Operators, in quantum mechanics and quantum optics, have been considered as an important characteristic by endowing Bose annihilation and creation operators with the basic commutative relation, such as [a, a^†] = 1. In order to deal with a lot of problems involved in quantum optics, say, calculating some expectation values of physical quality or various matrix elements, the operator ordering is often fallen back on due to its convenience. It is the main reason that the matrix elements of the normally ordered operator function :f(a, a^†) : in coherent states |z〉=exp(−12zz∗+za†)|0〉 yield^[1,2]

$$ \begin{eqnarray}\langle z|:f(a,{a}^{\dagger }):|{z}^{\prime}\rangle =f({z}^{\prime},{z}^{\ast })\exp \left(-\displaystyle \frac{1}{2}z{z}^{\ast }-\displaystyle \frac{1}{2}{z}^{\prime}{z}^{{\prime} \ast} +{z}^{\ast }{z}^{\prime}\right),\ \ \ \end{eqnarray}$$ (1)

$$ \begin{eqnarray}\hat{\rho }=\displaystyle \int \displaystyle \frac{{{\rm{d}}}^{2}z}{\pi }P(z,{z}^{\ast })|z\rangle \langle z|=\vdots P(a,{a}^{\dagger })\vdots,\end{eqnarray}$$ (2)

$$ \begin{eqnarray}\begin{array}{lll} & & \langle q|{\mathbb{Q}}g(Q,P){\mathbb{Q}}|p\rangle =g(q,p)\langle q|p\rangle,\\ & & \langle p|{\mathbb{P}}h(Q,P){\mathbb{P}}|q\rangle =h(q,p)\langle p|q\rangle,\end{array}\end{eqnarray}$$ (3)

In quantum optics, these operators such as (QP)^m and (QP)^−m keep coming up and It is troublesome in handling these operators due to a lack of proper mathematical tools. To our knowledge, there has been no report on a good solution to this problem. In this paper, we shall recast the quantum mechanical operators (a^†a)^m, with m being an arbitrary positive integer, into their normally ordered expressions by using Touchard polynomials. Also, we shall derive the anti-normally ordered expressions of (a^†a)^{± m} by using special functions as well as Stirling-like numbers together with the general mutual transformation rule between normal and anti-normal orderings of operators.^[4–11] Moreover, we shall obtain the ℚ- and ℙ-ordered forms of (QP)^±m by using an analogy method. Finally, some applications in the deduced operator ordering identities are discussed.

In this section, we deduce the normally and anti-normally ordered expansions of operators (a^†a)^m and (aa^†)^m, with m being an arbitrary positive integer. In this work we will make full use of a parameter differential method, special functions, and general mutual transformation rule between normal and anti-normal orderings of operators.^[4,11]

By using the completeness relation of Fock space ∑n=0∞|n〉〈n|=1 and a^†a|n 〉 = n | n 〉 and IWOP technique, we obtain

$$ \begin{eqnarray}\begin{array}{lll}{({a}^{\dagger }a)}^{m} & = & {({a}^{\dagger }a)}^{m}\displaystyle \sum _{n=0}^{\infty } |{n}\rangle \langle{n}| =\displaystyle \sum _{n=0}^{\infty }{n}^{m}\langle |{n}\rangle \langle{n}| \\ & = & :{{\rm{e}}}^{-{a}^{\dagger }a}\displaystyle \sum _{n=0}^{\infty }\displaystyle \frac{{n}^{m}}{n!}{({a}^{\dagger }a)}^{n}:,\end{array}\end{eqnarray}$$ (4)

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$$ \begin{eqnarray}\begin{array}{ll}{({a}^{\dagger }a)}^{m} & ={\left. \displaystyle \frac{{\partial }^{m}}{\partial {t}^{m}}:{{\rm{e}}}^{-{a}^{\dagger }a}\displaystyle \sum _{n=0}^{\infty }\displaystyle \frac{{({a}^{\dagger }a)}^{n}{{\rm{e}}}^{tn}}{n!}:\right|}_{t=0}\\ & ={\left. \displaystyle \frac{{\partial }^{m}}{\partial {t}^{m}}:{{\rm{e}}}^{-{a}^{\dagger }a}\displaystyle \sum _{n=0}^{\infty }\displaystyle \frac{{({a}^{\dagger }a{{\rm{e}}}^{t})}^{n}}{n!}:\right|}_{t=0}\\ & ={\left. \displaystyle \frac{{\partial }^{m}}{\partial {t}^{m}}:{{\rm{e}}}^{{a}^{\dagger }a({{\rm{e}}}^{t}-1)}:\right|}_{t=0}.\end{array}\end{eqnarray}$$ (5)

$$ \begin{eqnarray}{T}_{m}(\xi )={\left. \displaystyle \frac{{\partial }^{m}}{\partial {t}^{m}}{{\rm{e}}}^{\xi ({{\rm{e}}}^{t}-1)}\right|}_{t=0}.\end{eqnarray}$$ (6)

$$ \begin{eqnarray}{{\rm{e}}}^{\xi ({{\rm{e}}}^{t}-1)}=\displaystyle \sum _{m=0}^{\infty }\displaystyle \frac{{T}_{m}(\xi)}{m!}{t}^{m},\end{eqnarray}$$ (7)

$$ \begin{eqnarray}{T}_{m}(\xi)={{\rm{e}}}^{-\xi }{\left(\xi \displaystyle \frac{\partial }{\partial \xi }\right)}^{m}{{\rm{e}}}^{\xi }.\end{eqnarray}$$ (8)

$$ \begin{eqnarray}{({a}^{\dagger }a)}^{m}=:{T}_{m}({a}^{\dagger }a):,\end{eqnarray}$$ (9)

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To derive the anti-normally ordered form of (a^†a)^m, recall the general mutual transformation rules between normal and anti-normal orderings^[4,11]

$$ \begin{eqnarray}\begin{array}{lll} & & :F(a,{a}^{\dagger }):=\vdots \exp \left(-\displaystyle \frac{{\partial }^{2}}{\partial a\partial {a}^{\dagger }}\right)F(a,{a}^{\dagger })\vdots,\\ & & \vdots F(a,{a}^{\dagger })\vdots =:\exp \left(\displaystyle \frac{{\partial }^{2}}{\partial a\partial {a}^{\dagger }}\right)F(a,{a}^{\dagger }):,\end{array}\end{eqnarray}$$ (10)

$$ \begin{eqnarray}\begin{array}{lll}{({a}^{\dagger }a)}^{m} & = & {\left. \vdots \exp \left(-\displaystyle \frac{{\partial }^{2}}{\partial a\partial {a}^{\dagger }}\right)\displaystyle \frac{{\partial }^{m}}{\partial {t}^{m}}{{\rm{e}}}^{a{a}^{\dagger }({{\rm{e}}}^{t}-1)}\vdots \right|}_{t=0}\\ & = & {\left. \displaystyle \frac{{\partial }^{m}}{\partial {t}^{m}}\vdots \exp \left(-\displaystyle \frac{{\partial }^{2}}{\partial a\partial {a}^{\dagger }}\right){{\rm{e}}}^{a{a}^{\dagger }({{\rm{e}}}^{t}-1)}\vdots \right|}_{t=0}\\ & = & \displaystyle \frac{{\partial }^{m}}{\partial {t}^{m}}\vdots \exp \left(-\displaystyle \frac{{\partial }^{2}}{\partial a\partial {a}^{\dagger }}\right)\\ & & {\left. \times \displaystyle \int \displaystyle \frac{{{\rm{d}}}^{2}z}{\pi }{{\rm{e}}}^{-z{z}^{\ast }+za+{z}^{\ast }{a}^{\dagger }({{\rm{e}}}^{t}-1)}\vdots \right|}_{t=0}\\ & = & {\left. \displaystyle \frac{{\partial }^{m}}{\partial {t}^{m}}\vdots \displaystyle \int \displaystyle \frac{{{\rm{d}}}^{2}z}{\pi }{{\rm{e}}}^{-{{\rm{e}}}^{t}z{z}^{\ast }+za+{z}^{\ast }{a}^{\dagger }({{\rm{e}}}^{t}-1)}\vdots \right|}_{t=0}\\ & = & {\left. \displaystyle \frac{{\partial }^{m}}{\partial {t}^{m}}\vdots {{\rm{e}}}^{-t+a{a}^{\dagger }(1-{{\rm{e}}}^{-t})}\vdots \right|}_{t=0},\end{array}\end{eqnarray}$$ (11)

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$$ \begin{eqnarray}{X}_{m}(\xi )={\left. \displaystyle \frac{{\partial }^{m}}{\partial {t}^{m}}{{\rm{e}}}^{-t+\xi (1-{{\rm{e}}}^{-t})}\right|}_{t=0}.\end{eqnarray}$$ (12)

$$ \begin{eqnarray}{{\rm{e}}}^{-t+\xi (1-{{\rm{e}}}^{-t})}=\displaystyle \sum _{m=0}^{\infty }\displaystyle \frac{{X}_{m}(\xi)}{m!}{t}^{m}.\end{eqnarray}$$ (13)

$$ \begin{eqnarray}{({a}^{\dagger }a)}^{m}=\vdots {X}_{m}({a}^{\dagger }a)\vdots,\end{eqnarray}$$ (14)

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In this subsection we deduce the normally and anti-normally ordered expansions of operators (aa^†)^m, with m being an arbitrary positive integer. Considering [a^†,a] = −1 = [−a,a^†] and comparing formula (9), we can deduce that

$$ \begin{eqnarray}{(a{a}^{\dagger })}^{m}={(-)}^{m}{[(-a){a}^{\dagger }]}^{m}={(-)}^{m}\vdots {T}_{m}(-a{a}^{\dagger })\vdots .\end{eqnarray}$$ (15)

Moreover, by analogy with formula (14), we can obtain

$$ \begin{eqnarray}{(a{a}^{\dagger })}^{m}={(-)}^{m}{[(-a){a}^{\dagger }]}^{m}={(-)}^{m}:{X}_{m}(-a{a}^{\dagger }):,\end{eqnarray}$$ (16)

Now we deduce the normally and anti-normally ordered expansions of operators e^{λ a† a} and e^{λ aa†}. By using the power series expansion of e^{λ a† a} and Eqs. (9) and (7) we have

$$ \begin{eqnarray}\begin{array}{lll}{{\rm{e}}}^{\lambda {a}^{\dagger }a} & = & \displaystyle \sum _{n=0}^{\infty }\displaystyle \frac{{\lambda }^{n}}{n!}{({a}^{\dagger }a)}^{n}\\ & = & \displaystyle \sum _{n=0}^{\infty }\displaystyle \frac{{\lambda }^{n}}{n!}:{T}_{m}({a}^{\dagger }a):\,=\,:{{\rm{e}}}^{{a}^{\dagger }a({{\rm{e}}}^{\lambda }-1)}:.\end{array}\end{eqnarray}$$ (17)

$$ \begin{eqnarray}:{{\rm{e}}}^{\mu {a}^{\dagger }a}:\,={{\rm{e}}}^{{a}^{\dagger }aln(1+\mu)},\end{eqnarray}$$ (18)

On the other hand, using the power series expansion of e^{λ a† a} and Eqs. (14) and (13) turn into

$$ \begin{eqnarray}{{\rm{e}}}^{\lambda {a}^{\dagger }a}=\displaystyle \sum _{n=0}^{\infty }\displaystyle \frac{{\lambda }^{n}}{n!}\vdots {X}_{m}({a}^{\dagger }a)\vdots =\vdots {{\rm{e}}}^{-\lambda +{a}^{\dagger }a(1-{{\rm{e}}}^{-\lambda })}\vdots .\end{eqnarray}$$ (19)

$$ \begin{eqnarray}\vdots {{\rm{e}}}^{\mu {a}^{\dagger }a}\vdots =\displaystyle \frac{1}{1-\mu }{{\rm{e}}}^{-{a}^{\dagger }aln(1-\mu)},\end{eqnarray}$$ (20)

Similarly, by using Eqs. (15), (7), (16), and (13) we can obtain

$$ \begin{eqnarray}{{\rm{e}}}^{\lambda a{a}^{\dagger }}=\vdots {{\rm{e}}}^{{a}^{\dagger }a(1-{{\rm{e}}}^{-\lambda })}\vdots,\ \ \ \ \ {{\rm{e}}}^{\lambda a{a}^{\dagger }}={{\rm{e}}}^{\lambda }:{{\rm{e}}}^{{a}^{\dagger }a({{\rm{e}}}^{\lambda }-1)}:\,,\ \ \ \end{eqnarray}$$ (21)

In the following section, we deduce the ℚ- and ℙ-ordered expansions of operators (QP)^m and (PQ)^m, with m being an arbitrary positive integer.

In view of [Q,iℏP]=−1=[a†,a], by analogy with formula (9) we can deduce that

$$ \begin{eqnarray}\begin{array}{lll}{(QP)}^{m} & = & {(-{\rm{i}}\hslash)}^{m}{\left[Q\left(\displaystyle \frac{{\rm{i}}}{\hslash }P\right)\right]}^{m}\\ & = & {(-{\rm{i}}\hslash)}^{m}{\mathbb{Q}}{T}_{m}\left(\displaystyle \frac{{\rm{i}}}{\hslash }QP\right){\mathbb{Q}},\end{array}\end{eqnarray}$$ (22)

$$ \begin{eqnarray}\begin{array}{lll}{(QP)}^{m} & = & {(-{\rm{i}}\hslash)}^{m}{\left[Q\left(\displaystyle \frac{{\rm{i}}}{\hslash }P\right)\right]}^{m}\\ & = & {(-{\rm{i}}\hslash)}^{m}{\mathbb{P}}{X}_{m}\left(\displaystyle \frac{{\rm{i}}}{\hslash }QP\right){\mathbb{P}},\end{array}\end{eqnarray}$$ (23)

$$ \begin{eqnarray}\begin{array}{lll} & & {(PQ)}^{m}={({\rm{i}}\hslash)}^{m}{\left[\left(\displaystyle \frac{1}{{\rm{i}}\hslash }P\right)Q\right]}^{m}={({\rm{i}}\hslash)}^{m}{\mathbb{P}}{T}_{m}\left(\displaystyle \frac{1}{{\rm{i}}\hslash }QP\right){\mathbb{P}},\end{array}\end{eqnarray}$$ (24)

$$ \begin{eqnarray}\begin{array}{lll} & & {(PQ)}^{m}={({\rm{i}}\hslash)}^{m}{\left[\left(\displaystyle \frac{1}{{\rm{i}}\hslash }P\right)Q\right]}^{m}={({\rm{i}}\hslash)}^{m}{\mathbb{Q}}{X}_{m}\left(\displaystyle \frac{1}{{\rm{i}}\hslash }QP\right){\mathbb{Q}},\ \ \ \ \ \ \ \end{array}\end{eqnarray}$$ (25)

By using Eqs. (22)–(25), (7) and (13) or by analogy with formulas (17), (19), and (21), we can derive

$$ \begin{eqnarray}\begin{array}{lll}{{\rm{e}}}^{\lambda QP} & = & {\mathbb{Q}}\exp \left[\displaystyle \frac{{\rm{i}}}{\hslash }QP({{\rm{e}}}^{-{\rm{i}}\hslash \lambda }-1)\right]{\mathbb{Q}}\\ & = & {\mathbb{P}}\exp \left[{\rm{i}}\hslash \lambda +\displaystyle \frac{{\rm{i}}}{\hslash }QP(1-{{\rm{e}}}^{{\rm{i}}\hslash \lambda })\right]{\mathbb{P}},\end{array}\end{eqnarray}$$ (26)

$$ \begin{eqnarray}\begin{array}{lll}{{\rm{e}}}^{\lambda PQ} & = & {\mathbb{P}}\exp \left[-\displaystyle \frac{{\rm{i}}}{\hslash }QP({{\rm{e}}}^{{\rm{i}}\hslash \lambda }-1)\right]{\mathbb{P}}\\ & = & {\mathbb{Q}}\exp \left[-{\rm{i}}\hslash \lambda +\displaystyle \frac{{\rm{i}}}{\hslash }QP({{\rm{e}}}^{-{\rm{i}}\hslash \lambda }-1)\right]{\mathbb{Q}}.\end{array}\end{eqnarray}$$ (27)

Now we deduce the ordered expansions of operators such as (AB)^−m, with m being an arbitrary positive integer. We know that (a^†a)| n 〉 = n| n 〉, n = 0,1,2,3,…. To be precise, the eigenvalue of the number operator (a^†a) is discrete and contains zero. So the operator (a^†a) has no inverse, that is to say, (a^†a)⁻¹ does not exist.

From (a^†a)| n 〉 = n| n 〉 we obtain (aa^†)^m| n 〉 = (a^†a + 1)^m| n 〉 = (n + 1)^m| n 〉, n = 0, 1, 2, 3, …. Thus we see that the operator (aa^†)^m has inverse, denoted by (aa^†)^−m ≡ 1/(aa^†)^m. By using the completeness relation of the Fock space, we can obtain

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, we can obtain

$$ \begin{eqnarray}\begin{array}{lll}\displaystyle \frac{1}{{(a{a}^{\dagger })}^{m}} & = & :\displaystyle \sum _{l=0}^{\infty }\displaystyle \frac{{(-)}^{l}}{l!}{(a{a}^{\dagger })}^{l}\displaystyle \sum _{n=0}^{\infty }\displaystyle \frac{{(a{a}^{\dagger })}^{n}}{n!{(n+1)}^{m}}:\\ & = & :\displaystyle \sum _{n=0}^{\infty }\displaystyle \sum _{l=0}^{\infty }\displaystyle \frac{{(-)}^{l}{(a{a}^{\dagger })}^{n+l}}{l!n!{(n+1)}^{m}}:\\ & = & :\displaystyle \sum _{k=0}^{\infty }{(a{a}^{\dagger })}^{k}\displaystyle \sum _{l=0}^{k}\displaystyle \frac{{(-)}^{l}}{l!(k-l)!{(k-l+1)}^{m}}:.\end{array}\end{eqnarray}$$ (28)

In the last step of the above calculations, we have used the rearranging double summation formula

$$ \begin{eqnarray}\displaystyle \sum _{m=0}^{\infty }\displaystyle \sum _{l=0}^{\infty }{A}_{m}{B}_{l}=\displaystyle \sum _{k=0}^{\infty }\displaystyle \sum _{l=0}^{k}{A}_{k-l}{B}_{l}.\end{eqnarray}$$ (29)

Now we introduce a kind of special numbers here, defined by

$$ \begin{eqnarray}{S}_{k}(m)=\displaystyle \sum _{l=0}^{k}\displaystyle \frac{{(-)}^{l}}{l!(k-l)!{(k-l+1)}^{m}},\end{eqnarray}$$ (30)

then equation (28) can be rewritten as

$$ \begin{eqnarray}\displaystyle \frac{1}{{(a{a}^{\dagger })}^{m}}=\displaystyle \sum _{k=0}^{\infty }{S}_{k}(m){a}^{\dagger k}{a}^{k},\end{eqnarray}$$ (31)

which is just the compact normally ordered form of (aa^†)^−m. From the definition of S_k (m) we can easily derive that S_k (0) = δ_k,0, S₀ (m) = 1 and

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Then we have

$$ \begin{eqnarray}\displaystyle \frac{1}{a{a}^{\dagger }}=\displaystyle \sum _{k=0}^{\infty }\displaystyle \frac{{(-)}^{k}}{(k+1)!}{a}^{\dagger k}{a}^{k},\end{eqnarray}$$ (32)

which is just the normally ordered expansion of (aa^†)⁻¹. We can prove that

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That is to say, equation (32) is indeed the inverse operator of (aa^†). The more properties of the defined special number S_k (m) may be seen in Appendix C.

Moreover, by using Eq. (31) and the general mutual transformation rule between normal and anti-normal orderings of operators Eq.(10), we can obtain

$$ \begin{eqnarray}\begin{array}{lll}\displaystyle \frac{1}{{(a{a}^{\dagger })}^{m}} & = & \vdots \exp \left(-\displaystyle \frac{{\partial }^{2}}{\partial a\partial {a}^{\dagger }}\right)\displaystyle \sum _{k=0}^{\infty }{S}_{k}(m){a}^{\dagger k}{a}^{k}\vdots \\ & = & \vdots \displaystyle \sum _{k=0}^{\infty }{S}_{k}(m)\exp \left(-\displaystyle \frac{{\partial }^{2}}{\partial a\partial {a}^{\dagger }}\right){a}^{\dagger k}{a}^{k}\vdots \\ & = & \vdots \displaystyle \sum _{k=0}^{\infty }{S}_{k}(m){{\rm{H}}}_{k,k}(a,{a}^{\dagger })\vdots \\ & = & \vdots \displaystyle \sum _{k=0}^{\infty }{(-)}^{k}{S}_{k}(m){{\rm{L}}}_{k}(a{a}^{\dagger })\vdots,\end{array}\end{eqnarray}$$ (33)

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$$ \begin{eqnarray}\displaystyle \frac{1}{\hat{a}{\hat{a}}^{\dagger }}=\vdots \displaystyle \sum _{k=0}^{\infty }\displaystyle \frac{1}{(k+1)!}{{\rm{L}}}_{k}(a{a}^{\dagger })\vdots .\end{eqnarray}$$ (34)

Due to QP = (QP + PQ)/2 + i ℏ/2 and (QP + PQ) being a Hermitian operator, which implies the eigenvalue of (QP) does not contain zero, we know that the operator (QP) has its inverse operator. To be precise, (QP)⁻¹ ≡ 1/(QP) does exit.

In view of [Q,1iℏP]=1=[a,a†], by analogy with the formula (28) we can deduce that

$$ \begin{eqnarray}\begin{array}{lll}\displaystyle \frac{1}{{(QP)}^{m}} & = & \displaystyle \frac{1}{{({\rm{i}}\hslash)}^{m}}\displaystyle \frac{1}{{\left[Q\left(\displaystyle \frac{1}{{\rm{i}}\hslash }P\right)\right]}^{m}}\\ & = & \displaystyle \frac{1}{{({\rm{i}}\hslash)}^{m}}{\mathbb{P}}\displaystyle \sum _{k=0}^{\infty }{({\rm{i}}\hslash)}^{-k}{S}_{k}(m){Q}^{k}{P}^{k}{\mathbb{P}}\\ & = & \displaystyle \frac{1}{{({\rm{i}}\hslash)}^{m}}\displaystyle \sum _{k=0}^{\infty }{({\rm{i}}\hslash)}^{-k}{S}_{k}(m){P}^{k}{Q}^{k},\end{array}\end{eqnarray}$$ (35)

Similarly, by analogy with formula (33) we obtain

$$ \begin{eqnarray}\begin{array}{lll}\displaystyle \frac{1}{{(QP)}^{m}} & = & \displaystyle \frac{1}{{({\rm{i}}\hslash)}^{m}}\displaystyle \frac{1}{{\left[Q\left(\displaystyle \frac{1}{{\rm{i}}\hslash }P\right)\right]}^{m}}\\ & = & \displaystyle \frac{1}{{({\rm{i}}\hslash)}^{m}}{\mathbb{Q}}\displaystyle \sum _{k=0}^{\infty }{(-)}^{k}{S}_{k}(m){{\rm{L}}}_{k}\left(\displaystyle \frac{1}{{\rm{i}}\hslash }QP\right){\mathbb{Q}},\end{array}\end{eqnarray}$$ (36)

In a manner completely similar to the way of deriving Eqs. (35) and (36), we may derive that

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In the present section we discuss some applications of the above new identities. Since the operator (QP)^m is a complex one, with m being a big positive integer, it is tricky to calculate its matrix elements in phase space without Eqs. (22) and (23). Here we can see that it is easy to derive the matrix elements by using the new identity (22). According to Eq. (22) we can immediately obtain

$$ \begin{eqnarray}\begin{array}{lll}\langle q| {(QP)}^{m}|p \rangle & = & \displaystyle \frac{{(-{\rm{i}}\hslash)}^{m}}{\sqrt{2\pi \hslash }}{X}_{m}\left(\displaystyle \frac{{\rm{i}}}{\hslash }qp\right)\exp \left(\displaystyle \frac{{\rm{i}}}{\hslash }qp\right)\\ & = & \displaystyle \frac{{(-{\rm{i}}\hslash)}^{m}}{\sqrt{2\pi \hslash }}\exp \left(\displaystyle \frac{{\rm{i}}}{\hslash }qp\right)\\ & & \times \displaystyle \sum _{n=0}^{m}{\left(\displaystyle \frac{{\rm{i}}}{\hslash }qp\right)}^{n}\displaystyle \sum _{l=0}^{n}{(-)}^{l}\displaystyle \frac{{(n-l)}^{m}}{l!(n-l)!},\ \ \ \ \ \ \ \ \end{array}\end{eqnarray}$$ (37)

As another example, by using Eq. (34) we easily obtain the ℙ-representations of (aa^†)⁻¹ as follows:

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As the last example, we discuss the application of the general mutual transformation rules of normal and antinormal orderings Eq. (10). Using [aⁿ,a^†] = na^{n − 1} and [a,a^†n] = na^†n−1, we have

$$ \begin{eqnarray}\begin{array}{lll} & & {a}^{\dagger n+1}{a}^{m}={a}^{\dagger n}{a}^{m}{a}^{\dagger }-m{a}^{\dagger n}{a}^{m-1},\\ & & {a}^{\dagger n}{a}^{m+1}=a{a}^{\dagger n}{a}^{m}-n{a}^{\dagger n-1}{a}^{m}.\end{array}\end{eqnarray}$$ (38)

$$ \begin{eqnarray}\begin{array}{lll} & & {a}^{\dagger n+1}{a}^{m}=\vdots \exp \left(-\displaystyle \frac{{\partial }^{2}}{\partial a\partial {a}^{\dagger }}\right){a}^{\dagger n+1}{a}^{m}\vdots =\vdots {{\rm{H}}}_{n+1,m}({a}^{\dagger }a)\vdots,\\ & & {a}^{\dagger n}{a}^{m}=\vdots \exp \left(-\displaystyle \frac{{\partial }^{2}}{\partial a\partial {a}^{\dagger }}\right){a}^{\dagger n}{a}^{m}\vdots =\vdots {{\rm{H}}}_{n,m}({a}^{\dagger }a)\vdots,\\ & & {a}^{\dagger n}{a}^{m-1}=\vdots {{\rm{e}}}^{- \frac{{\partial }^{2}}{\partial a\partial {a}^{\dagger }}}{a}^{\dagger n}{a}^{m-1}\vdots =\vdots {{\rm{H}}}_{n,m-1}({a}^{\dagger }a)\vdots .\end{array}\end{eqnarray}$$ (39)

$$ \begin{eqnarray}\begin{array}{lll} & & {a}^{\dagger n}{a}^{m+1}=\vdots \exp \left(-\displaystyle \frac{{\partial }^{2}}{\partial a\partial {a}^{\dagger }}\right){a}^{\dagger n}{a}^{m+1}\vdots =\vdots {{\rm{H}}}_{n,m+1}({a}^{\dagger }a)\vdots,\\ & & {a}^{\dagger n}{a}^{m}=\vdots \exp \left(-\displaystyle \frac{{\partial }^{2}}{\partial a\partial {a}^{\dagger }}\right){a}^{\dagger n}{a}^{m}\vdots =\vdots {{\rm{H}}}_{n,m}({a}^{\dagger }a)\vdots,\\ & & {a}^{\dagger n-1}{a}^{m}=\vdots {{\rm{e}}}^{- \frac{{\partial }^{2}}{\partial a\partial {a}^{\dagger }}}{a}^{\dagger n-1}{a}^{m}\vdots =\vdots {{\rm{H}}}_{n-1,m}({a}^{\dagger }a)\vdots .\end{array}\end{eqnarray}$$ (40)

$$ \begin{eqnarray}\begin{array}{lll} & & \vdots {{\rm{H}}}_{n+1,m}({a}^{\dagger },a)\vdots =\vdots {{\rm{H}}}_{n,m}({a}^{\dagger },a)\vdots {a}^{\dagger }-m\vdots {{\rm{H}}}_{n,m-1}({a}^{\dagger },a)\vdots,\\ & & \vdots {{\rm{H}}}_{n,m+1}({a}^{\dagger },a)\vdots =a\vdots {{\rm{H}}}_{n,m}({a}^{\dagger },a)\vdots -n\vdots {{\rm{H}}}_{n-1,m}({a}^{\dagger },a)\vdots .\end{array}\end{eqnarray}$$ (41)

$$ \begin{eqnarray}\begin{array}{lll} & & {{\rm{H}}}_{n+1,m}(\eta,\xi)={{\rm{H}}}_{n,m}(\eta,\xi)\eta -m\vdots {{\rm{H}}}_{n,m-1}(\eta,\xi),\\ & & {{\rm{H}}}_{n,m+1}(\eta,\xi)=\xi {{\rm{H}}}_{n,m}(\eta,\xi)-n{{\rm{H}}}_{n-1,m}(\eta,\xi).\end{array}\end{eqnarray}$$ (42)

In this work, we recast the quantum mechanical operators (a^†a)^{± m} and (aa^†)^{± m}, with m being an arbitrary positive integer, into their normally ordered expansions by using Touchard polynomials and using special functions as well as a kind of special numbers (called Stirling-like numbers). Also, we derive their antinormally ordered expressions via the general mutual transformation rules between normal and antinomal orderings of operators. Moreover, the ℚ- and ℙ-ordered forms of (QP)^±m are also obtained by using an analogy method. Finally, some applications of these new identities are discussed.