### D.72 Electromagnetic commutators

The purpose of this note is to identify the two commutators of chapter 13.1; the one that produces the velocity (or rather, the rate of change in expectation position), and the one that produces the force (or rather the rate of change in expectation linear momentum). All basic properties of commutators used in the derivations below are described in chapter 4.5.4.

The Hamiltonian is

when the dot product is written out in index notation.

The rate of change in the expectation value of a position vector component is according to chapter 7.2 given by

so you need the commutator

Now the term can be dropped, since functions of position commute with each other. On the remainder, use the fact that each of the two factors comes out at its own side of the commutator, to give

and then again, since the vector potential is just a function of position too, the can be dropped from the commutators. What is left is zero unless is the same as , since different components of position and momentum commute, and when , it is minus the canonical commutator, (minus since the order of and is inverted), and the canonical commutator has value , so

Plugging this in the time derivative of the expectation value of position, you get

so the normal momentum is indeed given by the operator .

On to the other commutator! The -​th component of Newton’s second law in expectation form,

requires the commutator

The easiest is the term , since both and are functions of position and commute. And the commutator with is the generalized fundamental operator of chapter 4.5.4,

and plugging that into Newton’s equation, you can verify that the electric field term of the Lorentz law has already been obtained.

In what is left of the desired commutator, again take each factor to its own side of the commutator:

Work out the simpler commutator appearing here first:

the first equality because momentum operators and functions commute, and the second equality is again the generalized fundamental commutator.

Note that by assumption the derivatives of are constants, so the side of that this result appears is not relevant and what is left of the Hamiltonian becomes

Now let be the index following in the sequence and the one preceding it (or the second following). Then the sum above will have a term where , but that term is seen to be zero, a term where , and a term where . The total is then:

and that is

and the expression in brackets is the -​th component of and produces the term in Newton’s equation provided that .