Hamiltonian in an electromagnetic field

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SUMMARY

The discussion centers on the Hamiltonian operator in an electromagnetic field, specifically addressing its self-adjoint nature and the mathematical formulation used in quantum mechanics. Participants clarify that the Hamiltonian, represented as H, does not equate to its complex conjugate H*, and that self-adjointness is defined through the integral condition involving wave functions. The use of the operator (+iħ∇ - e/c A)² instead of its negative counterpart is justified mathematically, as it follows from the properties of complex conjugates in quantum mechanics.

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Lucy166 said:
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 \psi is any complex-valued function, and \vec{A} is any real vector-valued function, then [(-i\hbar \vec{\nabla} - e \vec{A})^2 \psi]^* = (+i \hbar \vec{\nabla} - e \vec{A})^2 \psi^*

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