Scattering Observables from One- and Two-body Densities: Formalism and Application to $\pmb \gamma$ ${}^3\hbox {He}$ Scattering

Abstract

We introduce the transition-density formalism, an efficient and general method for calculating the interaction of external probes with light nuclei. One- and two-body transition densities that encode the nuclear structure of the target are evaluated once and stored. They are then convoluted with an interaction kernel to produce amplitudes, and hence observables. By choosing different kernels, the same densities can be used for any reaction in which a probe interacts perturbatively with the target. The method therefore exploits the factorisation between nuclear structure and interaction kernel that occurs in such processes. We study in detail the convergence in the number of partial waves for matrix elements relevant in elastic Compton scattering on ${}^3\hbox {He}$. The results are fully consistent with our previous calculations in Chiral Effective Field Theory. But the new approach is markedly more computationally efficient, which facilitates the inclusion of more partial-wave channels in the calculation. We also discuss the usefulness of the transition-density method for other nuclei and reactions. Calculations of elastic Compton scattering on heavier targets like ${}^4\hbox {He}$ are straightforward extensions of this study, since the same interaction kernels are used. And the generality of the formalism means that our ${}^3\hbox {He}$ densities can be used to evaluate any ${}^3\hbox {He}$ elastic-scattering observable with contributions from one- and two-body operators. They are available at https://datapub.fz-juelich.de/anogga.

Appendices

Comment on Prior Compton Calculations on ${}^3\hbox {He}$

Our previous strategy for the computation of ${}^3\hbox {He}$ matrix elements of the Compton operators was based on the photodissociation calculation of ref. [56]. The analogous integrals for one-body and two-body operator contributions to the matrix elements were performed without splitting them into reaction-mechanism and density parts. While they were factorised into a piece involving the nucleons taking part in the reaction and the matrix element of the spectator $\delta$-distribution, the efficiency of defining densities that refer only to the quantum numbers of the active nucleons was not noticed; see refs. [24,25,26,27,28] for details.

In the course of this study, we found that refs. [24,25,26,27,28] contain a flaw in the reasoning leading to the original equations corresponding to Eq. (3.1), which in turn led to incorrect numerical implementations of the one-body part. The struck nucleon in the one-body part was considered to be not nucleon 3 but one of the nucleons of the (12) sub-system. Therefore, rather than ${{{\hat{O}}}}_{\lambda ^\prime \lambda }^{1B}(3)$ as in Sect. 2.3, the operator ${{{\hat{O}}}}_{\lambda ^\prime \lambda }^{1B}(1)$ was considered and replaced by $\frac{1}{2} [{{{\hat{O}}}}_{\lambda ^\prime \lambda }^{1B}(1) + {{{\hat{O}}}}_{\lambda ^\prime \lambda }^{1B}(2)]$ in the course of defining operators on the space of two-nucleon states. However, this replacement cannot be done at the level of the spin and isospin operators because the momentum assignment for the post-collision state differs depending on whether the struck particle is nucleon 1 or nucleon 2.

That error in refs. [24,25,26,27] was also present in our recent evaluation of $\gamma$ ${}^3\hbox {He}$ scattering [28]. It means that those works missed contributions at nonzero momentum-transfer, where the Compton-scattering collision induced a transition that changed either the spin or the isospin of the $\mathrm {N}\mathrm {N}$ state, but not both.

Fortunately, the numerical effect on observables is very small. For the neutron, this changes the matrix elements with insertions of $3\sigma _\mu$ by $\le 3\%$, except for about $16\%$ in $\sigma _x$ at the highest energy and momentum transfer we consider $(\omega =120\;\mathrm {MeV},\theta =165^\circ )$. The change is more pronounced for the proton, where it can amount to a factor of about 4.5 in $\sigma _x$ at that point and exceeds $10\%$ even at small $(\omega ,\theta )$. This might seem to imply big changes of the one-body amplitudes for the spin-polarisabilities. But the proton spins inside ${}^3\hbox {He}$ are mostly paired to spin-zero, so there is hardly any sensitivity to the mistake in matrix elements of the proton spin. The effect is also shielded for the neutron spin. Even at the “high” energy and momentum-transfer tested here, the effect of the neutron’s spin-polarisabilities is $\approx 10$% in the amplitude. Matrix elements of the neutron’s spin do not play a big role, and even a 16% error in them would only be a 1.6% error in matrix elements. This is associated with the fact that the biggest contributions to ${}^3\hbox {He}$ Compton scattering for $50\;\mathrm {MeV}\lesssim \omega \lesssim 120\;\mathrm {MeV}$ come from interactions with the two charged protons. These do not change the nucleon spin and are hence proportional to insertions of $\mathbb {1}\equiv \sigma _0$—and such matrix elements are changed by less than $0.2\%$ of the largest magnitude of all one-body matrix elements, see Eq. (3.9). Therefore, we were able to find the error only when we zoomed in on a detailed comparison between the “traditional” and “density” approach; see Sect. 3.4. In the “traditional” results quoted in the body of the paper this error is, of course, corrected.

In almost all cases, this mistake for the matrix elements with insertions $\sigma _\mu ^{(\mathrm {N})}$ only minimally alters the plots of both magnitudes and sensitivities of observables in ref. [28]. The cross section as well as the double-asymmetries $T_{11}^\mathrm {circ}\equiv \Sigma _{2x}$ and $T_{10}^\mathrm {circ}\equiv \Sigma _{2z}$ change by $<1\%$ at $50\;\mathrm {MeV}$, and by $\lesssim 3\%$ at $120\;\mathrm {MeV}$, where asymmetries exceed 0.1. For $\Sigma ^\mathrm {lin}\equiv \Sigma _{30}$, the thickness of the line is never exceeded ($<1\%$). To put this into perspective, the variation from using different ${}^3\hbox {He}$ wave functions is at all energies and angles at least a factor five bigger than the change from this error. In ref. [28], wave-function dependence was, in turn, estimated to be substantially smaller than the sum of all residual theoretical uncertainties. Therefore, we refrain from amending or updating the presentations of refs. [25,26,27,28]. Their conclusions are unchanged.

Symmetries of Matrix Elements

We now derive the symmetries that relate different matrix elements of the one- and two-nucleon operators in Sects. 3.4 and 3.3 by considering an insertion $\sigma _\mu \, \mathrm {e}^{\mathrm {i}\frac{2}{3} {\vec {q}} \cdot {\vec {r}}_3}$, which is independent of ${\vec {k}}$ (i.e.  $K=\kappa =0$). Note that we have employed the Jacobi-coordinate-space representation of the momentum-conservation relation (2.12) here to define the operator insertion. Hence, the plane wave deposits momentum into the Jacobi co-ordinate of nucleon 3. Also, $\sigma _\mu ^{(\mathrm {N})}$ acts not on the ${}^3\hbox {He}$ nucleus as a whole but only on the single “active” nucleon 3 at coordinate ${\vec {r}}_3$. Therefore, this set of operators cannot be represented by the standard Pauli matrices. However, their matrix elements do retain certain properties of a spin-$\frac{1}{2}$ representation.

We first prove relation (3.10). Under time reversal, ${\mathcal {T}} \sigma _i^{(\mathrm {N})} {\mathcal {T}}^{-1}=-\sigma _i^{(\mathrm {N})}$ is odd, while ${\mathcal {T}} \sigma _0^{(\mathrm {N})} {\mathcal {T}}^{-1}=\sigma _0^{(\mathrm {N})}$ is even. Since we wish to consider only operators with real matrix elements, we also note that ${\mathcal {T}} \mathrm {i}\sigma _y^{(\mathrm {N})} {\mathcal {T}}^{-1}=\mathrm {i}\sigma _y^{(\mathrm {N})}$ is even.

Now denoting the state ${\mathcal {T}}|\psi \rangle$ by $|{{\tilde{\psi }}}\rangle$, time reversal invariance gives for the matrix element of some operator ${\mathcal {Q}}$, see eg.,  ref. [57]:

$$\begin{aligned} \langle \phi | {\mathcal {Q}} |\psi \rangle =\langle {{\tilde{\psi }}} | {\mathcal {T}} {\mathcal {Q}}^\dagger {\mathcal {T}}^{-1} |{{\tilde{\phi }}}\rangle \;\;. \end{aligned}$$

(B.1)

If the matrix element is real, then $\langle \phi | {\mathcal {Q}} |\psi \rangle = \langle \psi | {\mathcal {Q}}^\dagger |\phi \rangle$, so

$$\begin{aligned} \langle \psi | {\mathcal {Q}}^\dagger |\phi \rangle =\langle {{\tilde{\psi }}} | {\mathcal {T}} {\mathcal {Q}}^\dagger {\mathcal {T}}^{-1} |{{\tilde{\phi }}}\rangle \;\;. \end{aligned}$$

(B.2)

This, of course, remains true if we replace ${\mathcal {Q}}^\dagger$ with ${\mathcal {Q}}$, which now serves as our starting point. Recalling Eq. (2.28)

$$\begin{aligned} {\mathcal {T}} |j,m_j\rangle =(-1)^{j+m_j} \;|j,-m_j \rangle \;\;, \end{aligned}$$

(B.3)

we obtain

$$\begin{aligned} \langle M'|\sigma _{x,z}^{(\mathrm {N})}|M \rangle&=(-1)^{2J+M'+M-1} \;\langle -M'|\sigma _{x,z}^{(\mathrm {N})}|-M \rangle \end{aligned}$$

(B.4)

$$\begin{aligned} \langle M'|\sigma _0^{(\mathrm {N})}|M \rangle&=(-1)^{2J+M'+M}\; \langle -M'|\sigma _0^{(\mathrm {N})}|-M \rangle \end{aligned}$$

(B.5)

$$\begin{aligned} \langle M'|\mathrm {i}\sigma _{y}^{(\mathrm {N})}|M \rangle&=(-1)^{2J+M'+M} \;\langle -M'|\mathrm {i}\sigma _{y}^{(\mathrm {N})}|-M \rangle \;\;. \end{aligned}$$

(B.6)

As $2(J+M)$ is even for both half-integer and integer quantum numbers, this proves Eq. (3.10). Replacing $\sigma _\mu ^{(\mathrm {N})}\rightarrow \sigma _\mu ^{(\mathrm {N})}\, \mathrm {e}^{\mathrm {i}\frac{2}{3} {\vec {q}} \cdot {\vec {r}}_3}$ does not alter this argument since the plane wave is time-reversal even (parity guarantees that only its real parts can contribute to the final result). Accounting for the multipolarity of ${\vec {k}}$ ($K\ne 0$), the relation is modified to:

$$\begin{aligned} A^{-M^\prime }_{-M}(\sigma _\mu ^{(\mathrm {N})};K,-\kappa ,{\vec {q}})= (-1)^{M^\prime -M+\mu +\kappa }\,A^{M^\prime }_{M}(\sigma _\mu ^{(\mathrm {N})};K,\kappa ,{\vec {q}})\;\;. \end{aligned}$$

(B.7)

This proves there are at most eight independent matrix elements of the $\sigma _\mu ^{(\mathrm {N})}\, e^{\mathrm {i}\frac{2}{3} {\vec {q}}\cdot {\vec {r}}_3}$ operator in the spin-$\frac{1}{2}$ basis $(M^\prime M)$. We now discuss symmetries that eliminate three more. First, since $\mathrm {i}\sigma _y^{(\mathrm {N})}$ is both real and anti-Hermitean, both its diagonal elements must be zero:

$$\begin{aligned} A^{M}_M(\mathrm {i}\sigma _y^{(\mathrm {N})};{\vec {q}})\equiv \langle M|\mathrm {i}\sigma _y^{(\mathrm {N})}\,\mathrm {e}^{\mathrm {i}\frac{2}{3} {\vec {q}} \cdot {\vec {r}}_3}|M \rangle =0\;\;. \end{aligned}$$

(B.8)

Furthermore $\sigma _0^{(\mathrm {N})}$ is symmetric, but its off-diagonal elements are equal and opposite by (B.5). Hence they must be zero:

$$\begin{aligned} A^{-M}_M(\sigma _0^{(\mathrm {N})};{\vec {q}})\equiv \langle -M|\sigma _0^{(\mathrm {N})}\,\mathrm {e}^{\mathrm {i}\frac{2}{3} {\vec {q}} \cdot {\vec {r}}_3}|M \rangle =0\;\;. \end{aligned}$$

(B.9)

These relations can also be established from the Lie algebra of the Pauli operators in a two-dimensional representation that is consistent with time reversal and in which $\sigma _x^{(\mathrm {N})}$ and $\sigma _z^{(\mathrm {N})}$ are real. They do not hold for densities which explicitly depend on ${\vec {k}}$ (i.e.  $K\ne 0$).

If the spin of the nucleon were always perfectly aligned with the spin of a $J=\frac{1}{2}$ nucleus, then off-diagonal matrix elements of $\sigma _z^{(\mathrm {N})}$ and diagonal matrix elements of $\sigma _x^{(\mathrm {N})}$ would also be zero. These matrix elements pick out densities for “wrong-spin” to “right-spin” transitions (or vice versa). Specifically:

$$\begin{aligned} \begin{aligned} \langle -\tfrac{1}{2}|\sigma _z^{(\mathrm {N})}\,\mathrm {e}^{\mathrm {i}\frac{2}{3} {\vec {q}} \cdot {\vec {r}}_3}|\tfrac{1}{2} \rangle&=\rho ^{00; m_3^t M_T,-\frac{1}{2}, +\frac{1}{2}}_{+\frac{1}{2}, +\frac{1}{2}} - \rho ^{00; m_3^t M_T, -\frac{1}{2}, +\frac{1}{2}}_{-\frac{1}{2}, -\frac{1}{2}} \;\;,\\ \langle \tfrac{1}{2}|\sigma _x^{(\mathrm {N})}\,\mathrm {e}^{\mathrm {i}\frac{2}{3} {\vec {q}} \cdot {\vec {r}}_3}|\tfrac{1}{2}\rangle&=\rho ^{00; m_3^t M_T,+\frac{1}{2}, +\frac{1}{2}}_{-\frac{1}{2}, +\frac{1}{2}}+ \rho ^{00; m_3^t M_T, +\frac{1}{2}, +\frac{1}{2}}_{+\frac{1}{2}, -\frac{1}{2}} \;\; . \end{aligned} \end{aligned}$$

(B.10)

Time-reversal alone is not enough to guarantee the equality of these two matrix elements. But the flipping symmetry of Eq. (2.38) means that the first terms of each line of Eq. (B.10) are equal. Using flipping symmetry in conjunction with time-reversal (2.31) shows that the second terms are equal, too. Therefore, the off-diagonal matrix elements of $\sigma _z^{(\mathrm {N})}$ and diagonal ones of $\sigma _x^{(\mathrm {N})}$ are identical:

$$\begin{aligned} A^{-M}_M(\sigma _z^{(\mathrm {N})};{\vec {q}})\equiv \langle -M|\sigma _z^{(\mathrm {N})}\,\mathrm {e}^{\mathrm {i}\frac{2}{3} {\vec {q}} \cdot {\vec {r}}_3}|M\rangle = \langle M|\sigma _x^{(\mathrm {N})}\,\mathrm {e}^{\mathrm {i}\frac{2}{3} {\vec {q}} \cdot {\vec {r}}_3}|M\rangle \equiv A^{M}_M(\sigma _x^{(\mathrm {N})};{\vec {q}}) \;\; . \end{aligned}$$

(B.11)

Ultimately then, there are five non-equal, non-zero ${}^3$He matrix elements of the one-body operators $\left\{ \sigma _0^{(N)},\sigma _x^{(N)},\sigma _y^{(N)},\sigma _z^{(N)}\right\}$, out of a possible 16. This matches the five independent transition densities after time-reversal, Hermitecity, and flipping symmetry have been applied.

The proof of the two-body relation (3.12) proceeds analogously to that of Eq. (B.7). Since the Compton two-body operator ${\hat{O}}_{12}$ is time-reversal even and the matrix element is real below all thresholds, one finds by inserting Eqs. (2.28) and (2.30):

$$\begin{aligned} \begin{aligned} \langle M'|\langle \lambda '| {\mathcal {T}} {\hat{O}}_{12} {\mathcal {T}}^{-1} |\lambda \rangle |M \rangle&=\langle - M'|\langle -\lambda '|{\hat{O}}_{12}|-\lambda \rangle |- M \rangle \\&= (-1)^{2J + M'+ M+ \lambda ^\prime + \lambda } \langle M'|\langle \lambda '| {\hat{O}}_{12} |\lambda \rangle |M \rangle \;\;. \end{aligned} \end{aligned}$$

(B.12)

Equation (3.12) follows directly because $2(J+M+\lambda )$ is even for both half-integer and integer quantum numbers.