I need your opinions. Why is the vector potential of a magnetic field important (or even necessary) to quantum mechanics? Why it has to be defined everywhere? Is there any fundamental reason you can think of?

My point of view:

  1. If one wants to have gauge invariance then he needs to have a vector potential. In fact if we construct the QED Lagrangian as being invariant under U(1), then the vector potential is naturally introduced by the covariant derivative. Additionally, the gauge field is a dynamical variable which needs to be defined everywhere.

  2. The Bohm-Aharonov effect shows that the vector potential (but not the gauge choice) is detectable and affect the probability distribution. Hence the vector potential is in some sense physical and need to be defined everywhere.

  • $\begingroup$ I can't comment on the mathematical requirements. I would certainly agree with your second point. The interesting question in my mind is, wether there is a classical effect that allows to detect the vector potential in the same way as the Bohm-Aharonov effect does? I am not aware of that at this moment. If there is none, then 2 is a unique requirement for the quantum case. If there is a classical effect, however, then the necessity for the vector potential just propagates into the quantum realm and the structural argument seems the stronger one. $\endgroup$
    – CuriousOne
    Dec 13, 2014 at 16:30

1 Answer 1


Even in Classical Mechanics the vector potential is defined in the whole space, i.e., it is a field, because it should be known wherever the charge in question goes. The charge may go everywhere we direct it with our initial conditions. In QM, where there is a wave function equation for a wave determined everywhere in space, the vector potential should also be defined within this space. A classical analogue to the Schroedinger equation is the Hamilton-Jacobi equation that determines a field S.

  • $\begingroup$ The Schrodinger equation is analog to the heat equation, not to the Hamilton-Jacobi equation. Look better at the latter. The derivative by time is taken on the action, not on the wave as in the Schrodinger equation. A good proof of the Schrodinger equation in analogy with the heat equation can be found in Gordon Baym "Lecture notes on quantum mechanics". $\endgroup$
    – Sofia
    Dec 13, 2014 at 21:32

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