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Different gauge in QM

  1. Aug 25, 2014 #1
    I have a problem concerning gauge invariance in QM.

    QM should be invariant of electromagnetic gauge. However, the following two physically equivalent vector potential:
    1. [itex]A = (-\frac{1}{2}By, \frac{1}{2}Bx, 0 )[/itex]
    2. [itex]A = (-By, 0 , 0 )[/itex]
    generates the following hamitonian:
    1. [itex]H = 1/{2M}[(P_x-\frac{eB}{2c}y)^2+(P_y+\frac{eB}{2c}x)^2 + {P_z}^2][/itex]
    2. [itex]H = 1/{2M}[(P_x-\frac{eB}{2c}y)^2+{P_y}^2 + {P_z}^2][/itex]

    For the first hamitonian, [itex]P_y[/itex] is a conserved quantity, but the second hamitonian yields the opposite result.

    So, is [itex]P_y[/itex] conserved or not?
  2. jcsd
  3. Aug 25, 2014 #2


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    The point is that in the presence of a magnetic vector potential, [itex] \hat{\vec{P}} [/itex] is not gauge invariant and so doesn't represent a physical quantity. Its place is taken by [itex] \hat{\vec{P}}-q\hat{\vec{A}} [/itex]
  4. Aug 25, 2014 #3


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    One must not forget that in quantum theory (as in Hamiltonian classical mechanics) [itex]p[/itex] denotes the canonical momentum rather than the mechanical momentum. The canonical momentum has not necessarily a direct physical meaning. As Shyan already said, you can see this in your case on the fact that it is a gauge-dependent quantity.

    Whatever you do, physical quantities are gauge independent. E.g., if you want to evaluate the Landau levels, i.e., the energy eigenvalues for a particle moving in a constant magnetic field, you'll always get the same, no matter which gauge you choose.

    In this case you should be able to do that in both gauges, because the part of the Hamiltonian in the transverse plane (transverse wrt. the magnetic field) map to the Hamiltonian of a harmonic oscillator. I've to check myself, whether I can solve the eigenvalue problem for the 1st gauge. In any case it's wise to use the 2nd gauge, because here you have a complete set of energy eigenstates in terms of the common eigenstates of [itex]H[/itex], [itex]P_x[/itex], [itex]P_z[/itex].

    Also, it's easy to show that the wave function changes under a gauge transformation only by a (space-time dependent) phase factor. So you get always the same energy eigenfunctions up to such a phase factor, but this has no physical meaning either!

    This is a very funny subject, including nice mind boggling quantum phenomena as the Aharonov Bohm effect.

    BTW: in the 2nd Hamiltonian [itex]P_x[/itex] is conserved, because (in the Heisenberg picture)
    [tex]\dot{P}_x=\frac{1}{\mathrm{i}} [H,P_{x}]=0,[/tex]
    but [itex]P_y[/itex] is not, because [itex]H[/itex] depends on [itex]y[/itex] and thus doesn't commute with [itex]P_y[/itex].
  5. Aug 25, 2014 #4
    Thanks all the above people. Now I am clear about this subject.
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