Hamiltonian in an electromagnetic field

[Lucy166]

I have a question connected with the problem:
https://www.physicsforums.com/threads/continuity-equation-in-an-electromagnetic-field.673312/

Why don’t we assume H=H*? Isn’t hamiltonian in magnetic field a self-adjoint operator? Why? Why do we use (+iħ∇-e/c A)² instead of (-iħ∇-e/c A)² two times?

 

[stevendaryl]

I have a question connected with the problem:
https://www.physicsforums.com/threads/continuity-equation-in-an-electromagnetic-field.673312/

Why don’t we assume H=H*? Isn’t hamiltonian in magnetic field a self-adjoint operator? Why? Why do we use (+iħ∇-e/c A)² instead of (-iħ∇-e/c A)² two times?

The second equation just mathematically follows from the first equation by taking complex conjugates. If [itex]\psi[/itex] is any complex-valued function, and [itex]\vec{A}[/itex] is any real vector-valued function, then [itex][(-i\hbar \vec{\nabla} - e \vec{A})^2 \psi]^* = (+i \hbar \vec{\nabla} - e \vec{A})^2 \psi^*[/itex]

Self-adjoint doesn't mean that [itex]H = H^*[/itex]. It means that [itex]\int \phi^*(\vec{x}) (H \psi(\vec{x})) d^3x = \int (H \phi(\vec{x}))^* \psi(\vec{x}) d^3x [/itex]