Hermite Polynomials As Part of the Solution to the Harmonic Oscillator

[Aero6]

Homework Statement

When trying to generate solutions to the harmonic oscillator, I'm trying to use hermite polynomials. I understand that there's a recursive relationship between the hermite polynomials but I'm confused in how each hermite polynomial is generated.

Homework Equations

Hn+1(p)=2*p*Hn(p)-2n*Hn-1(p) to generate each hermite polynomial
An+1,k = 2An,k-1 -2*n*an-1,k the formula for the coefficients of the hermite polynomials

The Attempt at a Solution

The solutions to the harmonic oscillator's energy levels are a combination of hermite polynomials*gaussian functions*normalization constant

There's a recursive relationship in trying to generate the hermite polynomials but I don't understand how to use the formula for the coefficients of the hermite polynomials. I have that H0=1 and H1=2p as the base cases for the hermite polynomials, but how do I make use of the coefficient formula to generate more hermite polynomials? Can't I just use recursive relationships and have H2 for example depend on H0 and H1? Also, I've read that there are odd and even solutions for hermite polynomials but I don't see how odd n values and even n values have to be treated separately.

Thanks.

 

[anathema]

Hello, thank you for your post. I understand your confusion regarding the generation of hermite polynomials. The recursive relationship you mentioned, Hn+1(p)=2*p*Hn(p)-2n*Hn-1(p), is indeed the key to generating each hermite polynomial. Let's break it down step by step:

1. The base cases for the hermite polynomials are H0=1 and H1=2p, as you correctly stated. These are the starting points for generating the rest of the polynomials.

2. To generate H2, we plug in n=1 into the recursive relationship. This gives us H2(p)=2p*H1(p)-2*H0(p). Since we know the values of H0 and H1, we can substitute them in and get H2(p)=2p*2p-2*1=4p^2-2. This is the second hermite polynomial.

3. To generate H3, we plug in n=2 into the recursive relationship. This gives us H3(p)=2p*H2(p)-2*2*H1(p). Again, we can substitute in the values we know and get H3(p)=2p*(4p^2-2)-4*2p=8p^3-12p. This is the third hermite polynomial.

4. We can continue this process to generate as many hermite polynomials as we need. Each polynomial will have a higher degree than the previous one, and the coefficients will follow the pattern given by the recursive relationship.

5. As for the odd and even solutions, this refers to the fact that for odd values of n, the hermite polynomial will be an odd function (symmetric about the origin) and for even values of n, it will be an even function (symmetric about the y-axis). This is why odd and even n values need to be treated separately.

I hope this explanation helps clarify the process of generating hermite polynomials. Let me know if you have any further questions. Good luck with your work!

 

[nlightner]

I can provide some clarification on the use of Hermite polynomials in solving the harmonic oscillator problem. The Hermite polynomials are a set of orthogonal polynomials that can be used to generate solutions to the Schrödinger equation for a harmonic oscillator. These polynomials are generated using a recursive relationship, as you have mentioned. This means that each polynomial in the set depends on the previous ones in the series.

To answer your question about the use of the coefficient formula, this is used to determine the specific coefficients for each term in the polynomial. The formula you have given is correct, and it can be used to generate the coefficients for any n value. However, as you have correctly stated, the base cases for H0 and H1 are 1 and 2p, respectively. So, to generate H2, you would use the formula with n=2 and plug in the values of H0 and H1. This will give you the coefficient for the H2 term, which you can then use to generate the full H2 polynomial.

Regarding the odd and even solutions, this refers to the fact that the Hermite polynomials can be separated into two sets - one with only even n values and one with only odd n values. This is because the solutions to the harmonic oscillator can be separated into even and odd solutions, and the Hermite polynomials reflect this property. So, when using the Hermite polynomials to solve the harmonic oscillator problem, it is important to consider both even and odd n values separately.

I hope this helps to clarify the use of Hermite polynomials in solving the harmonic oscillator problem. Keep in mind that these polynomials are just one part of the overall solution and must be combined with Gaussian functions and normalization constants to get the full solution.