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A Hamiltonian in an electromagnetic field

  1. Dec 3, 2017 #1
  2. jcsd
  3. Dec 3, 2017 #2

    stevendaryl

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    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]
     
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