N.14 The electric multipole problem

There is a big problem with electric multipole transitions in nuclei.
Electric multipole transitions arise from a matrix element of the form

Here

where

Under the approximation that the atomic or nuclear size is small
compared to the wave length of the emitted or absorbed photon, this
may be approximated. In that approximation. the single-particle
electric matrix element is commonly described as proportional to

where

Here

If the potential depends only on position, the commutator is zero, and
there is no problem. That is a valid approximation for the outer
electrons in atoms. But nuclear potentials include significant
momentum terms. These do not commute. One example is the spin-orbit
term in the shell model. Now consider the size of the commutator term
above compared to the first term. For a ballpark, note that it does
not make much difference whether the orbital angular momentum in the
spin-orbit term acts on the wave function

As a check on this ballpark, consider the simplest possible electric
multipole transition:

Here the part

Note also that the effect is especially counter-intuitive for electric dipole transitions. It would seem logical to think that such transitions could be approximately described by a straightforward interaction of the particles with the electric field. However, the commutator need not be zero. So the electric field could be dwarfed by a larger additional field. That field is then a consequence of the fact that quantum mechanics uses the vector potential rather than the classical electric field.

Which brings up the next problem. The commutator term will not be there if you use the gauge property of the electromagnetic field to approximate the leading order electric field through an electrostatic potential. So should the commutator be there or not to get the best solution? As far as this author can see, there is no way to tell. And certainly for higher multipole orders you cannot even ignore the problem by using the electrostatic potential.

(Note that the real problem is the fact that you get different answers
depending on how you select the gauge. If the nuclear Hamiltonian
(A.169) respected the quantum gauge property of
{A.19.5}, the commutator would be zero. That can be
seen by substituting in the gauge property: it shows that for any
particle the potential must commute with

This author knows not a single reference that gives a decent
description of the above issue. Many simply do not mention the
problem at all and just omit the commutator. Some simply state that
the potential is assumed to depend on the particle positions only,
like [33]. Surely it ought to be mentioned explicitly
that the leading electric multipole operator as listed may well be no
good at all? If it is listed, people will assume that is is
meaningful unless stated otherwise. As [33] notes,
“significant information regarding nuclear wave functions can
be obtained from a comparison of experimental

At least [11, p.9-172] and [5] can be read to say that there might be a nontrivial problem. The original relativistic derivation of Stech, [44], mentions the issue, but no ballpark is given. (It is however noted that the spin-orbit term might be significant for magnetic transitions. The purely nonrelativistic analysis used here does not show such an effect. The present author suspects that the difference is that the relativistic derivation of Stech inherently assumes that the gauge property is valid. Surely there must be ways to do the relativistic analysis such that, in say the electric dipole case, both the results with and without commutator are reproduced.)

To be sure, the author no longer believes that this is a potential
explanation why