A perturbative approach for the construction of the non-adiabatic nuclear kinetic energy operator for diatomic and triatomic systems

Abstract

In this work, we derive an effective non-adiabatic kinetic energy operator for nuclear motion in triatomic molecules on the basis of perturbation theory. For this purpose, we extended the approach of Herman and Asgharian (J Mol Spectr 19:305, 1966) originally developed for diatomic systems. General perturbative-type expressions for effective distance-dependent reduced nuclear masses have been obtained for a triatomic system. It is shown that in the diatomic limit our method reproduces correctly the previous known result of Herman and Asgharian.

Appendices

Appendices

A Derivation of the operators $\partial _{\alpha }$ and $\partial _{\alpha }\partial _{\beta }$ in the SF frame

In the notations of the variations we shall use the conventions adopted in I. The set of variations of the molecular (Jacobi) SF coordinates r, R, $\theta$, $\delta$, $\gamma$, and $\varphi$ are denoted by the symbol $\{\delta u\}_{\alpha \beta }$ and ordered in the same way as described by (I.B7). The subscripts $\alpha$ and $\beta$ denotes the Cartesian BF coordinates of the same Jacobi vector $\mathbf {r}$ or $\mathbf {R}$, i.e. $\alpha ,\beta =x,y,z$ or $\alpha ,\beta =X,Y,Z$ ($\alpha \ne \beta$); f is the dummy function. In the present work the Cartesian BF coordinates (in contrast to I) are not primed.

We shall demonstrate the transformation of the mixed derivatives $\partial _{\alpha } \partial _{\beta }$ from Cartesian BF coordinates into the SF molecular coordinates. Essentially, the procedure has been described in I [see (I.9)–(I.14), (I.77)]. Here, instead of (I.10) we use the finite difference representation for the mixed derivatives $\partial _{\alpha } \partial _{\beta }$

$$\begin{aligned} \partial _{\alpha } \partial _{\beta } f = \frac{ 1 }{\delta \alpha \delta \beta } \left[ \delta _2 f - \delta ^{(\alpha )}_1 f - \delta ^{(\beta )}_1 f \right] , \end{aligned}$$

(58)

with $\delta ^{(\alpha )}_1 f = f(\alpha + \delta \alpha ) - f(\alpha )$, $\delta ^{(\beta )}_1 f = f(\beta + \delta \beta ) - f(\beta )$ and $\delta _2 f = f(\alpha + \delta \alpha , \beta + \delta \beta ) -f(\alpha ,\beta )$. We treat $\delta _1 f$ and $\delta _2 f$ in exactly the same manner as described by (I.12) and (I.14), albeit for $\delta _2 f$ the variations of the molecular coordinates $\{ \delta u\}$ are represented by simultaneous variations of two independent BF Cartesian coordinates $\delta \alpha$ and $\delta \beta$. We start from the two Cartesian BF variations $\delta \alpha$ and $\delta \beta$, next we find the corresponding variations of the Cartesian SF variables with the help of the transformation (I.9). The obtained SF variations are then substituted into (I.13) that gives the following expressions for $\{ \delta u\}_{\alpha \beta }$

$$\begin{aligned} \{ \delta u \}_{xy}= & {} \left\{ \frac{\delta {x}^2}{2 r}+\frac{\delta {y}^2}{2 r}, \, 0, \, \frac{\delta {y}^2 \cot \theta }{2 r^2} - \frac{\delta {x}}{ r},\right. \nonumber \\&\frac{\cot \gamma \sin 2\varphi }{ 2 r^2 \sin \gamma }(\delta {y}^2 - \delta {x}^2) - \frac{\delta {x}\delta {y} \cot \gamma \cos 2\varphi }{ r^2 \sin \gamma } + \frac{\delta {x} \sin \varphi }{r \sin \gamma } +\frac{\delta {y} \cos \varphi }{r \sin \gamma },\nonumber \\&\frac{\cot \gamma }{2r^2} \bigg (\delta {x}^2 \sin ^2\varphi + \delta {x} \delta {y} \sin 2\varphi + \delta {y}^2 \cos ^2\varphi \bigg ) + \frac{1}{r} \bigg ({\delta {x} \cos \varphi } - {\delta {y} \sin \varphi }\bigg ),\nonumber \\&\frac{\sin 2\varphi }{8r^2 \sin ^2\gamma } \bigg ( 3\delta {x}^2 + \cos 2\gamma \big ( \delta {x}^2 - \delta {y}^2 \big ) - 3\delta {y}^2 \bigg ) + \frac{\delta {x}\delta {y} \big ( 3+\cos 2\gamma \big ) \cos 2 \varphi }{4 r^2 \sin ^2\gamma }\nonumber \\&\left. +\,\frac{\delta {x} \delta {y}}{2 r^2} -\frac{\delta {x} \delta {y} }{ r^2 \sin ^2\theta } - \frac{\cot \gamma }{r} \big ( {\delta {x} \sin \varphi } + {\delta {y} \cos \varphi } \big ) - \frac{\delta {y} \cot \theta }{ r}\right\} , \end{aligned}$$

(59)

$$\begin{aligned} \{\delta u\}_{xz}= & {} \left\{ \frac{ \delta x^2}{2 r}+ \delta z, \, 0, \, \frac{ \delta x \delta z}{ r^2} - \frac{ \delta x}{ r}, \, \frac{ \delta x \sin \varphi }{ r \sin \gamma } -\frac{ \delta x^2 \cot \gamma \sin 2\varphi }{2 r^2 \sin \gamma }\right. \nonumber \\&-\,\frac{ \delta x \delta z \sin \varphi }{r^2 \sin \gamma }, \, \frac{ \delta x^2 \cot \gamma \sin ^2\varphi }{2 r^2} +\frac{(r\delta x -\delta x \delta z) \cos \varphi }{ r^2},\nonumber \\&\left. \frac{ \delta x^2 [3 + \cos 2\gamma ] \sin 2\varphi }{8r^2 \sin ^2\gamma } + \frac{ (\delta x \delta z - r \delta x) \cot \gamma \sin \varphi }{r^2} \right\} , \end{aligned}$$

(60)

$$\begin{aligned} \{\delta u\}_{yz}= & {} \left\{ \frac{ \delta y^2}{2 r} + \delta z, \, 0, \, \frac{ \delta y^2 \cot \theta }{2 r^2}, \, \frac{ \cos \varphi \big ( \delta y^2 \cot \gamma \sin \varphi + r \delta y - \delta y \delta z \big ) }{r^2 \sin \gamma }, \right. \nonumber \\&\frac{ \delta y^2 \cot \gamma \cos ^2\varphi }{2 r^2} +\frac{ \delta y \delta z \sin \varphi }{ r^2} -\frac{ \delta y \sin \varphi }{ r}, \, -\frac{ \delta y^2 (3+\cos 2 \gamma ) \sin 2\varphi }{8r^2\sin ^2\gamma }\nonumber \\&\left. +\,\frac{ \delta y \delta z \cot \gamma \cos \varphi }{ r^2} +\frac{ \delta y \delta z \cot \theta }{ r^2} -\frac{ \delta y \cot \gamma \cos \varphi }{ r}-\frac{ \delta y \cot \theta }{ r}\right\} , \end{aligned}$$

(61)

and

$$\begin{aligned} \{\delta u\}_{XY}= & {} \left\{ 0, \, \frac{ \delta X^2 \cos ^2 \theta }{2 R} + \delta X \sin \theta + \frac{ \delta Y^2}{2 R}, \right. \nonumber \\&\left. -\,\frac{ \delta X^2 \sin 2\theta }{2R^2} +\frac{ \delta X \cos \theta }{R}+\frac{ \delta Y^2 \cot \theta }{2 R^2}, \, 0, \, 0, \, \frac{ \delta Y}{R \sin \theta } -\frac{ \delta X \delta Y}{R^2 \sin ^2\theta }\right\} , \end{aligned}$$

(62)

$$\begin{aligned} \{\delta u\}_{XZ}= & {} \left\{ 0, \, \frac{ \delta X^2 \cos ^2 \theta }{2 R}-\frac{ \delta X \delta Z \sin 2 \theta }{2 R} +\frac{\delta Z^2 \sin ^2 \theta }{2 R} \right. \nonumber \\&+\,\delta X \sin \theta + \delta Z \cos \theta , \, \frac{ (\delta Z^2-\delta X^2) \sin 2\theta }{2R^2}\nonumber \\&\left. -\,\frac{ \delta X \delta Z \cos 2\theta }{R^2} +\frac{ \delta X \cos \theta }{R} -\frac{ \delta Z \sin \theta }{R}, \, 0, \, 0, \, 0 \right\} , \end{aligned}$$

(63)

$$\begin{aligned} \{ \delta u\}_{YZ}= & {} \left\{ 0, \, \frac{ \delta Y^2 + \delta Z^2 \sin ^2 \theta }{2 R} + \delta Z \cos \theta , \right. \left. \frac{1}{2R^2} \big (\delta Y^2 \cot \theta + \delta Z^2 \sin 2\theta \big ) \nonumber \right. \\&\left. -\,\frac{ \delta Z \sin \theta }{R }, \, 0, \, 0, \, \frac{ \delta Y }{R \sin \theta } \right\} . \end{aligned}$$

(64)

With the help of (I.14) we compute $\delta _1 f$ [using the variations (I.B8), (I.B9)] and $\delta _2 f$ [using the variations (59)–(64)] and substitute the obtained results into (58). Taking the limit $\delta \alpha \rightarrow 0$, $\delta \beta \rightarrow 0$ in (58) we get the expression for the operator $\partial _\alpha \partial _\beta$ in the SF frame. Calculating the limit one should bear in mind that $\delta \alpha$ and $\delta \beta$ are of the same order of magnitude. As an example, consider the case of the $\partial _x \partial _y$ operator (obviously, $\partial _x \partial _y = \partial _y \partial _x$). Substituting all variations into (58) we reach the final result

$$\begin{aligned} \partial _x \partial _y= & {} \frac{\sin 2\varphi }{2r^2} \left( \cot \gamma \frac{\partial }{\partial {\gamma } } - \frac{\partial ^2}{\partial {\gamma }^2 } \right) + \frac{ \cos 2\varphi }{r^2 \sin \gamma } \frac{\partial ^2}{\partial {\delta }\partial {\gamma }} + \frac{\sin 2\varphi }{2 r^2 \sin ^2\gamma } \frac{\partial ^2}{\partial {\delta }^2 }\nonumber \\&+\,\left( \frac{\cos ^2 \varphi -\cos ^2 \gamma \sin ^2 \varphi }{r^2 \sin ^2\gamma } -\frac{1}{r^2 \sin ^2\theta }\right) \frac{\partial }{\partial {\varphi } } - \frac{\cot \gamma \cos 2\varphi }{r^2 \sin \gamma } \frac{\partial }{\partial {\delta } }\nonumber \\&-\,\left[ \frac{\cot \theta \sin \varphi }{r^2 \sin \gamma } + \frac{ \cot \gamma \sin 2\varphi }{r^2 \sin \gamma } \right] \frac{\partial ^2}{\partial {\delta }\partial {\varphi }} - \frac{\cos \varphi }{r^2 \sin \gamma } \frac{\partial ^2}{\partial {\theta }\partial {\delta }}\nonumber \\&+\,\frac{\cot \gamma \cos \varphi + \cot \theta }{r^2} \frac{\partial ^2}{\partial {\theta }\partial {\varphi }} -\left[ \frac{\cot \gamma \cos 2\varphi }{r^2} + \frac{\cot \theta \cos \varphi }{r^2} \right] \frac{\partial ^2}{\partial {\gamma }\partial {\varphi }}\nonumber \\&+\,\frac{\sin \varphi }{r^2} \frac{\partial ^2}{\partial {\theta }\partial {\gamma }} + \left( \frac{\cot \gamma \cot \theta \sin \varphi }{r^2} +\frac{\cot ^2 \gamma \sin 2\varphi }{2r^2}\right) \frac{\partial ^2}{\partial {\varphi }^2 }. \end{aligned}$$

(65)

Using the representations of the total angular momentum operators $\varPi _{x}$, $\varPi _{y}$ and $\varPi _{z}$ [see (I.A1)] the expression for $\partial _x \partial _y$ can be expressed in a more compact form:

$$\begin{aligned} \partial _x \partial _y = \frac{1}{\hbar ^2 r^2} \bigg [ \varPi _y \varPi _x + \cot \theta \varPi _y\varPi _z \bigg ] + \frac{i}{\hbar r^2 }\csc ^2\theta \, \varPi _z - \frac{i}{\hbar r^2} \bigg [ \varPi _x + \cot \theta \varPi _z \bigg ] \frac{\partial }{\partial {\theta } }. \end{aligned}$$

(66)

In the same manner we can compute all operators and finally arrive at the following representations for operators $\mathcal {O}_{\alpha \beta }$

$$\begin{aligned} \mathcal {O}_{xy}= & {} \frac{2}{\hbar ^2 r^2} \bigg [ \varPi _{y}\varPi _x + \cot \theta \varPi _{y}\varPi _z + \frac{i\hbar \, \varPi _z}{\sin ^2 \theta } - {i\hbar } \left( \varPi _x + \cot \theta \varPi _z \right) \frac{\partial }{\partial {\theta } } \bigg ], \end{aligned}$$

(67)

$$\begin{aligned} \mathcal {O}_{xz}= & {} \left( \frac{2}{r^2} - \frac{2}{r} \frac{\partial }{\partial {r} } \right) \left( \frac{i}{\hbar } \varPi _y + \frac{\partial }{\partial {\theta } } \right) , \end{aligned}$$

(68)

$$\begin{aligned} \mathcal {O}_{yz}= & {} \frac{2i}{\hbar } \left( \frac{1}{r} \frac{\partial }{\partial {r} } - \frac{1}{r^2} \right) \left( \varPi _x + \cot \theta \varPi _z \right) , \end{aligned}$$

(69)

$$\begin{aligned} \mathcal {O}_{XY}= & {} -\frac{2i}{\hbar } \left( \frac{1}{R}\frac{\partial }{\partial {R} } + \frac{\cot \theta }{R^2} \frac{\partial }{\partial {\theta } } - \frac{\csc ^2\theta }{R^2} \right) \varPi _z, \end{aligned}$$

(70)

$$\begin{aligned} \mathcal {O}_{XZ}= & {} \left( \frac{2}{R}\frac{\partial }{\partial {R} } - \frac{2}{R^2} \right) \cos 2\theta \frac{\partial }{\partial {\theta } } + \sin 2\theta \left( \frac{\partial ^2}{\partial {R}^2 } - \frac{1}{R} \frac{\partial }{\partial {R} } - \frac{1}{R^2} \frac{\partial ^2}{\partial {\theta }^2 } \right) , \end{aligned}$$

(71)

$$\begin{aligned} \mathcal {O}_{YZ}= & {} -\frac{2i}{\hbar } \left( \frac{\cot \theta }{R}\frac{\partial }{\partial {R} } - \frac{1}{R^2} \frac{\partial }{\partial {\theta } } \right) \varPi _z. \end{aligned}$$

(72)

The transformation of the 1st-order derivative operators $\partial _{\alpha }$ is considerably simpler and can be done within the same technique by employing variations (I.B8), (I.B9) and the standard definition for the 1st-order derivative operator [which we use now in place of (I.10)]

$$\begin{aligned} \partial _{\alpha } f = \lim _{\delta \alpha \rightarrow 0} \frac{ \delta ^{(\alpha )}_1 f }{\delta \alpha } , \end{aligned}$$

(73)

Obviously, for the purpose of the $\partial _{\alpha }$ transformation, one needs the linear parts (with respect to the $\delta \alpha$) of the expressions (I.13), (I.14), (I.B8) and (I.B9) only. One finally obtains the following representations for the $\partial _{\alpha }$ operators in the SF frame:

$$\begin{aligned} \partial _{x}= & {} - \frac{1}{r} \frac{\partial }{\partial {\theta } } - \frac{ i \varPi _y}{ \hbar \, r}, \quad \partial _{y} = \frac{i}{ \hbar \, r} \left( \varPi _x + \cot \theta \, \varPi _z \right) , \quad \partial _{z} = \frac{\partial }{\partial {r} },\nonumber \\ \partial _{X}= & {} \frac{\cos \theta }{R}\frac{\partial }{\partial {\theta } } + \sin \theta \frac{\partial }{\partial {R} }, \quad \partial _{Y} = -\frac{i}{\hbar }\frac{\varPi _z}{R\sin \theta }, \end{aligned}$$

(74)

$$\begin{aligned} \partial _{Z}= & {} \cos \theta \frac{\partial }{\partial {R} } - \frac{\sin \theta }{R}\frac{\partial }{\partial {\theta } }. \end{aligned}$$

(75)

The representations for the differential operators (65)–(72), (74) and (75) [as well as (48)–(50) for the diatomic molecule] are obtained under the geometrical constraints presented by the embedding $\big [ \mathbf {r} \, || \mathbf {z}\,\big ]$ of the Jacobi vectors $\mathbf {R}$ and $\mathbf {r}$ in the BF frame.

Finally we give the variations needed to transform the BF Cartesian derivative operators $\partial _{\alpha }$ and $\partial _{\alpha \beta }$ ($\alpha ,\beta =x,y,z$) into spherical SF coordinates $\{r,\theta ,\varphi \}$ used in Sect. 4 for the diatomic system:

$$\begin{aligned}&\{\delta u\}_{x} = \bigg \{ \frac{\delta x^2}{2r}, \, \frac{\delta x}{r}, \, 0 \bigg \}, \, \{\delta u\}_{y} = \bigg \{\frac{\delta y^2}{2r}, \, \frac{\delta y^2 \cot \theta }{2r^2}, \, \frac{\delta y }{r \sin \theta } \bigg \},\nonumber \\&\{\delta u\}_{z} = \big \{ \delta z, 0, 0 \big \}. \end{aligned}$$

(76)

$$\begin{aligned}&\{\delta u\}_{xy} = \bigg \{ \frac{\delta x^2 +\delta y^2}{2r}, \, \frac{2r \delta x + \cot \theta \delta y^2}{2r^2}, \, \frac{\delta y }{r \sin \theta } - \frac{\delta x \delta y \cos \theta }{r^2 \sin ^2 \theta } \bigg \},\nonumber \\&\{\delta u\}_{xz} = \bigg \{ \delta z + \frac{\delta x^2}{2r}, \, -\frac{\delta x \delta z}{r^2} + \frac{\delta x}{r}, \, 0 \bigg \},\nonumber \\&\{\delta u\}_{yz} = \bigg \{ \delta z + \frac{\delta y^2}{2r}, \, \frac{\delta y^2 \cot \theta }{2r^2}, \, -\frac{\delta y \delta z}{r^2 \sin \theta } + \frac{\delta y}{r \sin \theta } \bigg \}. \end{aligned}$$

(77)

B The recurrent relations between the matrix elements involving the associated Legendre polynomials

The general expression for the matrix elements $\left\langle n'm' | nm \right\rangle$, (I.C9), rather impractical for the large values of $n'$, n, $m'$ and m, because of the presence of numerous factorials and summations. Here we give the simple recurrent schemes for calculation of these matrix elements for some combinations of the quantum numbers. Namely, we consider the case $m'-m=2$. The mentioned restriction on the quantum numbers related to the practical needs in the calculation of the effective Hamiltonian [see (42) as an example].

We shall use the ideas used in the [34]. The associated Legendre polynomials $P_{n}^{m}(x)$ can be defined by the following two equivalent representations

$$\begin{aligned} P_{n}^{m}(x)= & {} (-1)^m \frac{(1-x^2)^{m/2}}{2^n n!} \partial _x^{n+m} (x^2-1)^n\nonumber \\= & {} \frac{(n+m)!}{2^n n! (n-m)!} {(1-x^2)^{-m/2}} \partial _x^{n-m} (x^2-1)^n, \end{aligned}$$

(78)

where (similarly as in the previous sections) the symbol $\partial _x^{\,k}$ denotes the k-th derivative on the variable x.

We shall use the following relation [see [35], (8.733-4)]

$$\begin{aligned} P_{n-1}^{m}(x) - P_{n+1}^{m}(x) = (2n+1) \sqrt{1-x^2} P_{n}^{m-1}(x). \end{aligned}$$

(79)

In the calculations of the integrals it is convenient to use the un-normalized expressions (78) for associated Legendre polynomials. The standard normalization coefficient $B_{nm}$ reads

$$\begin{aligned} \int _{-1}^{+1} \big [ B_{nm} P_{n}^{m}(x) \big ]^2 dx = 1, \quad B_{nm} = \sqrt{ \frac{(2n+1)}{2} \frac{(n-m)!}{(n+m)!} }. \end{aligned}$$

(80)

Let us introduce the following notation for the matrix elements with the polynomials $P_n^{m}(x)$

$$\begin{aligned} \int _{-1}^{+1} P_{n'}^{m'}(x) P_{n}^{m}(x) dx = \left\langle P_{n'}^{ m'} | P_n^{m} \right\rangle . \end{aligned}$$

(81)

We are going to derive: (i) the recurrent relations between the matrix elements $\left\langle P_n^{ m} | P_n^{m-2} \right\rangle$ and $\left\langle P_n^{ m-1} | P_n^{m+1} \right\rangle$ (type I); (ii) the matrix elements $\left\langle P_n^{2}|P_{n-k}^{0} \right\rangle$, $k=2,4,\ldots ,n$, where n is an even number (type II). Without loss of generality we assume that the quantum numbers m and $m'$ in (81) are positive, i.e. for the matrix elements of the type I, $m \ge 2$.

1.1 B.1 Matrix elements of the type I

Let us to introduce a short notation $I_{nm} = \left\langle P_n^{ m} | P_n^{ m-2} \right\rangle$. We shall derive the relation between the matrix elements $I_{nm}$ and $I_{nm+1}$, where $I_{nm+1}$ = $\left\langle P_n^{ m-1} | P_n^{m+1} \right\rangle$ = $\left\langle P_n^{ m+1} | P_n^{m-1} \right\rangle$. By using the representation (78) we have

$$\begin{aligned} I_{nm} = C_{n m} \int _{-1}^{+1} (1-x^2) \partial _x^{\, k} q_n(x) \partial _x^{\, p+2} q_n(x) dx, \quad q_n(x) =(x^2-1)^n, \end{aligned}$$

(82)

where $k=n+m$, $p=n-m$, and

$$\begin{aligned} C_{n m}=\frac{(-1)^m (n+m-2)!}{[2^n n!]^2(n-m+2)!}. \end{aligned}$$

(83)

Integrating by parts in (82) [$I_{nm} \equiv \int udv$, where $u=(1-x^2) \partial _x^{\, k} q_n(x)$] the $I_{nm}$ can be represented as follows:

$$\begin{aligned} I_{nm}= & {} -\,C_{n m} \int _{-1}^{+1} (1-x^2) \partial _x^{\,k+1} q_n(x) \, \partial _x^{\,p+1} q_n(x) dx\nonumber \\&+\,2 C_{n m} \int _{-1}^{+1} x \, \partial _x^{\,k} q_n(x) \, \partial _x^{\,p+1} q_n(x) dx = -\,C_{nm} I_{nm}^{(0)} + 2 C_{n m} I_{nm}^{(1)}. \end{aligned}$$

(84)

Consider (84) at $m=n$ (i.e. $k=2n$ and $p=0$), which give us the matrix element $I_{nn} \equiv \left\langle P_n^{ n} | P_n^{ n-2} \right\rangle$. Because of $\partial _x^{\,k+1} q_n(x) \equiv 0$, the first integral $I_{nm}^{(0)}$ in (84) is zero and $I_{nn}$ is represented solely by the second integral $I_{nm}^{(1)}$. Integrating the $I_{nm}^{(1)}$ by parts (with $u=x$) and using the relation

$$\begin{aligned} \int _{-1}^{+1} \partial _x^{\, k} q_n(x) \partial _x^{\, p} q_n(x) dx = K_{nm}, \quad K_{nm}= \frac{2(-1)^{m} \big ( 2^n n! \big ) ^2 }{2n+1} , \end{aligned}$$

(85)

we reach the final result for the matrix element $I_{nn}$

$$\begin{aligned} I_{nn}= -2 C_{nn} \int _{-1}^{+1} \partial _x^{\,k} q_n(x) \partial _x^{\,p} q_n(x) dx = \frac{-2(2n-2)!}{2n+1}. \end{aligned}$$

(86)

Consider now the general case, $n>m$. The integral $I_{nm}^{(0)}$ in the (84) represents (up to a constant) the matrix element $I_{nm+1}$ [one can see this immediately, by comparing the expression for $I_{nm}^{(0)}$ with the (82)], therefore we can write

$$\begin{aligned} I_{nm} = -\frac{C_{nm}}{C_{nm+1}} I_{nm+1} + 2C_{nm} I_{nm}^{(1)}. \end{aligned}$$

(87)

Integrating $I_{nm}^{(1)}$ by parts (with the $u= x \partial _x^{\,k} q_n(x)$) we obtain

$$\begin{aligned} I_{nm}^{(1)} = - K_{nm} - I_{nm}^{(2)}, \end{aligned}$$

(88)

where $I_{nm}^{(2)}$ can be treated in the same way and result is

$$\begin{aligned} I_{nm}^{(2)} = \int _{-1}^{+1} x \, \partial _x^{\,k+1} q_n(x) \partial _x^{\,p} q_n(x) dx = - K_{nm+1} -I_{nm}^{(3)}. \end{aligned}$$

(89)

We can write a general expression for $I_{nm}^{(r)}$ as follows

$$\begin{aligned} I_{nm}^{(r)} = - K_{n \, m+r-1} - I_{nm}^{(r+1)}, \end{aligned}$$

(90)

where

$$\begin{aligned} I_{nm}^{(r)} = \int _{-1}^{+1} x \, \partial _x^{\,k+r-1} q_n(x) \partial _x^{\,p-r+2} q_n(x) dx. \end{aligned}$$

(91)

The last non-zero integral $I_{nm}^{(N)}$ ($r=N$) can be determined from the condition $k+N-1=2n$ (i.e. $N=n-m+1$) and the result is $I_{nm}^{(N)} = K_{nn}$. Therefore, by using the recurrence relation (90) we can obtain the $I_{nm}^{(1)}$, namely

$$\begin{aligned} I_{nm}^{(1)} = -(n-m+1) K_{nm}. \end{aligned}$$

(92)

Substituting this expression for $I_{nm}^{(1)}$ into the (87) we finally reach the resulting recurrence relation

$$\begin{aligned} I_{nm} = \frac{I_{nm+1}}{(n+m-1)(n-m+1)} - \frac{4(n+1-m)}{(2n+1)}\frac{(n+m-2)!}{(n-m+2)!}. \end{aligned}$$

(93)

The recurrent relation (93) together with initial value for $I_{nn}$, (86), allows to calculate the all necessary matrix elements of the type $\left\langle P_n^{m} | P_n^{m-2} \right\rangle$.

1.2 B.2 Matrix elements of the type II

First we shall demonstrate that for $k > 0$ the auxiliary integral $J_k$ equal to zero

$$\begin{aligned} J_{k} \equiv \int _{-1}^{+1} (1-x^2) \partial _x^{n+2} (x^2-1)^n \bigg [ \partial _x^{n-k-2} (x^2-1)^{n-k-1} \bigg ] dx = 0. \end{aligned}$$

(94)

By means of the representations (78) we have

$$\begin{aligned}&(1-x^2) \partial _x^{n+2} (x^2-1)^n \propto P_n^{2}(x), \end{aligned}$$

(95)

$$\begin{aligned}&\partial _x^{n-k-2} (x^2-1)^{n-k-1} \propto \sqrt{1-x^2} P_{n-k-1}^{1}(x). \end{aligned}$$

(96)

Using now relation (79) the right-hand side of the (96) can be given as the linear combination of $P_{n-k-2}^{2}(x)$ and $P_{n-k}^{2}(x)$, therefore the integral $J_k$ can be represented as follows

$$\begin{aligned} J_{k} \propto \int _{-1}^{+1} P_n^{2}(x) \big ( a \, P_{n-k-2}^{2}(x) + b \, P_{n-k}^{2}(x) \big ) dx, \end{aligned}$$

(97)

The explicit expressions of the coefficients a and b can be obtained directly from (79). From the representation (97) it is obvious, that $J_k=0$ (for $k>0$) due to orthogonality properties of the associated Legendre polynomials.

Let us prove now the following relation

$$\begin{aligned} \left\langle P_n^{2} | P_{n-k}^{0} \right\rangle = \left\langle P_n^{2} | P_{n-k-2}^{0} \right\rangle = \cdots = \left\langle P_n^{2} | P_{0}^{0} \right\rangle . \end{aligned}$$

(98)

We shall use the induction method. Obviously, it is enough to show that relation

$$\begin{aligned} \left\langle P_n^{2} | P_{n-k}^{0} \right\rangle = \left\langle P_n^{2} | P_{n-k-2}^{0} \right\rangle \end{aligned}$$

(99)

holds for arbitrary $k>0$. Let us introduce the following notations: $\left\langle P_n^{2} | P_{n-k}^{0} \right\rangle \equiv J_A$ and $\left\langle P_n^{2} | P_{n-k-2}^{0} \right\rangle \equiv J_B$. Using the representation (78) we have

$$\begin{aligned} J_A= & {} C_k \int _{-1}^{+1} (1-x^2) \partial _x^{n+2} (x^2-1)^n \bigg [ \partial _x^{n-k} (x^2-1)^{n-k} \bigg ] dx, \end{aligned}$$

(100)

$$\begin{aligned} J_B= & {} C_{k+2} \int _{-1}^{+1} (1-x^2) \partial _x^{n+2} (x^2-1)^n \bigg [ \partial _x^{n-k-2} (x^2-1)^{n-k-2} \bigg ] dx, \end{aligned}$$

(101)

where

$$\begin{aligned} C_k = \bigg [ 2^{2n-k} n! (n-k)! \bigg ]^{-1}. \end{aligned}$$

(102)

Consider (100). We have,

$$\begin{aligned} \partial _x^{n-k} (x^2-1)^{n-k} = a_1 \, \partial _x^{n-k-2} (x^2-1)^{n-k-1} + a_2 \, \partial _x^{n-k-2} (x^2-1)^{n-k-2}, \end{aligned}$$

where $a_1=2(n-k)(2n-2k-1)$, $a_2=4(n-k)(n-k-1)$. Hence,

$$\begin{aligned} J_A= & {} C_k a_1 \int _{-1}^{+1} (1-x^2) \partial _x^{n+2} (x^2-1)^n \bigg [ \partial _x^{n-k-2} (x^2-1)^{n-k-1} \bigg ] dx\nonumber \\&\quad +\,C_k a_2 \int _{-1}^{+1} (1-x^2) \partial _x^{n+2} (x^2-1)^n \bigg [ \partial _x^{n-k-2} (x^2-1)^{n-k-2} \bigg ] dx.\nonumber \\ \end{aligned}$$

(103)

The first integral in (103) is zero, see (94). Taking into account that $C_k a_2 = C_{k+2}$, one can find immediately, that the second integral in (103) coincides with $J_B$. Hence, we obtain $J_A = J_B$, which completes the proof of the relation (99); by induction this also proves the relation (98) .

Finally, we need the explicit expression for any of the matrix elements in (98). For this purpose we shall obtain the expression for the simplest one $J_0 = \left\langle P_n^{2} | P_{0}^{0} \right\rangle$. The integral $J_0$ is quite trivial and can be obtained by sequential use of the relation (8.735) in [35]. However, to keep the representation self consistent we give below a short derivation. Integrating $J_0$ by parts twice we obtain

$$\begin{aligned} J_0= & {} C_n \int _{-1}^{+1} (1-x^2) \partial _x^{n+2} (x^2-1)^n dx\nonumber \\= & {} -\,4 C_n \int _{-1}^{+1} \partial _x^{n} (x^2-1)^n dx + 2 C_n uv \mid _{-1}^{+1}, \end{aligned}$$

(104)

where $u=x$ and $v= \partial _x^{n} (x^2-1)^n$. The integrand in the second line of (104) is the Legendre polynomial $P_n^{0}(x)$, therefore the corresponding integral is zero due to orthogonality properties. Hence $J_0=2 C_n u v \mid _{-1}^{+1}$ and can be calculated as follows

$$\begin{aligned} J_0= & {} 2 C_n \bigg [x \, \partial _x^{n} (x^2-1)^n \bigg ]_{-1}^{+1} \nonumber \\= & {} 4C_n \sum _{k=n/2}^{n} (-1)^{n-k} \frac{n!}{k! (n-k)!} \frac{(2k)!}{(2k-n)!}\nonumber \\= & {} C_n 2^{n+2} n! = 4. \end{aligned}$$

(105)