Energy levels in Hydrogen - derivation

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Homework Help Overview

The discussion revolves around the derivation of energy levels in the hydrogen atom, specifically using the expectation value of the Hamiltonian operator. Participants are exploring the mathematical framework involving wavefunctions and integrals in quantum mechanics.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the method of evaluating the integral for the expectation value of energy and question whether shortcuts exist for this process. There is a focus on the nature of the wavefunction and its relation to eigenstates.

Discussion Status

The conversation is active, with participants sharing insights about the evaluation of integrals and the use of tabulated values for certain expectation values. Some participants express concerns about the complexity of the integrals involved, while others challenge the standard methods being discussed.

Contextual Notes

There is an indication of differing opinions on the standard methods for deriving energy levels, with some participants emphasizing the need for integrals while others suggest alternative approaches. The discussion also highlights the potential for confusion regarding the application of eigenvalues and eigenfunctions in this context.

Grand
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Homework Statement


We know that:

E=<\psi|H|\psi>

where
H=-\frac{\hbar^2}{2m} \nabla^2 - \frac{Ze^2}{4\pi\epsilon_0 r}
and
\psi=R(r)Y(\theta, \phi)
with
R(r)=\frac{1}{\sqrt{(2n)!}}(\frac{2Z}{na_0})^{3/2}\left(\frac{2Zr}{na_0}\right)^{n-1}e^{-Zr/na_0}
If I want to find the energy, do I just evaluate this integral over all space? I started doing it, but it becomes very untidy at a point - I just wanted to ask if this is the method.


Homework Equations





The Attempt at a Solution

 
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Mod Note: moved to advanced physics. Please post such a question in that forum.
 
Essentially yes. However, there are shortcuts you can take to evaluate expectation values of the Hamiltonian. Is psi in an eigenstate (discrete portion of the spectrum), no integral necessary. Likewise for a weighted sum of eigenstates.
 
Psi is a wavefunction, and an eigenstate - but I believe it has to be an integral. However, can you tell me what these shortcuts are?
 
Tabulated (as a result of integrations on which you needn't worry) values for

\langle nlm |\frac{1}{r} | nlm \rangle(nlm are eigenvectors of the Hamiltonian) exist and ease your work.
 
Last edited:
I know these values - for inverse 1st, 2nd and 3rd powers - my concern is the first term - the one involving the differential - it is going to produce 2 integrals on its own.
 
NO, this is most definitely not the standard method. Do you know what eigenvalues and eigenfunctions are?
 
Yes, I do ofc. I know how to find the energy by requiring the raising operator to terminate the circular orbit wavefunction, I was just wondering is you can do it by directly evaluating this integral.
 
Grand said:
I know how to find the energy by requiring the raising operator to terminate the circular orbit wavefunction,

This is surely not the standard way. Please do demonstrate.
 
  • #10
Grand said:
I know these values - for inverse 1st, 2nd and 3rd powers - my concern is the first term - the one involving the differential - it is going to produce 2 integrals on its own.
So what specifically is the problem? These aren't particularly tricky integrals. They're straightforward but tedious. Show us what you've done so far.
 

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