I'm talking Quantum Mechanics here. Let's say that we have a constant magnetic field $\mathbf{B} = (0,0,B)$. Then I can pick two different vector potentials:

$\mathbf{A} = (\ -\tfrac{1}{2}By ,\ \tfrac{1}{2}Bx ,\ 0 \ )$

$\widetilde{\mathbf{A}} = ( \ 0 , \ Bx , \ 0 \ )$

Which both yield $\nabla \times \mathbf{A} = \nabla \times \widetilde{\mathbf{A}} = \mathbf{B}$. Then the way I understand it, I have two Hamiltonians:

$H = \tfrac{1}{2m} ( \mathbf{p} - \frac{e}{c} \mathbf{A} )^{2}$

$\widetilde{H} = \tfrac{1}{2m} ( \mathbf{p} - \frac{e}{c} \widetilde{\mathbf{A}} )^{2}$

These Hamiltonians are $different$, and yet they describe the same physical system!

How does this work? Why is this okay?

My thoughts are that it doesn't matter, since both $H$ and $\widetilde{H}$ will yield the same measurable quantities (energy eigenvalues). Is this correct?

  • $\begingroup$ Related to 263053. Did you work out the gauge transformation connecting the two As? $\endgroup$ Oct 19, 2016 at 23:06

1 Answer 1


Notice that the definition of the canonical momentum differs between the two Hamiltonians, so you should have a $\mathbf{p}$ and $\tilde{\mathbf{p}}$. You can see this when you move from the Hamiltonian to the canonical equations of motion, and apply the requirement that coordinate and its derivative, $\mathbf{x}$ and $\dot{\mathbf{x}}$, be gauge invariant.

You can get this same result by starting from the Lagrangian (linked for this formula): $$L = \frac{1}{2} m \mathbf{v}^2 - q\phi + \frac{q}{c} \mathbf{v} \cdot \mathbf{A},$$ which gives a gauge invariant action if you remember the identity $\frac{\operatorname{d} f}{\operatorname{d}t} = (\mathbf{v} \cdot \nabla) f + \frac{\partial f}{\partial t}$, and that surface terms (ie total time derivatives) are dropped because of the constraints. From that Lagrangian you get that the canonically conjugate momentum is: $$p_i \equiv \frac{\partial L}{\partial \dot{x}_i} = m v_i + \frac{q}{c} A_i.$$

  • $\begingroup$ Okay, but I thought the canonical momentum $\mathbf{p}$ is always the same, while the kinematical momentum $\Pi = m \frac{d\mathbf{x}}{dt} = \mathbf{p} - \frac{e}{c}\mathbf{A}$ would be different based on the vector potential used? I know that the gauge function connecting the two potentials I chose is $\Lambda = \tfrac{1}{2}B x y$, but I'm not sure what to do with this $\endgroup$ Oct 19, 2016 at 23:36
  • $\begingroup$ Nope, the physical momentum is always the same and the canonical momentum is gauge dependent. $\endgroup$ Oct 19, 2016 at 23:37
  • $\begingroup$ OH right! I just saw the edit to your post - the canonical momentum is defined by $\frac{\partial L}{\partial \dot{x}_{j}}$ $\endgroup$ Oct 19, 2016 at 23:39
  • $\begingroup$ My book (Sakurai), is defining $\mathbf{p}$ as the generator of translations...I'm not sure if this affects the above $\endgroup$ Oct 19, 2016 at 23:42
  • $\begingroup$ Not at all. The canonically conjugate momentum, $\mathbf{p}$, is the generator of translations in space. The link I linked to goes over this exact problem, FWIW. $\endgroup$ Oct 19, 2016 at 23:46

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