Graduate Hamiltonian in an electromagnetic field

Physics news on Phys.org
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
 
  • Like
Likes vanhees71 and Mentz114

Similar threads

  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 14 ·
Replies
14
Views
6K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K