20030214

8/11/2019 20030214

1/15

2003,23B(2):274-288

SOLVING SECOND ORDER DIFFERENTIAL

EQUATIONS IN QUANTUM MECHANICS BY

ORDER REDUCTION 1

C. Ted Chen

Grace Semiconductor Manufacturing Corporation

2nd Floor, 307 North Tun-Hwa Road, Taipei, Taiwan 10508, China

Abstract Solving the famous Hermite, Legendre, Laguerre and Chebyshev equations

requires different techniques of unique character for each equation. By reducing these

differential equations of second order to a common solvable differential equation of first

order, a simple common solution is provided to cover all the existing standard solutions

of these named equations. It is easier than the method of generating functions and more

powerful than the Frobenius method of power series.

Key words Second order differential equations, quantum mechanics, common solution

2000 MR Subject Classification 34A, 81Q

1 Introduction

An algebraic cubic equation is solved by reducing it to a quadratic equation. By using the

technique of order reduction for simplification of mathematical structure, we will show that the

common differential equation of second order

(a2x2 + a1x + a0)

dy2

dx2+ (b1x + b0)

dy

dx+ d0y= 0 (1.1)

has a particular solution

y(x) =cidi

dxiexp

(2a2i + 2a2 b1)x + a1i +a1 b0

a2x2 + a1x + a0dx

(1.2)

wherei is a non-negative integer determined by the condition which is called here the condition

of order reduction

d0 =

(i + 1)(a2i + 2a2

b1). (1.3)

This condition of order reduction will be shown useful for eigenvalues.

It is easy to verify the common solution numerically for any non-negative integer i by

substitutingd0 of the above condition of order reduction into Eq. (1.1).

1Received December 24, 2001

8/11/2019 20030214

2/15

No.2 Chen: SOLVING SECOND ORDER DIFFERENTIAL EQUATIONS 275

A mathematically beautiful common solution[1] of Eq. (1.1) has been derived from in-

conceivable definitions and axioms. So far, the formidable form of solution[1] has discouraged

physicists to appreciate it. The existing methods of generating functions[2] are powerful. How-

ever, there seem no traces of the path of how to arrive at the various generating functions to

start each solving procedure. Eq. (1.2) is an understandable simple common solution.

2 Method of Order Reduction

The Leibniz formula[3,4] for differentiating a producti times

di

dxi[V(x)W(x)] =

is=0

i!

s!(i s)!dsV

dxsdisW

dxis for i s 0 (2.1)

will be used to reduce Eq. (1.1) to a solvable equation of first order. Two useful examples of

this formula aredi+1

dxi+1(xW) = x

di+1

dxi+1W+ (i+ 1)

di

dxiW, (2.2)

di+1

dxi+1(x2W) = x2

di+2

dxi+2W+ 2(i + 1)x

di+1

dxi+1W+ i(i + 1)

di

dxiW. (2.3)

From Eqs (2.3) and (2.2), we have

di+1

dxi+1[a2x

2W] = a2x2 d

i+2

dxi+2W+ 2a2(i+ 1)x

di+1

dxi+1W+ a2i(i + 1)

di

dxiW,

di+1

dxi+1[a1xW

] = a1xdi+2

dxi+2W+ a1(i+ 1)

di+1

dxi+1W.

From definition, we havedi+1

dxi+1[a0W

] = a0di+2

dxi+2W.

With the substitution Q(x) = a2x2 +a1x+a0 and by adding the above three equations, we

obtain

di+1

dxi+1[QW] = Q

di+2

dxi+2W+ (i + 2)(2a2x + a1)

di+1

dxi+1W+ a2i(i + 1)

di

dxiW.

With P1 = b1 2(i + 1)a2, we have from Eq. (2.2)di+1

dxi+1[P1xW] = P1x

di+1

dxi+1W+ P1(i + 1)

di

dxiW.

With P0 = b0 (i + 1)a1, we havedi+1

dxi+1[P0W] = P0

di+1

dxi+1W.

With P=P1x + P0 andc0 =

(i +1)(a2i + 2a2

b1), we have from the above three equations

di+1

dxi+1[QW + P W] = Q

di+2

dxi+2W+ (b1x + b0)

di+1

dxi+1W+ c0

di

dxiW. (2.4)

With the definition

y(x) = di

dxiW(x) (2.5)

8/11/2019 20030214

3/15

276 ACTA MATHEMATICA SCIENTIA Vol.23 Ser.B

for transformation, Eq. (1.1) becomes

Qdi+2

dxi+2W+ (b1x + b0)

di+1

dxi+1W+ d0

di

dxiW = 0.

Under the condition of Eq. (1.3) or c0 = d0, we have from Eq. (2.4)

di+1

dxi+1 [QW

+ P(x)W] = 0

which can be written in the simplest form of first order differential equation as

QW + P W = 0

having the solution[5] W(x) =cie(P/Q)dx

. From Eq. (2.5), we have

y(x) =cidi

dxi(e

(P/Q)dx).

Thus, we have Eq. (1.2) as a particular solution for Eq. (1.1).

By slight modification of the definition in Eq. (2.5) from differentiation to integration, we

can find that i can also be a negative integer. In order to guarantee that we have finite termsin Eq. (1.2), we assume that i is a non-negative integer. It is physically incomprehensible but

of mathematical interest of special devotion[1] for the case that i is not an integer.

3 Applications

We will use the condition of order reduction to obtain discrete energy levels. We will take

a biographic approach to study individual differential equations and a sociological approach to

study the relations among different equations as a whole. Other power of the Leibniz differen-

tiation formula will also be studied.

3.1 Simple Harmonic Oscillator

3.1.1 The Hermite Equation

The Hermite equation has the form

y 2xy + 2ny = 0 (3.1)

where primes denote differentiation with respect to x.

3.1.1.1 The Standard Solution

This equation has the standard solution[2] y(x) = (1)nex2 dnex2

dxn

wheren is a non-negative integer.

3.1.1.2 Solutions by Order Reduction

From Eqs (1.2) and (1.3), we obtain y(x) =c1diex

2

dxi =c1dn1ex

2

dxn1

as the first particular solution for Eq. (3.1). In this solution, i orn 1 is a non-negativeinteger.

With simple transformation, we obtain the second particular solution as

y(x) = c2ex2d

iex2

dxi =c2e

x2dnex

2

dxn

8/11/2019 20030214

4/15

8/11/2019 20030214

5/15


We will show that the associated Laguerre equation to treat the radial part of hydrogen

atom can also treat the radial properties of the harmonic oscillator.

With the substitution z2 =w, Eq. (3.4) becomes

w d2

dw2H(w) +

w+1

2

d

dwH(w) +

E

2h 1

4

H(w) = 0 (3.6)

having the form of the associated Laguerre equation shown later in Eq. (5.1).From the third basic solution of Eq. (5.1), we obtain

EO = (2j+ 1 + 1/2)h and HO(w) =c3ew d

j

dwj(ewwj+1/2)

where the subscript O denotes that the vibrational quantum number 2j+ 1 is an odd integer.

From the fourth basic solution of Eq. (5.1), we obtain

EE= (2j+ 1/2)hv and HE(w) =c4eww1/2

dj

dwj(ewwj1/2)

where the subscript Edenotes that the vibrational quantum number 2j is an even integer.

Discarded solutions, obtained from the first and second basic solutions of Eq. (5.1), become

acceptable in terms of time reversal as we discussed before.It is interesting to note that theHE(z), having even quantum number, is an even function

ofz and HO(z), having odd quantum number, is an odd function ofz .

3.1.5 Solutions in Explicit Form and the Recursion Formula

From Eq. (2.1), we obtain

HO(w) =c3ew

js=0

j!(1)sews!(j s)!

djs

dwjswj+1/2

as a solution in explicit form with odd quantum number for Eq. (3.6). With w = x2, we have

HO(x) = c3

(j+ 3/2)j!

j

s=0

(

)s

(j s)!s!(s + 3/2) x2s+1

as a solution of odd function ofx for Eq. (3.3). The ratio of the (s + 1)th term to thesth term

is(s j)

(s+ l)(s + 3/2)x2.

The sth term ofHO(x) contains x2s+1. We want to write HO(x) by

H(x) =

k=0

ckxk

of which the (k+ 1)th term contains xk. Withk = 2s+ 1 and the odd vibrational quantum

number i = 2j+ 1 in EO, we have

ck+2 = 2(k i)

(k+ 1)(k+ 2)ck

which is in agreement with the two-term recursion relation obtained by the method of power

series[7].

8/11/2019 20030214

6/15


From Eq. (2.1), we obtain

HE(w) =c4eww1/2

js=0

j!(1)sew(j s)!s!

djs

dwjswj1/2

as a solution in explicit form with even quantum number for Eq. (3.6). Thus, we have

HE(x) = c4(j+ 1/2)j!

js=0

()s(j s)!s!(s+ 1/2) x

2s

as a solution of even function ofx for Eq. (3.3). With k = 2sand the even vibrational quantum

number i = 2j, we can have the same two-term recursion relation.

3.2 Angular Wave Equation

3.2.1 The Associated Legendre Equation

The associated Legendre equation has the form

(1 x2)y 2xy +

l(l+ 1) m2

1 x2

y= 0. (4.1)

3.2.1.1 The Standard SolutionThis equation has the standard solution[2,8]

y(x) = ci(1 x2)m/2 dl+m(x2 1)l

dxl+m . (4.2)

3.2.1.2 One Effective Solution by Order Reduction

Eq. (4.1) remains the same ifm is replaced withm or l is replaced withl 1 due tol(l+ 1) = (l+ 1/2)2 1/4. Four solutions are expected with such replacements. However, wewill show there is only one effective solution.

With the substitution y(x) = (1 x2)m/2F(x), Eq. (4.1) becomes

(1 x2

)

d2F

dx2 2(1 m)dF

dx (l m)(l+ 1 m)F= 0. (4.3)From Eqs (1.1) and (1.2), we obtain

y(x) = ci(1 x2)m/2 di

dxi(x2 1)im (4.4)

as a particular solution for Eq. (2.1) under the condition of

(i l m)(i+ l+ 1 m) = 0. (4.5)

Four solutions can be expected from this equation.

By solving the two roots ofi, we obtain from the previous equation

y(x) = ci(1 x2)m/2 dlmdxlm

(x2 1)l (4.6)

wherel m= 0, 1, 2, 3, . By solving another root and with L = l 1, we obtain

y(x) = ci(1 x2)m/2 dLm

dxLm(x2 1)L

8/11/2019 20030214

7/15


whereLm= 0, 1, 2, 3, . By replacingl with l1 in one of the above two solutions, we canhave another. Due to the similar form of these two solutions, the last one is thus a numerical

redundancy of Eq. (4.6).

By taking the upper and lower signs in front ofm, Eq. (4.6) yields two apparently different

solutions. By carrying out differentiation or by following the procedure to arrive at Eq. (4.9),

these two solutions have the same function ofxexcept the sign. Thus, there is only one effective

solution.3.2.2 The Schrodinger Equation

For any potential V(r) being a function of the radial coordinate only, there is the time

independent Schrodinger equation[8]

h2

82

1

r2

r(r2

r) +

1

r2 sin

(sin

) +

1

r2 sin2

2

2

+ V = E

where is the mass of a one-particle system or reduced mass of a two-particle system. With

m = 0,1,2,3, , it is shown[8] that by separating the radial and the angular parts of and substituting (r,,) = R(r)()eim, w = cos , and () = P(w), the above wave

equation becomes

(1 w2) d2

Pdw2

2w dPdw

+

m2

1 w2

P(w) = 0

whereis a constant of the procedure of the variable separation.

3.2.2.1 The Standard Solution

With = l(l+ 1), we have

(1 w2) d2P

dw2 2w dP

dw+

l(l+ 1) m

2

1 w2

P(w) = 0. (4.7)

The standard normalized solution[8] is

Pml (w) = N2ll!

(1 w2)m/2 dl+m(w2 1)l

dwl+m (4.8)

whereN = [(l + 1/2)(lm)!/(l + m)!]1/2. It has been shown[3] that, with the Leibniz formula,we can have

Pml (x) = (1)mPml (x) (4.9)indicating that Pml (x) is also a solution for Eq. (4.1). In other words, bothm and m arepermitted in the solution. Eq. (4.1) remains the same ifm is replaced withm.

Eq. (4.8) in explicit form [9, 10] is

P(w) = Nl!(1 w2)m/2lm2

k=0

(1)kk!(l k)!

(2l 2k)!(l m 2k)! w

lm2k. (4.10)

This solution is an even or odd function ofw if (l m) is an even or odd integer.3.2.2.2 Results from Order Reduction

From Eq. (4.6), we have

P(w) = ci(1 w2)m/2 dl+m

dwl+m(w2 1)l

8/11/2019 20030214

8/15


which is essentially the same as Eq. (4.8).

It has been shown[3] that, with the Leibniz formula, we can have 11

P(w)2dw= c2i

2

2l+ 1

(l+ m)!

(l m)! (2ll!)2.

From the definition of normalization constant 1

1P(w)2dw= 1, we have Eq. (4.8).

3.2.2.3 Solving the Associated Legendre Equation by the Hypergeometric Equa-tion

With the substitutions P(w) = (1 w2)m/2F(w) and z = w2, the associated Legendreequation shown by Eq. (4.7) becomes

z(1 z)d2

dz2F(z) (2m + 3)x 1)

2

d

dzF(z) (l+ m)(l+ 1 + m)

4 F(z) = 0 (4.11)

having the form of the hypergeometric equation shown by Eq. (6.1).

By solving Eq. (4.1) with the aid of the first solution of Eq. (6.1) and with c = 1/2,

a= (l+ m)/2 and b = (l+ 1 + m)/2, we have

PO(z) = c1(1 z)m/2 d

(l+m1)/2

dz(l+m1)/2 (1 z)(lm1)/2

z

(l+m+1)/2

where the subscriptO denotes that (l + m) is a positive odd integer, as a solution for Eq. (4.7).

From the second solution of Eq. (6.1), we have

PE(z) = c2(1 z)m/2z1/2 d(l+m)/2

dz(l+m)/2(1 z)(lm)/2z(l+m1)/2

where the subscriptEdenotes that (l + m) is a non-negative even integer. Two-term relation[11]

can easily be obtained from these two solutions. Withz = w2,we find that PE(w) orPO(w) is

an even or odd function ofw, respectively, depending on (l + m) being an even or odd integer.

The results are in agreement with Eq. (4.1) and similar to the even or odd quantum numbers

and solutions for the harmonic oscillator.

3.3 The Hydrogen Atom

3.3.1 The Associated Laguerre Equation

The associated Laguerre equation has the form

xy + (s+ 1 x)y + (q s)y = 0. (5.1)

3.3.1.1 Standard Solution

The above equation has the standard solution[2]

y(x) = ds

dxs

ex

dq

dxq(exxq)

wheres and qare non-negative integers due to the presence of ds

/dxs

and dq

/dxq

.3.3.1.2 Four Basic Solutions

With the transformation ofy = ekexxkpV(x) and under the conditions ofke(ke 1) = 0andkp(kp+ s) = 0, Eq. (5.1) becomes

xV + [(2ke 1)x + 2kp+ s + 1]V + (2kekp kp+ ske+ ke+ q s)V = 0.

8/11/2019 20030214

9/15


Thus, four particular solutions of Eq. (5.1) obtained from Eq. (1.2) are

y(x) =c1di

dxi(exxis)

under the condition ofi = s q 1;

y(x) = c2

xs di

dxi(exxi+s)

under the condition ofi = q 1;

y(x) = c3ex d

i

dxi(exxis)

under the condition ofi = qand

y(x) =c4exxs

di

dxi(exxi+s)

under the condition ofi = q s.In the standard solution, s and qshould be non-negative integers. In the second and third

solutions, it is not necessary to assume that s is an integer. In the first and fourth solutions,the difference ofs and qshould be an integer. Ifs is an integer, such as in the case of the radial

wave equation for the hydrogen atom shown later, then qshould be an integer.

3.3.1.3 Four Extended Solutions

With the transformation ofy(x) = dj V(x)/dxj , Eq. (1.1) becomes

(a2x2 + a1x + a0)

dj+2

dxj+2V + (b1x + b0)

dj+1

dxj+1V + d0

dj

dxjV = 0.

Differentiating

(a2x2 + a1x + a0)V

+ [(b1 2a2j)x + b0+ a1j]V + [j(a2+ a2j b1) + d0]V = 0

j times with respect tox, by following the procedure to arrive at Eq. (2.4), yields the previous

equation. For the case of Eq. (5.1), we have

xd2V

dx2 + (x + s j+ 1) dV

dx + (q s +j)V = 0.

Solving this equation by following the previous procedures for the four basic solutions, we obtain

y(x) = c5di+j

dxi+j(exxi+js)

under the condition ofi = s q 1 j;

y(x) = c6 d

j

dxj

xjs d

i

dxi (exxij+s)

under the condition ofi = q 1;

y(x) = c7dj

dxj

ex

di

dxi(exxi+js)

8/11/2019 20030214

10/15


under the condition ofi = qand

y(x) =c8dj

dxj

exxjs

di

dxi(exxij+s)

under the condition ofi = j + q s.Forj = 0, the above four extended solutions become the previous four basic solutions. For

j = s, we have the common special case for the seventh and eighth solutions

y(x) =c78ds

dxs

ex

dq

dxq(exxq)

(5.2)

which is the same as the standard solution. That for the third and fourth solutions is

y(x) =c56dsq1

dxsq1(exxq1). (5.3)

In this solution, s q 1 is a non-negative integer. In order to prevent this solution frombecoming divergent at x = 0,q 1 should be a non-negative integer.3.3.2 The Radial Wave Equation

The hydrogen atom has the potential V(r) = e2

/r wheree

represents the charge of theelectron. The Schrodinger radial wave equation[8] is

d2

dr2R(r) +

2

r

d

drR(r) +

l(l+ 1)

r2 +

82

h2 (E V(r))

R(r) = 0. (5.4)

With the substitutions and the change of variable [8]

2 = 82E/h2, (5.5)

= 42e2/h2, (5.6)

z= 2r, (5.7)

R(z) = ez/2zlL(z),

the radial wave equation becomes

z d2

dz2L(z) + (z+ 2 + 2l)d

dzL(z) + ( 1 l)L(z) = 0. (5.8)

With s = 2l+ 1 and q = + 1, the above equation and Eq. (5.1) become the same. The

coefficientss, qand are integers because l is an integer. With

= n (5.9)

and from Eqs (5.5) and (5.6), we have the Bohr formula

E= 22e4

n2h2 (5.10)

wheren can be an negative integer.

3.3.2.1 The Standard Radial Wave Solution

8/11/2019 20030214

11/15


The standard solution[8] of Eq. (5.8) is

L2l+1n+l (z) = d2l+1

dz2l+1

ez

dn+l

dzn+lezzn+l

.

The standard normalized solution[8] of Eq. (5.4) is

Rn,l(z) =

2na0

3(n l 1)!2n[(n + l)!]3

ez/2zlL2l+1n+l (z). (5.11)

From Eq. (5.2), we have

Rn,l(z) = c78ez/2zl

d2l+1

dz2l+1

ez

dn+l

dzn+1(ezzn+l)

which is essentially the same as the standard solution.

3.3.2.2 The Laguerre Polynomials in Explicit Form

The standard solution[8] of Eq. (5.8) can also be expressed as

L2l+1n+l (z) = [(n + l)!]2

nl1s=0

(1)s+1 zs

(2l+ 1 + s)!(n l 1 s)!s! . (5.12)

We will show how to derive it with the Leibniz differentiation formula.

Differentiating ezzn+l with respect to z , n + l times by using Eq. (2.1) yields

ez dn+l

dzn+lezzn+l =

n+lk=0

(1)k (n + l)!k!(n + l k)!

dn+lk

dzn+lkzn+l

.

Becaused2l+1

dz2l+1

dn+lk

dzn+lkzn+l

= 0 for k 2l,

we haved2l+1

dz2l+1

n+lk=0

dn+lk

dzn+lkzn+l

=

n+lk=2l+1

(n+ l)!

(k 2l 1)! zk2l1.

Thus, we have

d2l+1

dz2l+1

ez

d2l+1

dz2l+1ezzn+l

= [(n+ l)!]2

n+lk=2l+1

(1)k zk2l1

k!(n+ l k)!(k 2l 1)! .

With the substitution k = 2l+ 1 +s, we obtain Eq. (5.12) by shifting the range limits of the

summation in the above equation.

Two-term relation[9] obtained from the method of power series can easily generated from

Eq. (5.12).3.3.2.3 Simpler Radial Wave Solutions with Positive Principal Quantum Number

From the fourth basic solution of Eq. (5.1), we have

Rn,l(z) = c4ez/2zl1

dnl1

dznl1ezzn+l (5.13)

8/11/2019 20030214

12/15


as the fourth particular solution of Eq. (5.4). In this solution, there are n l 1 steps ofdifferentiation. In the standard solution shown by Eq. (5.11), there are n + 3l+ 1 steps. Thus

the above solution is simpler than the standard solution.

By following the same procedure to derive Eq. (5.12), we have

Rn,l(z) = c4(n l 1)!(n+ l)!ez/2zlnl1

s=0

(1)s zs

(2l+ 1 + s)!(n

l

1

s)!s!

.

Thus, the fourth particular solution and the standard solution shown by Eq. (5.11) are essen-

tially the same.

With L = l 1 and from the third basic solution of Eq. (5.1), we have

Rn,l(z) = c3ez/2zL1

dnL1

dznL1ezzn+L

which is the same as Eq. (5.13) by replacing l with L orl 1. This solution is a numericalredundancy.

3.3.3 Negative Principal Quantum Number

With the smallest Bohr radius a0 = h2/42e2, Eq. (5.4) becomes

d2

dr2R(r) + 2

rd

drR(r) +

l(l+ 1)r2

1n2a20

+ 2a0r

R(r) = 0.

Both this equation and Eq. (5.1) remain the same if n is replaced withn. The signs ofprincipal quantum number will be explained in terms of time reversal.

3.3.3.1 Negative Direction of Time and Principal Quantum Number

From the period of the Kepler motions[12], we have

2 = 2e4/2E3 (5.14)indicating that the total energy of hydrogen atomEshould have negative values and the period

of motionscan have both signs.

The quantum conditions [13] are pd= nhand prdr= nrhwhere means that theintegration is taken over one period of motions, p is the azimuthal momentum, pr is the radial

momentum, n is the azimuthal quantum number andnr is the radial quantum number. With

pd= (d/dt)2dt= 2Tdt,

prdr= (dr/dt)2dt= 2Trdt

where T is the azimuthal kinetic energy and Tr is the radial kinetic energy, the quantum

conditions can be rewritten as 20

Tdt= nh and 20

Trdt= nrh whereis the time spent

over one complete cycle of a periodic motion. Adding these two quantum conditions yields

2

0

Tdt= nh

where T =T+ Tr and n = n+ nr. The left side of this equation has the dimensions of the

action integral, energy time, in classical mechanics. From the result of virial theorem[14] forhydrogen atom, we have the time-averaged kinetic energy

T = ( 0

Tdt)/= E

8/11/2019 20030214

13/15


confirming that the total energy of hydrogen atom Eshould has negative values regardless the

signs of the period of motions . From the above two equations, we obtain

E= nh/2. (5.15)

By eliminatingfrom the above equation and Eq. (5.14), we can have the Bohr formula shown

by Eq. (5.15). In this approach, we have obtained the Bohr formula without having to know

the eccentricity of the orbit.In Eq. (5.14), we have E 0. Eq. (5.15) indicates that n and are of the same sign. In

other words, the principal quantum number and the direction of time are of the same sign.

3.3.3.2 Discarded Solutions and Negative Principal Quantum Number

From the first basic solution of Eq. (5.1), we have

Rn,l(z) = c1ez/2zl

dn+l

dzn+lezznl1

as the first particular solution of Eq. (5.4). This solution would be discarded due to unbound

physical conditions as z approaches infinity because of the presence of ez in it. However, it

becomes acceptable as z approaches negative infinity. As z changes the sign so will do

according to Eq. (5.7). Similarly, and n will change the signs according to Eqs (5.6) and(5.9). In Eq. (5.15), changes the sign as n changes the sign. Thus the above discarded

solution becomes acceptable for negative direction of time.

From Eq. (5.3), we obtain the same solution in form except the integration constants.

From the second basic solution of Eq. (5.10), we have the same solution except l is replaced

withl 1.3.3.4 The Confluent Hypergeometric and Associated Laguerre Equations

The confluent hypergeometric equation

xy + (x + c)y ay = 0 (5.16)

has the standard solution[15] y(x) = 1F1(a, c; x) = 1 + a

cx1! +

a(a+1)c(c+1)

x2

2! +

under the condition ofc= 0,1,2,3, . This solution does not guarantee to have finiteterms. Solutions of the confluent hypergeometric equation with finite terms are useful for the

relativistic hydrogen atom[16].

The confluent hypergeometric and associated Laguerre equations have the same form.

Solutions of Eq. (5.16) with finite terms can be obtained from the solutions of Eq. (5.1).

3.4 Symmetric Tops of Polyatomic Molecules

3.4.1 The Hypergeometric Equation

The hypergeometric equation

x(1 x)y + [c (a+ b + 1)x]y aby= 0 (6.1)

has the standard solution[2,17]

y(x) = 2F1(a,b,c; x) = 1 +ab

c

x

1!+

a(a + 1)b(b + 1)

c(c+ 1)

x2

2! + (6.2)

under the condition ofc= 0,1,2,3, . This solution does not guarantee to have finiteterms.

8/11/2019 20030214

14/15


Solutions with finite terms will be obtained. With the transformation of y = xkp (1 x)kmV(x) under the conditions ofkp(kp+ c 1) = 0 and km(km+ a+ b c) = 0, of Eq. (6.1)becomes in a form of Eq. (1.1). Four particular solutions of Eq. (6.1) are

y(x) = c1di

dxi[xic+1(1 x)iab+c]

under the condition of (

i + a

1)(

i + b

1) = 0;

y(x) = c2(1 x)1c di

dxi[xi+c1(1 x)iab+c]

under the condition of (i + a c)(i + b c) = 0;

y(x) =c3(1 x)ab+c di

dxi[xic+1(1 x)i+a+bc]

under the condition of (i a+ c 1)(i b + c 1) = 0 and

y(x) = c4x1c(1 x)xab+c d

i

dxi[xi+c1(1 x)i+a+bc]

under the condition of (i a)(i b) = 0.3.4.2 The Standard Solution for Symmetrical-Top Molecules

With Cto represent the moment of inertia about the symmetry axis of a symmetric top

andA the other two equal moments of inertia, the transformed rotational wave equation is[18]

x(1 x) d2F

dx2 + (x + ) dF

dx + F = 0 (6.3)

where

= |K M| + 1, = |K+ M| + |K M| + 2,

=82AE

h2 AK

2

C + K2 (/2 1)

2

and M andKhave integral values 0,1,2, . Substituting F(x) = =0

ax into Eq. (6.3)

yields the recursion formula

aj+1 =j(j 1) + j

(j+ 1)(j+ ) aj.

The condition for the series to break off after the j th term leads to the energy levels[18]

E= h2

82

J(J+ 1)

A + K2

1

C 1

A

(6.4)

whereJ=j + (|K+ M| + |K M|)/2. The wave solution[18] for Eq. (6.3) is

2F1(

1

J+ /2, J+ /2, 1 +|K

M|; x).

3.4.3 Results of Order Reduction

From the fourth solution, we obtain without having to solve for a or b

F(x) =cix|K+M|(1 x)|KM| d

j

dxj[xj+|K+M|(1 x)j+|KM|]

8/11/2019 20030214

15/15


and Eq. (6.4). This solution is easier to handle than the previous standard one.

3.4.4 The Hypergeometric and Confluent Hypergeometric Equations

With z = xb[19], the hypergeometric equation shown by Eq. (6.1) becomes

(z z2

b )

d2y

dz2+

c z zb az

b

dydz ay= 0.

As b approaches infinity, the above hypergeometric equation becomes the confluent hypergeo-

metric equation shown by Eq. (5.16).With lim

b(1 + z/b)b = ez, the four solutions of the hypergeometric equation shown by Eq.

(6.1) can be stretched into the four solutions of the confluent hypergeometric equation shown

by Eq. (5.16) correspondingly.

4 Conclusions

Important differential equations of second order in quantum mechanics can be classified

into two groups. The Hermite, associated Laguerre and confluent hypergeometric equations are

in one group for radial properties. The associated Legendre, Chebyshev and hypergeometric

equations are in another group for angular properties. Solutions of the confluent hypergeometric

equation are limiting cases of those of the hypergeometric equation. Discarded solutions of the

time-independent Schrodinger equations become acceptable in classical terms of time reversal.

Acknowledgment Discussions with Professor Shih-Tong Tu are inspiring.

References

1 Nishimoto K, Tu S -T. Fractional calculus method to extended linear ordinary differential equations of

Fuchs type (Examples in the case m= n= 2). Journal of Fractional Calculus, 1995,7: 17-28

2 Spiegel M R. Mathematical Handbook of Formulas and Tables. New York: McGraw-Hill, 1968. 146-160

3 Arfken G. Mathematical Methods for Physicists. 3rd ed. New York: Academic Press, 1985. 667-677

4 Spiegel M R. Mathematical Handbook of Formulas and Tables. New York: McGraw-Hill, 1968. 55

5 Spiegel M R. Mathematical Handbook of Formulas and Tables. New York: McGraw-Hill, 1968. 104

6 Pauling L, Wilson E B Jr. Introduction to Quantum Mechanics. New York: McGraw-Hill, 1935. 67-807 Levine I N. Quantum Chemistry. 3rd ed. Boston Mass: Allyn & Bacon, 1970. 58-62

8 Pauling L, Wilson E B Jr. Introduction to Quantum Mechanics. New York:McGraw-Hill,1935.Chapter 5

9 Mulder J J C. Closed-form spherical harmonics: Explicit polynomial expression for the associated Legendre

functions. Journal of Chemical Education, 2000,77: 244

10 Slater J C. The Quantum Theory Matters. New York: McGraw-Hill, 1968. 116

11 Levine I N. Quantum Chemistry. 3rd ed. Boston Mass: Allyn & Bacon, 1970. 89

12 Marion J B, Thornton S T. Classical Dynamics of Particles & Systems. 3rd ed. Orlando Florida: Harcourt

Brace Jovanovich, 1988. Chapter 7

13 Born M. Atomic Physics. 8th ed. New York: Dover, 1969. 384-391

14 Marion J B, Thornton S T. Classical Dynamics of Particles & Systems. 3rd ed. Orlando Florida: Harcourt

Brace Jovanovich, 1988. 236-238

15 Abramowitz M, Stegun I A. Handbook of Mathematical Functions. New York: Dover Publications, 1972.

332-336

16 Bethe H A, Salpeter E E. Quantum Mechanics of One- and Two-Electron Atoms. New York: Academic

Press, 1957. 65-69

17 Arfken G. Mathematical Methods for Physicists. 3rd ed. New York: Academic Press, 1985. 749

18 Pauling L, Wilson E B Jr. Introduction to Quantum Mechanics. New York: McGraw-Hill, 1935. 275-280

19 Wyld H W. Mathematical Methods for Physics. Reading Massachusetts: Benjamin/Cummings, 1976.

531-532

20030214

Documents