# Solution to the Schrodinger Equation

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1. Nov 23, 2014

1. The problem statement, all variables and given/known data

Consider the full 1-electron hydrogen wave function.

Prove that

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

is a solution to the Schrodinger 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.

2. Relevant equations

So as far as I understand the Schrodinger 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.

Last edited: Nov 23, 2014
2. Nov 23, 2014

I've had massive trouble trying to do powers and subscripts. Hopefully it's clear enough.

3. Nov 23, 2014

### BvU

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.

4. Nov 23, 2014

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

5. Nov 23, 2014

Hi,

After working on it all day, I haven't got what is needed.
A lot of the terms have cancelled 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

6. Nov 23, 2014

### vela

Staff Emeritus
You should make the effort to learn LaTeX so you can write mathematics. There's a good tutorial here:
$$\psi = A \left(6r-\frac{r^2}{a_0}\right) e^{-r^3/a_0}\sin\theta e^{i\phi} \\ -\frac{\hbar^2}{2m}\left[\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) + \frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta \frac{\partial}{\partial \theta}\right) + \frac{1}{r^2\sin^2\theta}\frac{\partial^2}{\partial\phi^2}\right]\psi - V\psi = E\psi.$$ I don't know if it was just a typo, but in the last term of $\nabla^2$, $\sin\theta$ should be squared.