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3rd and 4th hermite polynomials

  1. Feb 9, 2016 #1
    1. The problem statement, all variables and given/known data
    Calculate the third and fourth hermite polynomials

    2. Relevant 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))
    3. 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?
     
  2. jcsd
  3. Feb 9, 2016 #2

    SteamKing

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    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.
     
  4. Feb 9, 2016 #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?
     
  5. Feb 9, 2016 #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?
     
  6. Feb 9, 2016 #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.
     
  7. Feb 14, 2016 #6

    vela

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    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
    $$H_1(x) = e^{x^2/2} \left(x - \frac{d}{dx}\right)^1 e^{-x^2/2} = e^{x^2/2} (x e^{-x^2/2} + x e^{-x^2/2}) = 2x$$
     
  8. Feb 15, 2016 #7
    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!
     
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