How Do You Calculate 3rd and 4th Hermite Polynomials?

[nmsurobert]

Homework Statement

Calculate the third and fourth hermite polynomials

Homework Equations

(1/√n!)(√(mω/2ħ))ⁿ(x - ħ/mω d/dx)ⁿ(mω/πħ)^1/4 e^-mωx2/2ħ

a_k+2/a_k = 2(k-n)/((k+2)(k+1))

The Attempt at a Solution

i kind of understand how how to find the polynomials using the first equation up to n=1. I'm not sure i want to attempt to find it with n=3 because that will make (x - ħ/mω d/dx)ⁿ raised to the 3rd power. and then the 4th power.
we are provided with the second equation but i don't understand how to use it. there is an example in the book (townsend) and does the first 3 polynomials but the example makes no sense to me.

can anyone provide me with some insight?

 

[SteamKing]

Homework Statement

Calculate the third and fourth hermite polynomials

Homework Equations

(1/√n!)(√(mω/2ħ))ⁿ(x - ħ/mω d/dx)ⁿ(mω/πħ)^1/4 e^-mωx2/2ħ

a_k+2/a_k = 2(k-n)/((k+2)(k+1))

The Attempt at a Solution

i kind of understand how how to find the polynomials using the first equation up to n=1. I'm not sure i want to attempt to find it with n=3 because that will make (x - ħ/mω d/dx)ⁿ raised to the 3rd power. and then the 4th power.
we are provided with the second equation but i don't understand how to use it. there is an example in the book (townsend) and does the first 3 polynomials but the example makes no sense to me.

can anyone provide me with some insight?

It's not clear where you obtained this generating formula for Hermite polynomials, nor what the difficulty is with the alternate representation.

This article discusses other, simpler generating formulas for Hermite polynomials:

https://en.wikipedia.org/wiki/Hermite_polynomials

You shouldn't have to split the atom to come up with these.

 

[nmsurobert]

The difficulty with the alternate representation is I don't understand how to use it.
Apparently when n=1, the solution is X. Where does that come from?

 

[nmsurobert]

Even using one of the other formulas on Wikipedia like

(X - d/dx)ⁿ

For n=1, d/dx = 0 so the solution is X. I see that.

But for n=2 the solution is x² -1.

Why isn't it just x² don't all the d/dx go to zero?

 

[nmsurobert]

Even if I write it out as x^2 -d/dx(2x) + (d/dx)^2
I still don't get the correct solution for n=2.

 

[vela]

Even using one of the other formulas on Wikipedia like

(X - d/dx)ⁿ

For n=1, d/dx = 0 so the solution is X. I see that.

You shouldn't see that d/dx=0 because that's wrong. You need to apply the operator to $e^{-x^2/2}$ and then multiply the result by $e^{x^2/2}$. For $n=1$, you get
$$H_1(x) = e^{x^2/2} \left(x - \frac{d}{dx}\right)^1 e^{-x^2/2} = e^{x^2/2} (x e^{-x^2/2} + x e^{-x^2/2}) = 2x$$

 

[nmsurobert]

You shouldn't see that d/dx=0 because that's wrong. You need to apply the operator to $e^{-x^2/2}$ and then multiply the result by $e^{x^2/2}$. For $n=1$, you get
$$H_1(x) = e^{x^2/2} \left(x - \frac{d}{dx}\right)^1 e^{-x^2/2} = e^{x^2/2} (x e^{-x^2/2} + x e^{-x^2/2}) = 2x$$

ahhhh that makes sense. i spoke to the instructor and he cleared it up a bit for me too but he didn't explain it like that. thank you!