Expanding the electromagnetic hamiltonian

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SUMMARY

The Hamiltonian for a charged particle in a potential field A is defined as \(\hat{H} = \frac{1}{2m} (-i \hbar \nabla - q A)^{2}\). The expansion of this Hamiltonian results in \(\hat{H} = \frac{1}{2m} (-\hbar^{2}\nabla^{2} + q^{2}A^{2} + 2 q i \hbar A \cdot \nabla + q i \hbar (\nabla \cdot A))\). The factor of 2 in the third term arises from the application of the product rule when the Hamiltonian acts on a wavefunction. Properly applying the gradient operator is crucial to understanding this expansion.

PREREQUISITES
  • Understanding of quantum mechanics and Hamiltonians
  • Familiarity with vector calculus and the gradient operator
  • Knowledge of the Landau gauge in electromagnetic theory
  • Experience with wavefunctions in quantum mechanics
NEXT STEPS
  • Study the application of the product rule in quantum mechanics
  • Learn about the Landau gauge and its implications in electromagnetic theory
  • Explore the mathematical properties of the gradient operator in vector calculus
  • Investigate the role of Hamiltonians in quantum mechanics
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Students and researchers in quantum mechanics, physicists working with electromagnetic fields, and anyone seeking to deepen their understanding of Hamiltonians and their expansions.

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This isn't a homework problem - I can't understand a particular statement in my professor's notes. As such, I hope it's in the correct forum.

Homework Statement



The Hamiltonian for a charged particle in a potential field A is

[itex]\hat{H}[/itex] = (1/2m) ( -i [itex]\hbar[/itex] [itex]\nabla[/itex] - q A)[itex]^{2}[/itex]

The square bracket can be expanded.

Homework Equations



In my professor's notes, this expands to [itex]\hat{H}[/itex] = (1/2m) ( -[itex]\hbar[/itex][itex]^{2}[/itex][itex]\nabla[/itex][itex]^{2}[/itex] + q[itex]^{2}[/itex]A[itex]^{2}[/itex] + 2 q i [itex]\hbar[/itex] A[itex]\bullet \nabla[/itex] + q i [itex]\hbar[/itex] ( [itex]\nabla \bullet[/itex] A )

The Attempt at a Solution



When I attempt the expansion myself, I don't get the factor of 2 present in the 3rd term of the expansion. I know that it must be there - subsequent proofs using the Landau gauge don't work without it - but I don't understand where it came from.

Any help in understanding the reasoning behind this would be greatly appreciated.
 
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Always be careful when you have an operator like the gradient operator. It is not meaningful to just expand out the binomial product and write [itex]A\cdot \nabla[/itex], because that must be acting on something. So the way to expand the Hamiltonian is to write it acting on a wavefunction, [itex]\hat{H}f = \frac{1}{2m} (-i \hbar \nabla - q\vec{A})^2 f[/itex]; if you do that, you'll see where the factor of 2 came from (if you correctly apply the product rule).
 

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