Finding Hydrogen eigenfunction u(2,0)

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The discussion revolves around the calculation of the hydrogen eigenfunction u(2,0) using the operator A defined in the context of quantum mechanics. The initial attempt at deriving u_n^{l=n-2} led to a discrepancy with the expected result from Binney's text. It was clarified that the equation used was specific to the radial wavefunction for l = n - 1, and to find wavefunctions for l < n - 1, one must apply the raising operator A† to the known wavefunctions. After applying A† to u_n^{l=n-1}, the correct eigenfunction was successfully derived and normalized, resolving the initial confusion. The discussion highlights the importance of correctly applying quantum mechanical operators to derive the desired wavefunctions.
unscientific
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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:

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)

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

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

Thus, we want to solve:

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

Solving by integrating factor method, we obtain:

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

Normalizing,

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}

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?
 
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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."
 
Bill_K said:
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."

Ah, so by applying A to ##u_n^{l=n-1}## it will give ##u_n^{l=n-2}##?
 
unscientific said:
Ah, so by applying A to ##u_n^{l=n-1}## it will give ##u_n^{l=n-2}##?
That's my understanding. Try it and see!
 
Bill_K said:
That's my understanding. Try it and see!

It's mentioned right below the screenshot, I can't believe I missed it..

I tried it, and normalized and it came out right!
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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