HERMITE FUNCTIONS; LAGUERRE FUNCTIONS; LADDER OPERATORS

22. HERMITE FUNCTIONS; LAGUERRE FUNCTIONS; LADDER OPERATORS

In this section, we shall outline some of the important formulas for two more sets of named functions. Both Hermite and Laguerre functions are of interest in quantum mechanics where they arise as solutions of eigenvalue problems (see Problem 27, and Chapter 13, Problems 7.20 to 7.22). We shall also consider an operator method which is a useful alternative to series solution for some differential equations.

Hermite Functions The differential equation for Hermite functions is

(22.1) y ′′

2 n −x y n = −(2n + 1)y n , n = 0, 1, 2, · · · .

This equation can be solved by power series (Problem 5), but here we shall consider an operator method which is particularly efficient for this equation. Let’s use the operator D to mean d/dx; then (see Problem 5.31 of Chapter 8)

(D − x)(D + x)y = 2 −x (y ′ + xy) = y ′′ −x y + y, and similarly (22.2)

dx

(D + x)(D − x)y = y ′′

−x 2 y − y.

Using (22.2), we can write (22.1) in two ways: (22.3)

(D − x)(D + x)y n = −2ny n or

(D + x)(D − x)y n = −2(n + 1)y n .

Now let us operate on (22.3) with (D + x) and on (22.4) with (D − x), and change n to m for later convenience:

(22.5) (D + x)(D − x)[(D + x)y m ] = −2m[(D + x)y m ], (22.6)

(D − x)(D + x)[(D − x)y m ] = −2(m + 1)[(D − x)y m ].

(The brackets have been inserted to clarify our next step.) Now compare (22.3) and (22.6); if y n = [(D−x)y m ] and n = m+1, the equations are identical. We write

y m+1 = (D − x)y m

and we see that, given a solution y m of (22.1) for one value of n, namely n = m, we can find a solution when n = m + 1 by applying the “raising operator” (D − x) to y m . Similarly, from (22.4) and (22.5), we find that (Problem 1)

y m−1 = (D + x)y m .

We may call (D + x) a “lowering operator”; these operators are called creation and annihilation operators in quantum theory. Operators of this kind (see Problems 29,

30, and 23.27 for other examples) are called ladder operators since, like the rungs of a ladder, they enable us to go up or down in a set of functions. Now if n = 0, we find a solution of (22.3) [and therefore of (22.1)] by requiring

(D + x)y 0 = 0.

608 Series Solutions of Differential Equations Chapter 12

We solve this equation (Problem 2) to get (22.10)

y 0 =e −x 2 /2 .

Then, by (22.7), y n = (D − x) n

e 2 −x /2 . These are the Hermite functions; they can

be written in the simpler form y 2 n =e x 2 /2 (d n /dx n )e −x (Problem 3):

n = (D − x) e −x 2 /2 or

2 2 Hermite functions y n =e x /2 (d n /dx n )e −x .

n x If we multiply (22.11) by (−1) 2 e /2 , we obtain the Hermite polynomials; the fol- lowing equation may be called a Rodrigues formula for them:

Hermite polynomials

dx n

We find (Problems 4 and 5):

The Hermite polynomials satisfy the differential equation (Problem 6):

(22.14) y ′′ − 2xy ′ + 2ny = 0. Hermite equation

Using the differential equation, we can prove (Problem 7) that the Hermite poly- nomials are orthogonal on (−∞, ∞) with respect to the weight function e −x 2 . The normalization integral can be evaluated (Problem 10). Thus we have:

e −x H n (x)H m (x) dx = √

π2 n n! n = m.

Section 22 Hermite Functions; Laguerre Functions; Ladder Operators 609

The generating function for the Hermite polynomials is (Problem 8):

Φ(x, h) = e 2xh−h 2 = h H n (x) .

The generating function can be used to derive recursion relations for the Hermite polynomials. Two useful relations are (Problem 9):

(a) H ′ n (x) = 2nH n−1 (x), (22.17) (b) H n+1 (x) = 2xH n (x) − 2nH n−1 (x).

Laguerre functions The Laguerre polynomials may be defined by a Rodrigues formula:

Carrying out the differentiation (Problem 12), we find: x 2 x 3 n n(n − 1) n n(n − 1)(n − 2) (−1) x

(22.19) L n (x) = 1 − nx +

Laguerre polynomials

m m!

m=0

The symbol n m thors omit the 1/n! in (22.18); then the series in (22.19) is multiplied by n!. It is convenient to note that the series in (22.19) is like the binomial expansion of

(1 − x) n except that each power of x, say x m , is divided by an extra m!. We find (Problem 13):

(22.20) L 0 (x) = 1,

L 1 (x) = 1 − x,

2 (x) = 1 − 2x + x /2.

The Laguerre polynomials are solutions of the differential equation (Problems 14 and 15):

(22.21) xy ′′ + (1 − x)y ′ + ny = 0, y=L n (x).

610 Series Solutions of Differential Equations Chapter 12

Using the differential equation, we can prove (Problem 16) that the Laguerre polynomials are orthogonal on (0, ∞) with respect to the weight function e −x . In fact, we find (Problem 19) that, with the definition (22.18), the functions e −x/2 L n (x) are an orthonormal set on (0, ∞).

e −x L n (x)L k (x) dx = δ nk =

0 1, n = k.

The generating function for the Laguerre polynomials is (Problem 17):

e −xh/(1−h)

(22.23) Φ(x, h) =

L n (x)h n .

1−h

n=0

Using it, we can derive recursion relations; some examples are (Problem 18):

(a) L ′ n+1 (x) − L n ′ (x) + L n (x) = 0, (22.24)

(b) (n + 1)L n+1 (x) − (2n + 1 − x)L n (x) + nL n−1 (x) = 0, (c) xL ′ n (x) − nL n (x) + nL n−1 (x) = 0.

Warning: These formulas will be different if the factor 1/n! is omitted in the defi- nition (22.18), so check the notation of any reference you are using (computer, text, tables).

Derivatives of the Laguerre polynomials are called associated Laguerre polyno- mials; they may be found by differentiating (22.18), (22.19), or (22.20) (Problem 20). We define:

(22.25) L k

n (x) = (−1)

L n+k (x).

Associated Laguerre polynomials

dx k

Warning: The notation in various references may be confusing; some authors define L k n (x) as (d k /dx k )L n (x) [compare our definition in (22.25)], so read carefully the definition in the reference you are using. For example, associated Laguerre poly- nomials are used in the theory of the hydrogen atom in quantum mechanics. In various references you will find them denoted by L 2l+1 n−l−1 (x) and by L 2l+1 n+l (x); both these notations mean (except for sign) (d 2l+1 /dx 2l+1 )L n+l (x). (See Problems 26 to 28.)

By differentiating the Laguerre equation (22.21), we find the differential equation satisfied by the polynomials L k n (x) (Problem 21):

(22.26) xy ′′ + (k + 1 − x)y ′ + ny = 0, y=L k n (x).

Section 22 Hermite Functions; Laguerre Functions; Ladder Operators 611 The polynomials L k n (x) may also be found from the Rodrigues formula (Prob-

Note that in this form k does not have to be an integer; in fact, (22.27) is used to define L k n (x) for any k > −1.

Recursion relations for the polynomials L k n (x) may be found by differentiating recursion relations for the Laguerre polynomials. Some examples are (Problem 23):

n+1 (x) − (2n + k + 1 − x)L n (x) + (n + k)L k n−1 (x) = 0, (22.28)

Using the differential equation (22.26), we can show (Problem 24) that the func- tions L k n (x) are orthogonal on (0, ∞) with respect to the weight function x k e −x . We find (Problem 25):

x k e −x L k n (x)L k m (x) dx = (n+k)!

0 n!

, m = n.

The normalization integral needed in the theory of the hydrogen atom is not (22.29), but instead has the factor x k+1 . We find (see Problems 25 to 27):

x k+1 −x [L k

e n (x)] 2 dx = (2n + k + 1)

(n + k)!

0 n! Again warning: The formulas (22.28), (22.29), and (22.30) will be different in ref-

erences which omit the 1/n! in (22.18) and/or use a different definition of L k n (x) in (22.25).

PROBLEMS, SECTION 22

1. Verify equations (22.2), (22.3), (22.4), and (22.8). 2. Solve (22.9) to get (22.10). If needed, see Chapter 8, Section 2.

3. Show that e x 2 /2 D[e −x 2 /2 f (x)] = (D − x)f(x). Now set

D[e 2 −x f (x) = (D − x)g(x) = e /2 g(x)] to get

x 2 /2

2 −x (D − x) 2 g(x) = e D [e /2 g(x)]. Continue this process to show that

2 x 2 /2

F (x) = e 2 x (D − x) /2 D n [e −x 2 /2

F (x)]

for any F (x). Then let F (x) = e −x 2 /2 to get (22.11).

612 Series Solutions of Differential Equations Chapter 12

4. Using (22.12) find the Hermite polynomials given in (22.13). Then use (22.17b) to find H 3 (x) and H 4 (x).

5. By power series, solve the Hermite differential equation

y ′′ − 2xy ′ + 2py = 0

You should find an a 0 series and an a 1 series as for the Legendre equation in Section 2. Show that the a 0 series terminates when p is an even integer, and the a 1 series terminates when p is an odd integer. Thus for each integer n, the differential equation (22.14) has one polynomial solution of degree n. These polynomials with a 0 or a 1 chosen so that the highest order term is (2x) n are the Hermite polynomials. Find H 0 (x), H 1 (x), and H 2 (x). Observe that you have solved an eigenvalue problem (see end of Section 2), namely to find values of p for which the given differential equation has polynomial solutions, and then to find the corresponding solutions (eigenfunctions).

6. Substitute y n =e −x 2 /2 H n (x) into (22.1) to show that the differential equation satisfied by H n (x) is (22.14).

7. Prove that the functions H n (x) are orthogonal on (−∞, ∞) with respect to the weight function e −x 2 . Hint: Write the differential equation (22.14) as

e x 2 d (e −x 2 y ′ ) + 2ny = 0, dx

and see Sections 7 and 19. 8. In the generating function (22.16), expand the exponential in a power series and

collect powers of h to obtain the first few Hermite polynomials. Verify the identity

Substitute the series in (22.16) into this identity to prove that the functions H n (x) in (22.16) satisfy equation (22.14). Verify that the highest term in H n (x) in (22.16) is (2x) n . [You have then proved that the functions called H n (x) are really the Hermite polynomials since, by Problem 5, (22.14) has just one polynomial solution of degree n.]

9. Use the generating function to prove the recursion relations in (22.17). Hint for (a): Differentiate (22.16) with respect to x and equate coefficients of h n . Hint for (b): Differentiate (22.16) with respect to h and equate coefficients of h n .

10. Evaluate the normalization integral in (22.15). Hint: Use (22.12) for one of the H n (x) factors, integrate by parts, and use (22.17a); then use your result repeatedly.

11. Show that we have solved the following eigenvalue problem (see Problem 5 and end of Section 2): Given the differential equation y ′′ + (ǫ − x 2 )y = 0 [compare equation (22.1)]. find the possible values of ǫ (eigenvalues) such that the solutions y(x) of the given differential equation tend to zero as x → ±∞; for these values of ǫ, find the eigenfunctions y(x). What is ǫ, and what are the eigenfunctions?

12. Using Leibniz’ rule (Section 3), carry out the differentiation in (22.18) to obtain (22.19).

13. Using (22.19), verify (22.20) and also find L 3 (x) and L 4 (x). 14. Show that y = L n (x) given in (22.18) satisfies (22.21). Hint: Follow a method similar

to that used in Section 4. Let v = x n e −x and show that xv ′ = (n−x)v. Differentiate this last equation (n + 1) times by Leibniz’ rule, and use d n v/dx n = n! e −x L n (x) from (22.18).

Section 22 Hermite Functions; Laguerre Functions; Ladder Operators 613

15. Solve the Laguerre differential equation

xy ′′ + (1 − x)y ′ + py = 0

by power series. Show that the a 0 series terminates if p is an integer. Thus for each integer n, the differential equation (22.21) has one solution which is a polynomial of degree n. These polynomials with a 0 = 1 are the Laguerre polynomials L n (x). Find L 0 (x), L 1 (x), L 2 (x), and L 3 (x). (This is an eigenvalue problem—compare Problem 5 and Section 2.)

16. Prove that the functions L n (x) are orthogonal on (0, ∞) with respect to the weight function e −x . Hint: Write the differential equation (22.21) as

e x d (xe −x y ′ ) + ny = 0, dx

and see Sections 7 and 19. 17. In (22.23), write the series for the exponential and collect powers of h to verify the

first few terms of the series. Verify the identity

Substitute the series in (22.23) into this identity to show that the functions L n (x) in (22.23) satisfy Laguerre’s equation (22.21). Verify that the constant term is 1 by putting x = 0 in the generating function. [You have then proved that the functions called L n (x) in (22.23) are really Laguerre polynomials since, by Problem 15, (22.21) has just one polynomial solution of degree n.]

18. Verify the recursion relations (22.24) as follows: (a)

Differentiate (22.23) with respect to x to get hΦ = (h − 1)(∂Φ/∂x); equate coefficients of h n+1 .

(b) Differentiate (22.23) with respect to h to get (1 − h) 2 (∂Φ/∂h) = (1 − h − x)Φ; equate coefficients of h n .

(c) Combine (a) and (b) to get x(∂Φ/∂x) + hΦ − h(1 − h)∂Φ/∂h = 0. Substitute the series for Φ and equate coefficients of h n .

19. Evaluate the normalization integral in (22.22). Hint: Use (22.18) for one of the L n (x) factors; integrate by parts n times. Use (22.19) to find (d n /dx n R ∞ − )L n (x) and

Chapter 11, Section 3, to evaluate 0 x n e x dx. 20. Using (22.25), (22.20), and Problem 13, find L k n (x) for n = 0, 1, 2, and k = 1, 2. 21. Verify that the polynomials L k n (x) in (22.25) satisfy (22.26). Hint: Write (22.21)

with n replaced by n + k and differentiate k times by Leibniz’ rule. 22. Verify that the polynomials given by (22.27) are the same as the L k n (x) defined in

(22.25). Hints: Show that the functions in (22.27) satisfy (22.26) as follows. Let − v=e x x n+k

and show that xv ′ = (n+k−x)v. (Compare Problem 14.) Differentiate this equation n + 1 times by Leibniz’ rule, and use d − n v/dx n = n! e x x k L k

n (x) from (22.27). Also show that the coefficient of x n

in both (22.25) and (22.27) is (−1) n /n! [Thus, assuming that (22.26) for one k has only one polynomial solution of degree

n (which can be shown by series solution), (22.27) gives the same polynomials as (22.25) for integral k.]

614 Series Solutions of Differential Equations Chapter 12

23. Verify the recursion relation relations (22.28) as follows: (a)

In (22.24b), replace n by n + k and differentiate k times by Leibniz’ rule; in (22.24a), replace k by n + k and differentiate k − 1 times. Subtract k times the second result from the first.

(b) In (22.24c), replace n by n + k and differentiate k times. 24. Show that the functions L k

n (x) are orthogonal on (0, ∞) with respect to the weight

function x e −x . Hint: Write the differential equation (22.26) as

x −k e x d (x k+1 e −x y ′ ) + ny = 0 dx

and see Sections 7 and 19. 25. Evaluate the normalization integrals (22.29) and (22.30). Hints: Use (22.27) for

one of the L n k (x) factors in (22.29); integrate by parts n times. Use (22.25) and then (22.19) to evaluate d n /dx n L k n (x). Compare Problem 19. To evaluate (22.30),

multiply (22.28a) by x k e −x

26. Solve the following eigenvalue problem (see end of Section 2 and Problem 11): Given the differential equation

where l is an integer ≥ 0, find values of λ such that y → 0 as x → ∞, and find the corresponding eigenfunctions. Hint: let y = x l+1 e −x/2 v(x), and show that v(x) satisfies the differential equation

xv ′′ + (2l + 2 − x)v ′ + (λ − l − 1)v = 0. Compare (22.26) to show that if λ is an integer > l, there is a polynomial solution

v(x) = L 2l+1 λ−l−1 (x). 27. The functions which are of interest in the theory of the hydrogen atom are

f n (x) = x l+1 e −x/2n L 2l+1 n−l−1

where n and l are integers with 0 ≤ l ≤ n − 1. (Note that here k = 2l + 1, and we 3 have replaced n by n − l − 1; in this problem L 2 , say, means l = 1, n = 4.) For l = 1, show that „

« f 2 (x) = x 2 e −x/4

, f 3 (x) = x 2 e −x/6

, f 4 (x) = x 2 e −x/8

Hint: Find the polynomials L 3 0 ,L 3 1 ,L 3 2 as in Problem 20 (with k = 3) and then replace x by x/n. The functions f n (x) are very different from those in (22.29) since x/n changes from one function to the next. However, it can be shown (Prob- lem 23.25) that for one fixed l, the set of functions f n (x), n ≥ l + 1, is an orthogonal set on (0, ∞). Verify this for these three functions. Hint: The integrals are Γ functions—see Chapter 11, Section 3.

28. Repeat Problem 27 for l = 0, n = 1, 2, 3. 29. Show that R p = p x − D and L p = p x + D where D = d/dx, are raising and lowering operators for Bessel functions, that is, show that R p J p (x) = J p+1 (x) and L p J p (x) =

J p−1 (x). Hint: Use equations (15.5). Note that these operators depend on p as well as x, so they are not as simple as the Hermite function raising and lowering operators (22.7) and (22.8). If you want to operate, say, on J p+1 , you must change p in R or L to p + 1, etc. Making this adjustment, show that the equations LRJ p =J p and RLJ p =J p both give Bessel’s equation.

30. Find raising and lowering operators (see Problem 29) for spherical Bessel functions. Hint: See problems 17.15 and 17.16.

Section 23 Miscellaneous Problems 615