Derivation process of Selection Rule of hydrogenic atom

Click For Summary
The discussion focuses on the selection rules for hydrogenic atoms, particularly the lack of regulation on the principal quantum number n in selection rules. It highlights the integral of the radial part of the wavefunctions, which is non-zero when n and n' differ but l and m remain the same. The participants explore the use of orthogonality in solving the integral involving associated Laguerre polynomials. A specific integral is presented to illustrate the relationship between wavefunctions with different n values but identical l and m quantum numbers. The conversation emphasizes the importance of understanding these integrals in quantum mechanics.
smjchris
Messages
1
Reaction score
0
Homework Statement
Prove that the radial integral is always non-zero
Relevant Equations
Selection rule, wavefunction of hydrogeninc atom.
This page is Quantum mechanics by bransden. My homework is explain why there is no regulation of quantum number n in selection rule. Also explain that by solving that integral of radial part is always non-zero.

∫∞0[rRnl(r)]Rn′l′(r)r2dr

(n is different with n')

I tried to solve it by just calculate it, but I can't calculate the associated laguerre polynomial. I think the answer is using orthogonality, but how can I solve it?

pdfcoffee.com_quantum-mechanics-bransden-pdf-2-pdf-free.png
 
Last edited by a moderator:
Physics news on Phys.org
smjchris said:
Also explain that by solving that integral of radial part is always non-zero.
I think always is too strong. Consider two wavefunctions with different ##n## but the same ##l## and ##m## quantum numbers. What should the following integral give you?$$I=\int_0^{\infty}R_{n'l}R_{nl}~r^2dr\int_0^{2\pi}d\phi\int_0^{\pi}Y^*_{lm}Y_{lm}~\sin\theta d\theta.$$
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
2K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K