Solution to the Schrodinger Equation

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Homework Help Overview

The discussion revolves around proving that a specific wave function for a one-electron hydrogen atom is a solution to the Schrödinger equation. The wave function is given, and participants are tasked with demonstrating its validity in relation to the Hamiltonian operator and determining its energy and principal quantum number.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to substitute the wave function into the Schrödinger equation and differentiate as needed. They express uncertainty about what constitutes proof of a solution and how to derive the energy value.
  • Some participants question whether the left-hand side of the equation should yield a constant times the wave function, indicating a need for clarification on the steps taken.
  • Others suggest that the poster should show their work in detail to identify any potential mistakes in earlier calculations.

Discussion Status

Participants are actively engaging with the problem, with some providing guidance on the expected outcome of the substitution process. There is an acknowledgment of the need for clarity in the original poster's calculations, and the discussion is focused on exploring the reasoning behind the steps taken.

Contextual Notes

There are indications of formatting challenges with mathematical expressions, and one participant notes a potential typo in the original equation setup. The original poster expresses frustration over the complexity of the calculations and the cancellation of terms.

bradx888
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Homework Statement



Consider the full 1-electron hydrogen wave function.
[/B]
Prove that

ψ =A(6r –r2/a0) exp[-^r3/a0] sinθ exp[+iφ],

is a solution to the Schrödinger equation H|ψ> = E|ψ>, where H is the Hamiltonian operator.

Hence show that it's energy E= -1.51 eV and its principle quantum number n = 3.

Homework Equations


[/B]
So as far as I understand the Schrödinger equation is

-ħ2/2m[1/r2 δ/δr(r2 δ/δr) + 1/(r2sinθ) ∂/∂θ(sinθδ/δθ) + 1/(r2sinθ) δ2/δφ2]ψ - Vψ = Eψ

3. The Attempt at a Solution

I have subbed the solution into the equation, differentiating where needed and so on. I'm unsure as to what actually proves it is a solution, do I need to get one side equal to zero?
I also have no idea how to show that its E = -1.51 eV. I would be able to get from E to n though.
 
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I've had massive trouble trying to do powers and subscripts. Hopefully it's clear enough.
 
Hello Brad, welcome to PF :)

I have subbed the solution into the equation
Good. You are to find that your left hand side yields a constant times ##\Psi##. The constant is E.

Did you find that ? If not, show what you did find.
 
BvU said:
Hello Brad, welcome to PF :)Good. You are to find that your left hand side yields a constant times ##\Psi##. The constant is E.

Did you find that ? If not, show what you did find.
Ok I'm in the process of trying to get it like that. I'm struggling to factorise out the 'r's so may be a little while :P
 
BvU said:
Hello Brad, welcome to PF :)Good. You are to find that your left hand side yields a constant times ##\Psi##. The constant is E.

Did you find that ? If not, show what you did find.

Hi,

After working on it all day, I haven't got what is needed.
A lot of the terms have canceled out. I may have done it wrong but have checked thoroughly.

On the left hand side I am left with,

Ae-r/3aesinθ(-r2/9a3 + ((3a + 8)/3a2)r - (12ħ2 + 6m)/aħ2

Please please help :(
 
bradx888 said:
ψ =A(6r –r2/a0) exp[-^r3/a0] sinθ exp[+iφ],

-ħ2/2m[1/r2 δ/δr(r2 δ/δr) + 1/(r2sinθ) ∂/∂θ(sinθδ/δθ) + 1/(r2sinθ) δ2/δφ2]ψ - Vψ = Eψ
You should make the effort to learn LaTeX so you can write mathematics. There's a good tutorial here:
https://www.physicsforums.com/threads/physics-forums-faq-and-howto.617567/#post-3977517
LaTeX is pretty straightforward. If you quote this post, you can see the markup used to generate the equations below:
$$\psi = A \left(6r-\frac{r^2}{a_0}\right) e^{-r^3/a_0}\sin\theta e^{i\phi} \\
-\frac{\hbar^2}{2m}\left[\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) +
\frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta \frac{\partial}{\partial \theta}\right) +
\frac{1}{r^2\sin^2\theta}\frac{\partial^2}{\partial\phi^2}\right]\psi - V\psi = E\psi.$$ I don't know if it was just a typo, but in the last term of ##\nabla^2##, ##\sin\theta## should be squared.

You need to show your work. Just posting what you end up with isn't very helpful since your mistakes, if any, occurred in earlier steps. Unless we can see them, we don't know where you went wrong.
 

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