E.
I. Du Pont De Nernours and Co. Wilmington, Delaware 19899
,
1
I
An introduction
Some topics in the study of science are made unnecessarily difficult by the need to learn simultaneously unfamiliar physical concepts and unfamiliar mathematical methods. Chemists feel the difficulty most acutely, perhaps, in the study of quantum chemistry or quantum mechanics. A prior acquaintance with the mathematical methods makes study of these subjects far easier, leads to a firmer understanding, and saves time in the long run. The solutions to a number of key problems in quantum mechanics occur as polynomials, or power series of finite length. The purpose of this paper is to explain thoroughly, but in easy stages, how such solutions are found. The subject is easy to understand when studied for itself and will even be entertaining to those who like mathematical games. The material presented here will equip the reader to develop new polynomial solutions of appropriate equations. The method will first be explained with the aid of two examples. The principal polynomials used in quantum chemistry will then be taken up in turn. Example 1
Suppose we want to solve the following equation
Equation (I), being linear, is easily solved by standard methods; it is used here as a convenient example to illustrate the basic points of the polynomial method. I n the notation to be used throughout the paper eqn. (1) will be written y' zy' - 2y = 0 (1) We must first decide what we mean by the phrase "to solve eqn. (I)." In practice eqn. (1) will no doubt describe some physical system; for example, y might denote position and x might denote time. Why do we not simply use eqn. (1) as the sole and sufficient description of the systcm? The answer is that, if the symbols have the meaning given above, eqn. (1) contains velocity (y') as well as position and time. The result is that eqn. (1) does not enable us to visualize the workings of the system nor to answer simple questions such as "what is the position at a specified time?" Clearly, we wish to get rid of y' and establish a relationship between position and time only. It will then be easy to write relationships between velocity and time, velocity and position, acceleration and time, etc. The same considerations apply regardless of the quantities denoted by y and x. To solve the equation, then, means to establish a
+
functional relation between y and x. The relation must he a continuous function throughout' the range of values to which the equation applies. The latter will be dictated by the physical content of the problem; for example, if y is the concentration of a chemical species, the solution must usually apply over the whole rangeyo>y>Oor07y7y,. As the first step in finding a solution we assume that an answer exists in the form in which the a,'s are constants to be determined. From eqn. (2) we obtain After substituting eqns. (2) and (3) into eqn. (1) we have
I t will be convenient to group eqn. (4) according to powers of x,as follows (a, - 2 0+ ) (2a. - adz (3ar)za . . . [(v 1)ay+, ( Y - 2)ar1z1 . . . = 0 (5) Now, how may we solve eqn. (5)? It has an unknown number of terms, but even if we arbitrarily limit the number of terms we cannot solve eqn. (5) as one solves an equation with known values of the coefficients. For example, suppose we had reason to limit eqn. (5) to the first two terms. We could then write
++
+ +
z
=
+
a, -2ao -
(6)
a, - 2-
If the physical system dictated values for ao, a,, and an, we would insert these values into eqn. (6) and solve for x. That value of x, together with the values of the three coefficients could then he substituted into eqn. (2). We should then have a single value of x and a single value of y for which eqn. (1) is true. For example, if aa = 2 and a, = aa = 1, it follows that x = 3, y = 14, and y' = 7. As was stated earlier, this is not the kind of solution that is usually required; y must be a continuous function of x over the whole range of y and x values that apply to the physical problem. (In the problem a t issue let us say the range is 0 to for both x and y.) Only one method is available to attain the kind of solution that is desired: in eqn. (5) we must set the coefficient of each power of x equal to zero. It is obvious that eqn. (5) will then he true for any value of x whatsoever. Volume 46, Number 3, March 1969
/
151
After carrying out that operation, we have
examples. I n such instances it is clear that a polynomial solution exists only when the coefficient = b,, b2, etc. It happens that eqn. (14) is equivalent to
from which we find Equation (15) is a solution of eqn. (12), but the right,hand side is not a polynomial. T h e Hermite Polynomials'
Since aa is proportional to as, as is proportional to a,, etc., the vanishing of aa causes all the a , ( u >3) to vanish, too. Our solution is, then y = a.
+ 2aoz + a
d
(9)
Differentiation and substitution into eqn. (1) yields an equation true for all values of x. Equation (10) was written out to show that a. appears in all terms. Hence, the value of a. is arbitrary as far as the algebra is concerned, but it will usually be determined by some boundary condition of the problem. Freedom to take on arbitrary values is almost always enjoyed by the coefficient of lowest index. We shall not refer to this feature again hut shall uniformly assign the value 1. Our solution is then Substitution will show that eqn. (11) is indeed a solution of eqn. (1). Example 2 We wish to solve Y'
+ zy' + 2y = 0
(12)
We proceed as before and find a,=--2ao =-2 az = -3a1/2 = 3 as = -4a9/3 = - 4 a, = - 5 4 4 = 5
(13)
ctc. The solution is an infinite series, of which the first few terms are Since an unending series is not an acceptable answer, we conclude that the polynomial method does not work for . eqn. (12). At this point we should recall what we did in finding an unacceptable solution for eqn. (1) ( x = 3, y = 14). We first obtained equ. (6) by setting a3 and all coefficients of higher index equal to zero, and we then chose values for ao, al, and az. I n other words, we arbitrarily set values for all the coefficients. I n the polynomial method the values of the coefficients (except for ao, and in some instarices al as well) are set ultimately by the equation to be solved. We are not a t liberty to break off the scrics a t will; if it does not terminate itself the answer is an infinite series, as in eqn. (14). Some series terminate themselves absolutely as in Example 1. Other series terminate when a coefficient in the original differential equation takes on certain values, bl, br, ba, . . . , as we shall see in all the following 152 / Journal of Chemical Education
Let H be a function of x. tion H"
- 22H'
We wish to solve the equa-
+ 2nH
=
0
(16)
As before, we assume that H is a polynomial in x, and we find the first and second derivatives in the usual way. After substituting these quantities into eqn. (16) and collecting terms, we have
Setting the coefficient of each power of x equal to zero gives a. = -ma.
The solution produced by eqns. (18) has odd and i n t e r e s h g properties. I t mill go into an infinite series if we do not make a careful choice of ao,al, and n. First note that the coefficients of even index form u series in which each term is proportional to the preceding one. If any term vanishes, all the following t.erms will vanish, too. For example, if al = 0, it will follow that a,, as, ax, et,c. all equal zero. The point a t which the series ao,ai, a,, . . breaks off is determined by the value of n. If n = 0, only a. is left. If n = 4, thc series consists only of ao,a%,and ad. Unfortunately, when the series of coefficientsof ever1 indcx is terminated in this way, the coefficients of odd index still form a u infinite series. The series of coefficients of odd index can be terminated by proper choice of n (e.g., n = 1 or 3), but this leaves an infinite series of even coefficients. The only way out is to set either a. or a, equal to zero. No sleight of hand is involved here. I t was stated above that a polynomial once started must terminate itself. That is true: oue cannot select a few members out of a series of terms and tell the remaining terms to plcase go away. On the other hand, one is not obliged to start a series that would turn out to be embarrassing. If we set al = 0, we shall have H as a series of even powers of x . Certain values of n will cause that series to break off. With each such value of 7~ we shall have a solution of eqn. (16). Similarly, if a. = 0, H d l consist of a series of odd pol\-ers of x , and certaiii values of 71. will terminate the series. These results are summarized in Table 1. Each polynomial in Table 1 is a solutiou of eqn. (lG), and an infinite uumhcr of additional solutions (for higher values of n) may be found in the same way.
' Charles Hermite (1822-1901)
m ++3-;-:?" ~ a b f e1. aa
a,
'
a
,.st!y;?
. ' 9. -9 ", :
" 3 . ' "
Solutions of Equation (1 6)
2.
Solutions of Equation'
H
n
T.
Furthermore, any of the polynomials may be multiplied through by any constant whatsoever, and the resulting polynomial will be a solution of eqn. (16); thus an infinite number of solutions exists for each value of n. If we multiply any of the polynomials in Table 1 by the number in brachets, we obtain the corresponding Hemzte polynomzal. The number in brackets is, of
-,
-"
h"r-rinlnno nf n. nv n. AiLnL ~ n l n r rtI ~ " 1
Table 3. R
I
1 2 2 3 3 6
0 1 0 2 1 0
Solutions of Equation (21)
L 1 1
,[-I1
1 - x/2
1 1 - z/4 1-x xz/6
+
.[-61 .l-41
.[-l%~]
.[-!XI .1-181
tY hI I oU Unrm;tn nnl~innm;nl I L ~ L I I I I Y C p"LJLL"L.IIU..
These values arise because the entire series of Hermite polynomials is derived from a single "generating function." Generating functions for these and other polynomials are given in Appendix I. The Hermite polynomials occur in solutions of the Schroedinger equation for harmonic oscillators. I n such problems n is directly related t'o the energy of the system. The finding that solutions exist only for certain values of n means that solutions exist only for certain values of the energy. The Laguerre Polynomials1
We wish to solve the following equation
Each coefficient is proportional to the preceding one. Hence, if any coefficient vanishes, all the succeeding coefficients vanish. I t follows that the simple step of setting 1. equal to u will break off the polynomial following the term in x". The first few polynomials are given in Table 2. Multiplying each solution by the number in square brackets yields the corresponding Laguerre polynomial. We have taken these up in preparation for studying the polynomials of the next section; it is the latter that will be used in quantum mechanics. The Associated Laguerre Polynomials
The following equation Again we suppose that L may be a polynomial in x . With the experience of solving two equations back of us we might as well try a short cut: Instead of writing out the polynonlial and its derivatives, let us simply find the eocfficient of x'. The term in L" must contain x'-' before it is multiplied by x. The exponent, 1) before the two difthen, must have been ( u ferentiations. Therefore, we start with a"+, xV+', differentiate twice to obtain u(u 1) a"+, x"-', and mull ) a , + ~a'. The steps intiply by x to obtain u ( u volved in dealing with the other terms of eqn. (19) are given in the following table.
+
+
Original term First derivative Final term
+
L' -zL' rL autlxY+' avzY avzv ( v f l)a,+,zv ~ a , 2 ~ - ~ ... (P l ) a Y + ~ x Y - m.zV ra,xv
+
As before, the coefficient of each power of x must vanish. The coefficient of x' found above is ( u a.+~ (r - ")a,. Setting this coefficient equal to zero leads to
+
+
We may now substitute values of u ( = 0, 1, 2, . . . ) into eqn. (20) to obtain the following relations among the coefficients of the polynomial L Y
0 1 2 3
Equation ( 2 0 )
XL"
+ (21 + 2 - z)L' + (n - l - 1)L
=
0
(21)
is identical to eqn. (19) except that the coefficients are more complicated. The reader will be able to solve eqn. (21) unassisted. H e will find that the series a d breaks off with the uth term provided 1 = n - u - 1. I t follows that L will have only one term (ao) for numberless combinations of n and u, for example, 1 and 0, 2 and 1, 3 and 2, etc. Similarly, L will consist only of (ao alx) for the n and u combinations 2/0, 3/1,4/2, etc. The first few polynomials follow in Table 3. Multiplication of each polynomial by the number in square brackets yields the corresponding associated Laguerre polgnomial. These appear in the radial part of solutions of the Schroedinger equation for atomic orbitals. The symbols n and 1stand for two important quantum numbers.
+
The Legendre Functions3
We take up the simple Legendre functions (or Legendre polynomials) because the results will be needed later. The equation to be solved is (1 - z2)P,"- 2zP,'
+ 1(1 + 1 ) P , = 0
(22)
The subscript 1 is used here to distinguish these functions from the associated functions to come later. The latter are denoted by Pm,.
a, = -rao an = or = a, =
(1 - r)a1/4 ( 2 - r)as/9 (3 - r)an/l6
Edmond Laguerre (1834-1886). Adrien Marie Legendre (1752-1833). Volume 46, Number 3, March 1969
/
153
so unless a. and all the higher coefficientsare to be zero, we must have (27)
s = 3
Equation (27) is called the indicial equation, because it gives us the value of s, the iudcx. We now rewrite eqn. (26) as F
=
(28)
xa (G)
If we differentiate and substitute into eqn. (23) we obtain A solution of eqn. (22) is found easily if one assumes that PIis a polynomial in x. The first six polynomials are given in Table 4, in two forms, (1) with a. = a, = 1 and (2) with a. and a, as determined by the generating function. Singular Points and Indicia1 Equations
The solution is C = 1 - x, and hence the solution of eqn. (23) is F = x3 - x4. The reader might like to work out the indicial equation for a problem having adjustable constants. We take
Before taking up the last problem we must look briefly at singular points and indicial equations. Suppose we wish to solve
A few solutions are given in Appendix I1 The Associated Legendre Functions
If we develop the values of the coefficients in the usual way, we find
The equation to be solved is 1
d
[Sin B
21 +
[1(1
+ 1) - smW -1m'
0 = 0
(31)
in which e is a function of the angle 8. If we carry out the second differentiation, we have All the remaining coefficients are infinite. The troublc with eqn. (23) is that it has a singular point a t x = 0. (At a singular point the solution is either discontinuous, is not unique, or does not exist.) To see what happens to eqn. (23) a t x = 0, we rearrange it and drop the (1 - x)
That is, the slope becomes infinite, and hence the function itself does not have a defined value. Multiplying eqn. (23) through by x does not remove t,he problem but merely hides it, for we then have a t x = 0, 0.F' - 3F
=
0
(25)
Equation (25) does contain some useful information: F must be zero when x is zero. This can be arranged, as will be shown below. Some singular points permit the existence of a polynomial solution, others do not. The distinctions are discussed in many treatises.' The subject was introduced here because singular points occur in our final problem. We now take up the method that will be used to eliminate them. Assume that F in eqn. (23) is a polynomial from which the lower powers of x have been removed. That is, if G(x) is a polynomial starting with a constant term F(z) z8 G(x) = a& + a,z8+' + . . . + a,zSfY+ . . . (26) Differentiate eqn. (26), substitute into eqn. (23) and isolate the term having the lowest power. I t turns out to be As before, the coefficient of each power of x must vanish; 154
/
Journal of Chemicol Edumtion
As solutions we could try polynomials in 8, sin 8, tan (A polynomial in 8 would not be compatible with the sines and cosines already present. The derivatives of tan 8, cot 8, sec 8, and csc 8 are complicated, and they bring in functions other than the one being differentiated. A polynomial in either sin 8 or cos 8 is strongly suggested by the presence of sines and cosines in eqn. (32) and by the fact that sines and cosines transform into each other on being differentiated.) I t happens that the answer will consist of products of sines and cosines, and the easiest route to it will start with a polynomial in cos 8. We first make two transformations of the starting equation, the first for convenience and the second of necessity. Virst, we note that the function Q in eqn. (31) is to be differentiated with respect to the angle 8. This step produces products of sin 8 and cos 8 to various powers. I t is perhaps a little more convenient to have a single variable and to differentiate with respect to that variable. The way to attain this state of affairs is to differentiate the polynomial in cos 8 with respect to cos 8 (which we denote by x). This is easily done as follows. The function Q remains unchanged but we give it the new symbol P be8, or other functions.
' PAULING,L.,A N D WILSON,E. B., "Introduction to Quantum IMeehmics," MeGraw-Hill Book Co., New Yark, 1935, pp. 109-10,118-19. INCE,E. L., "Ordinary Differential Equations," reprint of 1st ed., Dover Publications, New York, 1956, pp. 69-71, H., AND MURPHY,G. "The 160-1, 3 6 5 9 . MARGENAU, Mathematics of Physics and Chemistry," (2nd ed.), D. Van Nostrand Co., New York, 1956, p. 61, 69-72.
cause P' and I"' will differ from 8' and 0". It should be obvious that
After similarly finding P" we have the following to substitute into eqn. (32) 8
= P
8' = -sin OP' 9" = sinz OP" - cos
OP'
The result is the following
x2)"I2,we shall have the polynomials e (8) to satisfy eqn.
\--,.
(21)
We should look upon eqn. (34) as simply an equation in x, though we shall eventually put the solutions back into trigonometric terms.s We must now make a second transformation, because eqn. (34) has singular points a t x = + l , - 1 . (The red flag appears in the term m2P/(1 - x2).) I t is usual in such instances to make in turn the substitutions, y = 1 - x and z = 1 x, so as to bring the singular points to the origin. If we let Y(y) = P ( x ) ,the first substitution yields
+
Equation (38) can be solved in the usual way, but a short cut is a~ailable.~Note that eqn. (38) is identical to eqn. (22) except for the constants. If we differentiate eqn. (22) m times with respect to x we obtain eqn. (38) except that G" in eqn. (38) is replaced by the (m 2)nd derivative of P , G' is replaced by the (m l)st, derivative, and G by the (m)thderivative. We previously found some of the Pl that satisfy eqn. (22); they are given in Table 4. The mth derivative of a P Lwill satisfy eqn. (22) after the whole equation has been differentiated m times. Hence the solutions of eqn. (38) are
+
+
The indicial equation reveals that 4sP = m%or s
=
&m/2
(36)
Negative powers in our solution would compound our difficulties, so we take only the positive root. That is, whether m is positive or negative, m2 is positive and s2 is positive; we can make sure of taking the positive root of s2 by writing s = lm1/2. For convenience we shall use simply the symbol m, but it will be understood that m takes on positive values, only. The substitution z = 1 x, Z(z) = P(x) yields the same value for s. Hence, we are instructed by the indicial equations to consider P ( x ) to be made up as follows
+
Substituting eqn. (37) into eqn. (34) gives
Equation (38) can now be solved by the polynomial method. Each polynomial solution, multiplied by (1 - x ~ ) " / ~will , yield P(x) to solve eqn. (34). If we then substitute cos 9 for each x and (sin 8)"' for ( 1 All the texts consulted state that "it is convenient" to make the substitution leading to eqn. (34). The convenience is slight in the author's opinion. The indicial equation can be reached in one step from eqn. (32). The trigonometric equivalent of wn. (381 is then easilv solved.
We can now obtain values for G by differentiating the P I in Table 4. A few solutions obtained in this way are given in Table .5. After making the substitutions discussed above we shall have a series of e (8) to solve eqn. (31). These are given in Table 5 , also. It is amusing to find that eqn. (31)-an equation that looked forbiddingly difficulthas innumerable solutions. We now revert to the choice of a polynomial in cos 8 [see paragraph following eqn. (32)l. One can treat 8 as a polynomial in sin 8 , but then the equation corresponding to eqn. (34) has three singular points. Nonetheless, the equation can be solved. The solutions are equivalent to those in Table 5 , that is, one set of solutions can be converted into the other through use of the relationship sin28 cos28 = 1. The functions e (8) occur in the angular part of solutions of the Schroedinger equation. They play an important part in describing the shapes of electronic orbitals.
+
Summary
The solutions of certain differential equations are polynomials, or power series of finite length. The basic method of finding these solutions was explained with the aid of two simple examples. The principal polynomials occurring in quantum mechanics were then taken up in turn. Special mathematical points are associated with each of these. The polynomials treated here are used widely in science. Appendix I. Generating Functions for the Polynomials7
Thepolynomial breaksooff withx"', where u' = 1 - m. ' d'/dz" denotes the vth derivative; theother superscriptsare exponents. v in this footnote represents all the symbols that denote the order of a. derivative in the text: 1, m, VL, and 1.
Hermite
Volume 46, Number 3, March 1969
/
155
Laguerre
Associated Legendro d' L,(z) = e" - (zre-=) dz'
Appendix ll.
Associated Laguenx
Legendre Pdz)
156
/
P,"(z)
=
1 d' - - (2%- 111 29! d d
Journal o f Chemical Educofion
=
(1
- z')*'Z
dm dzm P42) [m 2 01
Solutions of Equation (30)
