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Finding Hydrogen eigenfunction u(2,0)

  1. Feb 23, 2014 #1
    Taken from Physics of Quantum Mechanics, by James Binney.

    dnmrr9.png

    I try to calculate ##u_{n}^{l=n-2}##, something goes wrong:

    Starting, we define operator A by:

    [tex]A_{n-2} = \frac{a_0}{\sqrt 2}\left(\frac{i}{\hbar}p_r + \frac{1-n}{r} + \frac{Z}{(n-1)a_0}\right)[/tex]

    Substituting ##p_r = -i\hbar (\frac{\partial}{\partial r} + \frac{1}{r})##:

    [tex]A_{n-2} = \frac{a_0}{\sqrt 2}\left( \frac{\partial}{\partial r} + \frac{2-n}{r} + \frac{Z}{(n-1)a_0}\right)[/tex]

    Thus, we want to solve:

    [tex]\left(\frac{\partial}{\partial r} + \frac{2-n}{r} + \frac{Z}{(n-1)a_0} \right) u_{n}^{l=n-2} = 0[/tex]

    Solving by integrating factor method, we obtain:

    [tex]u_n^{l=n-2} = A r^{n-2} e^{\frac{Z}{(n-1)a_0}r}[/tex]

    Normalizing,

    [tex]u_n^{l=n-2} = \frac{1}{\sqrt{[2(n-1)]!}} \left(\frac{2Z}{(n-1)a_0}\right)^{\frac{3}{2}} \left(\frac{2Z}{(n-1)a_0}\right)^{n-2} e^{-\frac{Z}{(n-1)a_0}r} [/tex]

    This is similar to ##u_n^{l=n-1}##, simply replace n by n-1:

    ajx85d.png

    But when I substitute n = 2, so l = 0, I get ##u_2^0 = \frac{1}{\sqrt 2} \left(\frac{2Z}{a_0}\right)^{\frac{3}{2}} e^{-\frac{Z}{a_0}r}##

    I get a completely different result from the book:

    1ftl6a.png

    I'm not sure what's wrong with my derivation?
     
  2. jcsd
  3. Feb 23, 2014 #2

    Bill_K

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    Science Advisor

    The copy of Binney I'm looking at is slightly different from yours, so bear with me, but the equation you're trying to use,

    An-1|n, n-1> = 0

    is only for (as he says) "the radial wavefunction of the circular orbit with angular momentum ℓ = n - 1." That is, it gives the wavefunctions along the diagonal in Table 8.1. To get the other wavefunctions you need to start with one of these wavefunctions and one or more times apply A. This will step you up in the table, along one of the columns.

    "Applications of A should generate the wavefunctions for ℓ < n - 1."
     
  4. Feb 23, 2014 #3
    Ah, so by applying A to ##u_n^{l=n-1}## it will give ##u_n^{l=n-2}##?
     
  5. Feb 23, 2014 #4

    Bill_K

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    That's my understanding. Try it and see!
     
  6. Feb 23, 2014 #5
    It's mentioned right below the screenshot, I can't believe I missed it..

    I tried it, and normalized and it came out right!
     
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