Different Hamiltonians because of different choice of magnetic vector potential

Question

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?

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: $$p_i \equiv \frac{\partial L}{\partial \dot{x}_i} = m v_i + \frac{q}{c} A_i.$$