Gaussian Integrals Depending by a Quantum Parameter in Finite

Home , Gaussian function, Gaussian integral

arXiv:2107.06874v1 [math.GM] 14 Jul 2021 Keywords. q solving of practice the integrals. regarding Gaussian analogies and examples methods, h asinitga in integral Gaussian The 2 Introduction 1 Contents Classification. functions. Subject Boys AMS number, Grassmann mean, geometric lz atclrcaso asinitgasta eed ytequ the by s sc depends we that of notes integrals branches these Gaussian In different of parameter in class interesting. used particular and a are actual alyze always they T topic that integrals. the Gauss fact makes of evaluation the the with is mathematics coupled in theme common A h asinitga in integral Gaussian The 3 asinitgasadhmgnosfntos14 functions homogeneous and integrals Gaussian 4 asinitgasdpnigb quantum a by depending integrals Gaussian ∗ . asinitga rGusterm 11 7 ...... theorem? . Gauss . or integral integral Gaussian Gaussian the write to 2.2 Ways 2.1 e-mail [email protected] auz (CN) Saluzzo [email protected], : ~ trigfo lsia eut,w ilpeeta vriwon overview an present will we results, classical from Starting . asinitga,qatzto,seilfntos arithmetic– functions, special quantization, integral, Gaussian aaee nfiiedimension finite in parameter ioeCamosso Simone 1 n Abstract 31,0A5 75,8S0 78,92E10. 97I80, 81S10, 97I50, 00A05, 33B15, dmnin3 –dimension dmnin 12 –dimensions 1 ∗ alan- hall uantum antum ience, his, 2 5 Gaussian matrix integrals 15

6 The symplectic structure 17

7 The quantum postulates 18

8 Gaussian integrals depending by a quantum parameter 19

9 Quantum integrals 23

10 On a Gaussian integral in one dimension and the Lemniscate 24

11 The Berezin–Gaussian integral: a generalization of the Gaus- sian integral 27 11.1 On the Laplace method and a “mixed Gaussian Integral” . . . 29

12 Quantum Gaussian integrals with a perturbative term 30

13 Gaussian integrals and the Boys integrals 31

Bibliography 34

1 Introduction

Let ~ be a quantum parameter, an important practice in quantum mechanics, is the evaluation of Gaussian integrals. We will focus our attention on a group of Gaussian integrals that come from quantum mechanics and quantum ﬁeld theory (QFT). A general Gaussian integral in this class depends by a 1 ~ quantum parameter k = ~ with “ 0 ”. In our calculations we shall treat 1 → k = ~ as a purely formal parameter. Summarising, we will treat integrals of the following form:

i ψ(v,w) A (v,w) e− ~ dv, (1) Rn Z where A is the “amplitude”, ψ is generally quadratic and v,w Rn. This paper is based on previous discussions of [Zee], [Wi], [GS1]∈ and [GS2] on diﬀerent Gaussian integrals of the form (1). A basic text is the work of [H¨o] and, for the asymptotic analysis we treat only the Laplace method described in [Wo]. In order to get the result of a Gaussian integral, sometimes it is

2 necessary to introduce a special function (a good reference for the properties of special functions is [A]). Instead on the techniques used, an exhaustive source is the book of Nahin [Na]. Regarding the different examples discussed in these pages, they have been inspired by the following works concerning the geometric quantization, they are [GP], [P1], [P2], [P3] and [Cam]. At the beginning of the paper we will review basic facts concerning Gaus- sian integrals in 1 and n–dimension. We proceed with the definition of Gaus- sian matrix integrals and with different examples coming from geometric quantization. A special section of this work is dedicated to the connection between Gaussian integrals and elliptic functions. In this special section old results of number theory illustrated by [S1] and [S2], have been used in order to express certain Gaussian integrals in term of the arithmetic–geometric mean AGM. The next section is dedicated to the Berezin–Gaussian integral. The Berezin integral is defined for G–numbers (Grassmannian numbers) that are noncommutative quantities. The generalized version of the Gaussian integral for these variables presents interesting mathematical properties (details are in [Zee]). The last section concerns the relation between Gaussian integrals and Boys integrals. Boys integrals are used in quantum chemistry and they are of fundamental importance for many electron systems ([Bo],[Ki]). In the end we want to specify that all examples and discussion has been considered only for the finite dimensional case. There are generalizations of the Gaussian integral in infinite dimension. For example it is possible consider Gaussian integrals in Hilbert spaces, but the discussion of this subject is not part of these notes (examples of interesting articles on this suject are [Ne] and [Dr]). The reader can think about this paper as a manual (sure not complete!) or a sort of digest for common use by both, students and teachers.

2 The Gaussian integral in 1–dimension

We start with the Gauss integral:

3 + ∞ x2 e− dx = √π. (2) Z−∞ The result follows taking the square of the left–hand side and using the polar coordinates:

+ 2 + + ∞ x2 ∞ ∞ x2 y2 e− dx = e− − dxdy (3) −∞ −∞ −∞ Z 2π Z + Z + ∞ ρ2 ∞ ρ2 = ρe− dρdϑ = 2π ρe− dρ 0 0 0 Z Z + Z t ∞ s 1 = π e− ds = π lim = π. · t + −es Z0 → ∞ 0

In his article ([Con]) Conrad gives eleven proofs of the integral (2) using diﬀerent methods. A quick method consist to use the non elementary formula: π Γ(s) Γ(1 s) = , (4) · − sin (πs) + ∞ x s 1 where Γ(s) = 0 e− x − dx (for s< 0). Using (4) we have that: R + 2 + 2 ∞ x2 ∞ x2 e− dx = 2 e− dx (5) 0 Z−∞ Z 1 1 π = Γ Γ = = π. 2 · 2 sin π 2 The curious reader can read the interesting article of [I] where the author treat the integrals as a “puzzle” to solve. Another Gaussian integral is the following:

+ ∞ ax2 π e− dx = , (6) a Z−∞ r for every a > 0. We can modify the original integral in order to obtain diﬀerent versions. For example adding the term bx we ﬁnd that: −

4 + 2 ∞ ax2 bx b π e− − dx = e 4a . (7) a Z−∞ r 2 We can prove the formula completing the square ax2 bx = √ax + b + − − − 2√a b2 4a . In this case:

2 + + b b2 ∞ ax2 bx ∞ √ax+ + e− − dx = e− 2√a 4a dx (8)

Z−∞ Z−∞ 2 2 b 2 + b + 2 b ∞ √ax+ e 4a ∞ s2 b π = e 4a e− 2√a dx = e− ds = e 4a . √a a Z−∞ Z−∞ r Alternatives of the Gaussian integral are in the book of Nahin [Na], an example is:

+ ∞ 2n x2 (2n)!√π x e− dx = , (9) 2n!4n Z0 for n 0, where for n = 0 we reﬁnd (2) between 0 and + . Another version≥ of (9) is the following: ∞

+ ∞ n ax2 1 3 5 (n 1)√π x e− dx = · · ···n n+1− , n =2, 4,... (10) 2 2 a 2 Z−∞ where a> 0, or this:

+ px2 qx2 ∞ e− e− − dx = √π (√q √p) , (11) x2 − Z0 for q>p 0. A complex≥ modiﬁcation of the Gaussian integral (7) is the following:

+ 2 ∞ ax2 ηx η π e− − dx = e 4a , (12) a Z−∞ r where η C and the integral converges uniformly in any compact region. Now, the∈ integral deﬁnes an analytic function that may be evaluated by taking η to be real and then using analytic continuation. To prove this we use the same trick of the integral (7) and the result is true for all η C. As observed in [GS1] when η = iξ we get that: ∈

5 + 2 ∞ ax2 iξx π ξ e− − dx = e− 4a , (13) a · Z−∞ r 2 ax2 ξ or, in other words, the Fourier transform of e− is the Gaussian e− 4a . Considering the properties of the Fourier transform we can prove the following result.

Proposition 2.1 Let m be a positive integer and k be a real positive constant, then

+ m ∞ m iξx 1 kx2 ( i) 1 ξ2 x e− − 2 dx = √2π − Pm(ξ)e− 2k , (14) m+ 1 k 2 Z−∞ m m 2j where Pm(ξ) = ξ + j 1 pmjξ − is a monic polynomial in ξ of degree m and parity ( 1)m. ≥ − P Proof .

By a property of the Fourier transform:

m m 1 kx2 m d 1 kx2 x e− 2 = i e− 2 . F dξm F 2 1 2 π ξ 2 kx 2 2k Now the Fourier transform of e− is equal to k e− and deriving m 1 ξ2 times e− 2k we find the polynomial Pm where the principalq term has coeffi- ( 1)m cient −km . We must collect this coefficient in order to find the polynomial Pm. 2 In one dimension we can do much more using special functions as the gamma function and its inductive property Γ(n+1) = n Γ(n). For example · we have the cubic Gauss integral:

+ ∞ x3 4 e− dx = Γ . (15) 3 Z0 This can be proved by a simple substitution y = x3:

+ + ∞ x3 1 ∞ y 2 1 1 4 e− dx = e− y− 3 dy = Γ = Γ . (16) 3 3 · 3 3 Z0 Z0 6 This last argument can be generalized for integrals of the form:

+ ∞ xm 1+ m e− dx = Γ , (17) m Z0 where m 1. We observe that for the case m = 2 the gamma function is 3 √≥π Γ 2 = 2 and we recover (2) between 0 and + . Another approach involves the use of the Taylor∞ expansion. In order to explain the method we consider the following integral:

+ 2 ∞ e x e− − 1 dx. (18) − Z−∞ In this case we can work “formally” and write that:

2 e x x2 1 2x2 1 3x2 e− − 1 = e− + e− e− + , − − 2! − 3! ··· where we use the Mc. Laurin expansion of ex 1+ x + x2 + . Now ∼ 2! ··· integrating each term we have that:

+ + + + 2 ∞ k ∞ k ∞ e x ( 1) ∞ kx2 ( 1) π e− − 1 dx = − e− dx = − . (19) − k! k! k k=1 k=1 r Z−∞ X Z−∞ X So we have:

+ + 2 ∞ k ∞ e x ( 1) π e− − 1dx = − . (20) − k! k k=1 r Z−∞ X 3 + e x The reader can try with ∞ e− − 1 dx, have a nice exercise! 0 − R 2.1 Ways to write the Gaussian integral There are diﬀerent ways to write the Gaussian integral (2), for example with the substitution x2 = ln t, we have:

+ ∞ 1 dt = √π. (21) 0 2√ln t Z Another substitution is to put ex = s, so:

7 + ∞ ln s 1 s− − ds = √π. (22) Z0 A no obvious way to write the Gaussian integral (2) is the following:

2 π + x ∞ 1 + sin arctan e− 2 − dx = √π, (23) x2 π cos arctan e− Z−∞ − 2 x where the Gudermannian function gd(x) = 0 sechwdw is used in order to prove the integral (see [GR] for useful mathematical relations). R Another way to express the Gaussian integral is through the error function:

+ ∞ t2 e− dt = √π erf( ), (24) ∞ Z−∞ where

x 2 t2 erf(x) = e− dt, (25) √π Z0 and it has interesting properties (see [L]). For example we can express the “ Plasma dispersion function ” in terms of the erf function. The “ Plasma dispersion function ” is deﬁned as:

+ t2 1 ∞ e D(x) = − dt, (26) √π t x Z−∞ − for (x) > 0, and its analytic continuation to the rest of the complex x plane.ℑ Instead of the function (27) we can consider the “ incomplete Plasma dispersion function”:

+ t2 1 ∞ e D(ν, x) = − dt, (27) √π t x Zν − for (x) > 0. We obtain D(x) as limit of D(ν, x) for ν . The function ℑ → −∞ D(ν, x) satisfy the following diﬀerential equation:

d 1 e ν2 D(ν, x)+2xD(ν, x) = − 1+erf(ν). dx √π ν x − −

8 For references see [Ba]. For ν , from erf( ) = 1, we have that: → −∞ −∞ −

D′(x)+2xD(x) = 2, − that is a linear diﬀerential equation with solution:

x x2 t2 D(x) = e− c 2 e dt . − Z0 In order to determine c we observe that in the positive part of the complex plane for x 0 we have that: → + t2 1 ∞ e D(x) = pp − dt + i√π = i√π, √π t "Z−∞ # where pp is the principal part and the argument of the integral is an odd function, so the integral is zero. In the end we have that:

x x x2 t2 x2 2i t2 x2 D(x) = e− i√π 2 e dt = i√πe− 1+ e dt = i√πe− [1 + erf(ix)] , − √π Z0 Z0 that is the relation searched. The last way to write the Gaussian integral, is given by Ramanujan in [Ha] using the following continued fraction for the integral:

a a2 x2 √π e− e− dx = (28) 2 1 0 − 2a + a+ 2 Z 2a+ 3 a+ 4 2a+ 5 ··· We give a little sketch of the proof. First we observe that the continued fraction (28) is equivalent to:

+ a2 ∞ x2 1 e e− dx = 1 , (29) a 2a + a+ 2 Z 2a+ 3 a+ 4 2a+ 5 ··· √π + 2 a 2 + 2 + 2 ∞ x x ∞ x ∞ x in fact 2 = 0 e− dx so 0 e− dx 0 e− dx = a e− dx, now 2 − − we change the sign and multiply by ea in order to obtain (28). R R R R 9 Second we consider the generating function:

+ x2 ∞ x2 ψ(x) = e e− dx. Zx The function ψ has interesting properties, for example the n–th derivative is given by:

ψ(n)(x) = P (x)ψ(x) Q (x), n − n where Pn and Qn are given by the following recurrence relations:

P0(x)=1, P1(x)=2x;

Q0(x)=0, Q1(x) = 1;

Pn+1(x)=2xPn(x)+2nPn 1(x); − Qn+1(x)=2xQn(x)+2nQn 1(x). − Furthermore we have the following identity proved by induction:

Q (x)P (x) P (x)Q (x)=( 2)nn! n+1 n − n+1 n − and the following expression for Pn(x):

n x2 d x2 P (x) = e− e . n dxn Third and ﬁnal step we have to prove that:

Q (x) lim n = ψ(x), n + → ∞ Pn(x) for all x > 0. The limit can be showed using upper and lower bounds on ψ(x) in a similar way of [Ko] that in its article used a diﬀerent generating x2 + x2 2 ∞ 2 function φ(x) = e x e− dx instead our ψ. Now it is clear that Qn(x) is the n–th convergent of the continued fraction: R Pn(x) 1 2 2a + 2a+ 4 2a+ 6 2a+ 8 2a+ 10 ··· that is equivalent simplifying by 2 to:

10 1 1 2a + a+ 2 2a+ 3 a+ 4 2a+ 5 ··· and the result (28) is proved. Passing to the limit:

+ a2 ∞ x2 √π e− e− dx = 2 lim   . (30) a + 2 1 → ∞ − 2a + a+ 2 Z−∞ a 3  2 + 4   a+ 5   2a+   ···  2.2 Gaussian integral or Gauss theorem? The reader that is familiar with physics knows the statement of the Gauss theorem for the flux of the electric field E~ generated by a finite number of charges qi confinated in a closed surface. This statement can be summarized by the formula:

n q Φ E~ = i=1 i , (31) ε0 P where ε0 is called the dielettric constant. It is curious that the formula can be expressed in this other complicated way:

n + πε0 y2 2 ∞ − Pn q Φ E~ = sgn q e i=1 i dy . (32) i · | | i=1 ! X Z−∞ This is a way to express the Gauss theorem with a Gaussian integral! This formulation seems not very useful, may be or may be not! The reader can imagine heuristically to write the charges qi as an integer multiplied by the 19 fundamental charge e = 1, 6 10− C. Now e is related to ~ and it is possible to substitute inside the integral· the parameter ~. Somebody can Taylor expand the integral in powers of ~, can this provides a quantization for the ﬂux? The interested reader can think about these as formal manipulations, or simply let it go and move on to the next section.

11 3 The Gaussian integral in n–dimensions

In n dimension we have that:

x 2 n e−k k dx = √π , (33) Rn Z 2 2 2 x 2 where x = x1 + +xn. The result follows observing that Rn e−k k dx = k k n ··· + t2 ∞ e− dt and using the polar coordinates: R −∞ R + n + 1 ∞ t2 1 ∞ ρ2 n 1 e− dt = e− ρ − c dρ (34) n n n √π √π 0 Z−∞ Z + cn 1 ∞ n 1 s 1 cn n = s 2 − e− ds = Γ = 1, 2 √πn √πn 2 2 Z0 n 2π 2 where cn = n is the area of the unit sphere. The result has been gener- Γ( 2 ) alized by [Hö] for a symmetric positive definite n n–matrix A: × n Ax,x √π e−h idx = , (35) Rn √det A Z where , is the inner product on Rn. This result is the analogue of (6) in oneh· dimension.·i Always in [Hö] we can find another result for the Fourier transform. We recall here the theorem.

Theorem 3.1 (Hörmander) If A is a non–singular symmetric matrix and 1 Ax,x A 0 the Fourier transform of u(x) = e− 2 h i is a Gaussian function ℜ ≥ n 1 1 Bξ,ξ 1 u(ξ) = (2π) 2 (det B) 2 e− 2 h i where B = A− and the square root is well defined. If A = iA0 where A0 is real, symmetric and non-singular then n 1 πi sgnA0 1 1 − i A− ξ,ξ ub(ξ) = (2π) 2 det A0 − 2 e 4 − 2 h 0 i. | | In the previous theorem the term sgnA is called the signature of A . b 0 0 The Gaussian integral (35) admits different generalizations, for example we can consider the problem to evaluate the integral:

1 xT Ax I = xixje− 2 · dx, (36) Rn Z with A a real symmetric n n matrix, T denotes the transpose and the ordinary product of matrices.× ·

12 A general procedure in order to solve Gauss integrals as (36) consists to introduce a generating function (J) depending by a parameter J: Z

J1 . J =  .  , Jn   where  

1 xT Ax+xT J (J) = e− 2 · · . Z Rn Z Now we have that:

∂2 (J) I = Z . ∂J ∂J i j J=0

We will use this version of the Wick’s theorem.

Theorem 3.2

n ∂ 1 T 1 2 J A− J 1 1 e · = A− i i A− i i , (37) ∂J ∂J p1 p2 ··· pn 1 pn i1 ··· in J=0 − X where the sum is taken over all pairings: (ip1 , ip2 ),... (ipn 1 , ipn ), of i1,...,in. −

We will use this theorem in the case where i1 = i and i2 = j. We observe that there is another way to write (J): Z

1 (x A 1J)T A(x A 1J)+ 1 JT A 1J (J) = e− 2 − − − − 2 − . Z Rn Z 1 If we substitute y = x A− J we have that: − n 1 JT A 1J 1 yT Ay (2π) 2 1 JT A 1J (J) = e 2 − e− 2 dy = e 2 − . Z Rn √det A Z Reconsidering the integral I we have that:

n 2 1 JT A 1J (2π) 2 ∂ e 2 − I = . √det A ∂Ji∂Jj J=0

After calculations:

13 n 2 1 (A 1) JkJl (2π) 2 ∂ e 2 − kl I = √det A ∂Ji∂Jj J=0 1 1 a 1 1 b 1 (A 1) JkJl n 2 − kl (2π) 2 ∂ ( 2 (A− )aiJ + 2 (A− )ibJ )e = √ h ∂J i det A j J=0 1 a 1 (A 1) JkJl n 2 − kl (2π) 2 ∂ ((A− )iaJ )e = √ h ∂J i det A j J=0 n (2π) 2 1 1 1 1 k l 1 1 a 1 b 1 c (A− )klJ J = (A− ) +(A− ) J (A− ) J + (A− ) J e 2 √ ij ia 2 bj 2 jc det A J=0 n (2π) 2 1 1 k l 1 1 a 1 b (A− )klJ J = (A− )ij +(A− )iaJ (A− )bjJ e 2 √ · det A J=0 n (2 π) 2 1 = (A− )ij. √det A (38) n (2π) 2 1 We may call = so the result is I = (A− ) . This work can Z0 √det A Z0 ij be generalized and, a good reference is [F].

4 Gaussian integrals and homogeneous functions

A real function ϕ : Rn [0, + [, at least upper semicontinuous is α– → ∞ homogeneous, with α = (α1,...,αn) and αi > 0 if

α1 αn ϕ (t x1,...,t xn) = tϕ (x1,...,xn) (39) for each x Rn and t> 0. ∈ We may consider the sphere for the α–homogeneous function ϕ deﬁned for each r > 0:

B (r) = x Rn : ϕ(x)

14 In [Ve] the following theorem has been proved.

Theorem 4.2 Let ϕ : Rn [0, + [ be an α–homogeneous function, then → ∞

ϕ(x) e− dx = Leb(Bϕ)p! (40) Rn Z where p = α + + α is the weight of ϕ. 1 ··· n The next proposition is an application of a study on homogeneous functions and Gaussian integrals.

Proposition 4.3 Let c > 0, p > 0, and let A be an n n positive deﬁnite symmetric matrix, then ×

n Γ n +1 (c Ax,x )p π 1 2p e− h i dx = n . (41) Rn c √det A Γ +1 Z r 2 Observation 4.4 From the previous proposition we reﬁnd that:

+ + + 3 1 ∞ x3 1 ∞ x 3 1 ∞ (x2) 2 Γ 3 +1 4 e− dx = e−| | dx = e− dx = √π 1 = Γ . 0 2 2 Γ +1 3 Z Z−∞ Z−∞ 2 5 Gaussian matrix integrals

Gaussian integrals may be also deﬁned on the space of Hermitian matrices. Let be the space of all Hermitian matrices H = h HN { ij}i=1,...,N,j=1,...,N where h C and h = h . Every matrix H may be represented as: ij ∈ ij ji ∈ HN x (x + iy ) (x + iy ) 11 12 12 ··· 1N 1N .  (x12 + iy12) x22 .  H = . . ···......      (x1N + iy1N ) xNN   ······  where xij,yij R for 1 i < j N and xii R for i =1,...,N. We have that RN∈2 and we≤ denote by:≤ ∈ HN ∼

15 N

dLeb(H) = dxii dxijdyij, i=1 i

2 Tr(H ) = hijhji = hijhij (42) i,j ,...,N i,j ,...,N =1X =1X 2 2 2 = xii + xij + yij i=j i=j X X6 N 2 2 2 = xii +2 xij + yij = (Bh, h) . i=1 i

x11 .  .  xNN    x12   .  h =  .   .     xN 1,N   −   y12   .   .     yN 1,N   −  and  

16 1 ..  .  1 B =    2   .   ..     2    2 2  N 2 N  is an N N matrix, with det B = 2 − . We may deﬁne the Gaussian measure: ×

2 N N 1 2 1 − Tr(H ) dµ(H) = 2 2 e− 2 dLeb(H), (43) (√2π)N 2 · and the Gaussian matrix integral:

N 2 1 2 (√2π) 2 Tr(H ) e− dLeb(H) = N2 N . (44) − N 2 ZH 2 A complete and depth discussion on the topic can be found in [Zv].

6 The symplectic structure

In this section we consider Rd with d = 2n. We assume on Rd the presence of J : Rd Rd such that J 2 = id, where id is the identity map on Rd → − and J(v) = iv for all v Rd. We say that (Rd,J) is a complex vector space Rd ∈ Rd Rd C also denoted by J . Now if we assume to have H : J J be a positive definite hermitian product then g = (H) and ω ×= →(H) define ℜ −ℑ respectively a real symmetric scalar product and a symplectic structure on Rd . The reverse is also true. In fact given g : Rd Rd R a real J–invariant J J × J → scalar product if we define ω( , ) = g(J , ) and H = g iω then H is a positive definite hermitian product.· · · · − Furthermore if we define g(v,w) = vT w where the exponent T is the Rd · transpose and v,w J , we have that g is a real J–invariant scalar product and the structure ω∈can be described in term of g. Recalling that J T = J − we have:

ω(v,w) = g(Jv,w), ω(v,Jw) = g(v,w), ω(Jv,Jw) = g(Jv,w), (45)

17 and also ω is J–invariant.

7 The quantum postulates

In his work [Di], P. Dirac deﬁnes the quantum Poisson bracket [ , ] of any two variables u and v as · ·

u v vu= i~[u, v], (46) − 34 where ~ = 6, 626070040(81) 10− J s is the Planck constant. The formula · · (46) is one of the basic postulates of quantum mechanics. We can summarize these postulates as follows. To start we ﬁx a symplectic manifold (M,ω) of dimension dM , with ω the corresponding symplectic structure and an Hilbert space . The quantization is a “way” to pass from the classical system to the quantumH system. In this case the classical system, or phase space, is described by the symplectic manifold M and the Poisson algebra of smooth functions on M denoted by ( ∞(M), , ). The quantum system is described by . We deﬁne quantizationC a map{·Q·}from the subset of the commutative H algebra of observables ∞(M) to the space of operators in . Let f ∞(M) we have that Q(f): C is the corresponding quantumH operator.∈ CWe can summarize the quantumH → axioms H in this scheme:

1. linearity: Q(αf + βg) = αQ(f)+ βQ(g), for every α, β scalar and f,g observables;

2. normality: Q(1) = I, where I is the identity operator;

3. hermiticity: Q(f)∗ = Q(f);

4. (Dirac) quantum condition: [Q(f), Q(g)] = i~Q( f,g ); − { }

5. Irreducibility condition: for a given set of observables fj j I , with the { } ∈ property that for every other g ∞(M), such that f ,g = 0 for ∈ C { j } all j, then g is constant. We can associate a set of quantum operators Q(fj) j I such that for every other operator Q that commute with all ∈ of{ them} is a multiple of the identity.

The last postulate says that in the case we consider a connected Lie group G we say that it is a group of symmetries of the physical system if we have the

18 two following irreducible representation: one as symplectomorphisms acting on (M,ω) and another as unitary transformations acting on . For many H details about these postulates see [E].

2n Example 7.1 (Schr¨odinger quantization) Let M = R and (qj,pj) the canonical coordinates of position and momentum. In this case Q(qj) = qj ∂ 2 n that acts as multiplication and Q(pj) = i~ . The space = L (R , dqj) − ∂qj H is the quantum space and there are the following relations of commutations:

[Q(qk), Q(qj)] = [Q(pk), Q(pj)] = 0, [Q(qk), Q(pj)] = i~δkjI.

Quantization models in the past years have been studied from diﬀerent point of view. We cite as examples the case of geometric quantization and deformation quantization but the literature is very wide on the topic. We will consider the basic case on R2n presented in the above example as quantization scheme for the next sections.

8 Gaussian integrals depending by a quantum parameter

In this section we study a particular kind of Gaussian integrals deriving from quantum mechanics. The general form of these integrals is the following:

i g(v,w) e− ~ dv, (47) Rn Z where the function g is usually quadratic in its variables and (v,w) R2n. Here we are interested to particular forms of g. ∈

Proposition 8.1 We have the following cases:

1) if g = g (v,w) = ω(v,w) i v w 2 then: 1 − 2 k − k

i n 1 2 g1(v,w) w e− ~ dv = (2π~) 2 e− 2~ k k ; (48) Rn Z

19 2) if g = g (v,w,u) = ω(v,w + u) i v 2 then: 2 − − 2 k k

i n 1 2 g2(v,w,u) w+u e− ~ dv = (2π~) 2 e− 2~ k k ; (49) Rn Z 3) if g = g (v,w,u) = ω(v,w)+ ω(w,u) i v w 2 i w u 2 then: 3 − 2 k − k − 2 k − k

i n i g3(v,w,u) g1(v,u) e− ~ dw = (π~) 2 e− ~ , (50) Rn Z for all v,w,u Rn. ∈ Proof .

We start with the ﬁrst Gaussian integral and we have that:

i i 1 2 g1(v,w) ω(v,w) v w e− ~ dv = e− ~ − 2~ k − k dv (51) Rn Rn Z Z w 1 n ig J(β), β 2 n 1 w 2 = ~ 2 e− √~ − 2 k k dβ = (2π~) 2 e− 2~ k k , Rn Z 1 where β = √~ (v w) is a new variable used in the integration. For the second− case we proceed in a similar way:

i i 1 2 g2(v,w,u) ω(v,w+u) v e− ~ dv = e ~ − 2~ k k dv (52) Rn Rn Z Z 2 2 i g(Jv,w+u) 1 v ig J v , w+u 1 v ~ 2 √~ √~ √~ 2 √~ = e− − dv = e− − dv Rn Rn Z Z w+u 1 2 n ig Jβ, β n 1 w+u 2 = ~ 2 e− √~ − 2 k k dβ = (2π~) 2 e− 2~ k k , Rn Z 1 where we used β = √~ v as a new variable. i In the last case we consider the function g3(v,w,u) = ω (v,w) 2 v w 2 + ω (w,u) i w u 2 where v,w,u Rn. − k − k − 2 k − k ∈

i i 1 2 2 ~ g3(v,w,u) ~ [ω(v,w)+ω(w,u)] ~ [ v w + w u ] e− dw = e− − 2 k − k k − k dw. (53) Rn Rn Z Z 20 We set v w = t so: −

i i i 1 2 2 ~ g3(v,w,u) ~ ω(v,u) ~ [ω(t,v u)] ~ [ t + v u t ] e− dw = e− e− − − 2 k k k − − k dt. (54) Rn Rn Z Z v u A second variable − t = z permits to write: 2 −

2 2 i i i 1 v u v u g3(v,w,u) ω(v,u) ~ [ω(z,u v)] ~ − z + − +z e− ~ dw = e− ~ e− − − 2 hk 2 − k k 2 k idz. Rn Rn Z Z (55) After calculations we ﬁnd that:

i i 1 2 i 1 2 g3(v,w,u) ω(v,u) v u [ω(z,u v)] z e− ~ dw = e− ~ − 4~ k − k e− ~ − − ~ k k dz. (56) Rn Rn Z Z Now the last integral is an ordinary Gaussian integral of simple estimation:

i n i 1 2 n i g3(v,w,u) ω(v,u) v u g1(v,u) e− ~ dw = (π~) 2 e− ~ − 2~ k − k = (π~) 2 e− ~ . (57) Rn Z 2 i 2 A variation of the exponent is the function g4(v,w) = ω(v,Jw) 2 v i (1+2i cot ϑ) w 2. This function appears in [P1] as the function ψ (−v ,k eiϑkv −). 2 k k 2 0 1 Proposition 8.2 Let ϑ [0, π], v,w Rn, then ∈ ∈

i n n π n ~ g4(v,w) n i( ϑ) e− dvdw = 2 2 (~π) sin 2 ϑe 2 − 2 . (58) R2n Z Proof . In this case we have that:

i i 1 2 1 2 g4(v,w) ω(v,Jw) v (1+2i cot ϑ) w e− ~ dvdw = e− ~ − 2~ k k − 2~ k k dvdw (59) R2n R2n Z Z ~n n n 1 (2+2i cot ϑ) w 2 (2π) = (~2π) 2 e− 2~ k k dw = , Rn det(2+2i cot ϑ)I Z n 21 p 2i where In is the identity matrix. Observing that 2+2i cot ϑ = sin ϑ [ i sin ϑ + cos ϑ] we have that: −

n n i ~ (2π) n n π n ~ g4(v,w) n i( ϑ) e− dvdw = = (~π) 2 2 sin 2 (ϑ)e 2 − 2 . (60) iϑ R2n 2ie− Z det sin(ϑ) In q 2

1 2 2 Rn Proposition 8.3 Let g5(u,v,w) = 2 ( w v + v u ) and v,w,u , then k − k k − k ∈

2 i g u,v,w n i w u iπn ~ 5( ) ~ 2 2~ −2 4 e dv = (2π ) e k k e . (61) Rn Z Proof .

From the integral:

i w v 2+ v u 2 i (w u) t 2+ t 2 e 2~ (k − k k − k )dv = e 2~ (k − − k k k )dt (62) Rn Rn Z Z

(w u) − where we used the substitution v u = t. A second substitution t = 2 z bring to the integral: − −

2 2 2 i (w u) (w u) i (w u) 2 ~ − +z + − z ~ − + z = e 2 k 2 k k 2 − k dz = e 2 k 2 k k k dz (63) Rn Rn Z Z i (w u) 2 i n i (w u) 2 i − z 2 − q 2 2~ 2 2~ ~ 2 2~ 2 2 = e k k e k k dz = e k k e k k dq Rn Rn Z Z n i (w u) 2 iπn − ~ 2 2~ 2 4 = (2π ) e k k e , 1 2 where the last substitution was √~ z = q.

22 9 Quantum integrals

Now we shall consider a slight variation in the original integral adding a multiplicative factor A in front of the exponential:

i ψ(v,w) A(v,w)e− ~ dv. (64) Rn Z What is the general approach to this kind of integrals? Without specifying the form of the functions A, ψ it is difficult to find a general strategy that works always. We can observe that the integral (64) is an oscillating integral 1 with parameter ~ . So the classical approach is the method of stationary phase. The idea consists to expand the function ψ(v,w) near its critical points in order to have a “good” approximation of the integral. References are [Hö] and [GS2]. Another very flexible method is the Laplace method. It consists to Tay- lor expand A, ψ around the critical points of ψ and estimate the integral. Let A, ψ be infinitely differentiable functions in Rn. In our case assume to i 2 consider a specific form of ψ(v,w) = 2 v w . Thus we must evaluate the integral: − k − k

1 v w 2 I = A(v,w)e− 2~ k − k dv. (65) Rn Z 1 Assuming that I converges absolutely for ~ suﬃciently large, we can start changing variable v w = u: −

1 u 2 I = A(u + w,w)e− 2~ k k du. (66) Rn Z Now the function ψ(u) = u 2 has u = 0 as critical point. Expanding A(u + w,w) around zero we havek k that:

1 2 1 2 ~ u ~ u I = A(u+w,w)e− 2 k k du = (A(w,w)+ DA(u + w,w) u=0 u + ...) e− 2 k k du, Rn Rn | k k Z Z (67) that integrated term by term gives:

23 1 u 2 n I = A(u + w,w)e− 2~ k k du = A(w,w) (2π~) 2 + (68) Rn Z n+1 n n 2 2 +1 2 π Γ 2 n+1 + DA(u + w,w) u=0 ~ 2 + | Γ n 2 n+2 n n 2 2 +2 2 2 π Γ 2 n+2 n +1 + D A(u + w,w) u ~ 2 + ~ 2 . | =0 Γ n O 2 We used the Laplace approximation formula in [Wo] for J(k) as k + : → ∞

kψ(x) kψ(ˆx) n 1 n J(k) = A(x)e− dx = A (ˆx) e− (2π) 2 Σ 2 k− 2 + (1/k), (69) Rn | | O Z 2 1 where Σ = D ψ(ˆx)− is the Hessian in the critical pointx ˆ of ψ. Weaker assumptions can be made on the functions A, ψ requiring to be A continuous and ψ has continuous second order derivarites in a neighborhood ofx ˆ.

10 On a Gaussian integral in one dimension and the Lemniscate

Proposition 10.1 Let ~ be the Plank constant then:

+ 1 1 3 ~ 2 2 2 ∞ 1 x4 1 x2 2 π 4~ + 2~ ~ e− dx = 1(n, ) 1 . (70) C · AGM(p √2, 1) 2 Z−∞ where

+ + 1 + ∞ ∞ ∞ 1 (2n)! k 2 (2n +4k 1) 2l(4l 3) ~ n − − 1(n, ) = 3n − . C ~ 2 n!v2 n! k(2n +4k 3) (2l 1)(4l 1) n=0 u k=1 l=1 X u Y − Y − − t (71)

Proof .

24 + n + + ∞ 2n ∞ 1 x4+ 1 x2 ∞ 1 x 1 x4 e− 4~ 2~ dx = e− 4~ dx (72) 2~ n! Z−∞ Z−∞ n=0 ! X + n + ∞ 2n ∞ 1 x 1 x4 = 2 e− 4~ dx. 2~ n! 0 n=0 ! Z X x4 We substitute the variable 4~ = u, then we have:

+ + + n 3 ∞ 1 4 1 2 ∞ 1 1 ∞ 1 u 2 − 4 ~ x + ~ x u e− 4 2 dx = 2− 2 ~ 4 n e− du, (73) ~ 2 n! n=0 0 Z−∞ X Z 1 ~ 4 + 1 n 1 that we can write as ∞ n Γ + . Now using some properties of √2 n=0 ~ 2 n! 2 4 the Γ function and the formula (2.8) of [Ka] we ﬁnd an expression denoted ~ P 1 by 1(n, ) multiplied by Γ 4 . It is a fact from the number theory that Γ 1C is related to the arithmetic–geometric mean denoted by AGM, details 4 are in [Vi] and [AAR]. 2

Proposition 10.2 Let ~ be the Plank constant then:

+ 1 2 ∞ 1 x3 1 x2 π 3 2− 9 3~ + 2~ ~ e− dx = 2(n, ) 1 . (74) 0 C · 5 3 Z 3− 12 AGM 2, 2+ √3 p where

+ + 2 ∞ ∞ 1 2 n 2 n k 3 n + k 1 2 2(n, ~) = 3 3 ~ 3 − Γ n . (75) C (2~)nn! k 1 2 n + k 2 · 3 n=0 k=1 3 3 3 X Y − − Proof .

+ n + + ∞ 2n ∞ 1 x3+ 1 x2 ∞ 1 x 1 x3 e− 3~ 2~ dx = e− 3~ dx. (76) 2~ n! 0 0 n=0 ! Z Z X x3 We substitute the variable 3~ = u, then we have

25 + ∞ 1 x3+ 1 x2 e− 3~ 2~ dx (77) 0 Z + ∞ + 1 2 n 2 2 n+ 1 ∞ ( 2 n+ 1 ) 1 u = 3 3 − 3 ~ 3 3 u 3 3 − e− du (2~)nn! n=0 Z0 X + ∞ 1 2 n 2 2 n+ 1 2 1 = 3 3 − 3 ~ 3 3 Γ n + . (2~)nn! 3 3 n=0 X 2 1 Now by the proposition 2.1 of [Ka] with x = 3 n and b = 3 we have an expression for the gamma function:

+ 2 2 2 1 ∞ k n + k 1 Γ n Γ n + = 3 − 3 (78) 3 3 k 1 2 n + k 2 · Γ 2 k=1 3 3 3 3 Y − − + 2 2 1 ∞ k n+ k 1 Γ n Γ 3 = 3 − 3 3 . k 1 2 n + k 2 · √ k=1 3 3 3 2π 3 Y − − ~ 1 We ﬁnd an expression denoted by 2(n, ) multiplied by Γ 3 . As in the previous proposition also the factorC Γ 1 is related to the arithmetic– 3 geometric mean denoted by AGM, details are in [Vi] and [AAR]. 2 Proposition 10.3 Let 1 1 ψ (v,w) = v w 4 + v w 2, 2,4 −4~k − k 2~k − k we have that:

n n 2m 2 n 4~ −4 ψ2,4(v,w) π 2 − n +2m e dv = n Γ . (79) Rn m! 4 Γ 2 +1 m=0 Z X Proof . Setting v w = tand using the spherical coordinates we have that: −

26 eψ2,4(v,w)dv (80) Rn Z 2π π π + ∞ 1 r4+ 1 r2 n 1 = e− 4~ 2~ r − φn 1=0 φn 2=0 ··· φ1=0 0 Z − Z − Z Z n 2 n 3 sin − φ1 sin − φ2 sin φn 2drdφ1 dφn 1 − − · ··· +··· ∞ 1 4 1 2 n 1 ~ r + ~ r n 1 4 2 = dS − V r − e− dr. n 1 ZS − Z0 n π 2 All angular integrals are equal to Vn 1(1) = n and for the second, − Γ( 2 +1) we use the Taylor expansion of the exponential in this way:

+ + ∞ 2m ∞ n 1 r 1 r4 r − e− 4~ dr, (81) (2~)mm! 0 m=0 ! Z X now substituting r4 = 4~u and, after some calculations, we reﬁnd the Gamma function:

+ + m n n 2m ∞ 2 4~ −4 ∞ n 1 r 1 r4 2 − n +2m r − e− 4~ dr = Γ . (82) (2~)mm! m! 4 0 m=0 ! m=0 Z X X 2

11 The Berezin–Gaussian integral: a generalization of the Gaussian integral

In QFT (quantum ﬁeld theory) Gaussian integrals are treated in the case of anticommutative quantities:

ψ ψ = ψ ψ , ψ2 = ψ2 = 0, (83) 1 2 − 2 1 1 2 called Grassmann numbers or G-numbers. We want to treat an integral of the form:

ψ1∗rψ1 e− dψ1∗dψ1, (84) ZΛ 27 where ψ1∗ is the complex conjugation of a complex Grassmann number with the convention that (ψ ψ )∗ = ψ∗ψ∗ = ψ∗ψ∗, r is a real exponent and Λ is 1 2 2 1 − 1 2 the exterior algebra of polynomials in anticommuting elements. In order to treat integrals of the form (84) we observe that the Taylor expansion of a function is f(ψ)= g1 + g2ψ where g1,g2 can be both ordinary or Grassmann numbers. The integral, called the Berezin integral, is a linear functional with the properties:

∂f 1) Λ ∂ψ dψ = 0; R 2) Λ ψdψ = 1; for everyR f Λ where ∂ is a left or right derivative. We can treat the n ∈ ∂ψ dimensional case ψ =(ψ1,...,ψn) in the same manner:

f(ψ)dψ = f(ψ)dψ1 dψn. (85) n ··· ··· ZΛ ZΛ ZΛ The differentiation is defined as: ∂ dψ, (86) ∂ψ ≡ · ZΛ and this is consistent with previous definitions. Now using the properties of the Berezin integral we can treat the integral r (84) with 2 real:

r r r ψ1∗ 2 ψ1 e− dψ1∗dψ1 = 1 ψ1∗ ψ1 dψ1∗dψ1 = . (87) Λ Λ − 2 2 Z Z The result in n dimension with an hermitian matrix A = Aij is given by: { }

28 n 2 1 ψi∗Aij ψj e− 2 dψk∗dψk n Λ k ! Z Y=1 1 = [ ψi∗1 Ai1j1 ψj1 ] ... [ ψi∗n Ainjn ψjn ]dψ1∗dψ1 ...dψn∗dψn n! n − − ZΛ 1 = ψj∗1 ...ψj∗n Ai1j1 ...Ainjn (ψjn ...ψj1 dψ1 ...dψn) dψn∗ ...dψ1∗ n! n ZΛ 1 = ǫ ǫ A A = det A, n! i1...in j1...jn i1j1 ··· injn (88) where ǫi1...in = Λn ψi1 ...ψin dψn ...dψ1 and where there is only one non–zero term in the expansion. So R n 1 2 ψi∗Aij ψj e− dψk∗dψk = √det A, (89) n Λ k Z Y=1 that diﬀer from the usual Gaussian integral essentially because √det A is in the numerator instead denominator (details can be founded in [Zee]). This integral is also called the fermionic Gaussian integral.

11.1 On the Laplace method and a “mixed Gaussian Integral” Now we look if we can say something on integrals of the following form:

k g(ψ,ψ ) f(ψ, ψ∗)e− 2 ∗ dψ∗dψ, (90) 2 ZΛ where we consider the case with n = 2. We can apply a Laplace method? k We can choose g = ψ∗ 2 ψ an “heuristic” approach shows that if f = − 1 f1 + f2ψ + f3ψ∗ and k = ~ the integral can be computed:

1 g(ψ,ψ ) 1 f(ψ, ψ∗)e− 2~ ∗ dψ∗dψ = . (91) 2 2~ ZΛ Always with an “heuristic” point of view we can consider the following special “mixed Gaussian integral”:

29 n k 1 2 2 ψi∗ idij ψj 2 k x e− − k k dx1 dxn dψl∗dψl, (92) n Rn ··· Λ l Z × Y=1 where ψi are anticommutative variables and xi are commutative. In this case 1 Aij = idij is the identity matrix and k = ~ is the “quantum parameter”. A combination of tools of previous sections shows that:

n k 1 2 n 2 ψi∗ idij ψj 2 k x 2 e− − k k dx1 dxn dψl∗dψl = (2π) , (93) n Rn ··· Λ l Z × Y=1 independently from k!

12 Quantum Gaussian integrals with a perturbative term

In this subsection we consider the presence of a little perturbation term P (x) of the original Gaussian integral. We examine the case where the term P (x) is a polynomial that consists of a single term: λ P (x) = xk, (94) k! where k N and λ> 0. We have the following result in one dimension. ∈ Proposition 12.1 Let P (x) be a perturbation term with the form of (94), then

1 + 2 k 2 ∞ 1 ax2+P (x)+Jx 2π λ ( d ) J e− 2 dx = e k! dJ e 2a , (95) a Z−∞ where a> 0 and Jx is a source term.

Proof . We start with the identity:

1 + 2 2 ∞ 1 ax2+Jx 2π J e− 2 dx = e 2a , (96) a Z−∞ that follows by completing the square and changing coordinates in the integral. If we consider in addition the perturbative term, then:

30 + + + ∞ n ∞ 1 ax2+P (x)+Jx ∞ P (x) 1 ax2+Jx e− 2 dx = e− 2 dx. (97) n! n=0 ! Z−∞ Z−∞ X Let us consider this other formula :

+ k + ∞ k 1 ax2+Jx d ∞ 1 ax2+Jx x e− 2 dx = e− 2 dx. (98) dJ Z−∞ Z−∞ Using (96) the equation (98) may be written as:

1 k + 2 2 ∞ k 1 ax2+Jx 2π d J x e− 2 dx = e 2a . (99) a dJ Z−∞ Now the proposition follows because:

+ ∞ 1 ax2+P (x)+Jx e− 2 dx Z−∞ + + k n ∞ + ∞ + 1 ∞ n 1 ax2+Jx 1 λ d ∞ 1 ax2+Jx = P (x) e− 2 dx = e− 2 dx n! n! k! dJ n=0 Z−∞ n=0 ! Z−∞ X + Xk n 1 1 ∞ 1 λ d 2π 2 J2 2π 2 λ d k J2 = e 2a = e k! ( dJ ) e 2a . n! k! dJ a a n=0 ! X (100) 2 Further information on the perturbative methods in quantum ﬁeld theory can be found in the book of [M].

13 Gaussian integrals and the Boys integrals

In this last section we will consider a class of Gaussian functions of type:

l m n ar2 x y z e− A , Z 2 2 2 2 where l, m, n +, a> 0, rA = xA + yA + zA, xA = x Ax, yA = y Ay, z = z A ∈and A , A , A R. We observe that this− class contains− the A − z x y z ∈ complete system of hermitian functions around any point.

Proposition 13.1 Let a,b > 0 and A , A , A , B , B , B R, then x y z x y z ∈ 31 1) 3 2 ar2 br2 π abS e− A e− B dxdydz = e− a+b ; (101) R3 a + b Z 2)

3 2 2 2 ar2 1 br2 π 3ab 2Sa b abS − A − B − a+b e ∆ e dxdydz = 2 e ; R3 −2 a + b a + b − (a + b) Z (102)

3) 2 2 arA brB e− e− 2π 2 dxdydz = F0 pRCP ; (103) R3 r p Z C where S = (A B )2, ∆ is the Laplacian in R3, C is the center i=x,y,z i − i position of the nucleus, p = a + b, P = aAi+bBi for i = x, y, z called the P i p 2 2 1 xt2 2n center of mass, R = (C P ) and F (x) = e− t dt is called CP i=x,y,z i − i n 0 the “Boys function”. P R Proof . The ﬁrst integral can be written as

ar2 br2 e− A e− B dxdydz R3 Z 2 2 2 2 2 2 a(x Ax) a(y Ay ) a(z Az) b(x Bx) b(y By) b(z Bz) = e− − − − − − − − − − − − dxdydz, R3 Z (104) we consider only the ﬁrst of the three integrals and it is equal to:

2 2 2 (a+b)x +2(bBx+aAx)x aA bB e− − x− x dx, (105) R Z the other two are the same integral but with y and z instead x. The integral (105) can be easily solved completing the square and ﬁnding that is equal to:

2 (Ax Bx) ab π − e− a+b , a + b r the result for the ﬁrst integral follows considering the other two integrals. For the second we reduce the problem in the estimation of:

32 2 2 2 a(x Ax) 1 ∂ b(x Bx) − − − − e 2 e dx, (106) R −2 ∂x Z that gives two integrals:

2 2 2 2 a(x Ax) b(x Bx) 2 2 a(x Ax) b(x Bx) = b e− − − − dx 2b (x Bx) e− − − − dx. R − R − Z Z (107) The ﬁrst is the same of (105) and the second gives:

2 ab(Ax Bx) 2 2 2 2 − 1 (2A + b +2B a + a 4Ba A)e− a+b π − . 2 (a + b)2 · a + b r After algebraic sempliﬁcations and recollecting, we have that:

2 2 2 a(x Ax) 1 ∂ b(x Bx) e− − 2 e− − dx R −2 ∂x Z 2 2 2 2 ab(Ax Bx) ab 2a b (Ax Bx) − = − e− a+b . a + b − (a + b)2 (108) We remember the initial reduction and, considering the other two integrals in y and z, the result follows. The last integral is a “Coulomb integral” that can be solved using a formula for the product of two Gaussians. We recall here the formula:

ax2 bx2 ab X2 px2 e− A e− B = e− a+b AB e− P , · · aAx+bBx where p = a+b (is the total exponent), XAB =(Ax Bx) and Px = p . ab − The quantity a+b is called the reduced exponent, we will use this formula after. Now before to start we consider another identity very easy to show using the classical Gaussian integral:

+ 1 1 ∞ r2 t2 = e− C dt. (109) rC √π Z−∞ We transform the original 3–dimensional integral in the following 4-dimensional integral:

33 ar2 br2 e A e B − − dr R3 r Z C 1 pr2 t2r2 = e− P e− C drdt √π R4 Z 2 2 2 pt 2 1 (p+t )r 2 RCP = e− dre− p+t dt, √π R R3 Z Z (110) where the last passage follows from the product formula for Gaussians. Now π the integral in dr is equal to p+t2 and the initial integral in 3–dimension is now reduced to be an integral in 1–dimension:

ar2 br2 + 2 e− A e− B 2 ∞ π pt R2 − p+t2 CP dr = 2 e dt. (111) R3 r √π p + t Z C Z0 2 t2 3 3 We change variable setting w = p+t2 , so pt− dt = w− dw and the integral reduces to

ar2 br2 1 e− A e− B 2π 2 2 2π pw RCP 2 dr = e− dw = F0 pRCP . (112) R3 r p p Z C Z0 2 In [Bo] the author proves that all Schrödinger integrals can be estimated for this class of functions, moreover general molecular integrals can be reduced to these integrals. A Schrödinger integral is a solution of the many– electron Schrödinger equation for stationary states expressed in atomic units. The atomic units have been chosen such that the foundamental electron properties are all equal to one atomic unit. This is the reason because the quantum parameter doesn’t appears in this section. An example of four–center integral is treated in the following letter [Ki].

References

[A] E.Artin, “Einf¨uhrung in die Theorie der Gamma funktion ”, Teubner, Leipzig (1931), English transl. “The Gamma Function”, New York: Holt, Rinehart and Winston, 1964.

34 [AAR] G.E.Andrews, R.Askey, R.Roy, “Special Functions”, Cambridge Univ. Press, Cambridge, 1999. [Ba] S.D.Baalrud, “The incomplete plasma dispersion function: Proper- ties and application to waves in bounded plasmas”, Physics of Plasmas, 20(1):012118, 2013. doi: 10.1063/1.4789387. [Bo] S.F.Boys, “Electronic wave functions I. A general method of calculation for the stationary states of any molecular system”, Proceedings of the Royal Society A, 200, 542–554. [Cam] S. Camosso, “Scaling asymptotics of Szego kernels under commuting Hamiltonian actions”, (English summary) Ann. Mat. Pura Appl. (4) 195 (2016), no. 6, 2027–2059. [Con] K. Conrad, “The Gaussian Integral”, https://kconrad.math.uconn.edu/blurbs/analysis/gaussianintegral.pdf. [Di] P.A.M.Dirac, “The principles of quantum mechanics”, Oxford at the clarendon press (fourth edition 1958). [Dr] B.Driver, “Path integral notes”, online notes on the site: http://www.math.ucsd.edu/, January 7 (2013). [E] A.Echeverr´ıa–Enriqhéz, M. C. Muñoz–Lecanda, N. Román-Roy, C. Victoria–Monge, “Mathematical Foundations of Geometric Quantiza- tion”, Departament de Matemática Aplicada y Telemática Edificio C- 3, Campus Norte UPC C/ Jordi Girona 1 E-08034 Barcelona Spain, preprint (1991). [F] A.Friberg, “Matrix Integrals: Calculating Matrix Inte- grals Using Feynman Diagrams ”, Dissertation, retrived from http://urn.kb.se/resolve?urn=urn:nbn:se:uu:diva 227928. [GP] A.Galasso, R.Paoletti, “Equivariant Asymptotics of Szegökernels under Hamiltonian U(2) actions”, eprint arXiv:1802.05644 (2018). [GR] I.S.Gradshteyn, I.M.Ryzhik, “Tables of Integrals, Series and Products ”, Academic Press, 7th ed., (2007) New York. [GS1] V.Guillemin, S.Sternberg, “Symplectic techniques in physics”, Cam- bridge University Press 1984, pp. 51–54.

35 [GS2] V.Guillemin, S.Sternberg, “Semi–classical analysis”, on–line notes, http://www.math.harvard.edu/~shlomo/docs/Semi_Classical_Analysis_Start.pdf. [Ha] G.H.Hardy, “Ramanujan: Twelve Lectures on subjects suggested by his life and work”, 3rd ed. New York: Chelsea, 1999, pp. 8–9. [Hö] L.Hörmander, “The analysis of Linear Partial Differential Operators I”, Springer, Classics in Mathematics, Reprint of the 2nd Edition 1990. [I] H.Iwasawa, “Gaussian Integral Puzzle”, Math. Intelligencer 31 (2009), 38–41. [Ka] D.Karayannakis, “An algorithm for the evaluation of the gamma function and ramifications”, arXiv:0712.0292[math.CA] (2007). [Ki] R.Kikuchi, “ Gaussian Functions in Molecular Integrals ”, The Journal of Chemical Physics 22, 148 (1954), p. 148. [Ko] O.Kouba, “Inequalities related to the error function”. Preprint: http://arxiv.org/abs/math/0607694, 2006. [L] N.G.Lehtinen, “ Error function”, Stanford University Webpage, Web- page: http://nlpc.stanford.edu/nleht/Science/reference/errorfun.pdf, April 2010. [M] M.Marcolli, “Feynman Motives”, World Scientific (2009), https://doi.org/10.1142/7245. [Na] P.J.Nahin, “Inside Interesting Integrals”, Springer, Undergraduate Lec- ture Notes in Physics 2015. [Ne] A.Neagu, “Path Integrals I”, The West University of Timisoara Semi- nar of Theoretical Physics Faculty of Physics Nr.1/1995, pp. 16–22. [P1] R.Paoletti, “Szegökernels and asymptotic expansions for Legendre polynomials”, preprint, arXiv:1612.05009. [P2] R.Paoletti, “Scaling limits for equivariant Szegökernels”, J. Symplectic Geom. 6 (2008), no. 1, 9-32. [P3] R.Paoletti, “Asymptotics of Szegökernels under Hamiltonian torus actions”, Israel Journal of Mathematics 2011, DOI: 10.1007/s11856-011- 0212-4.

36 [S1] R.Sridharan, “From linteria to lemniscate II: Gauss and Landen’s work”, Resonance 9(6), 11–20 (2004).

[S2] R.Sridharan, “Physics to Mathematics: from Litearia to Lemniscate–I”, Resonance 9(4), 21–29 (2004).

[Ve] S.Venturini, “ Gaussian Integrals, Multinomial Coeﬃcients, and Ho- mogeneous Functions ”, Advances in Applied Mathematics 22, 297–302 (1999).

[Vi] R.Vidunas, “Expressions for values of the gamma function”, Kyushu J. Math., 59 (2005), pp. 267–283.

[Wi] W.O.Straub, “A Brief Look at Gaussian Integrals”, article: http://www.weylmann.com/gaussian.pdf, Pasadena, California (2009).

[Wo] R.Wong, “Asymptotic Approximations of Integrals”, Society for Indus- trial and Applied Mathematics, Philadelphia, pp. 494–498.

[Zee] A.Zee, “Quantum Field Theory in a Nutshell”, Princeton University Press, 2003.

[Zv] A.Zvonkin, “Matrix Integrals and Map Enumeration: An Accessible In- troduction”, Math. Comput. Modelling 26 281-304 (1997).