Expanding the electromagnetic hamiltonian

In summary, the conversation is about the expansion of the Hamiltonian for a charged particle in a potential field A and the confusion surrounding the factor of 2 in the third term of the expansion. The expert advises to be careful when expanding the binomial product and to remember that it must be acting on something. The factor of 2 can be understood by correctly applying the product rule when writing the Hamiltonian acting on a wavefunction.
  • #1
xahdoom
3
0
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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  • #2
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).
 

1. What is the electromagnetic Hamiltonian?

The electromagnetic Hamiltonian is a mathematical representation of the total energy of a system that includes both electric and magnetic fields. It is used in quantum mechanics to describe the behavior of particles in the presence of electromagnetic fields.

2. How is the electromagnetic Hamiltonian expanded?

The electromagnetic Hamiltonian is expanded by considering the contributions of different types of interactions, such as the electric and magnetic dipole moments of particles, as well as their spin and orbital angular momentum.

3. Why is expanding the electromagnetic Hamiltonian important?

Expanding the electromagnetic Hamiltonian allows for a more accurate description of the behavior of particles in electromagnetic fields, which is crucial in many areas of physics, including quantum mechanics and particle physics.

4. What are some applications of the expanded electromagnetic Hamiltonian?

The expanded electromagnetic Hamiltonian has many applications, including in the study of atomic and molecular systems, as well as in the development of new technologies such as quantum computing and magnetic resonance imaging (MRI).

5. Are there any limitations to expanding the electromagnetic Hamiltonian?

While expanding the electromagnetic Hamiltonian can provide a more complete description of particle behavior, it can also lead to more complex calculations and may not always accurately predict experimental results. Additionally, it may not be applicable in extreme conditions such as in the presence of strong gravitational fields.

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