- #1

unscientific

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- 13

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:

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:

I'm not sure what's wrong with my derivation?