Paramagnetic term of the hamiltonian

[dd331]

The Hamiltonian for particle in an EM field is

H = 1/2m (p - qA)^2 + q phi

If we take the cross-terms, which corresponds to the paramagnetic term, we have
$$H para = -q/2m * (p.A + A.p ) = iqh/2m * (\nabla .A + A.\nabla)$$

What I do not understand is how this simplifies into $$iqh/m * A.\nabla$$?

assuming that $$\nabla .A = 0$$ (i.e. Coulomb gauge). Why does the factor of 1/2 disappears? I'm only a first year undergraduate and I'm learning this on my own. I will appreciate it if you give a fuller answer. Thank you.

 

[Meir Achuz]

In QM, H is assumed to act on a wave function \psi.
This means that del.A really means del.(A \psi)=(del.A)\psi + (A.del)\psi, so the (A.del) comes in twice.

 

[Entropia]

The paramagnetic term in the Hamiltonian represents the interaction between the particle's momentum and the electromagnetic (EM) field. In this case, the EM field is described by the vector potential, A, and the scalar potential, phi. The cross-terms in the Hamiltonian, H para, involve the dot product between the momentum, p, and the vector potential, A. This can be rewritten as p.A = A.p, as dot products are commutative.

Now, in order to simplify this further, we need to consider the Coulomb gauge, where the divergence of the vector potential is equal to zero, \nabla .A = 0. This means that the vector potential only has transverse components and does not have any longitudinal components. In this case, the dot product of A with the gradient operator, \nabla, will also be equal to zero, A.\nabla = 0.

Therefore, in the Coulomb gauge, the paramagnetic term simplifies to H para = iqh/2m * (\nabla .A + A.\nabla) = iqh/2m * (0 + 0) = 0. This is why the factor of 1/2 disappears, as it is multiplied by zero.

In summary, the factor of 1/2 disappears in the Coulomb gauge because the dot product between the momentum and the vector potential is equal to zero. This simplification only holds in the Coulomb gauge, as in other gauges the dot product may not be equal to zero and the factor of 1/2 would not disappear. I hope this explanation helps you understand the concept better. Keep up the good work in your studies!