Differential equations II: Hermite equation and Hermite polynomials
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Introduction to power series method:
Solutions of second order differential equations through power series method:
We know the solution of the above equation already; it is
y(x) = Acosωx + Bsinωx
as can be verified by direct substitution in the differential equation.
Power series solution: Let
We assume that the power series on the right hand side converges in a certain domain of values for x. Our solution will then be valid in that domain.
Also we assume that the derivatives are all well defined.
This is NOT a pure mathematics course for us to worry about convergence, validity of such representations etc. However, whenever we face problems, we agree to go back to the mathematician!
Note that the “.” (dot, period) between numbers is actually the multiplication symbol. Substitute the above in the equation
and equate coefficients of every power of x to zero.
i.e., coefficient of x0 = 0, coefficient of x1 = 0 etc. Then
(ω2a4 + 6.5a6) = 0
Also ω2a1 + 3.2a3 = 0
ω2a3 + 5.4a5 = 0
Thus we get two recurrence relations
The two constants which are left undefined are a0 and a1. This is true for all second order differential equations. They require two conditions to arrive at a particular solution.
Thus the solution is
The above example provides an illustration for a new method to solve an old problem with known solutions. The harmonic oscillator equation will be solved using this method.
The Schrödinger differential equation is
and to is the domain
Rewrite the above equation:
x has the dimension of length L . λ and αhave the dimension of L − 2.
where ψ = ψ(x) and
What is the role of y?
note What is the dimension of ?
It is the same as
Verify that the dimension is
Since x has the dimension of length, we have introduced y so that it is dimensionless variable. Therefore in solving the differential equation we can drop all the excess baggage of constants.
Thus the differential equation that we want to solve is
Note that the equation is valid for all
Asymptotic (large y) solution:
What does the solution look for large values of y? To do this, we can try the “asymptotic [or, limiting] equation for large y, namely,
(take y sufficiently large so that the term we dropped out, is small and can be neglected). Try the solution Then
For to vanish for large values of choose only
and y being very large,
for very large
It is not acceptable. Hence the asymptotic solution must be for large values of y
Recurring coefficients for Hermite equation:
The complete solution be chosen as
where the function is arbitrary and must satisfy the equation
for all values of y.
Substituting for Ψ as above, we get,
This is known as the Hermite’s differential equation.
Problem: Show that Hermite’s equation is obtained when Eq. (12) is substituted in Eq.(10).
Propose a series solution as follows:
Substitute this in the differential equation:
Rearrange the above in increasing powers of y to get the following,
Considering even n’s
Repating this for all even n’s
Considering odd n’s:
The equations for are known as recurrence relations.
The power series has the following asymptotic behaviour.
What is the behaviour of for large values of n?
Thus the absolute covergence property of the power series is the same as that of as n tends to infinity; Then the wave function Ψ will be such that may not be finite for all the values of y if is an infinite series. However, if can be truncated at some finite n, then
That is an + 2 = 0 for some arbitrary n, but not an.
Or for some n.
n can be any positive integer, but must be the largest n value, for a given λ and α. With the substitution
The power series for becomes
For n even, the series multiplying a0 terminates so it is the appropriate result for Hn(y). a1 is necessarily zero.
For n odd, the series multiplying a1 terminates so it is the appropriate result for Hn(y). a0 is necessarily zero.
The traditional method of writing Hermite polynomials is such that the coefficient of highest power of y, (yn), is 2n.
Therefore for odd n,
For even n,
Harmonic oscillator wave functions. Summary:
These wave functions are NOT normalised. Hence to get the harmonic oscillator wave functions given in the lecture on harmonic oscillator, we must multiply by the normalization constant.
The relation is
A derivation of this relation is found in some standard quantum mechanics text books, for example, “Quantum Mechanics” by Eugen Merzbacher, John Wiley & Sons, 1999 which is an excellent text book on this subject. Refer to page 87 of the text book, in particular equations 5.36-5.38.
For the harmonic oscillator wave functions recall:
The harmonic oscillator wave functions are normalized as follows:
Transform the variable and do the integral.
A few examples of normalized harmonic oscillators wave functions are:
The Hermite differential equation is
In this section we want to obtain a relation satisfied among Hermite polynomials, other than the differential equation, namely,
Therefore the differential equation above is satisfied for each polynomial so that
The above can be verified trivially for
To derive the recursion formula, we will first verify the following identity
It must be kept in mind that when n is even, (n-1) is odd and vice versa. Consider Hn(y) for even n and therefore Hn − 1(y) for odd (n-1) as
Likewise one can prove the relation
starting with odd n and even (n-1).
The above is important in the derivation of a relation between Hermite polynomials since
Therefore the Hermite differential equation becomes
This is valid for all n’s Replacing n by n +1,
Thus the recursion formula for Hermite polynomials is
Using the above formula,
which we have obtained earlier. All other expression such as H3(y), H4(y), H5(y) etc. can be obtained by using this expression as well.
Generating function for Hermite polynomials:
Hermite polynomials can be constructed using a generating function technique. (i.e the function generates the polynomials upon substituting the right values of n)
Let us define a new function which collects all Hermite polynomials in a series expansion as:
Integrate the above equation to get
All the with n being odd are zero. All the with n being even are given by
From the fact that
and expanding as a Taylor series in s, we get
The last step is verified below for n=1 and 2 to give a feel for the equation. Try to do this generally for yourself.
Likewise one can verify the general result.
Thus the final result:
The right hand side is known as the generating function as, by successive differentiation, one can generate Hermite polynomials of various orders.
Energies of harmonic oscillator:
The boundary conditions for the wave function require that
is the expression for energy.
has the dimension of angular frequency 2πυ, where
Thus the energy of the harmonic oscillator is quantized with n = 0, 1, 2…….. etc as possible values.