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

\begin{displaymath}
H_{21} =
\sum_i \big\langle\psi_{\rm {L}}\big\vert \skew...
...ehat{\skew{-.5}\vec p}}_i \big\vert\psi_{\rm {H}}\big\rangle
\end{displaymath}

Here $\psi_{\rm {L}}$ is the final atomic or nuclear state and $\psi_{\rm {H}}$ the initial one. The sum is over the atomic or nuclear particles, with $i$ the particle index. Also

\begin{displaymath}
{\skew 4\widehat{\skew{-.5}\vec p}}_i = \frac{\hbar}{{\rm ...
...c r}_i\times\nabla_i
j_\ell(kr_i) Y_\ell^m(\theta_i,\phi_i)
\end{displaymath}

where $j_\ell$ is a spherical Bessel function and $Y_\ell^m$ a spherical harmonic.

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

\begin{displaymath}
\big\langle\psi_{\rm {L}}\big\vert r_i^\ell Y_{\ell,i}^m \...
...angle
\qquad Y_{\ell,i}^m \equiv Y_\ell^m(\theta_i,\phi_i)
\end{displaymath}

where $r_i^{\ell}Y_{\ell,i}^m$ is a harmonic polynomial. However, the correct inner product is

\begin{displaymath}
\big\langle\psi_{\rm {L}}\big\vert
r_i^\ell Y_{\ell,i}^m...
...,r_i^\ell Y_{\ell.i}^m]
\big\vert\psi_{\rm {H}}\big\rangle
\end{displaymath}

Here $\hbar\omega$ is the energy of the photon. The additional commutator term is not mentioned in any other basic textbook on nuclei that this author knows of. A derivation is in {D.43}.

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 $\psi_{\rm {H}}$ or on the harmonic polynomial. So the relative size of the commutator term ballparks to the ratio of the spin-orbit energy to the photon energy. That ratio can be big. For example, in the so-called “islands of isomerism” transitions, one state has enough spin-orbit energy to cross a major shell boundary. But the energy release $\hbar\omega$ stays within the same shell and could be quite small.

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

\begin{displaymath}
\psi_{\rm {L}} = R_{\rm {L}}(r_i) Y_{0,i}^0{\uparrow}\qqua...
...} {\skew 4\widehat{\vec L}}_i\cdot{\skew 6\widehat{\vec S}}_i
\end{displaymath}

Here the part $V_0$ of the potential $V$ represents terms that only depend on position, or do not involve the position coordinates of particle $i$. This part commutes with the harmonic polynomial. Also $f$ is a function of radial position of order 1. Then the constant $V_{\rm {so}}$ gives the magnitude of the spin orbit energy. In the above case, only the $z$ components of the angular momentum operators give a contribution. Then it is easily seen that the commutator term produces a contribution that is larger in magnitude than the other term by a factor of order $V_{\rm {so}}\ell$$\raisebox{.5pt}{$/$}$$\hbar\omega$.

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 $e^{-{{\rm i}}q_i\chi_i/\hbar}$, and the exponential is a completely arbitrary function of the particle coordinates. But the fact that the commutator should be zero does not take away the fact that it is not zero. Presumably, if you described the nucleus in terms of the individual quarks, you could write down a potential that respected the gauge property. But using quarks is not an option. The reality is that you must use protons and neutrons. And in those terms, a potential that does a decent job of describing the nucleus will simply not respect the gauge property. Yes, this does suggest that describing gamma decay using protons and neutrons instead of quarks is an inherently fishy procedure.)

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 [32]. 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 [32] notes, “significant information regarding nuclear wave functions can be obtained from a comparison of experimental $\gamma$-​decay transition probabilities with theoretical values calculated on basis of specific models of the nucleus.” If one is not aware of the possibility of the additional commutator as a leading order effect, one might incorrectly conclude that a nuclear wave function is poor where the real problem is the ignored commutator.

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, [43], 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 ${\rm {E}}1$ transitions are so much slower than the Weisskopf estimate. The deviations in chapter 14.20.5 figures 14.61 and 14.62 seem much too big and systematic to be explained by this mechanism. And it does not seem to address the concerns about mixed transitions that are mentioned there.