20030214
Transcript of 20030214
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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
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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)
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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
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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].
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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
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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
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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.
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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)
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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
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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)
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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
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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.
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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|]
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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.
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