数理解析研究所講究録
第 1797 巻 2012 年 144-166

144

Feynman’s Operational Calculi for Noncommuting Operators
By

Byoung Soo KIM *Abstract
We survey the author’s recent development of Jefferies and Johnson‘s theory of Feynman’s operational calculi. Extraction of a linear factor in Feynman‘s operational calculi can simplify disentangling
under various conditions. We also introduce a first order infinitesimal calculus for a function of noncommuting operators. Further, we consider a measure permutation formula and its applications in Feynman’s
operational calculi.
$n$\S 1. Introduction
It is important in several areas of mathematics and its applications to be able to form functions of operators. If one has a single self-adjoint operator or several commuting self-adjoint
operators, the spectral theorem provides an extremely rich functional calculus. However, assoon as we have two or more noncommuting operators, the functional calculus becomes much
more complicated even if the operators are self-adjoint.
Feynman‘s 1951 paper [3] on the operational calculus for noncommuting operators arose outof his ingenious work on quantum electrodynamics and was inspired in part by his earlier workon the Feynman path integral. Indeed, Feynman thought of his operational calculus as a kindof generalized path integral. Much surprisingly varied work on the subject has been done sinceby mathematicians and physicists. References can be found in the recent books of Johnson and
Lapidus [11] and Nazaikinskii, Shatalov and Stemin [13].
Recently, Jefferies and Johnson [4], [5], [6] introduced a mathematically rigorous approach
to Feynman‘s operational calculi. The central objects of this theory are the disentangling algebra, a commutative Banach algebra, and the disentangling map which carnies the commutativestmcture into the noncommutative algebra of operators.
2010 Mathematics Subject Classification(s): $47A60,46J32,28M60$
Key Words: Feynman‘s operational calculi, disentangling, extraction ofa linear factor, derivation formula, Newton’s
formula with remainder, Taylor’s formula with remainder, measure permutation formula, commutation formula,Pauli matrices*School of Liberal Arts, Seoul National University of Science and Technology, Seoul 139-743, Korea

145
This paper is a survey article on the author’s recent development of the Jefferies and Johnson’s theory of Feynman‘s operational calculi. We summarize the results in [1], [2], [7], [9],[10]. Sections 2 and 3 are concemed with the formulas which simplify disentangling in Feynman’s operational calculi under various conditions. In Section 4, we explores the differential
calculus associated with the disentangled operators arising from Feynman‘s operational calculi.
Sections 5 and 6 are concemed with a measure permutation formula, which correspond to theindex permutation formula in Maslov’s discretized version of Feynman’s operational calculus.
We will not give the detailed proof and just state the properties.
We tum now to reviewing the basic definitions of the Jefferies and Johnson‘s theory of
Feynman‘s operational calculi.
Given a positive integer and positive numbers, let
be the space ofcomplex-valued functions of complex variables $f(z1, \ldots,z_{n})$, which are analytic at $(0, \ldots,0)$ ,and are such that their power series expansion
$n$

$n$

$\gamma_{l},$$A(\gamma_{1}, \ldots, r_{n})$$\ldots,\gamma_{n}$

$n$

(1.1)

$f(Z1, \ldots,Z_{n})=\sum^{\infty}c_{m_{1},\ldots,m_{n}}z_{1}^{m_{1}}\cdots d_{n}^{n_{n}}m_{1},\ldots,m_{n}=0$converges absolutely, at least on the closed polydisk $|zl|\leq rl,$$\ldots,$$|z_{n}|\leq r_{n}$.

For $f\in A(r_{1}, \ldots, r_{n})$ given by (1.1), let(1.2)$\Vert f\Vert=\Vert f\Vert_{A(r_{1},\ldots,r_{n})}=\sum^{\infty}|C_{m_{1},\ldots,m_{n}}m_{1},\ldots,m_{n}=0|r_{1}^{m_{1}}\cdots r_{n}^{m_{n}}$.

The functions on
defined by (1.1) and (1.2) makeinto a Banach
algebra with identity under pointwise multiplication of the functions involved [4].
Let be a Banach space and, the space of boundednonzero operators fromlinear operators acting on . Except for the numbers, which will serve as
weights, we ignore for the present the nature ofas operators and introduce a commutative Banach algebra consisting of“analytic functions”, wherearetreated as purely formal commuting objects.
Now consider the collectionof all expressions of the form
$A(r_{1}, \ldots,r_{n})$$X$$A(r_{1}, \ldots,r_{n})$$A_{1},$ $\ldots,A_{n}$$\mathcal{L}(X)$$X$$\Vert A_{1}\Vert,$$\Vert A_{n}\Vert$$\ldots,$$A_{1},$ $\ldots,A_{n}$$f(\tilde{A}_{1}, \ldots,\tilde{A}_{n})$$D(A_{1}, \ldots,A_{n})$

(1.3)

where

$f( \tilde{A}_{1}, \ldots,\tilde{A}_{n})=\sum^{\infty}c_{m_{1},\ldots,m_{n}}\tilde{A}_{1}^{m_{1}}\cdots\tilde{A}_{n}^{m_{n}}m_{1},\ldots,m_{n}=0$$c_{m_{1},\ldots,m_{n}}\in \mathbb{C}$for all $m1,$ $\ldots,m_{n}=0,1,$$\ldots$, and$\Vert f(\tilde{A}_{1}, \ldots,\tilde{A}_{n})\Vert=\Vert f(\tilde{A}_{1}, \ldots,\tilde{A}_{n})\Vert_{D(A_{1},\ldots A_{n})}$(1.4)$= \sum^{\infty}|c_{m_{1},\ldots,m_{n}}|\Vert A_{1}\Vert^{m_{1}}\cdots\Vert A_{n}\Vert^{m_{n}}<\infty m_{1},\ldots,m_{n}=0^{\cdot}$$\tilde{A}_{1},$$\ldots,\tilde{A}_{n}$146
The space $D(A_{1}, \ldots,A_{n})$ equipped with pointwise operations and the norm (1.4) is a commuta, thenfor $j=1,$
tive Banach algebra with identity. In fact, if we takethe two Banach algebras are the same except for a renaming of the indeterminants. We refer toof boundedas the disentangling algebra associated with the n-tuple
linear operators acting on .
continuous probability meaand
be nonzero operators fromLet
, the Borel class of $[0, T]$ (such measures are continuous prosures defined at least on
vided that each single point set has measure ). We wish to define the disentangling mapping$\Vert\tilde{A}_{j}\Vert=\Vert A_{j}\Vert=r_{j}$$D(A_{1}, \ldots,A_{n})$$\ldots,n$$(A_{1}, \ldots,A_{n})$$X$$\mathcal{L}(X)$$A_{1},$ $\ldots,A_{n}$$\mu_{1},$$\ldots,\mu_{n}$$\mathcal{B}[0, T]$

$0$

(1.5)

$\mathcal{T}_{\mu_{1},\ldots,\mu_{n}}:D(A_{1}, \ldots,A_{n})arrow \mathcal{L}(X)$. Putting it another way, given any
according to the mle determined by the measuresof, we wish to form the function
analytic function.
as directed by
not necessarily commuting operators, let
Given nonnegative integers$\mu_{1},$$\ldots,\mu_{n}$$f\in A(\Vert A_{1}\Vert, \ldots, \Vert A_{n}\Vert)$$f_{\mu_{1},\ldots,\mu_{n}}(A_{1}, \ldots,A_{n})$$A_{1},$ $\ldots,A_{n}$$\mu_{1},$$\ldots,\mu_{n}$$m_{1},$ $\ldots,m_{n}$(1.6)$P^{m_{1},\ldots,m_{n}}(z1, \ldots,z_{n})=z_{1}^{m_{1}}\cdots d_{n}^{n_{n}}$so that(1.7)$P^{m_{1},\ldots,m_{n}}(\tilde{A}_{1}, \ldots,\tilde{A}_{n})=\tilde{A}_{1}^{m_{1}}\cdots\tilde{A}_{n}^{m_{n}}$For each $m=0,1,$

given$\pi\in S_{m}$$\ldots$, let$S_{m}$.

denote the set of all permutations of the integers 1,$\{$$\ldots$

,

$m\}$

, and

, we set(1.8)$\Delta_{m}(\pi)=\{(sl, \ldots,S_{m})\in[0,T]^{m};0<s_{\pi(1)}<\cdots<s_{\pi(m)}<T\}$

We now define the mapping(1.9)$\mathcal{T}_{\mu_{1},\ldots,\mu_{n}}$. For $j=1,$$\ldots,n$.

and all $s\in[0,T]$ , let$A_{j}(s)=A_{j}$and for $i=1,$$\ldots,m$(1.10), we define$C_{i}(s):=\{\begin{array}{l}A_{1}(s) if i\in\{1, \ldots,ml\},A_{2}(s) if i\in\{m1+1, \ldots,m_{1}+m_{2}\},:A_{n}(s) if i\in\{ml+\cdots+m_{n-1}+1, \ldots,m\},\end{array}$for all $0\leq s\leq T$ .
in (1.9) is time independent. This implies that the integrands can be pulled
Note thatoutside of the integral in all of the cases involved. The measures effect the weights on thevarious terms as usual but, even more centrally for this theory, the measures effect the orderingof the operators (the time dependent setting has been developed elsewhere, for example, in [8],$A_{j}$[14], [15] .
$)$

147
Definition 1.1. For any nonnegative integers$m_{1},$ $\ldots,m_{n}$and $m=m_{1}+\cdots+m_{n}$ ,$\mathcal{T}_{\mu_{1},\ldots,\mu_{n}}P^{m_{1},\ldots,m_{n}}(\tilde{A}_{1}, \ldots,\tilde{A}_{n})$(1.11)$= \sum_{\pi\in S_{m}}\int_{\Delta_{m}(\pi)}C_{\pi(m)}(s_{\pi(m)})\cdots C_{\pi(1)}(s_{\pi(1)})(\mu_{1}^{m_{1}}\cross\cdots\cross\mu_{n}^{m_{n}})(ds\iota, \ldots,ds_{m})$Then, for$f(\tilde{A}_{1}, \ldots,\tilde{A}_{n})\in D(A_{1}, \ldots,A_{n})$(1.12).

given by$f( \tilde{A}_{1}, \ldots,\tilde{A}_{n})=\sum^{\infty}c_{m_{1},\ldots,m_{n}}\tilde{A}_{1}^{m_{1}}\cdots\tilde{A}_{n}^{m_{n}}m_{1},\ldots,m_{n}=0$’

we set(1.13)$\mathcal{T}_{\mu_{1},\ldots,\mu_{n}}f(\tilde{A}_{1}, \ldots,\tilde{A}_{n})=\sum^{\infty}c_{m_{1},\ldots,m_{n}}\mathcal{T}_{\mu_{1},\ldots,\mu_{n}}P^{m_{1},\ldots,m_{n}}(\tilde{A}_{1}, \ldots,\tilde{A}_{n})m_{1},\ldots,m_{n}=0^{\cdot}$

We will often use the altemative notation:(1.14)$P_{\mu_{1},,\mu_{n}}^{m_{1}.’.\cdot.\cdot\cdot,m_{n}}(A_{1}, \ldots,A_{n})=\mathcal{T}_{\mu l,\ldots,\mu_{n}}P^{m_{1},\ldots,m_{n}}(\tilde{A}_{1}, \ldots,\tilde{A}_{n})$and.

(1.15)$f_{\mu_{1},,\mu_{n}}(A_{1}, \ldots,A_{n})=\mathcal{T}_{\mu_{1},\ldots,\mu_{n}}f(\tilde{A}_{1}, \ldots,\tilde{A}_{n})$For elementary properties of the disentangling algebra $D(A_{1}, \ldots,A_{n})$ and the disentangling, see the introductory papers [4], [5], [6], [8].
map$\mathcal{T}_{\mu_{1},\ldots,\mu_{n}}$\S 2. Extraction of a Linear Factor
The operation of ’disentangling’ is the key to Feynman’s operational calculi for noncommuting operators. Hence, formulas which simplify this procedure under various conditions arecentral to the subject. The main results of this section make it possible to carry out the disentangling in an iterative manner. In this section we summarize the results in [9].
Let
, wherebelong tois a Banach space andbe continuous$A_{1},$ $\ldots,A_{n}$probability measures on$\mathcal{L}(X)$$\mathcal{B}[0, T]$$X$$\mu_{1},$$\ldots,\mu_{n}$.

Theorem 2.1. Suppose that the probability measures
are supported by $[a,b]\subset$$[0, T]$ and that the probability measures
are supported by $[0,a]\cup[b, T]$ . Let
be nonnegative integers. Let$\mu_{1},$$\ldots,\mu_{k}$$\mu_{k+1},$ $\ldots,\mu_{n}$

(2.1)

and

$\mu 0$(2.2)$m_{1},$ $\ldots,m_{n}$$K_{m_{1},\ldots,m_{k}}=P_{\mu l,,\mu_{k}}^{m_{1}.’.\cdot.\cdot\cdot,m_{k}}(A_{1}, \ldots,A_{k})$any continuous probability measure supported by $[a,b]$ . Then.

$P_{\mu_{1},,\mu_{n}}^{m_{1}.’.\cdot.\cdot\cdot,m_{n}}(A_{1}, \ldots,A_{n})=P_{\mu\circ,\mu_{k}+1\mu_{n}}^{1,m_{k+\cdots,m_{n}}}1,,\ldots,(K_{m_{1},\cdots,m_{k}},A_{k+1}, \ldots,A_{n})$148
Theorem 2.2 (Extraction of a Linear Factor).
andsume that

Let
Then

be given as in Theorem 2.1. As-$h(\tilde{A}_{k+1}, \ldots,\tilde{A}_{n})\in D(A_{k+1}, \ldots,A_{n})$$g(\tilde{A}_{1}, \ldots,\tilde{A}_{k})\in D(A_{1}, \ldots,A_{k})$(2.3)$Let\mu_{1},$ $\ldots,\mu_{n}$$f(z1, \ldots,z_{n})=g(z,.,z)h(z_{k+1}, \ldots,z_{n})$$K=\mathcal{T}_{\mu_{1},\ldots,\mu_{k}}g(\tilde{A}_{1}, \ldots,\tilde{A}_{k})$and$f(\tilde{A}_{1}, \ldots,\tilde{A}_{n})\in D(A_{1}, \ldots,A_{n})$

. Let

.

any continuous probability measure supported by $[a,b]$ .
and$\mu_{0}$

,

(2.4)

$\mathcal{T}_{\mu_{1},\ldots,\mu_{n}}f(\tilde{A}_{1}, \ldots,\tilde{A}_{n})=\mathcal{T}_{\mu_{0},\mu_{k+1},\ldots,\mu_{n}}F(\tilde{K},\tilde{A}_{k+1}, \ldots,\tilde{A}_{n})$where $F(zo,z_{k+1}, \ldots,z_{n})=zoh(z_{k+1}, \ldots,z_{n})$ .

Theorem 2.2 extends to Jefferies and Johnson‘s theory of Feynman’s operational calculi acomputational technique which is due to Maslov and which is used many times both in his book[12] and in the book by Nazaikinskii, Shatalov and Stemin [13]. The technique is referred toin [13] as “the extraction of a linear factor” and is used in conjunction with the “autonomousbracket” notation of Maslov.
The technique of extracting a linear factor is used in [12], [13] to establish equalities while
doing calculations with noncommuting operators. Of course, such calculations can be organized
so that the goal is to show that a related operator expression is equal to the zero operator. Our
first corollary below will make it easy to apply Theorem 2.2 in the manner which we have justdiscussed.
Corollary 2.3. Let the hypotheses of Theorem 2.2 be satisfied and suppose that

.

(2.5)

Then

$K=\mathcal{T}_{\mu_{1},\ldots,\mu_{k}}g(\tilde{A}_{1}, \ldots,\tilde{A}_{k})=O$$\mathcal{T}_{\mu_{1},\ldots,\mu_{n}}f(\tilde{A}_{1}, \ldots,\tilde{A}_{n})$also equals the zero opemtor

Next we give an example illustrating the use of Theorem 2.2 and Corollary 2.3. This exam.
ple depends on a special relationship between$A_{1},$ $\ldots,A_{k}$be given and suppose that $g(Z1,Z2)=z_{1}^{3}-$Example 2.4. Let bounded operators
are continuous. Further, assume thatandbeing subsets of $[a,b]$ and theprobability measures on $[0, T]$ with the supports of andbeing subsets of $[0,a]\cup[b, T]$ where $0<a<b<T$ . Finally, we takeandsupports ofin the$f(Z1,Z2,Z3,Z4)=g(Z1,Z2)h(Z3,Z4)$ and ask for the computation of$A_{2}=A_{1},$Hence,by Proposition 3.1
certainly commute.
andcase where $A_{2}=A_{1}$ . Since
agree with the usual commutative
acting onof [4], all of the functional calculi
functional calculus. Thus,$A_{1},$ $\ldots,A_{4}$$Z_{1}^{2}Z2+z\iota z_{2}^{2}-z_{2}^{3}$$h(Z3,Z4)\in A(\Vert A_{3}\Vert, \Vert A_{4}\Vert)$$\mu_{1},$$\mu_{1}$$\mu_{3}$$\ldots,\mu_{4}$$\mu_{2}$$\mu_{4}$$\mathcal{T}_{\mu_{1},\ldots,\mu_{4}}f(\tilde{A}_{1},\tilde{A}_{2},\tilde{A}_{3},\tilde{A}_{4})$$A_{1}$$A_{2}$$(\tilde{A}_{1},\tilde{A}_{2})$$\mathcal{T}_{v_{1},v_{2}}$$\mathcal{T}_{\mu_{1},\mu_{2}}g(\tilde{A}_{1},\tilde{A}_{2})=g(A_{1},A_{2})=A_{1}^{3}-A_{1}^{2}A_{2}+A_{1}A_{2}^{2}-A_{2}^{3}=O$

where the last equality comes ffom the fact that $A_{2}=A_{1}$ . It follows immediately from Corollaryequals the zero operator.
2.3 that$\mathcal{T}_{\mu_{1},\ldots,\mu_{4}}f(\tilde{A}_{1}, \ldots,\tilde{A}_{4})$149
Corollary 2.5. Suppose that the probability measures
are supported by $[0,a]$ andthat the probability measures
are supported by $[a, T]$ . Let $f,g$ and be given as in
Theorem 2.2. Then$\mu_{1},$ $\ldots,\mu_{k}$$h$$\mu_{k+1},$ $\ldots,\mu_{n}$

.

(2.6)

$\mathcal{T}_{\mu_{1},\ldots,\mu_{n}}f(\tilde{A}_{1}, \ldots,\tilde{A}_{n})=\mathcal{T}_{\mu_{k+1},\ldots,\mu_{n}}h(\tilde{A}_{k+1}, \ldots,\tilde{A}_{n})\mathcal{T}_{\mu_{1},\ldots,\mu_{k}}g(\tilde{A}_{1}, \ldots,\tilde{A}_{k})$Results in this section can be combined with various results in [4], [5], [6] to yield further
corollaries.

\S 3. Methods for Iterative Disentangling
The results of this section permit us, under appropriate assumptions, to disentangle in an
iterative manner. The main result ’extraction of a multilinear factor’, Theorem 3.3, is a multilinear version of Theorem 2.2. Theorem 3.7 also produces a linear term at each stage of theiteration but, unlike Theorem 3.3, the new linear term depends on the previous one. In thissection we summanze the results in [7].
Let be a positive integer. For each $j=1,$, let
be the nonempty subset of $I=$$\{1, \ldots,n\}$ such that$d$$\ldots,d$$I_{j}$$I_{j}=\{i_{j-1}+1, \ldots,i_{j}\}$where $i_{0}=0$ and let$I_{0}=I-(I_{1}\cup\cdots\cup I_{d})$.

Now we introduce the abbreviated notation.

Notation. We write$P_{\mu_{i},i\in I}^{n_{i},i\in l}(A_{i}, i\in I)=P_{\mu_{1},,\mu_{n}}^{m_{1}.’.\cdot.\cdot\cdot,m_{n}}(A_{1}, \ldots,A_{n})$,as well as,$f(\tilde{A}_{i}, i\in I)=f(\tilde{A}_{1}, \ldots,\tilde{A}_{n})$$f(z_{i},i\in I)=f(z_{1}, \ldots,z_{n})$,$D(A_{i},$ $i\in

D=D(A_{1}, \ldots,A_{n})$$\mathcal{T}_{\mu_{i},i\in I}=\mathcal{T}_{\mu_{1},\ldots,\mu_{n}}$,.

Theorems 3.3 and 3.7 below can be established via induction and Theorem 2.2. The twomonomial cases Theorems 3.1 and 3.5 could be obtained as special cases of Theorems 3.3 and3.7 respectively. However the monomial cases are easier to understand, and so we will statethem as separate theorems.
Theorem 3.1. Let $a_{j},b_{j},j=1,$(3.1)$\ldots,d$be real numbers such that$0\leq a_{1}<b_{1}\leq a2<b_{2}\leq\cdots\leq a_{d}<b_{d}\leq T$

.

150
. Let
have supports contained within
Suppose thatfor $j=1,$, be any continuous probability measures having supports contained within, letGiven nonnegative integers$\ldots,d$$[a_{j},b_{j}]$$\mu_{i},i\in I_{j}$$1,$ $\ldots,d$$v_{j},$$j=$$[a_{j},b_{j}]$.

$m_{1},$ $\ldots,m_{n}$(3.2)$K_{j}=P_{\mu_{i},i\in I_{j}}^{m_{i},i\in l_{j}}(A_{i},i\in I_{j})$for $j=1,$$\ldots,d$(3.3). Then$P_{\mu_{1},,\mu_{n}}^{m_{1}.’.\cdot.\cdot\cdot,m_{n}}(A_{1}, \ldots,A_{n})=P_{v_{1},\ldots,v_{d};\mu_{i},i\in I_{0}}^{1,\ldots,1;m_{i},i\in I_{0}}(K_{1}, \ldots,K_{d};A_{i}, i\in I_{0})$be given as in Theorem 3.1. Suppose
Corollary 3.2. Let,. For any nonnegative integers.

$I_{0}=\emptyset$$\mu_{1},$ $\ldots,\mu_{n}$$I_{1}\cup\cdots\cup I_{d}$

(3.4)

,

that is, $I=$

$m_{1},$ $\ldots,m_{n}$$P_{\mu_{1},,\mu_{n}}^{m_{1}.’.\cdot.\cdot\cdot,m_{n}}(A_{1}, \ldots,A_{n})=K_{d}\cdots K_{1}=P_{\mu_{i},i\in I_{d}}^{m_{i},i\in l_{d}}(A_{i},i\in I_{d})\cdots P_{\mu_{i},i\in l_{1}}^{m_{i},i\in l_{1}}(A_{i},i\in I_{1})$.

Now we come to the theorem which allows us to iteratively disentangle a multilinear factor
where $f(z1, \ldots,z_{n})$ is an appropriately factorable analytic function offromdefined below can be computed independently. Note that the linear factorsof one another; this will not be true of Theorem 3.7.
$\mathcal{T}_{\mu_{1},\ldots,\mu_{n}}f(\tilde{A}_{1}, \ldots,\tilde{A}_{n})$$z_{1},$$K_{1}’,$$\ldots,z_{n}$$\ldots,K_{d}’$andTheorem 3.3 (Extraction of a Multilinear Factor). Let
in Theorem 3.1. Assume thatfor $j=1,$. Let$\mu_{1},$$\ldots,\mu_{n}$$g_{j}(\tilde{A}_{i}, i\in I_{j})\in D(A_{i},i\in I_{j})$$v_{1},$$\ldots,d$$\ldots,$and$\nu_{d}$be given as$h(\tilde{A}_{i},i\in I_{0})\in$$D(A_{i},i\in I_{0})$(3.5)$f(z \iota, \ldots,z_{n})=[\prod_{j=1}^{d}g_{j}(z_{i},i\in I_{j})]h(z_{i},i\in I_{0})$

and
(3.6)

for $j=1,$$K_{j}’$$\ldots,d$(3.7). Then$:=\mathcal{T}_{\mu_{i},i\in I_{j}}g_{j}(\tilde{A}_{i},i\in I_{j})$$f(\tilde{A}_{1}, \ldots,\tilde{A}_{n})\in D(A_{1}, \ldots,A_{n})$and$\mathcal{T}_{\mu_{1},\ldots,\mu_{n}}f(\tilde{A}_{1}, \ldots,\tilde{A}_{n})=\mathcal{T}_{v_{1},\ldots,v_{d};\mu_{i},i\in l_{0}}F(\tilde{K}_{1}’, \ldots,\tilde{K}_{d}’;\tilde{A}_{i},i\in I_{0})$,where $F(w1, \ldots,w_{d};z_{i},i\in I_{0})=w1\ldots w_{d}h(z_{i},i\in I_{0})$.
Corollary 3.4. Let. Then$\mu_{1},$$\ldots,\mu_{n}$and$f(z_{1}, \ldots,z_{n})$be given as in Theorem 3.3. Suppose $I=$

$I_{1}\cup\cdots\cup I_{d}$

(3.8)

.

$\mathcal{T}_{\mu_{1},\ldots,\mu_{n}}f(\tilde{A}_{1}, \ldots,\tilde{A}_{n})=K_{d}’\cdots K_{1}’=\mathcal{T}_{\mu_{i},i\in I_{d}}g_{d}(\tilde{A}_{i},i\in I_{d})\cdots \mathcal{T}_{\mu_{i},i\in I_{1}}g_{1}(\tilde{A}_{i},i\in I_{1})$The results in the rest of this section are consequences of the main results of Section 2.
However, the supports of the clusters of measures beyond the first are in nonabutting pairs of151intervals arranged so that Theorem 2.2 is applicable and there is linearity at each stage of thedisentangling but not multilinearity of the processes as a whole. The difference in the results
is perhaps most clearly seen by comparing the multilinear expression
in formula (3.4)
of Corollary 3.2 with (3.13) of Corollary 3.6 below. Perhaps the most cogent point is that theoperatorsin Corollary 3.2 can be calculated independently of one another and in any
order whereas this is not tme ofin Corollary 3.6.
$K_{d}\cdots K_{1}$$K_{1},$ $\ldots,K_{d}$$L_{1},$

Theorem 3.5. Let $a_{j},b_{j},j=1,$(3.9)$\ldots,L_{d}$$\ldots,d$be real numbers such that$0\leq a_{d}\leq\cdots\leq a_{2}\leq a_{1}<b_{1}\leq b_{2}\leq\cdots\leq b_{d}\leq T$.

Suppose thathave supports contained within $[a_{j},a_{j-1}]\cup[b_{j-1},b_{j}]$ for $j=1,$
where $a_{0}=b_{1}$ and $b_{0}=a_{1}$ . Let, be any continuous probability measures having$j=1,$
suppons contained within. Given nonnegative integers, let$\mu;,i\in I_{j}$$\ldots,d$$\ldots,d$$\eta_{j},$$[a_{j},b_{j}]$(3.10)$m_{1},$ $\ldots,m_{n}$$L_{j}=P_{\eta_{j-1};\mu_{i},i\in I_{j}}^{1;m_{i},i\in I_{j}}(L_{j-1};A_{i},i\in I_{j})$for $j=1,$

where is the identity operator andhaving suppon contained within. Then$\ldots,d$$L_{0}$$\eta 0$is any continuous probability measure$[a_{1},b_{1}]$(3.11)$P_{\mu_{1},,\mu_{n}}^{m_{1}.’.\cdot.\cdot\cdot,m_{n}}(A_{1}, \ldots,A_{n})=P_{\eta_{d};\mu_{i},i\in I_{0}}^{1;m_{i},i\in I_{0}}(L_{d};A_{i},i\in I_{0})$Corollary 3.6. Let
and. For any nonnegative integers$\mu_{1},$$\ldots,\mu_{n}$$\cup I_{d}$(3.12)$\eta_{1},$$m\iota,$$\ldots,$$\eta_{d}$$\ldots,m_{n}$.

be given as in Theorem 3.5. Suppose $I=I_{1}\cup$,$P_{\mu_{1},,\mu_{n}}^{m_{1}.’.\cdot.\cdot\cdot,m_{n}}(A_{1}, \ldots,A_{n})=L_{d}=P_{\eta_{d}-1,\mu_{i},i\in I_{d}}^{1m_{i},i\in J_{d}}(L_{d-1};A_{i},i\in I_{d})$whereis given inductively by the formula (3.10). Equation (3.12) can be expressed more
explicitly by the formula$L_{d-1}$$P_{\mu_{1},,\mu_{n}}^{m_{1}.’.\cdot.\cdot\cdot,m_{n}}(A_{1}, \ldots,A_{n})=P_{\eta_{d-1};\mu_{i},i\in I_{d}}^{1;m_{i},i\in I_{d}}(P_{\eta_{d-2};\mu_{i},i\in I_{d-1}}^{1;m_{i},i\in J_{d-1}}(\cdots$(3.13)$(P_{\eta_{1};\mu_{i},i\in I_{2}}^{1;m_{i},i\in l_{2}}(P_{\mu_{i},i\in I_{1}}^{m_{i},i\in I_{1}}(A_{i},i\in I_{1});A_{i},i\in I_{2})$$);A_{i},i\in I_{d-1});A_{i},i\in I_{d})$;.

Now we come to one of the main result of this section.

Theorem 3.7. Let
andgiven as in Theorem 3.3. For each $j=1,$$\mu_{1},$(3.14)$\ldots,\mu_{n}$$\eta_{1},$be given as in Theorem 3.5 and $f(z1, \ldots,z_{n})$ be, let$\ldots,\eta_{d}$$\ldots,d$$F_{j-1}(w_{j-1};z_{i},i\in I_{j})=w_{j-1}g_{j}(z_{i},i\in I_{j})$and(3.15)$L_{j}’=\mathcal{T}_{\eta_{j-1};\mu_{i},i\in I_{j}}F_{j-1}(\tilde{L}_{j-1}’;\tilde{A}_{i}, i\in I_{j})$152
is the identity operator and. Thensupport contained withinwhere$w_{0}=\iota,$$h’$$\eta_{0}$is any continuous probability measure having$[a_{1},b_{1}]$(3.16)$\mathcal{T}_{\mu_{1},\ldots,\mu_{n}}f(\tilde{A}_{1}, \ldots,\tilde{A}_{n})=\mathcal{T}_{\eta_{d};\mu_{i},i\in 0}F(\tilde{L}_{d}’;\tilde{A}_{i},i\in I_{0})$where $F(w_{d};z_{i},i\in I_{0})=w_{d}h(z_{i},i\in I_{0})$ .
Corollary 3.8.
pose
. Then we have$\ t\mu_{1},$ $\ldots,\mu_{n},$$\eta_{1},$$\ldots,\eta_{d}$and $f(z1, \ldots,z_{n})$ be given as in Theorem 3.7. Sup-

$I=I_{1}\cup\cdots\cup I_{d}$(3.17)

where$\mathcal{T}_{\mu_{1},\ldots,\mu_{n}}f(\tilde{A}_{1}, \ldots,\tilde{A}_{n})=L_{d}’=\mathcal{T}_{\eta_{d-1};\mu,i\in l_{d}}F_{d-1}(\tilde{L}_{d-1}’;\tilde{A}_{i},i\in I_{d})$$F_{d-1}$

and$L_{d-1}’$

are given inductively by the formulas (3.14) and (3.15), respectively.

We use the linearity of the disentangling mapping again to extend Theorem 3.3 to finite
sums of functions of the form (3.5). In the language of tensor products, the extension is fromelementary tensors to algebraic tensors.

Corollary 3.9. Let theTheorem 3.3. Further, let(3.18)$a_{j}$’s and$b_{j}’ s,$$A_{1},$ $\ldots,A_{n},\mu_{1},$ $\ldots,\mu_{n}$and$v_{1},$$\ldots,v_{d}$be exactly as in$f(z1, \ldots,z_{n})=\sum_{l=1}^{N}f_{l}(z\iota, \ldots,z_{n})$where each$f_{l}$

has the $fom$(3.19)$f_{l}(z \iota, \ldots,z_{n})=\prod_{j=1}^{d}g_{jl}(z_{i},i\in I_{j})h_{l}(z_{i},i\in I_{0})$and satisfies the conditions of Theorem 3.3. For any $l\in\{1, \ldots,N\}$ , let(3.20)$K_{jl}’$for $j=1,$$\ldots,d$. Then$:=\mathcal{T}_{\mu_{i},i\in l_{j}}g_{jl}(\tilde{A}_{i},i\in I_{j})$$f(\tilde{A}_{1}, \ldots,\tilde{A}_{n})\in D(A_{1}, \ldots,A_{n})$and$\mathcal{T}_{\mu_{1},\ldots,\mu_{n}}f(\tilde{A}_{1}, \ldots,\tilde{A}_{n})=\sum_{l=1}^{N}\mathcal{T}_{v_{1},\ldots,v_{d};\mu_{i},i\in l_{0}}F_{l}(\tilde{K}_{1l}’, \ldots,\tilde{K}_{dl}’;\tilde{A}_{i},i\in I_{0})$(3.21)$= \sum_{l=1}^{N}\mathcal{T}_{v_{1},\ldots,v_{d};\mu_{i},i\in I_{0}}[\tilde{K}_{1l}’\cdots\tilde{K}_{dl}’h_{l}(\tilde{A}_{i},i\in I_{0})]$where$F_{l}(w\iota, \ldots,w_{d};z_{i},i\in I_{0})=w_{1}\cdots w_{d}h_{l}(z_{i},i\in I_{0}),$$l=1,$ $\ldots,N$

.

,

153
A variety of further corollaries could be written down using the results of this section either
by themselves or in conjunction with earlier results.

\S 4. The derivation formula and higher-order exapnsions
In this section we summanize the results in [10]. We explores the differential or derivational
calculus associated with the disentangled operators arising from Feynman‘s operational calculi
for noncommuting operators. We deal with a first order infinitesimal calculus for a function of
noncommuting operators. Here the first derivatives are replaced by the first order derivational
derivatives. The derivational derivatives of the first and higher order have been useful in, for example, operator algebras, noncommutative geometry and Maslov’s discrete form of Feynman ’s
operational calculus.
Let be a nonzero operator inand letbe an entire function. Then
$n$

$A$

$f(x)= \sum_{m=0}^{\infty}c_{m}x^{m}$$\mathcal{L}(X)$for any continuous probability measure on$\mu$$\mathcal{B}[0, T]$,$\mathcal{T}_{\mu}f(\tilde{A})=\sum_{m=0}^{\infty}c_{m}A^{m}$and we let$f(A)= \sum_{m=0}^{\infty}c_{m}A^{m}$.

In this section we are interested in the term of order in the expression$\epsilon$

(4.1)

where

$f(A+\epsilon B)=f(A)+\epsilon C_{1}+\epsilon^{2}C_{2}+\cdots+O(\epsilon^{n})$$B$is an operator in$\mathcal{L}(X)$(4.2). The coefficient of can be obtained as$\epsilon$$C_{1}= \frac{d}{d\epsilon}f(A+\epsilon B)|_{\epsilon=0}$.

We will find an expression for a more general “derivative”, of which this is a particular case.
First order derivations will play the central role here.
A derivation ofis an arbitrary linear mapping$\mathcal{L}(X)$$D:\mathcal{L}(X)arrow \mathcal{L}(X)$satisfying the Leibniz mle(4.3)$D(AB)=D(A)B+AD(B)$ ,$A,B\in \mathcal{L}(X)$

Derivations of the form
(4.4)

$D(A)=D_{B}(A)=BA-AB$ ,.

154
where is an arbitrary element of
Now we claim that
$B$

(4.5)

$\mathcal{L}(X)$, are called inner derivations of$D[f(A)]=\mathcal{T}_{\mu,\mu}f_{1}(\tilde{A},\overline{D(A}))$$\mathcal{L}(X)$.

,and $f_{1}(x,y)=f’(x)y$ . In fact we will
where is any continuous probability measure onshow a more general result in Theorems 4.1 and 4.3 below.
$\mathcal{B}[0, T]$$\mu$

Theorem 4.1. For each
, we have$f(\tilde{A}_{1}, \ldots,\tilde{A}_{n})\in D(A_{1}, \ldots,A_{n})$and for an inner derivation

$D$

of

$\mathcal{L}(X)$$D[\mathcal{T}_{\mu_{1},\ldots,\mu_{n}}f(\tilde{A}_{1}, \ldots,\tilde{A}_{n})]$

(4.6)

,

$= \sum_{j=1}^{n}\mathcal{T}_{\mu_{1},\ldots,\mu_{j},\mu_{j},\mu_{j+}l,\ldots,\mu_{n}}F_{j}(\tilde{A}_{1}, \ldots,\tilde{A}_{j},\overline{D(A_{j})},\tilde{A}_{j+1}, \ldots,\tilde{A}_{n})$

where
(4.7)

$F_{j}(x_{1}, \ldots,x_{j},y,x_{j+1}, \ldots,x_{n})=\frac{\partial}{\partial x_{j}}f(x_{1}, \ldots,x_{n})y$Lemma 4.2. Let$D$

be a derivation of. Define$\mathcal{D};\mathcal{L}(\mathcal{L}(X))arrow \mathcal{L}(\mathcal{L}(X))$by$D(T)=[D, T]=DT-TD$ .

(4.8)

Then

$\mathcal{L}(X)$.

$\mathcal{D}$is an inner derivation of$\mathcal{L}(\mathcal{L}(X))$(4.9). Furthermore, for any$D(L_{A})=L_{D(A)}$$A\in \mathcal{L}(X)$,.

Theorem 4.3. Fomula (4.6) is valid for an arbitmry derivation of$\mathcal{L}(X)$.

Remark We have considered disentangling maps and derivations associated with the al, whereis a Banach space. In fact there is no change in the definitions of the
gebra
by an arbitrary algebra . Moreover
disentangling map and the derivation if we replace
Theorem 4.1 through Theorem 4.3 remain tme in this new setting.
$\mathcal{L}(X)$$X$$\mathcal{A}$$\mathcal{L}(X)$whose elements are (infinitely differentiable) families of eleConsider the algebra
depending on a numerical parameter . Clearly, the mapping
ments of$\mathcal{L}(X)_{\{t\}}$$t$$\mathcal{L}(X)$(4.10)

taking each family$f(x)= \sum_{m=0}^{\infty}c_{m}x^{m}$(4.11)$\frac{d}{dt}$:$\mathcal{L}(X)_{\{t\}}arrow \mathcal{L}(X)_{\{t\}}$. Let
into its t-derivative is a derivation of the algebrabe an entire function. Then by Theorem 4.3 and its remark, we obtain$\mathcal{L}(X)_{\{t\}}$$A_{t}\in \mathcal{L}(X)_{\{t\}}$$\frac{d}{dt}f(A_{t})=\mathcal{T}_{\mu,\mu}f_{1}(\tilde{A}_{t},\overline{\frac{dA_{t}}{dt}})$,

155
where is any continuous probability measure on
and $f_{1}(x,y)=f’(x)y$ .
From (4.11) we easily derive a formula for the coefficient (4.2). Namely, takethenand$\mathcal{B}[0, T]$$\mu$$A_{\epsilon}=A+\epsilon B$$\frac{dA}{d\epsilon}\subseteq=B$(4.12)$C_{1}=\mathcal{T}_{\mu,\mu}f_{1}(\tilde{A},\tilde{B})$

,

i.e.,
(4.13)$f(A+\epsilon B)=f(A)+\epsilon \mathcal{T}_{\mu,\mu}f_{1}(\tilde{A},\tilde{B})+O(\epsilon^{2})$.

We will develop here some special cases of higher order expansions.

Theorem 4.4. Let(4.14)

where$f(x)= \sum_{m=0}^{\infty}c_{m}x^{m}$be an entirefunction. For each $n=1,2,$$\frac{d^{n}}{dt^{n}}f(A+tB)=\mathcal{T}_{\mu,\mu}f_{n}(\overline{A+tB},\tilde{B})$$\mu$

is any continuous probability measure on$\mathcal{B}[0, T]$$\ldots$,,

and $f_{n}(x,y)=f^{(n)}(x)y^{n}$ .

Thus we can write down the usual Maclaurin expansion.

(4.15)$f(A+ \epsilon B)=f(A)+\sum_{n=1}^{N}\frac{\epsilon^{n}}{n!}\mathcal{T}_{\mu,\mu}f_{n}(\tilde{A},\tilde{B})+O(\epsilon^{N+1})$

Our next theorem will give an expansion of arbitrarily high order accompanied by an explicit
remainder term.

Theorem 4.5. (Newton’s formula with remainder) Let $A,B$ be nonzero opemtors inand let. Let be a continuous probability measure on. For any positive integer$N$, wehave$\mathcal{L}(X)$$\epsilon>0$(4.16)$\mathcal{B}[0, T]$$\mu$$f(A+ \epsilon B)-f(A)=\sum_{n=1}^{N}\frac{\epsilon^{n}}{n!}\mathcal{T}_{\mu,\mu}f_{n}(\tilde{A},\tilde{B})+Q_{N}$,where $f_{n}(x,y)=f^{(n)}(x)y^{n}$ for $n=1,2,$ $\ldots,N$ and QN is given by the formula(4.17)$Q_{N}= \frac{\epsilon^{N+1}}{(N+1)!}\mathcal{T}_{\mu,\mu}g_{N}(\tilde{A},\tilde{B})$where $g_{N}(x,y)=f^{(N+1)}(x+\epsilon_{1}y)y^{N+1}$ for some

$0<\epsilon_{1}<\epsilon$,.

Theorem 4.6. (Taylor’s formula with remainder) Let $A,C$ be nonzero opemtors inand letbe continuous probability measures on. For any positive integer $N$, wehave$\mathcal{L}(X)$$\mu_{1},\mu_{2}$(4.18)$\mathcal{B}[0, T]$$f(C)-f(A)= \sum_{n=1}^{N}\frac{1}{n!}\mathcal{T}_{\mu_{1},\mu_{2}}h_{n}(\tilde{A},\tilde{C})+R_{N}$

,

,

156
where $h_{n}(x,y)=f^{(n)}(x)(y-x)^{n}$ for $n=1,2,$ $\ldots,N$ and(4.19)$R_{N}$

is given by the formula$R_{N}= \frac{1}{(N+1)!}\mathcal{T}_{\mu_{1},\mu_{2}}k_{N}(\tilde{A},\tilde{O}$,where $k_{N}(x,y)=f^{(N+1)}(x+t(y-x))(y-x)^{N+1}$ for some $0<t<1$ .

Remark Unlike Newton’s formula with remainder in (4.16), Taylor’s formula with remainder in (4.18) does not generally provide an expression in power of if $C=A+\epsilon B$ .
$\epsilon$\S 5. Measure permutation formulas, whereis
In this section we summarize the results in [2]. Let and belong toa Banach space. Let and be continuous probability measures and let $f(x,y)=x^{m}y^{m}$ for
nonnegative integer . Then, as one can expect, we have$A$$\mu 1$$B$$\mathcal{L}(X)$$X$$\mu_{2}$

$m$

(5.1)

$\mathcal{T}_{\mu_{1},\mu_{2}}f(\tilde{A},\tilde{B})=\mathcal{T}_{\mu_{2},\mu_{1}}f(\tilde{B},\tilde{A})$,that is, the measures and operators can be exchanged simultaneously for $f(x,y)=x^{m}/n$ . Proposition 2.11 of [4] is a generalization of the above identity. But we can not expect that
(5.2)$\mathcal{T}_{\mu_{1},\mu_{2}}f(\tilde{A},\tilde{B})=\mathcal{T}_{\mu_{2},\mu_{1}}f(\tilde{A},\tilde{B})$in general. To understand this, let us consider a simple example. Let $f(x,y)=xy$ . Then$\mathcal{T}_{\mu_{1},\mu 2}f(\tilde{A},\tilde{B})=AB(p_{1}\cross\mu_{2})[s2<s_{1}]+BA(\mu_{l}\cross\mu_{2})[s1<s2]$and$\mathcal{T}_{\mu_{2},\mu_{1}}f(\tilde{A},\tilde{B})=AB(\mu\cross)[s<s_{1}]+BA(/12\cross\mu_{1})[s_{1}<s_{2}]$where, for example,$(\mu_{1}\cross\mu_{2})[s_{2}<s_{1}]$denotes$(\mu_{1}\cross\mu_{2})\{(s_{1},s_{2}):s_{2}<s_{1}\}$,. Hence the equalityin (5.2) can not be tme unless$(\mu_{1}\cross\mu_{2})[s2<s_{1}]=(\mu_{2}\cross\mu_{1})[s_{2}<s_{1}]$.

Moreover, the conditions to ensure (5.2) are much more complicate if, for example, $f(x,y)=$or $f(x,y)=\mathscr{H}+x^{2}y^{2}$ , etc.
In this section we are interested in the difference between the two operators. Our first lemma below is easy to obtain but it is useful to simplify the proof
andof the Measure permutation formula I in Theorem 5.2.

$y^{3}$$\mathcal{T}_{\mu_{1},\mu_{2}}f(\tilde{A},\tilde{B})$$\mathcal{T}_{\mu_{2},\mu_{1}}f(\tilde{A},\tilde{B})$

Lemma 5.1. Let$m_{1}$

and $m2$ be positive integers and let be a continuous probability mea$\mu$

sure on $B[0,T]$ . Let(5.3)$K_{m_{1}}=m{}_{1}P_{\mu,\mu}^{n_{1}-1,1}(A, [A,B])$.

157
Then the identity(5.4)$A^{m_{1}}B^{m_{2}}-B^{m_{2}}A^{m_{1}}=B^{m_{2}-1}K_{m_{1}}+B^{m_{2}-2}K_{m_{1}}B+\cdots+K_{m_{1}}B^{m_{2}-1}$

holds
Now we are ready to prove measure permutation formulas, the main results in this section.

Theorem 5.2. (Measure permutation formula I) Let andmeasures on. Assume that is supported by $[0,a]$ andsome $a\in[0, T]$ . Then for any positive integers and $m2$ ,$\mu\iota$$\mathcal{B}[0, T]$$\mu_{1}$$\mu_{2}$$\mu_{2}$be continuous pmbability
is supported by $[a, T]$ for$m\iota$(5.5)$P_{\mu_{2},\mu l}^{m_{1},m_{2}}(A,B)=P_{\mu_{1},\mu_{2}}^{m_{1},m_{2}}(A,B)+m2P_{\mu,\mu}^{m_{2}-1,1}(B,K_{m_{1}})$,is given by (5.3) and is any continuous pmbability measure on
assume thatand letwhere$K_{m_{1}}$$\mu$$\mathcal{B}[0,T]$. Further$f(\tilde{A},\tilde{B})\in D(A,B)$(5.6)$f(x,y)= \sum_{m_{1},m_{2}=0}^{\infty}c_{m_{1},m_{2}}x^{m_{1}}y^{m_{2}}$,then we have
(5.7)$\mathcal{T}_{\mu_{2},\mu_{1}}f(\tilde{A},\tilde{B})=\mathcal{T}_{\mu_{1},\mu_{2}}f(\tilde{A},\tilde{B})+\sum_{m_{1},m_{2}=1}^{\infty}c_{m_{1},m_{2}}m_{2}P_{\mu,\mu}^{m_{2}-1,1}(B,K_{m_{1}})$Let$\frac{\delta f}{\delta y}(x;y,y_{1})$denote the difference derivative of $f(x,y)$ with respect to
,

(5.8)

$\frac{\delta f}{\delta y}(x;y,y_{1})=\{$

if

$y$

.

, that is,$y\neq y_{1}$$\frac{f(x,y)-f(x,y_{1})}{\mathcal{T}_{\mathcal{Y}}^{f_{(x,y)}}\partial y-y_{1}}$

if $y=y1$ .

The infinite series of operators on the right hand side of (5.7) can be expressed as a disentangling of a function which is related with $f$ . We present it as Measure permutation formula II in
the following theorem. The technique of ’extraction of a linear factor’ [7, 9] plays an importantrole in the next theorem.

Theorem 5.3. (Measure permutation formula II) Under the assumptions of Theorem 5.2
one has
(5.9)

where

$\mathcal{T}_{\mu_{2},\mu_{1}}f(\tilde{A},\tilde{B})=\mathcal{T}_{\mu_{1},\mu_{2}}f(\tilde{A},\tilde{B})+\mathcal{T}v_{2},v_{2},v_{2},v_{1},v_{3^{\wedge}}[A,B]\frac{\delta^{2}f}{\delta x\delta y}(\tilde{A},\tilde{A};\tilde{B},\tilde{B})$and$[b_{2},T]$for any$v_{1},$$\gamma_{2}$$v3$$b_{1},b_{2}$andare continuous probability measures supported on $[0,b_{1}],$
with $0<b_{1}<b_{2}<T$ , respectively. Moreover (5.9) can be simplified$[b_{1},b_{2}]$$as$(5.10)$\mathcal{T}_{\mu_{2},\mu_{1}}f(\tilde{A},\tilde{B})=\mathcal{T}_{\mu_{1},\mu_{2}}f(\tilde{A},\tilde{B})+\mathcal{T}_{v_{2},v_{2},v_{1},v_{3}}[A,B]\frac{\delta f_{x}}{\delta y}(\tilde{A};\tilde{B},\tilde{B})\sim$.

158
Now we give some corollaries and an example of Measure permutation formulas I and II.
For $f(\tilde{A})\in D(A)$ , let us denote$f(A)=f_{\mu}(A)=\mathcal{T}_{\mu}f(\tilde{A})$. Our first corollary, called commutation
for any continuous probability measure on
formula, gives a relationship between $f(A)B$ and $Bf(A)$ . It is an easy consequence of Theorems$\mu$$\mathcal{B}[0,T]$

5.2 or 5.3.
Corollary 5.4. (Commutation formula I) Let $f(\tilde{A})\in D(A)$ and let(5.11)$f(x)= \sum_{m=0}^{\infty}c_{m}x^{m}$.

Then we have
(5.12)

where$f(A)B=Bf(A)+\mathcal{T}_{\mu,\mu}f_{1}(\tilde{A},[A,B])\sim$$\mu$

is any continuous probability measure on$\mathcal{B}[0,T]$,

and $f_{1}(x,z)=f’(x)z$.

The following is a generalized commutation formula. If we take $g(y)=y$ in Corollary 5.5,
then Corollary 5.4 is immediate from CorollaIy 5.5.

Corollary 5.5. (Commutation formula II) Let $f(\tilde{A})\in D(A)$ and $g(\tilde{B})\in D(B)$ . Then wehave,(5.13)

where$f(A)g(B)=g(B)f(A)+\mathcal{T}_{v_{2},v_{2},v_{1},v_{3}}(f\otimes g)_{1}(\tilde{A},[A,B];\tilde{B},\tilde{B})\sim$$\nu_{1},v2,v_{3}$are the measures as in Theorem 5.3 and(5.14)$(f \otimes g)_{1}(x,z;y,y_{1})=f’(x)z\frac{\delta g}{\delta y}(y,y_{1})$.

Derivation formula (4.6) in Section 4 plays a key role in a differential calculus for Feynman‘soperational calculi. (For details, see Sections 3 and 4 of [10].) In the following corollary weobtain the derivation formula (4.6) as a corollary of the commutation formula (5.13) in Corollary
5.5.
Corollary 5.6. Let(5.15)

where$D$

be a derivation of$\mathcal{L}(X)$and let $f(\tilde{A})\in D(A)$ . Then we have$D[f(A)]=\mathcal{T}_{\mu,\mu}f_{1}(\tilde{A},\overline{D(A}))$$\mu$

is any continuous probability measure on$\mathcal{B}[0, T]$,

and $f_{1}(x,z)=f’(x)z$.

In our next example we illustrate the measure permutation formula for a function of thePauli matrices.

159
Example 5.7. Let’s consider the simplest non-commuting example of the Pauli matrices$\sigma_{1}=(\begin{array}{l}0110\end{array})$Let$f(\tilde{\sigma}_{1},\tilde{\sigma}_{2})\in D(\sigma_{1},\sigma_{2})$and let,$\sigma_{2}=(\begin{array}{l}0-ii0\end{array})$,$\sigma_{3}=(\begin{array}{ll}1 00-1 \end{array})$.

be the measures given as in Theorem 5.2. Now let us
in the form
so that the matrix
acts first$\mu_{1},\mu_{2}$rewrite the operator
in every terms of the expression. Since$\mathcal{T}_{\mu_{2},\mu_{1}}f(\tilde{\sigma}_{1},\tilde{\sigma}_{2})$$\mathcal{T}_{\mu l,\mu_{2}}f(\tilde{\sigma}_{1},\tilde{\sigma}_{2})$$[\sigma_{1},\sigma_{2}]=2i\sigma_{3}$$\sigma_{1}$,$\mathcal{T}_{\mu_{2},\mu 1}f(\tilde{\sigma}_{1},\tilde{\sigma}_{2})=\mathcal{T}_{\mu_{1},\mu_{2}}f(\tilde{\sigma}_{1},\tilde{\sigma}_{2})+2i\mathcal{T}_{v_{2},v_{2},v_{2},\nu_{1},v_{3}}\tilde{\sigma}_{3}\frac{\delta^{2}f}{\delta x\delta y}(\tilde{\sigma}_{1},\tilde{\sigma}_{1};\tilde{\sigma}_{2},\tilde{\sigma}_{2})$

,

where
are the measures given as in Theorem 5.3. Now let us investigate the second
term on the right hand side of the above equation. Assume that$v_{1,2}v,v_{3}$$f(x,y)= \sum_{m_{1},m_{2}=0}^{\infty}c_{m_{1},m_{2}}x^{m_{1}}y^{m_{2}}$.

Then we have$\mathcal{T}_{v_{2},v_{2},v_{2},v_{1},v_{3}}\tilde{\sigma}_{3}\frac{\delta^{2}f}{\delta x\delta y}(\tilde{\sigma}_{1},\tilde{\sigma}_{1};\tilde{\sigma}_{2},\tilde{\sigma}_{2})$$= \sum_{m_{1},m_{2}=1}^{\infty}c_{m_{1},m_{2}}m_{1}(\sum_{l=0}^{m_{2}-1}P_{v_{2},v_{2},v_{1},v_{3}}^{1,m_{1}-1,l,m_{2}-l-1}(\tilde{\sigma}_{3},\tilde{\sigma}_{1};\tilde{\sigma}_{2},\tilde{\sigma}_{2}))$.

We can extract a linear factor to obtain,$P_{v_{2},\nu_{2},v_{1},v_{3}}^{1,m_{1}-1,l,m_{2}-l-1}(\tilde{\sigma}_{3},\tilde{\sigma}_{1};\tilde{\sigma}_{2},\tilde{\sigma}_{2})=P_{v_{2},v_{1},v_{3}}^{1,l,m_{2}-l-1}(\tilde{H}_{m_{1}};\tilde{\sigma}_{2},\tilde{\sigma}_{2})$where$H_{m_{1}}=P_{vv_{2}}^{1,m_{1}-1}(\tilde{\sigma}_{3},\tilde{\sigma}_{1})2,\cdot$

Note that the Pauli matrices$\sigma_{1}$$\sigma_{1}^{2}=id$,and$\sigma_{3}$satisfy$\sigma_{1}\sigma_{3}+\sigma_{3}\sigma_{1}=O$,$\sigma_{3}+\sigma_{1}\sigma_{3}\sigma_{1}=O$.

Hence if $m1$ is odd, then$H_{m_{1}}= \frac{1}{m_{1}}(\sigma_{1}^{m_{1}-1}\sigma_{3}+\sigma_{1}^{m_{1}-2}\sigma_{3}\sigma_{1}+\cdots+\sigma_{1}\sigma_{3}\sigma_{1}^{m_{1}-2}+\sigma_{3}\sigma_{1}^{m_{1}-1})=\frac{1}{m_{1}}\sigma_{3}$.

Similarly, if $m1$ is even, then we know that $H_{m_{1}}=O$ . Hence$m{}_{1}P_{v_{2},v_{2},v_{1},v_{3}}^{1,m_{1}-1,l,m_{2}-l-1}(\tilde{\sigma}_{3},\tilde{\sigma}_{1};\tilde{\sigma}_{2},\tilde{\sigma}_{2})=\{\begin{array}{ll}P_{v_{2},v_{1^{\mathcal{V}}3}}^{1,l,m_{2}-l-1}(\tilde{\sigma}_{3};\tilde{\sigma}2,\tilde{\sigma}_{2}), if m_{1} is oddO, if m1 is even\end{array}$160
and so$\mathcal{T}_{v_{2)}v_{2},v_{2},v_{1},v_{3}}\tilde{\sigma}_{3}\frac{\delta^{2}f}{\delta x\delta y}(\tilde{\sigma}_{1},\tilde{\sigma}_{1};\tilde{\sigma}_{2},\tilde{\sigma}_{2})$$= \sum_{m_{1},m_{2}=1;m_{1}=odd}^{\infty}c_{m_{1},m_{2}}(\sum_{l=0}^{m_{2}-1}P_{v_{2},\nu_{1},v_{3}}^{1,l,m_{2}-l-1}(\tilde{\sigma}_{3};\tilde{\sigma}_{2}, ei 2))$$= \mathcal{T}_{v_{2},\nu_{1},\nu_{3}}\tilde{\sigma}_{3}\frac{\delta^{2}f}{\delta x\delta y}(1, -1;\tilde{\sigma}2,\tilde{\sigma}2)$.

Finally, we have.

$\mathcal{T}_{\mu_{2},\mu_{1}}f(\tilde{\sigma}_{1},\tilde{\sigma}_{2})=\mathcal{T}_{\mu_{1},\mu_{2}}f(\tilde{\sigma}_{1},\tilde{\sigma}_{2})+2i\mathcal{T}_{v_{2},v_{1},v_{3}}\tilde{\sigma}_{3}\frac{\delta^{2}f}{\delta x\delta y}(1, -1;\tilde{\sigma}_{2},\tilde{\sigma}_{2})$\S 6. Generalized measure permutation formulas
In this section we summarize the results in [1]. Let and be bounded linear operators
acting on , where is a Banach space. Let and be continuous probability measures onand. In Section 5, we introduced relationships between the two operators, and called the relationships as measure permutation formulas. But in Section 5
we considered the case where the numbers of operators and measures involved are exactly two.
In this section we generalize the results for more than three operators and measures.
The technique of ‘extraction of a linear factor’ [7, 9] plays an important role not only in
Theorem 6.2 below but also in the rest of this paper. Let’s start with a simple lemma.
$A$

$X$

$X$

$\mu_{1}$$B$$\mu_{2}$$\mathcal{B}[0,T]$$\mathcal{T}_{\mu_{1},\mu_{2}}f(\tilde{A},\tilde{B})$$\mathcal{T}_{\mu_{2},\mu_{1}}f(\tilde{A},\tilde{B})$

Lemma 6.1. Let and be continuous probability measures ontive integer , we have
$\mu$

$v$

$\mathcal{B}[0, T]$. For any nonnega-

$n$

(6.1)

$P_{\mu,v}^{1,n}(A+B,C)=P_{\mu,v}^{1,n}(A,C)+P_{\mu,\nu}^{1,n}(B,C)$.

Of course this lemma is an easy consequence as anyone can expect. But (6.1) may not be
tme if the power associated with the operator $A+B$ is greater than 1. Here is a simple example,
where we take the operator to be the identity operator and to be 1. Now it is obvious to
see that
$I$

$C$

$n$

$P_{\mu,v}^{m,1}(A+B,D=(A+B)^{m}$and$P_{\mu,v}^{m,1}(A,D+P_{\mu,v}^{m,1}(B,D=A^{m}+B^{m}$and these two operators may not be equal unless $m=1$ .

,161Theorem 6.2. (Generalized measure permutation formula I) Let
be real numbers such that $0\leq a_{1}<a_{2}<a_{3}\leq T$. Supposeand be continuous probability measuresonhaving supports contained inand $[a_{2},a3]$ , respectively. Let be any continuous pmbability measure having support contained in $[0,a_{1}]\cup[a_{3}, T]$ . Then for any positive
integers
and nonnegative integer , we have$a_{1},a_{2},a3$$that\mu_{1}$$\mathcal{B}[0, T]$$\mu_{2}$$[a_{1},a_{2}]$$m_{1},m_{2}$(6.2)$\lambda$$n$,$P_{\mu_{2},\mu_{1},\lambda}^{m_{1},m_{2},n}(A,B,C)=P_{\mu_{1},\mu_{2},\lambda}^{m_{1},m_{2},n}(A,B,C)+m_{2}P_{\mu,\mu,\lambda}^{m_{2}-1,1,n}(B,K_{m_{1}},C)$

where is any continuous pmbability measure having supports contained in $[a_{1},a3]$ andis given as in (5.3). Further assume thatand let$\mu$$K_{m_{1}}$$f(\tilde{A},\tilde{B},\tilde{C})\in D(A,B,C)$(6.3)$f(x,y,z)= \sum_{m_{1},m_{2},n=0}^{\infty}c_{m_{1},m_{2},n}x^{m_{1}}y^{m_{2}}z^{n}$.

Then we have$\mathcal{T}_{\mu 2,\mu_{1},\lambda}f(\tilde{A},\tilde{B},\tilde{C})$

(6.4)

.

$= \mathcal{T}_{\mu l,\mu_{2},\lambda}f(\tilde{A},\tilde{B},\tilde{C})+\sum_{m_{1},m_{2}=1;n=0}^{\infty}c_{m_{1},m_{2},n}m{}_{2}P_{\mu,\mu,\lambda}^{m_{2}-1,1,n}(B,K_{m_{1}},C)$

By introducing additional operatorsand measureslowing corollary. It is obtained simply by extracting a linear factor.
$C_{1},$ $\ldots,C_{k}$$1_{1},$$\ldots,\lambda_{k}$we obtain the fol-

Corollary 6.3. Letandbe given as in Theorem 6.2. Let, be any
continuous probability measures having suppons contained in $[0,a_{1}]\cup[a_{3}, T]$ . For any positive
integers $m1,m_{2}$ and nonnegative integers, we have$a_{1},a_{2},a_{3},\mu l$$l_{1},$$\mu_{2}$$n_{1},$$\ldots$ノ$k$$\ldots,n_{k}$$P_{\mu_{2},\mu_{1},\lambda_{1},,\lambda_{k}}^{m_{1},m_{2},n_{1}.’.\cdot.\cdot,n_{k}}(A,B,C_{1}, \ldots,C_{k})$(6.5)$=P_{\mu_{1},\mu_{2},\Lambda_{1,.\prime}}^{m_{1},m_{2},n_{1}\prime}.\cdot.\cdot,\cdot|_{k}^{n_{k}}(A,B,C_{1}, \ldots,C_{k})+m_{2}P_{\mu,\mu,\lambda_{1^{j}},.,t_{k}}^{m_{2}-1,1..’

where andand let$\mu$$K_{m_{1}}$(6.6)

are given as in (5.3). Further assume thatn_{1},\ldots,n_{k}}(B,K_{m_{1}},C_{1}, \ldots,C_{k})$,$f(\tilde{A},\tilde{B},\tilde{C}_{1}, \ldots,\tilde{C}_{k})\in D(A,B,C_{1}, \ldots,C_{k})$$f(x,y_{Z1}, \ldots,z_{k})=\sum_{m_{1},m_{2},n_{1},.n_{k}=0}^{\infty}..,c_{m_{1},m_{2},n_{1},\ldots,n_{k}}x^{m_{1}}y^{m_{2}}z_{1}^{n_{1}}\cdots z_{k}^{n_{k}}$.

Then we have$\mathcal{T}_{\mu_{2},\mu i}\iota,’\iota_{1},\ldots,\iota_{k}f(\tilde{A},\tilde{B},\tilde{C}_{1}, \ldots,\tilde{C}_{k})$$=\mathcal{T}_{\mu_{1},\mu_{2},\lambda_{1},\ldots,\lambda_{k}}f(\tilde{A},\tilde{B},\tilde{C}_{1}, \ldots,\tilde{C}_{k})$(6.7)$+ \sum_{m_{1},m_{2}=1;n_{1},\ldots,n_{k}=0}^{\infty}c_{m_{1},m_{2},n_{1},\ldots,n_{k}}m{}_{2}P_{\mu,\mu,\lambda_{1},.,\lambda_{k}}^{m_{2}-1,1..’n_{1},\ldots,n_{k}}(B,K_{m_{1}},C_{1}, \ldots,C_{k})$.

162
Measure permutation formula II in Theorem 5.3 can also be generalized as follows. That is,
the infinite series of operators on the right hand side of (6.4) can be expressed as a second orderdifference derivative of $f$ .

Theorem 6.4. (Generalized measure permutation formula II) Under the assumptions ofTheorem 6.2 one has
(6.8)

where

$\mathcal{T}_{\mu_{2},\mu_{1},,t}f(\tilde{A},\tilde{B},\tilde{C})=\mathcal{T}_{\mu_{1},\mu_{2},\lambda}f(\tilde{A},\tilde{B},\tilde{\fbox{}}+\mathcal{T}_{\nu_{2},v_{2},v_{2},v_{1},\nu_{3},1^{\sim}}[A,B]\frac{\delta^{2}f}{\delta x\delta y}(\tilde{A},\tilde{A};\tilde{B},\tilde{B};\tilde{C})$$v_{1},$$[b_{2},a_{3}]$

$v2$

and

for any

(6.9)

$v3$

$b_{1},b_{2}$are continuous probability measures supported onwith$a_{1}<b_{1}<b_{2}<a_{3}$ ,$[a_{1},b_{1}],$,$[b_{1},b_{2}]$andrespectively. Moreover (6.8) can be simplified as$\mathcal{T}_{\mu_{2},\mu_{1},\lambda}f(\tilde{A},\tilde{B},\tilde{C})=\mathcal{T}_{\mu_{1},\mu_{2},\lambda}f(\tilde{A},\tilde{B},\tilde{C})+\mathcal{T}v_{2},v_{2},v_{1},\nu_{3},\lambda^{\wedge}[A,B]\frac{\delta f_{x}}{\delta y}(\tilde{A};\tilde{B},\tilde{B};\tilde{C})$,

Corollary 6.5. Under the assumptions of Corollary 6.3 one has$\mathcal{T}_{\mu_{2},\mu_{1},\lambda_{1},\ldots,\lambda_{k}}f(\tilde{A},\tilde{B},\tilde{C}_{1}, \ldots,\tilde{C}_{k})$$=\mathcal{T}_{\mu_{1},\mu_{2},\lambda_{I},\ldots,\lambda_{k}}f(\tilde{A},\tilde{B},\tilde{C}_{1}, \ldots,\tilde{C}_{k})$$+ \mathcal{T}_{\nu}2^{\nu_{2},\nu_{2},v_{1},\nu_{3^{J}},!_{1,\ldots,\prime}\iota_{k}^{\sim}}[A,B]\frac{\delta^{2}f}{\delta x\delta y}(\tilde{A},\tilde{A};\tilde{B},\tilde{B};\tilde{C}_{1}, \ldots,\tilde{C}_{k})$(6.10)$=\mathcal{T}_{\mu_{1},\mu_{2},\lambda_{1},\ldots,\lambda_{k}}f(\tilde{A},\tilde{B},\tilde{C}_{1}, \ldots,\tilde{C}_{k})$,$+ \mathcal{T}_{v_{2},\nu_{2},\nu_{1},v_{3’},1_{1},\ldots,\lambda_{k}}[A,B]\frac{\delta f_{x}}{\delta y}(\tilde{A};\tilde{B},\tilde{B};\tilde{C}_{1}, \ldots,\tilde{C}_{k})\sim$where$v_{1},$ $v_{2}$

and

$v3$

are the measures given as in Theorem 6.4.

In Theorem 6.2 through Corollary 6.5 we permute and where these are the first and thesecond measures in the definition of the disentangling map. Theorem 6.6 below says that wecan permute the second and the third, or the first and the third measures to get similar formulas$\mu_{2}$$\mu_{1}$as in Theorem 6.4.
Theorem 6.6. (Generalized measure permutation formula II) Under the assumptions ofTheorem 6.4 one has$\mathcal{T}_{\lambda,\mu_{2},\mu_{1}}f(\tilde{A},\tilde{B},\tilde{\fbox{}}=\mathcal{T}_{\lambda,\mu_{1},\mu_{2}}f(\tilde{A},\tilde{B},\tilde{C})+\mathcal{T}_{\nu_{2},\lambda,\nu_{2},v_{2},v_{1},\nu_{3}}^{\sim}[B,C]\frac{\delta^{2}f}{\delta y\delta z}(\tilde{A};\tilde{B},\tilde{B};\tilde{C},\tilde{C})$(6.11)$= \mathcal{T}_{\lambda,\mu_{1},\mu_{2}}f(\tilde{A},\tilde{B},\tilde{C})+\mathcal{T}_{v_{2},\lambda,v_{2},v_{1},v_{3}}[B,C]\frac{\delta f_{y}}{\delta z}(\tilde{A};\tilde{B};\tilde{C},\tilde{C})\sim$and$\mathcal{T}_{\mu_{2},\lambda,\mu 1}f(\tilde{A},\tilde{B},\tilde{\fbox{}}=\mathcal{T}_{\mu_{1},\lambda,\mu_{2}}f(\tilde{A},\tilde{B},\tilde{C})+\mathcal{T}_{v_{2},\nu_{2},v_{2},\lambda,v_{1},v_{3}}^{\sim}[A,C]\frac{\delta^{2}f}{\delta x\delta z}(\tilde{A},\tilde{A};\tilde{B};\tilde{C},\tilde{C})$(6.12)$= \mathcal{T}_{\mu_{1},/l,\mu_{2}}f(\tilde{A},\tilde{B},\tilde{C})+\mathcal{T}_{\nu_{2},\nu_{2},\lambda,\nu_{1},v_{3}}[A,C]\frac{\delta f_{x}}{\delta z}(\tilde{A};\tilde{B};\tilde{C},\tilde{C})\sim$where$v_{1},$ $v_{2}$

and

$v3$

are given as in Theorem 6.4.

163
By extracting a linear factor, the following corollary for additional operators and measuresare easy consequences of Theorem 6.6.
Corollary 6.7. Under the assumptions of Comllary 6.3 one has$\mathcal{T}_{t_{1},\ldots,\lambda_{k},\mu_{2},\mu_{1}}f(\tilde{C}_{1}, \ldots,\tilde{C}_{k},\tilde{A},\tilde{B})$$=\mathcal{T}_{\lambda_{1},\ldots,\lambda_{k},\mu_{1},\mu_{2}}f(\tilde{C}_{1}, \ldots,\tilde{C}_{k},\tilde{A},\tilde{B})$$+ \mathcal{T}_{v_{2},\lambda_{1},\ldots,\lambda_{k},\nu_{2},\nu_{2},v_{1},v_{3}}[A,B]\frac{\delta^{2}f}{\delta x\delta y}(\tilde{C}_{1}, \ldots,\tilde{C}_{k}, ;\tilde{A},\tilde{A};\tilde{B},\tilde{B})\sim$(6.13)$=\mathcal{T}_{\lambda_{1},\ldots,\lambda_{k},\mu_{1},\mu_{2}}f(\tilde{C}_{1}, \ldots,\tilde{C}_{k},\tilde{A},\tilde{B})$$+ \mathcal{T}_{v\lambda\lambda vvv}2,1,\ldots,k,2,1,3[A,B]\frac{\delta f_{x}}{\delta y}(\tilde{C}_{1}, \ldots,\tilde{C}_{k}, ;\tilde{A};\tilde{B},\tilde{B})\sim$and$\mathcal{T}_{\mu_{2},\lambda_{1},\ldots,\lambda_{k},\mu_{1}}f(\tilde{A},\tilde{C}_{1}, \ldots,\tilde{C}_{k},\tilde{B})$$=\mathcal{T}_{\mu_{1^{l}},i}t_{1},\ldots,!_{k},\mu_{2}f(\tilde{A},\tilde{C}_{1}, \ldots,\tilde{C}_{k},\tilde{B})$(6.14)$+ \mathcal{T}_{v_{2},v_{2},v_{2},\lambda_{1},\ldots,\lambda_{k},v_{1},\nu_{3}}[A,B]\frac{\delta^{2}f}{\delta x\delta y}(\tilde{A},\tilde{A};\tilde{C}_{1}, \ldots,\tilde{C}_{k}, ;\tilde{B},\tilde{B})\sim$$=\mathcal{T}_{\mu_{1},\lambda_{1},\ldots,t_{k},\mu_{2}}f(\tilde{A},\tilde{C}_{1}, \ldots,\tilde{C}_{k},\tilde{B})$$+ \mathcal{T}_{v_{2},\nu_{2},\lambda_{1},\ldots,\lambda_{k},v_{1},v_{3}}[A,B]\frac{\delta f_{x}}{\delta y}(\tilde{A};\tilde{C}_{1}, \ldots,\tilde{C}_{k};\tilde{B},\tilde{B})\sim$where$v_{1},v_{2}$and$v_{3}$,are given as in Theorem 6.4.

Until now in this section, we permute two measures in the disentangling map when the two, respectively, and none of the other measures
measures have supports inandhave supports in. But in our next theorem we permute two measures when anothermeasure have support in between the supports of the two measures.
$[a_{1},a_{2}]$$[a_{2},a_{3}]$$[a_{1},a_{3}]$Theorem 6.8. (Iterative measure permutation formula) Suppose that
andarecontinuous pmbability measures having supports contained in $[0,a_{1}],$ $[a_{1},a2]$ and $[a2, T]$ with
$0<a_{1}<a2<T$, respectively. Let. Then we have$\mu_{1},\mu_{2}$$\mu_{3}$$f(\tilde{A},\tilde{B},\tilde{C})\in D(A,B,C)$$\mathcal{T}_{\mu_{3},\mu_{2},\mu_{1}}f(\tilde{A},\tilde{B},\tilde{C})=\mathcal{T}_{\mu_{1},\mu_{2},\mu_{3}}f(\tilde{A},\tilde{B},\tilde{C})$$+ \mathcal{T}_{v_{2},v_{2},v_{2},v_{1},v_{3},\mu_{1}}^{\sim}[A,B]\frac{\delta^{2}f}{\delta x\delta y}(\tilde{A},\tilde{A};\tilde{B},\tilde{B};\tilde{C})$(6.15)$+ \mathcal{T}_{v_{2},v_{2},v_{2},\mu_{3},v_{1},v_{3}}[A,C]\frac{\delta^{2}f}{\delta x\delta z}(\tilde{A},\tilde{A};\tilde{B};\tilde{C},\tilde{C})\sim$$+\mathcal{T}_{v,\nu_{2^{V}2^{V}1^{V}3}}^{\sim}2,\mu\iota^{[B,C]\frac{\delta^{2}f}{\delta y\delta z}(\tilde{A};\tilde{B},\tilde{B};\tilde{C};\tilde{C})}$

’

where
and
$[b_{2},a2]$
andfor$v_{1},v2$$v_{3}$are continuous pmbability measures having suppons on$b_{1},b_{2}$with $a_{1}<b_{1}<b_{2}<a_{2}$ , respectively.

$[a_{1},b_{1}],$$[b_{1},b_{2}]$164
Here are some comments on the above theorem.

directly because the support of lies in beand
Remark 1. We can not permute
tween the supports of and .
2. In Theorem 6.8, we permute and in the first step, and then and , and final. Of course this procedure is notfromand to obtainin the first step, and thenand , andandmandatory. That is, we can permute
finally and . But this new steps induce the same identity as (6.15).
$\mu_{1}$$\mu_{3}$$\mu_{2}$$\mu_{2}$$\mu_{3}$$\mu_{2}$$\mu_{2}$$Iy\mu_{3}$$\mu_{1}$$\mathcal{T}_{\mu 3,\mu 2,\mu[}f(\tilde{A},\tilde{B},\tilde{C})$$\mathcal{T}_{\mu_{1},\mu 2,\mu_{3}}f(\tilde{A},\tilde{B},\tilde{C})$$\mu_{2}$$\mu_{2}$$\mu_{3}$$\mu_{1}$$\mu_{3}$$\mu_{1}$$\mu_{2}$$\mu_{1}$In Example 5.7 we applied measure permutation formula to derive an identity for a functionof the Pauli matrices. Our final example is a further investigation of the example.

Example 6.9. Consider the simplest noncommuting example of the Pauli matrices,$\sigma_{1}=(\begin{array}{l}0110\end{array})$,$\sigma 2^{=}(\begin{array}{l}0-ii0\end{array})$,$\sigma_{3}=(\begin{array}{l}010-1\end{array})$.

and letbe thebe given by
$[a,
T]$
5.7,
$[0,a]$, respectively. In Example
we permute the measures
andmeasures supported on
acts first in every term of the expression, and obtained
and so that the matrixLet$f(x,y)= \sum_{m_{1},m_{2}=0}^{\infty}c_{m_{1},m_{2}}x^{m_{1}}y^{m_{2}}$$f(\tilde{\sigma}_{1},\tilde{\sigma}_{2})\in D(\sigma_{1},\sigma_{2})$$\mu_{1}$$\mu_{1},\mu_{2}$$\sigma_{1}$$\mu_{2}$,(6.16)$\mathcal{T}_{\mu_{2},\mu_{1}}f(\tilde{\sigma}_{1},\tilde{\sigma}_{2})=\mathcal{T}_{\mu_{1},\mu_{2}}f(\tilde{\sigma}_{1},\tilde{\sigma}_{2})+2i\mathcal{T}_{v_{2},v_{1},\nu_{3}}\tilde{\sigma}_{3}\frac{\delta^{2}f}{\delta x\delta y}(1, -1;\tilde{\sigma}_{2},\tilde{\sigma}_{2})$whereand are continuous probability measures supported on $[0,b_{1}],[b_{1},b_{2}]$ and $[b_{2},T]$for $0<b_{1}<b_{2}<T$ , respectively. Let us permute the measures and in the second term on
and , respectively, so
the right hand side of (6.16), which are associated with the matricesthat the matrix
acts first in every term of the expression. By Theorem 6.6, we have$v_{l},v2$

$v3$

$v2$

$v3$

$\sigma_{3}$$\sigma_{2}$$\sigma_{2}$$\mathcal{T}_{v_{2},\nu_{l},\nu_{3}}\tilde{\sigma}_{3}\frac{\delta^{2}f}{\delta x\delta y}(1, -1;\tilde{\sigma}_{2},\tilde{\sigma}_{2})$(6.17),$= \mathcal{T}_{\nu_{3},v_{1},v_{2}}\tilde{\sigma}_{3}\frac{\delta^{2}f}{\delta x\delta y}(1, -1;\tilde{\sigma}_{2},\tilde{\sigma}_{2})-\mathcal{T}_{\eta_{2},\eta_{2},\eta_{2},v_{1},\eta_{1},\eta_{3}}^{\sim}[\sigma_{3},\sigma_{2}]\frac{\delta^{2}g}{\delta z\delta y_{1}}(\tilde{\sigma}_{3},\tilde{\sigma}_{3};\tilde{\sigma}_{2};\tilde{\sigma}_{2},\tilde{\sigma}_{2})$where$[c_{2}, T]$$\eta_{1},\eta_{2}$and$\eta_{3}$are continuous probability measures supported on

for any $b_{1}<c_{1}<c2<T$ , respectively, and$g$$[b_{1,Cl}],$ $[c_{1},c_{2}]$andis given by$g(z,y,y_{1})=z \frac{\delta^{2}f}{\delta x\delta y}(1, -1;y,y_{1})$.

Now let us investigate each terms on the right hand side of (6.17). By extracting a linear factor,
we know that the first term on the right hand side of (6.17) is reduced towhere$\mathcal{T}_{v_{3},v_{2}}\tilde{\sigma}_{3}\tilde{K}=\sigma_{3}K$$K= \mathcal{T}_{\nu_{1},v_{2}}\frac{\delta^{2}f}{\delta x\delta y}(1, -1;\tilde{\sigma}_{2},\tilde{\sigma}_{2})$.

165
But since,$\frac{\delta^{2}f}{\delta x\delta y}(1, -1;y,y_{1})=\sum_{m_{1},m_{2}=1,m_{1}=odd}^{\infty}c_{m_{1},m_{2}}(\sum_{l=0}^{m_{2}-1}y^{l}y_{1}^{m_{2}-l-1})$

we have$K= \sum_{m_{1},m_{2}=1;m_{1}=odd}^{\infty}c_{m_{1},m_{2}}m_{2}\sigma_{2}^{m_{2}-1}=\mathcal{T}_{v_{2}}\frac{\delta f_{y}}{\delta x}(1, -1;\tilde{\sigma}_{2})$and so.

$\mathcal{T}_{v_{3},v_{1},v_{2}}\tilde{\sigma}_{3}\frac{\delta^{2}f}{\delta x\delta y}(1, -1;\tilde{\sigma}_{2},\tilde{\sigma}_{2})=\mathcal{T}_{v_{3},v_{2}}\tilde{\sigma}_{3}\frac{\delta f_{y}}{\delta x}(1, -1;\tilde{\sigma}_{2})=\mathcal{T}_{\mu_{2},\mu_{1}}\tilde{\sigma}_{3}\frac{\delta f_{y}}{\delta x}(1, -1;\tilde{\sigma}_{2})$For the second term on the right hand side of (6.17), note thatand$[\sigma_{3},\sigma_{2}]=-2i\sigma[$$\frac{\delta^{2}g}{\delta z\delta y_{1}}(z,z1;y;y_{1},y_{2})=\frac{\delta^{3}f}{\delta x\delta y\delta_{\mathcal{Y}1}}(1, -1;y;y\iota,y_{2})$$= \sum_{m_{1},m_{2}=1;m_{1}=odd}^{\infty}c_{m_{1},m_{2}}(\sum_{l=0}^{-2m}\sum_{r=0}^{-l-2}y^{l}y_{1}^{r}y_{2}^{m_{2}-l-r-2})m_{22}$.

Hence the term is reduced to$m_{2}-2m_{2}-l-2$$M \equiv 2i\sum_{m_{1},m_{2}=1;m_{1}=odd}^{\infty}c_{m_{1},m_{2}}$$( \sum_{l=0}$$P_{\eta_{2,1},\eta_{1},\eta_{3}}^{1,l,r,m_{2}-l-r-2})’(\sigma_{1}, \sigma_{2}, \sigma_{2}, \sigma_{2}))$$\sum_{r=0}$

We extract a linear factor to obtain$P_{\eta_{2},v_{1},\eta_{1}^{2},\eta_{3}}^{1,l,r,m-l-r-2}(\sigma_{1},\sigma_{2},\sigma_{2},\sigma_{2})=P_{\eta_{2},\eta_{1},\eta_{3}}^{1,l+r,m_{2}-l-r-2}(\sigma_{1},\sigma_{2},\sigma_{2})$and so,$M=2i\mathcal{T}_{\eta_{2},\eta_{1},\eta_{3}}k(\tilde{\sigma}_{1},\tilde{\sigma}_{2},\tilde{\sigma}_{2})=2i\mathcal{T}_{v_{2},v_{1},v_{3}}k(\tilde{\sigma}_{1},\tilde{\sigma}_{2},\tilde{\sigma}_{2})$where$k(z,y,y_{2})= \sum_{m_{1},m_{2}=1;m_{1}=odd}^{\infty}c_{m_{1},m_{2}}(\sum_{l=0}^{-2m}\sum_{r=0}^{-l-2}zy^{l+r}y_{2}^{m_{2}-l-r-2})m_{22}$$=z \sum_{m_{1},m_{2}=1,m_{1}=odd}^{\infty}c_{m_{1},m_{2}}(\sum_{s=1}^{m_{2}-1}s\gamma_{y_{2}^{m_{2}-s-1})}^{-1}$$=z( \frac{\delta^{2}f}{\delta x\delta y})_{y}(1, -1;y,y_{2})$.

Finally from (6.16) we have$\mathcal{T}_{\mu_{2},\mu_{1}}f(\tilde{\sigma}_{1},\tilde{\sigma}_{2})=\mathcal{T}_{\mu_{1},\mu_{2}}f(\tilde{\sigma}_{1},\tilde{\sigma}_{2})+2i\mathcal{T}_{\mu_{2},\mu_{1}}\tilde{\sigma}_{3}\frac{\delta f_{y}}{\delta x}(1, -1;\tilde{\sigma}_{2})$$(6.18)$$+(2i)^{2} \mathcal{T}_{vvv}\tilde{\sigma}_{1}2,1,3(\frac{\delta^{2}f}{\delta x\delta y})_{y}(1, -1;\tilde{\sigma}_{2},\tilde{\sigma}_{2})$

.

.

166
One can proceed with these manipulations. That is, using Theorem 6.4 we can permute
in the third term of the right hand side of (6.18).

$v_{2}$

and$\nu_{1}$References
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[2] K.S. Chang, B.S. Kim and Y.H. Park, Measure permutation formulas in Feynman’s operational calculi, Russian J. Math. Phys. 17 (2010), 26-34.
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