In general, in quantum mechanics we can prove probability current or the Schrodinger equation and other quantities are gauge invariant. However, the Hamiltonian isn't gauge invariant. Under a gauge transformation, the Hamiltonian operator changes(or have i understood wrong?) Does this mean that the Hamiltonian doesn't describes a true physical quantity like in classical mechanics?Closing, if the above are correct, do they have any affect on the principle of least action?
Thank you.
Note: The Hamiltonian is: $$H_f = {1 \over 2m} [P- qA(R,t)]^2 +qU(R,t) $$After a gauge transformation: $$H_g = {1 \over 2m} [P- qA'(R,t)]^2 +qU'(R,t) $$. Thus, we have $$H_f \neq H_g $$
Best Answer
Even in classical mechanics, Hamiltonian for one particle in external field EM is function
$$ H(\mathbf r,\mathbf p) = \frac{(\mathbf p - \frac{q}{c}\mathbf A(\mathbf r, t))^2}{2m} + q\phi(\mathbf r,t) $$
where $\mathbf A,\phi$ are any of the valid functions that describe the same external EM field.
The form of the Hamiltonian (dependence on the potentials and $\mathbf r,\mathbf p) is unique, but its value is not; it depends on the choice of the above two functions ("choice of gauge").
This non-uniqueness is no big deal, as Hamiltonian function is mostly a theoretical concept that is useful to formulate the laws and derive other laws; it non-uniqueness is not necessary for that use.
In classical physics, the laws and their consequences can be formulated also in a gauge-independent way with EM fields $\mathbf E,\mathbf B$ only. The potentials and the Hamiltonian can be avoided.
The situation with gauge-dependence is similar to the one with kinetic energy having value that depends on the inertial system it is evaluated for. The value is frame-dependent, but it poses no problem for its use.