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Hermite functions,Ladder operators

  1. May 6, 2007 #1
    1. The problem statement, all variables and given/known data

    It is possibly not a homework problem.However,to do a homework problem,I require this:
    Boas writes the effect of Ladder operators on y_n that satisfies
    y"_n-x^2y_n=-(2n+1)y_n,n=0,1,2,3...

    (D-x)(D+x)y_n=-2ny_n
    (D+x)(D-x)y_n=-2(n+1)y_n

    Then,she proved y_(m-1)=(D+x)y_m and the other raising operator eqn.


    So far there is no problem...

    Now she says,if n=0,we find a solution of (D-x)(D+x)y_n=-2ny_n by requiring (D+x)y_0=0

    My question is if n=0,we have (D-x)(D+x)y_0=0.
    Does that mean (D+x)y_0=0 necessarily?


    2. Relevant equations



    3. The attempt at a solution

    treating (d+x)y_0=t,I saw that we have (D-x)t=0 or,t=c exp[x^2/2]
    So,they are treating c=0?...why?

    I am toatlly confused.Please help.
     
  2. jcsd
  3. May 15, 2007 #2
    The last point (if you take c = 0 or t = 0) you get

    [tex]\frac{dy}{dx} + xy = 0[/tex]

    which gives

    [tex]y(x) = Ae^{-x^{2}/2}[/tex]

    But what you're asking is the converse:

    for n = 0 do we always have

    [tex](D-x)(D+x)y = y \implies (D+x)y = 0[/tex]

    Is this what you're asking?
     
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