3.4 Series solutions and orthogonal polynomials

You should all be familiar with this from the Legendre polynomials discussed in the second year math course (or see http://walet.phy.umist.ac.uk/2C1).

These functions arise naturally in the problem of the one-dimensional quantum-mechanical harmonic oscillator.

3.4.1 The quantum-mechanical oscillator and Hermite polynomials

The quantum-mechanical Harmonic oscillator has the time independent Schrödinger equation

Solutions to such equations are usually required to be normalisable,

i.e., ψ ∈ {ℒ}^{2}(ℝ).

Mathematical functions other than simple polynomials always act on pure numbers (since otherwise the result of the function would contain a mixture of quantities of different dimensions, as we can see by Taylor expanding). This holds here as well, and we must be able to define “dimensionless variables”. We combine all the parameters of the problem to define two scales, a harmonic oscillator length

and a scale for energy {E}_{ω} = ℏω. We can then define a dimensionless coordinate and energy

In these variables the Schrödinger equation reads

Functions in {ℒ}^{2}(ℝ) must decay sufficiently fast at infinity: not all solutions to (3.12) have that property! Look at large z, where λ ≪ {z}^{2}, and we find that a function of the form ψ(z) =\mathop{ exp}\nolimits (±{z}^{2}∕2) satisfies exactly the equation

for large z. Since we have neglected a constant to obtain this result, we can conclude that any behaviour of the form {z}^{a}\mathop{ exp}\nolimits (±{z}^{2}∕2) is allowed (since the pre-factor gives subleading terms–please check). Since the wave function must vanish at infinity, we find the only acceptable option is a wave function of the form

where f(z) does not grow faster than a power as z →∞.

The easiest thing to do is substitute this into (3.12), and find an equation for f,

This is of Sturm-Liouville form; actually we can multiply if with p(z) = {e}^{−{z}^{2} } to find

This is a Sturm-Liouville problem, with eigenvalues 2λ − 1. The points z = ±∞ are singular, since p vanishes. Thus [a,b] is actually (−∞,∞), as we would expect.

So how do we tackle Hermite’s equation (3.13)? The technique should be familiar: we substitute a Taylor series around z = 0,

and collect the coefficient of terms containing the same power of z, and equate all these coefficients to zero

This recurrence relation can be used to bootstrap our way up from {c}_{0} or {c}_{1}. It never terminates, unless (2λ − 1) is an even integer. It must terminate to have the correct behaviour at infinity (like a power). We are thus only interested in even or odd polynomials, and we only have non-zero c’s for the odd part (if λ − 1∕2 is odd) or even part (when λ − 1∕2 is even).

If we call λ = n + 1∕2, with n integer, the first solution is {H}_{0}(z) = 1, {H}_{1}(z) = z, {H}_{2}(z) = 1 − {z}^{2}, …. These are orthogonal with respect to the weighted inner product

This shows that the eigenfunctions of the Harmonic oscillator are all of the form

with eigenvalue (n + {1\over 2})ℏω.

3.4.2 Legendre polynomials

A differential equation that you have seen a few times before, is Legendre’s equation,

Clearly x = ±1 are singular points of this equation, which coincides with the fact that in most physically relevant situations x =\mathop{ cos}\nolimits θ, which only ranges from − 1 to 1. As usual, we substitute a power series around the regular point x = 0, y(x) ={\mathop{ \mathop{∑ }}\nolimits }_{n}{c}_{n}{x}^{n}. From the recurrence relation for the coefficients,

we see that the solutions are terminating (i.e., polynomials) if λ = n(n + 1) for n ∈ ℕ. These polynomials are denoted by {P}_{n}(x). Solutions for other values of λ diverge at x = 1 or x = −1.

Since Eq. (3.15) is of Sturm Liouville form, the polynomials are orthogonal,

As for all linear equations, the {P}_{n}’s are defined up to a constant. This is fixed by requiring {P}_{n}(1) = 1.

Generating function

A common technique in mathematical physics is to combine all the solutions in a single object, called a “generating function”, in this case

We shall now prove that

and show that we can use this to prove a multitude of interesting relations on the way. The calculation is rather lengthy, so keep in mind where we do wish to end up: The coefficients of {t}^{n} in Eq. (3.16) satisfy Eq. (3.15).

1. First we differentiate (3.16) w.r.t. to x and t,
   \begin{eqnarray} t{(1 − 2tx + {t}^{2})}^{−3∕2}& =& {\mathop{∑ }}_{n}{t}^{n}{P'}_{ n}(x), %&(3.17) \\ (x − t){(1 − 2tx + {t}^{2})}^{−3∕2}& =& {\mathop{∑ }}_{n}n{t}^{n−1}{P}_{ n}(x).%&(3.18) \\ \end{eqnarray}

2. We then replace {(1 − 2tx + {t}^{2})}^{−1∕2} on the l.h.s. of (3.17) by a {\mathop{\mathop{∑ }}\nolimits }_{n}{t}^{n}{P}_{n}(x), multiplying both sides with (1 − 2tx + {t}^{2}),
   \mathop{∑ }{t}^{n+1}{P}_{ n}(x) ={ \mathop{∑ }}_{n}{P'}_{n}(x)({t}^{n} − 2x{t}^{n+1} + {t}^{n+2}).

   Equating coefficients of the same power in t, we find

   {P}_{n}(x) = {P'}_{n+1}(x) − 2x{P'}_{n}(x) + {P'}_{n−1}(x).

   (3.19)

3. Since (x − t) times the l.h.s. of (3.17) equals t times the l.h.s. of (3.18), we can also equate the right-hand sides,
   (x − t){\mathop{∑ }}_{n}{t}^{n}{P'}_{ n}(x) = t{\mathop{∑ }}_{n}n{t}^{n−1}{P}_{ n}(x),

   from which we conclude that

   x{P'}_{n}(x) − {P'}_{n−1}(x) = n{P}_{n}(x).

   (3.20)

4. Combine (3.19) with (3.20) to find

   (n + 1){P}_{n}(x) = {P'}_{n+1}(x) − x{P'}_{n}(x).

   (3.21)

5. Let n go to n − 1 in (3.21), and subtract (3.20) times x to find
   (1 − {x}^{2}){P'}_{ n}(x) = n\left ({P}_{n−1}(x) − x{P}_{n}(x)\right ).

6. Differentiate this last relation,
   \begin{eqnarray*} \left [(1 − {x}^{2}){P'}_{ n}(x)\right ]'& =& n{P'}_{n−1}(x) − n{P}_{n}(x) − nx{P'}_{n}(x)%& \\ & =& −n(n + 1){P}_{n}(x), %& \\ \end{eqnarray*}

   where we have applied (3.20) one more time.

This obviously completes the proof.

We can now easily convince ourselves that the normalisation of the {P}_{n}’s derived from the generating function is correct,

i.e., {P}_{n}(1) = 1 as required.

Expansion of |{r}_{1} −{r}_{2}{|}^{−1}.

One of the simplest physical applications is the expansion of |{r}_{1} −{r}_{2}{|}^{−1} in orthogonal functions of the angle between the two vectors.

Let us first assume {r}_{2} > {r}_{1},

where we have used the generating function with t = {r}_{1}∕{r}_{2}.

Since the expression is symmetric between {r}_{1} and {r}_{2}, we find the general result

where {r}_{<,>} is the smaller (larger) or {r}_{1} and {r}_{2}.

Normalisation

When developing a general “Legendre series”, f(x) ={\mathop{ \mathop{∑ }}\nolimits }_{n}{c}_{n}{P}_{n}(x), we need to know the normalisation of {P}_{n}(x). This can be obtained from the generating function, using orthogonality,

Substituting the generating function, we find

Thus

Electrostatic potential due to a ring of charge

Figure 3.5: A homogeneously charged ring in the xy plane.

As a final example we discuss the case of a homogeneously charged ring of radius a in the xy plane, see fig. 3.5.

The equation to solve is ΔV = 0, apart from on the ring itself. The problem can easily be tackled by separation of variables in polar coordinates, and we see from symmetry that the potential can only depend on r and θ. The angular equation gives Legendre polynomials, and the radial equation is trivial to solve (it has a power of r as solution), resulting in the expansion

where we have imposed the boundary condition V (∞) = 0. Actually, we can be slightly more specific and use the fact that from far away the ring looks like a point charge, V → q∕(4π{ϵ}_{0}r) for r →∞.

Now how do we determine the coefficients {c}_{n} in (3.24)? The simplest technique is based on a calculation of the potential on the positive z axis. You should have derived the result before (it is a standard example in basic electrostatics)

This can easily be expanded in powers of a∕z, and if we use \sqrt{{z}^{2}} = z we get

Since on the positive z axis r = z and {P}_{n}(\mathop{cos}\nolimits θ) = {P}_{n}(1) = 1, we conclude that

3.4.3 Bessel functions and the circular drum

Bessel’s equation of order ν takes the form

This equation has a regular singular point at x = 0, and the point x = ∞ is regular. It is thus not of Sturm-Liouville form, without additional boundary conditions (see below).

The solutions can be found in many places: we substitute a generalised power series around x = 0,

From the index equation (lowest power in x) we find γ = ±ν; this leads to two independent solutions if ν is not a half-integer. The recurrence relations are

The main result are the Bessel functions (regular solutions) for ν ≥ 0,

The simplest use of these regular solutions is for example in the caculation of the modes in a circular drum. With u(r,ϕ) = R(r){e}^{imϕ} we find that

with the explicit boundary condition y(a) = 0 and the implicit boundary condition y(0) is finite. With these conditions we have an Sturm-Liouville problem!

We can move λ into the variable by using the transformation

which turns the equation into Bessel’s equation of order m. Thus

with the boundary condition

If we are given the zeroes {x}_{n} of {J}_{m}(x), we find that

We tabulate the zeroes of {J}_{m}(x) in Tab. 3.2.