Determinant/permanent division and direct mappings in Fock space

Abstract

The standard mapping between different N-particle states in Fock space is very abstract and consists of combinations of creation and annihilation operators being formally applied to the vacuum state. The resulting formalism provides us with the correct symmetries in the case of fermions and bosons, but the permanents and Slater determinants that are formally joined by the action of these operators must be individually constructed from product states using symmetrizers and antisymmetrizers. Surely, this abstract approach does not correspond to a direct mapping between different N-particle states in Fock space. In the following, we will examine direct mappings in Fock space, which are generalizations of an old determinant division scheme by Schweins (1825). The suggested new scheme is based on a suitable definition of isolated states. We will show that all the classical results obtained by Schweins may be recovered. Furthermore, the new division scheme may easily be extended to permanents. This leads to direct mappings in Fock space, which correspond to creation and annihilation operators.

Appendices

Appendix A

In this Appendix, we will show how to work out the brackets of Eqs. (21) and (25). This is a very tedious bookkeeping exercise, and we start our discussion by tackling the first bracket in Eq. (21):

$$\begin{aligned} M_{3}M_{4}^{'}-M_{3}^{'}M_{4}=\det {(h_{1}b_{2}c_{4}})\, \det {(a_{1}b_{2}c_{3}})-\det {(a_{1}b_{2}c_{4}})\det {(h_{1}b_{2}c_{3}}) \end{aligned}$$

(A1)

with the following Laplace expansions:

$$\begin{aligned} \det {(h_{1}b_{2}c_{4})}= & {} h_{1}\det {\left( b_{2}c_{4} \right) }-h_{2}\det {\left( b_{1}c_{4} \right) +h_{4}\det \left( b_{1}c_{2} \right) } ,\nonumber \\ \det {(a_{1}b_{2}c_{3})}= & {} \, a_{1}\det {\left( b_{2}c_{3} \right) }-a_{2}\det {\left( b_{1}c_{3} \right) +a_{3}\det \left( b_{1}c_{2} \right) } ,\nonumber \\ \det {(a_{1}b_{2}c_{4})}= & {} \, a_{1}\det {\left( b_{2}c_{4} \right) }-a_{2}\det {\left( b_{1}c_{4} \right) +a_{4}\det \left( b_{1}c_{2} \right) } ,\nonumber \\ \det {(h_{1}b_{2}c_{3})}= & {} \, h_{1}\det {\left( b_{2}c_{3} \right) }-h_{2}\det {\left( b_{1}c_{3} \right) +h_{3}\det \left( b_{1}c_{2} \right) }. \end{aligned}$$

(A2)

Next, we collect terms according to minors of the determinant $\det {(h_{1}b_{2}c_{4}})$, which leads to the following result:

$$\begin{aligned}{} & {} M_{3}M_{4}^{'}-M_{3}^{'}M_{4} \nonumber \\{} & {} \quad =\det {\left( b_{2}c_{4} \right) \left\{ \det {\left( h_{1}a_{1} \right) \det {\left( b_{2}c_{3} \right) -\det {\left( h_{1}a_{2} \right) \det {\left( b_{1}c_{3} \right) +\det {\left( h_{1}a_{3} \right) \det \left( b_{1}c_{2} \right) }}}}} \right\} } \nonumber \\{} & {} \qquad -\det {\left( b_{1}c_{4} \right) \left\{ \det {\left( h_{2}a_{1} \right) \det {\left( b_{2}c_{3} \right) -\det {\left( h_{2}a_{2} \right) \det {\left( b_{1}c_{3} \right) +\det {\left( h_{2}a_{3} \right) \det \left( b_{1}c_{2} \right) }}}}} \right\} } \nonumber \\{} & {} \qquad + \det {\left( b_{1}c_{2} \right) \left\{ \det {\left( h_{4}a_{1} \right) \det {\left( b_{2}c_{3} \right) -\det {\left( h_{4}a_{2} \right) \det {\left( b_{1}c_{3} \right) +\det {\left( h_{4}a_{3} \right) \det \left( b_{1}c_{2} \right) }}}}} \right\} } \nonumber \\{} & {} \quad =\det \left( b_{2}c_{4} \right) \det {\left( \det {\left( h_{1}a_{1} \right) b_{2}c_{3}} \right) -}\det \left( b_{1}c_{4} \right) \det \left( \det {\left( h_{2}a_{1} \right) b_{2}c_{3}} \right) \nonumber \\{} & {} \qquad +\, \det \left( b_{1}c_{2} \right) \det {\left( \det {\left( h_{4}a_{1} \right) b_{2}c_{3}} \right) =\, A_{1}\det {\left( b_{2}c_{4} \right) -A_{2}\det {\left( b_{1}c_{4} \right) +A_{4}\det \left( b_{1}c_{2} \right) }}}. \end{aligned}$$

(A3)

From the definitions of the $A_{i}$ in Eq. (17), we further note that $\det {\left( h_{1}a_{1} \right) =\det {\left( {\begin{array}{*{20}c} h_{1} &{} h_{1}\\ a_{1} &{} a_{1}\\ \end{array} } \right) =0}}$, and that $\det {\left( h_{2}a_{2} \right) =0.\, }$ On the other hand, we also find that $\det {\left( h_{1}a_{2} \right) =\det {\left( {\begin{array}{*{20}c} h_{1} &{} h_{2}\\ a_{1} &{} a_{2}\\ \end{array} } \right) =}}\, h_{1}a_{2}-h_{2}a_{1}$.

The other brackets in Eq. (21) may be worked out in a similar fashion, and those results may be summarized as follows:

$$\begin{aligned} M_{3}M_{4}^{'}-M_{3}^{'}M_{4}= & {} A_{1}\det {\left( b_{2}c_{4} \right) -A_{2}\det {\left( b_{1}c_{4} \right) +A_{4}\det \left( b_{1}c_{2} \right) }} ,\nonumber \\ M_{2}M_{4}^{'}-M_{2}^{'}M_{4}= & {} A_{1}\det {\left( b_{3}c_{4} \right) -A_{3}\det {\left( b_{1}c_{4} \right) +A_{4}\det \left( b_{1}c_{3} \right) }} ,\nonumber \\ M_{3}M_{4}^{'}-M_{3}^{'}M_{4}= & {} A_{2}\det {\left( b_{3}c_{4} \right) -A_{3}\det {\left( b_{2}c_{4} \right) +A_{4}\det \left( b_{2}c_{3} \right) }}. \end{aligned}$$

(A4)

To work out the brackets for the permanents in Eq. (25), we will proceed in the same fashion as we did for the determinants in this section. The first bracket in Eq. (25) was:

$$\begin{aligned} L_{3}L_{4}^{'}-L_{3}^{'}L_{4}=\text {perm}~{(h_{1}b_{2}c_{4}})\, \text {perm}~{(a_{1}b_{2}c_{3}})-\text {perm}~{(a_{1}b_{2}c_{4})} \text {perm}~{(h_{1}b_{2}c_{3}}) \end{aligned}$$

(A5)

and it involved the following Laplace expansions:

$$\begin{aligned} \text {perm}~{(h_{1}b_{2}c_{4}})= & {} h_{1}\text { perm}~{\left( b_{2}c_{4} \right) }+h_{2}\text { perm}~{\left( b_{1}c_{4} \right) +h_{4}\text { perm}~\left( b_{1}c_{2} \right) } ,\nonumber \\ \text {perm}~{(a_{1}b_{2}c_{3}})= & {} \, a_{1}\, \text { perm}~{\left( b_{2}c_{3} \right) }+a_{2}\text { perm}~{\left( b_{1}c_{3} \right) +a_{3}\text { perm}~\left( b_{1}c_{2} \right) } ,\nonumber \\ \text {perm}~{(a_{1}b_{2}c_{4}})= & {} \, a_{1}\, \text { perm}~{\left( b_{2}c_{4} \right) }+a_{2}\text { perm}~{\left( b_{1}c_{4} \right) +a_{4}\text { perm}~\left( b_{1}c_{2} \right) } ,\nonumber \\ \text {perm}~{(h_{1}b_{2}c_{3}})= & {} \, h_{1\, }\text { perm}~{\left( b_{2}c_{3} \right) }+h_{2}\text { perm}~{\left( b_{1}c_{3} \right) +h_{3}\text { perm}~\left( b_{1}c_{2} \right) }. \end{aligned}$$

(A6)

Once again, we start to collect terms according to the minors of the permanent $\text {perm}~{(h_{1}b_{2}c_{4}})$, which leads us to the following result:

$$\begin{aligned}{} & {} L_{3}L_{4}^{'}-L_{3}^{'}L_{4} \nonumber \\{} & {} \quad =\text {perm}~{\left( b_{2}c_{4} \right) \left\{ \det {\left( h_{1}a_{1} \right) \text {perm}~{\left( b_{2}c_{3} \right) +\det {\left( h_{1}a_{2} \right) \text {perm}~{\left( b_{1}c_{3} \right) +\det {\left( h_{1}a_{3} \right) \text {perm}~\left( b_{1}c_{2} \right) }}}}} \right\} } \nonumber \\{} & {} \qquad +\,\text {perm}~{\left( b_{1}c_{4} \right) \left\{ \det {\left( h_{2}a_{1} \right) \text {perm}~{\left( b_{2}c_{3} \right) +\det {\left( h_{2}a_{2} \right) \text {perm}~{\left( b_{1}c_{3} \right) +\det {\left( h_{2}a_{3} \right) \text {perm}~\left( b_{1}c_{2} \right) }}}}} \right\} } \nonumber \\{} & {} \qquad +\,\text {perm}~{\left( b_{1}c_{2} \right) \left\{ \det {\left( h_{4}a_{1} \right) \text {perm}~{\left( b_{2}c_{3} \right) +\det {\left( h_{4}a_{2} \right) \text {perm}~{\left( b_{1}c_{3} \right) +\det {\left( h_{4}a_{3} \right) \text {perm}~\left( b_{1}c_{2} \right) }}}}} \right\} } \nonumber \\{} & {} \quad =\text {perm}~\left( b_{2}c_{4} \right) \text {perm}~{\left( \det {\left( h_{1}a_{1} \right) b_{2}c_{3}} \right) }+\text {perm}~\left( b_{1}c_{4} \right) \text {perm}~\left( \det {\left( h_{2}a_{1} \right) b_{2}c_{3}} \right) \nonumber \\{} & {} \qquad +\,\text {perm}~\left( b_{1}c_{2} \right) \text {perm}~{\left( \det {\left( h_{4}a_{1} \right) b_{2}c_{3}} \right) =\, B_{1}~\text {perm}~{\left( b_{2}c_{4} \right) +B_{2}~\text {perm}~{\left( b_{1}c_{4} \right) +B_{4}~\text {perm}~\left( b_{1}c_{2} \right) }}}. \end{aligned}$$

(A7)

The final results of Eq. (A7) are based on the definitions of the $B_{i}$ from Eq. (18). The other brackets in Eq. (25) can be worked out in a similar fashion, and those results may be summarized as follows:

$$\begin{aligned} L_{3}L_{4}^{'}-L_{3}^{'}L_{4}= & {} B_{1}~\text {perm}~{\left( b_{2}c_{4} \right) }+B_{2}~\text {perm}~{\left( b_{1}c_{4} \right) }+B_{4}~\text {perm}~{\left( b_{1}c_{2} \right) } ,\nonumber \\ L_{2}L_{4}^{'}-L_{2}^{'}L_{4}= & {} B_{1}~\text {perm}~{\left( b_{3}c_{4} \right) }+B_{3}~\text {perm}~{\left( b_{1}c_{4} \right) }+B_{4}~\text {perm}~{\left( b_{1}c_{3} \right) } ,\nonumber \\ L_{3}L_{4}^{'}-L_{3}^{'}L_{4}= & {} B_{2}~\text {perm}~{\left( b_{3}c_{4} \right) }+B_{3}~\text {perm}~{\left( b_{2}c_{4} \right) }+B_{4}~\text {perm}~{\left( b_{2}c_{3} \right) }. \end{aligned}$$

(A8)

Appendix B

In this Appendix, we will point out a very particular property of the Schweinsian division schemes and of our general division scheme for determinants and permanents, which Aitken calls the extra shutter property [4]. We will see that from the comparison of different division schemes for determinants and permanents of dimensions 1–4, we may easily derive the corresponding division schemes for all higher dimensions, simply by systematically extending all terms lower-dimensional division schemes!

This extra shutter property may look like a conjecture rather than a sound mathematical result, which is based an elegant mathematic proof. We nevertheless think that for readers with a background in physics rather than in mathematics, this Appendix will allow for a very intuitive approach to the various expansion schemes, which might be more useful than a difficult mathematical proof.

Let us now start with the first Schweinsian expansion, where we list results for dimensions 1–4 in the form of divisions with remainder:

• 1. $N=1$ (trivial):

  $$\begin{aligned} h_{1}=\frac{h_{1}}{a_{1}}a_{1}. \end{aligned}$$

  (B1)

• 2. $N=2:$

  $$\begin{aligned} \det \left( h_{1}b_{2} \right) =\frac{h_{1}}{a_{1}}\det \left( a_{1}b_{2} \right) +\det \left( h_{1}a_{2} \right) \, \frac{b_{1}}{a_{1}}. \end{aligned}$$

  (B2)

• 3. $N=3:$

$$\begin{aligned} \det \left( h_{1}b_{2}c_{3} \right) =\left( \frac{h_{1}}{a_{1}}+\frac{\det \left( h_{1}a_{2} \right) }{\det \left( a_{1}b_{2} \right) }\, \frac{b_{1}}{a_{1}} \right) \det \left( a_{1}b_{2}c_{3} \right) +\det \left( h_{1}a_{2}b_{3} \right) \frac{\det \left( b_{1}c_{2} \right) }{\det \left( a_{1}b_{2} \right) }. \end{aligned}$$

Or:

$$\begin{aligned} \det \left( h_{1}b_{2}c_{3} \right) =\frac{\det \left( h_{1}b_{2} \right) }{\det \left( a_{1}b_{2} \right) }\det \left( a_{1}b_{2}c_{3} \right) +\det \left( h_{1}a_{2}b_{3} \right) \frac{\det \left( b_{1}c_{2} \right) }{\det \left( a_{1}b_{2} \right) }. \end{aligned}$$

(B3)

• 4. $N=4:$

$$\begin{aligned}{} & {} \det \left( h_{1}b_{2}c_{3}d_{4} \right) =\left( \frac{h_{1}}{a_{1}}+\frac{\det \left( h_{1}a_{2} \right) }{\det \left( a_{1}b_{2} \right) }\, \frac{b_{1}}{a_{1}}+\frac{\det \left( h_{1}a_{2}b_{3} \right) }{\det \left( a_{1}b_{2}c_{3} \right) }\frac{\det \left( b_{1}c_{2} \right) }{\det \left( a_{1}b_{2} \right) } \right) \det \left( a_{1}b_{2}c_{3}d_{4} \right) \\{} & {} \quad +\det \left( h_{1}a_{2}b_{3}c_{4} \right) \frac{\det \left( b_{1}c_{2}d_{3} \right) }{\det \left( a_{1}b_{2}c_{3} \right) }. \end{aligned}$$

Or:

$$\begin{aligned} \det \left( h_{1}b_{2}c_{3}d_{4} \right) =\frac{\det \left( h_{1}b_{2}c_{3} \right) }{\det \left( a_{1}b_{2}c_{3} \right) }\det \left( a_{1}b_{2}c_{3}d_{4} \right) +\det \left( h_{1}a_{2}b_{3}c_{4} \right) \frac{\det \left( b_{1}c_{2}d_{3} \right) }{\det \left( a_{1}b_{2}c_{3} \right) }. \end{aligned}$$

(B4)

Next, we list the second Schweinsian expansions of dimensions 1–4:

• 1. $N=1$ (trivial):    

  $$\begin{aligned} a_{1}=a_{1}\cdot 1. \end{aligned}$$

  (B5)

• 2. $N=2:$    

  $$\begin{aligned} \det \left( a_{1}b_{2} \right) ={\left( a_{1}-\, \frac{a_{2}b_{1}}{b_{2}} \right) }b_{2}=a_{1}b_{2}-a_{2}b_{1}. \end{aligned}$$

  (B6)

• 3. $N=3:$    

  $$\begin{aligned} \det \left( a_{1}b_{2}c_{3} \right) =\left( a_{1}-\, \frac{a_{2}b_{1}}{b_{2}}-\frac{\det \left( a_{2}b_{3} \right) }{b_{2}}\, \frac{\det \left( b_{1}c_{2} \right) }{\det \left( b_{2}c_{3} \right) } \right) \det \left( b_{2}c_{3} \right) . \end{aligned}$$

Or:

$$\begin{aligned} \det \left( a_{1}b_{2}c_{3} \right) =\frac{\det \left( a_{1}b_{2} \right) }{b_{2}}\det \left( b_{2}c_{3} \right) -\frac{\det \left( a_{2}b_{3} \right) }{b_{2}}\, \det \left( b_{1}c_{2} \right) . \end{aligned}$$

(B7)

• 4. $N=4:$

$$\begin{aligned}{} & {} \det \left( a_{1}b_{2}c_{3}d_{4} \right) =\left( a_{1}-\, \frac{a_{2}b_{1}}{b_{2}}-\frac{\det \left( a_{2}b_{3} \right) }{b_{2}}\, \frac{\det \left( b_{1}c_{2} \right) }{\det \left( b_{2}c_{3} \right) }-\frac{\det \left( a_{2}b_{3}c_{4} \right) }{\det \left( b_{2}c_{3} \right) }\, \frac{\det \left( b_{1}c_{2}d_{3} \right) }{\det \left( b_{2}c_{3}d_{4} \right) } \right) \\{} & {} \quad \cdot \det \left( b_{2}c_{3}d_{4} \right) . \end{aligned}$$

Or:

$$\begin{aligned} \det \left( a_{1}b_{2}c_{3}d_{4} \right) =\frac{\det \left( a_{1}b_{2}c_{3} \right) }{\det \left( b_{2}c_{3} \right) }\det \left( b_{2}c_{3}d_{4} \right) -\frac{\det \left( a_{2}b_{3}c_{4} \right) }{\det \left( b_{2}c_{3} \right) }\, \det \left( b_{1}c_{2}d_{3} \right) . \end{aligned}$$

(B8)

Now, we come to our general expansion scheme for determinants of dimensions 1–4:

• 1. $N=1:$    

  $$\begin{aligned} h_{1}=\frac{h_{1}}{a_{1}}a_{1}+A_{1}=\frac{h_{1}}{a_{1}}a_{1} \quad \text {as } A_{1}=0. \end{aligned}$$

  (B9)

• 2. $N=2:$

  $$\begin{aligned} \det \left( h_{1}b_{2} \right) =\frac{h_{1}}{a_{1}}\det \left( a_{1}b_{2} \right) +\frac{\det {(A_{1}0)}}{a_{1}} \end{aligned}$$

with:

$$\begin{aligned} \det {(A_{1}0)}=\det \left( {\begin{array}{*{20}c} A_{1} &{} A_{2}\\ b_{1} &{} 0\\ \end{array} } \right) . \end{aligned}$$

(B10)

• 3. $N=3:$    

  $$\begin{aligned} \det \left( h_{1}b_{2}c_{3} \right) =\frac{\det \left( h_{1}b_{2} \right) }{\det \left( a_{1}b_{2} \right) }\det \left( a_{1}b_{2}c_{3} \right) +\frac{\det {(A_{1}b_{2}0)}}{\det \left( a_{1}b_{2} \right) } \end{aligned}$$

with:

$$\begin{aligned} \det {(A_{1}b_{2}0)}=\det \left( {\begin{array}{*{20}c} A_{1} &{} A_{2} &{} A_{3}\\ b_{1} &{} b_{2} &{} b_{3}\\ c_{1} &{} c_{2} &{} 0\\ \end{array} } \right) . \end{aligned}$$

(B11)

• 4. $N=4:$    

  $$\begin{aligned} \det \left( h_{1}b_{2}c_{3}d_{4} \right) =\frac{\det \left( h_{1}b_{2}c_{3} \right) }{\det \left( a_{1}b_{2}c_{3} \right) }\det \left( a_{1}b_{2}c_{3}d_{4} \right) +\frac{\det {(A_{1}b_{2}c_{3}0)}}{\det \left( a_{1}b_{2}c_{3} \right) } \end{aligned}$$

with:

$$\begin{aligned} \det {(A_{1}b_{2}c_{3}0)}=\det \left( {\begin{array}{*{20}c} A_{1} &{} A_{2} &{} {\begin{array}{*{20}c} A_{3} &{} A_{4}\\ \end{array} }\\ b_{1} &{} b_{2} &{} {\begin{array}{*{20}c} b_{3} &{} b_{4}\\ \end{array} }\\ {\begin{array}{*{20}c} c_{1}\\ d_{1}\\ \end{array} } &{} {\begin{array}{*{20}c} c_{2}\\ d_{2}\\ \end{array} } &{} {\begin{array}{*{20}c} {\begin{array}{*{20}c} c_{3}\\ d_{3}\\ \end{array} } &{} {\begin{array}{*{20}c} c_{4}\\ 0\\ \end{array} }\\ \end{array} }\\ \end{array} } \right) . \end{aligned}$$

(B12)

The meaning of the $A_{i}$ is the same as in Eq. (17).

Last but not least, we also list our general expansion scheme for permanents of dimensions 1–4:

• 1. $N=1:$    

  $$\begin{aligned} h_{1}=\frac{h_{1}}{a_{1}}a_{1}+B_{1}=\frac{h_{1}}{a_{1}}a_{1}\quad \text {as }B_{1}=0. \end{aligned}$$

  (B13)

• 2. $N=2:$    

  $$\begin{aligned} \text {perm}\left( h_{1}b_{2} \right) =\frac{h_{1}}{a_{1}}\text {perm}\left( a_{1}b_{2} \right) +\frac{\text {perm}~{(B_{1}0)}}{a_{1}} \end{aligned}$$

with:

$$\begin{aligned} \text {perm}~{(B_{1}0)}=\text {perm}\left( {\begin{array}{*{20}c} B_{1} &{} B_{2}\\ b_{1} &{} 0\\ \end{array} } \right) . \end{aligned}$$

(B14)

• 3. $N=3:$    

  $$\begin{aligned} \text {perm}\left( h_{1}b_{2}c_{3} \right) =\frac{\text {perm}~\left( h_{1}b_{2} \right) }{\text {perm}\left( a_{1}b_{2} \right) }\text {perm}\left( a_{1}b_{2}c_{3} \right) +\frac{\text {perm}~{(B_{1}b_{2}0)}}{\text {perm}\left( a_{1}b_{2} \right) } \end{aligned}$$

with:

$$\begin{aligned} \text {perm}~{(B_{1}b_{2}0)}=\text {perm}\left( {\begin{array}{*{20}c} B_{1} &{} B_{2} &{} B_{3}\\ b_{1} &{} b_{2} &{} b_{3}\\ c_{1} &{} c_{2} &{} 0\\ \end{array} } \right) . \end{aligned}$$

(B15)

• 4. $N=4:$    

  $$\begin{aligned} \text {perm}\left( h_{1}b_{2}c_{3}d_{4} \right) =\frac{\text {perm}\left( h_{1}b_{2}c_{3} \right) }{\text {perm}\left( a_{1}b_{2}c_{3} \right) }\text {perm}\left( a_{1}b_{2}c_{3}d_{4} \right) +\frac{\text {perm}~{(B_{1}b_{2}c_{3}0)}}{\text {perm}\left( a_{1}b_{2}c_{3} \right) } \end{aligned}$$

with:

$$\begin{aligned} \text {perm}~{(B_{1}b_{2}c_{3}0)}=\text {perm}\left( {\begin{array}{*{20}c} B_{1} &{} B_{2} &{} {\begin{array}{*{20}c} B_{3} &{} B_{4}\\ \end{array} }\\ b_{1} &{} b_{2} &{} {\begin{array}{*{20}c} b_{3} &{} b_{4}\\ \end{array} }\\ {\begin{array}{*{20}c} c_{1}\\ d_{1}\\ \end{array} } &{} {\begin{array}{*{20}c} c_{2}\\ d_{2}\\ \end{array} } &{} {\begin{array}{*{20}c} {\begin{array}{*{20}c} c_{3}\\ d_{3}\\ \end{array} } &{} {\begin{array}{*{20}c} c_{4}\\ 0\\ \end{array} }\\ \end{array} }\\ \end{array} } \right) . \end{aligned}$$

(B16)

The meaning of the $B_{i}$ is the same as in Eq. (18).

Finally, we would like to remark that Aitken gives a mathematical proof of the two Schweinsian division schemes [4]. Furthermore, the key ideas behind our general division scheme are so simple and so general, that they also hold in any dimension. The procedure is always the same: first use a Laplace expansion of the two determinants or the two permanents, which are the denominators and the numerators of a given division scheme. Then apply a division scheme with remainder and carry out a systematic re-ordering of terms, and finally arrive at more compact expressions like the ones itemized above.