3rd and 4th hermite polynomials

In summary, the conversation discusses the calculation of the third and fourth Hermite polynomials using two equations provided. The difficulty lies in understanding how to use the second equation. After some discussion and clarification, it is determined that the operator must be applied to the exponential function in the equation and then multiplied by the exponential function again to get the correct solution.
  • #1
nmsurobert
288
36

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

ak+2/ak = 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?
 
Physics news on Phys.org
  • #2
nmsurobert said:

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

ak+2/ak = 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.
 
  • #3
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?
 
  • #4
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?
 
  • #5
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.
 
  • #6
nmsurobert said:
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
$$H_1(x) = e^{x^2/2} \left(x - \frac{d}{dx}\right)^1 e^{-x^2/2} = e^{x^2/2} (x e^{-x^2/2} + x e^{-x^2/2}) = 2x$$
 
  • #7
vela said:
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!
 

1. What are 3rd and 4th Hermite polynomials?

The 3rd and 4th Hermite polynomials are mathematical functions that are used in physics and engineering to describe the behavior of physical systems. They are part of a family of orthogonal polynomials known as Hermite polynomials, which are solutions to a specific type of differential equation.

2. How are 3rd and 4th Hermite polynomials different from each other?

The main difference between the 3rd and 4th Hermite polynomials lies in their degree, or order. The 3rd Hermite polynomial is a cubic function, while the 4th Hermite polynomial is a quartic function. This means that the 4th Hermite polynomial has an extra term compared to the 3rd Hermite polynomial, making it slightly more complex.

3. What are the applications of 3rd and 4th Hermite polynomials?

The 3rd and 4th Hermite polynomials have many applications in physics and engineering. They are commonly used to describe the behavior of quantum mechanical systems, such as the energy levels of atoms and molecules. They are also used in signal processing and image analysis, as well as in statistics for data analysis and curve fitting.

4. How are 3rd and 4th Hermite polynomials calculated?

The 3rd and 4th Hermite polynomials can be calculated using recursive formulas or by using the Rodrigues' formula. These methods involve using the coefficients of the previous Hermite polynomial in the series to calculate the next one. Alternatively, they can also be calculated using specialized software or mathematical libraries.

5. What is the significance of the Hermite polynomials in mathematics?

The Hermite polynomials hold a special place in mathematics due to their numerous applications in physics and engineering. They are also important in the study of special functions and mathematical analysis. Additionally, they have connections to other areas of mathematics, such as combinatorics and number theory, making them a versatile and valuable tool in various fields of study.

Similar threads

  • Advanced Physics Homework Help
Replies
10
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
11
Views
2K
  • Differential Equations
Replies
1
Views
1K
  • General Math
Replies
2
Views
2K
  • Advanced Physics Homework Help
Replies
7
Views
2K
  • Advanced Physics Homework Help
Replies
10
Views
2K
  • Linear and Abstract Algebra
Replies
8
Views
996
  • Differential Equations
Replies
1
Views
3K
  • Set Theory, Logic, Probability, Statistics
Replies
13
Views
2K
Back
Top