained for a perfect periodic arrangements of atoms and it correspond to electron stationary states. In reality atoms are not arranged perfectly; there are defects in crystals plus thermal movement of atoms (here is another chunk of solid state physics: phonons).
    These imperfection lead to scattering, hence resistance.
    Once you have all of this figure out, you can consider electrical properties. Electrical properties of a given solid depends on three factors
    • number of free carriers (electrons and holes)
    • their effective mass (how easily they can be accelerated by an electric field)
    • scattering (rate of momentum loss due to collision with lattice vibrations and impurities)
    Some materials, like gold, silver, copper will have a large number of electrons and low scattering rate. Other metals like lead, tungsten, etc. will have much higher scattering rate, hence, lower resistivity. Semiconductors have usually carrier concentrations billions times smaller than metals and, therefore, much higher resistivity. Everything depends on type of atoms and their arrangement.
    Take carbon, for example. If it forms a cubic structure (diamond), the energy gap is about 5 eV and it is a very good insulator. Another allotrope of carbon is graphite that has layered structure. In this case, there is a slight overlap between the bands which give it some conductivity but not as much as that of metal.
    Carbon can also exist as amorphous with conductivity even lower than that of graphite.
    I hope I gave you an idea but you really have to get a good solid state physics book.
     
  4. Mar 26, 2017 #3
    Thanks Henryk. I appreciate your detailed reply.
    What a good book would you suggest to me?
     
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