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Charge In Schroedinger's Equation

  1. Feb 28, 2008 #1
    This is a simple question which I'm sure has a simple explanation. While mass of the particle is explicitly included in the Schroedinger Equation, the charge is not. Why isn't it? Is it included along with the potential energy or....?
  2. jcsd
  3. Feb 28, 2008 #2
  4. Feb 28, 2008 #3
    Thank you.
  5. Feb 29, 2008 #4


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    Also, you can write down a SE for systems that do NOT involve mass. A good example would be tunneling from one state to another in a superconducting component called Josephson junctions. If you write down the SE for such a junction you will find that what would be "mass" in the equation for a e.g. a travelling electron is now actually the effective capacitance [itex]m=(\hbar/(2e) C[/itex] of the junction (or, equivalently, m is the effective mass of a "phase particle" in phase space).

    Hence, Schroedinger equations that involve a "real mass" are just special cases of a more general formalism.
  6. Feb 29, 2008 #5
    I have a question...what would this "more general formalism" look like, mathematically ?
  7. Feb 29, 2008 #6


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    Last edited: Feb 29, 2008
  8. Feb 29, 2008 #7


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    Maybe I was a bit unclear. I only meant that it is possible to write down a SE for systems where there is no real "mass" (i.e. nothing that has the unit of kg unless it is multipled by some constant), in this case the SE is for a "phase particle" with an effective "mass" given by the capacitance of the junction which is moving in phase space with the "coordinates" of the particle given by the electric phase across the junction.

    Of course nothing is really moving in real space, the SE is merely describing the state of the junction (if the current through the junction is constant problem is just a simple "particle in a well" type of problem).

    Hence, my point was simply that the SE is in no way limited to describing e.g. electrons and other "real" particles with properties like mass, charge etc as one could be lead to believe by the fact that there is a "m" in the equation. As the OP was asking about including charge in the SE I assumed that he/she might not know this.
  9. Feb 29, 2008 #8
    Unfortunately with all those nasty Quantum equations charge is hard to work around, because if you think of an electron, it is surrounded by virtual photons mediating the electromagnetic force around it that are continutally undergoing pair prodution causing positrons to come nearer the electron. Theoretically there are infinite positrons there, making the charge of an electron infinite.

    As physicists we have no problem making infinity - infinity = a finite sum. Those weird mathematician types tend to snub their noses at us though.
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