PHYS 424 Notes

The solutions of the differential equations of physics are analytic functions, and every analytic function has a Taylor Series, so the most straightforward way to seek solutions to differential equations is to attempt to find the power series. Power series is not so strange a way to define a function; after all cos and sin are algebraically defined in exactly that way. Hence we could approach our present problem of finding the solutions of Hy = Ey where

y(x) = S_n b_n xⁿ .

We would quickly drown in algebra. There are a couple of standard life preservers. First, get rid of a few constants, so that the algebra is as compact as possible, by defining

in terms of which

Second, find out how y behaves at large x [or x] and hope that this asymptotic behavior determines the most complicated part of the power series. At large x,

d²y/dx² = x² y . We can easily guess the solution of this equation (compare y₀ from the last section):

y(x) = A e ^{- x2/2} + B e ^{+ x2/2}   The second of these is not normalizable and cannot be used, so we take B=0 so that, at large x at least,

y = A e ^{- x2/2}

The ghost of the second term will return to haunt us in a moment. We don't have the full solution, of course, since the differential equation was exact only for large x. However, we can use this result to suggest parametrizing the exact solution as

If y has a power series, then so does h. However, h is hopefully simpler than y, since its asymptotic behavior must not upset the known asymptotic behaviour of y. It turns out to be a lot simpler.

Now we apply the power-series idea, by seeking a solution h(x) in power-series form,

(d²h/dx²) e ^{- x2/2} - 2 x (dh/dx) e ^{- x2/2} + h (x² -1) e ^{- x2/2} = (x² - K) h e ^{- x2/2}

(d²h/dx²) - 2 x (dh/dx) + (K -1)h = 0

Now the power series:

S_n=0 ^¥ a_n [n(n-1)x^n-2 -2n xⁿ +(K-1)xⁿ] = 0

In the first term, let q=n-2, and let q=n in the other terms. Then all the powers of x will be the same, except that there will be terms for x^{-1, -2}:

S_q=0 ^¥ { a_q+2[(q+2)(q+1)x^q] +a_q[ -2q x^q +(K-1)x^q]} + a_-2[(0)(-1)x^-2] + a_-1[(1)(0)x^-1]} = 0

and the two extra terms have zero coefficients so that they don't mess up the algebra. The only way that this equation can vanish for all x is for each coefficient to vanish separately:

Hence we have a recursion relation. Given a₀ we know all the even a_q, and given a₁ we know all the odd a_q. In other words, we have two solutions for h, each multiplied by an arbitrary constant [a₀ and a₁]. However, for any given E [K] we should be able to discard one solution as not being normalizable, since that is what happened with y when we were exploring the asymptotic behavior of the wave function. Therefore one or the other solution for h must be too large at ¥ to normalize. How can we check for this? Try looking at the recursion relation when x is large. The behavior will be dominated by the terms with large q [at least if there are terms with large q], so we have approximately

a_q+2 = a_q[2q/q²]
        = (2/q) a_q

Hence a_q = a₀(2/2)(2/4)(2/6)...(2/q)
                = a₀/[1*2*3*...*(q/2)]
            a_q = a₀/(q/2)!

or a_q = a₁(2/3)(2/5)(2/7)...(2/q)
          > a₁(2/4)(2/6)(2/8)...2/(q+1)
          = a₀/[2*3*...*(q+1)/2]
    a_q-1 > a₁/(q/2)!

Therefore h(x) is -- or is greater than --

h(x) = S_{q=0, 2, 4, ...} Cx^q /(q/2)!

= S_{q=0, 1, 2, ...} C(x²)^q /q!

= exp( x² )

which implies that

y(x) ³ exp( x²/2 ) for large x

and is therefore not normalizable. We cannot allow the series to reach large values of q. Therefore there must be some value of q, which we can call n, for which

a_n+2 = 0.

In this case, the asymptotic behavior of y is xⁿ exp[-x²/2] which is normalizable. Examining 7-4, a_n+2 can vanish only if

2n+1-K=0,

K = 2n+1

implying that [see 7-0b]

which we had previously obtained by algebraic methods.

Note the following:

• There is a normalizable solution for every n.
• If n is even, the series based on a₀ terminates and the series based on a₁ is not normalizable. If n is odd, the series based on a₁ terminates and the series based on a₀ is not normalizable. Hence for each n we have one normalizable solution only.
• y is either even or odd.
• The polynomials h, properly normalized, are called the Hermite polynomials. The lowest few are given in the text.
• y can be nonzero in the classically-forbidden regions of x.

Show a few examples of y_n superimposed on the potential and the value of the total energy.