# Finding a constant from eigenfunction

1. Oct 20, 2014

### terp.asessed

1. The problem statement, all variables and given/known data
Since Hamiltonian operator is:

Ĥ = - (ħ2/(2m))(delta)2 - A/r
where r = (x2+y2+z2)
(delta)2 = ∂2/∂x2 + ∂2/∂y2 + ∂2/∂z2
A = a constant

from Ĥg(r) = Eg(r) form, where:

g(r) = D e-r/b(1-r/b)

with b, D as constants, is an EIGENFUNCTION of Ĥ, find the correct b and give the eigenvalue E.

2. Relevant equations
g(r) = D e-r/b(1-r/b)

Ĥ = - (ħ2/(2m))(delta)2 - A/r
where r = (x2+y2+z2)
(delta)2 = ∂2/∂x2 + ∂2/∂y2 + ∂2/∂z2
D = a constant

3. The attempt at a solution
g(r) = D e-r/b(1-r/b)
g'(r) = -De-r/b/b - De-r/b/b + Dre-r/b/b2
= -2De-r/b/b + Dre-r/b/b2
g''(r) = 3De-r/b/b2 - Dre-r/b/b3

[delta_g(r)]2 = 5De-r/b/b2 -Dre-r/b/b3 - 4De-r/b/(br)

Ĥg(r) = Eg(r)
- (ħ2/(2m))(5De-r/b/b2 -Dre-r/b/b3 - 4De-r/b/(br)) - A/r * (D e-r/b(1-r/b)) = E*D e-r/b (1-r/b)
..which is then reduced to:

-5ħ2/(2mb2) + ħ2r/(2mb3) + 4ħ2/(2mbr) - A/r + A/b = E (1-r/b)....I managed to cancel out constant D and e-r/b, but I am at loss how how to eliminate r?

Should I equivelate:
ħ2r/(2mb3) + 4ħ2/(2mbr) - A/r = -E(r/b) ? I tried this way, and still haven't managed to find a way to cancel out r....to make b an independent number...
With regard to E, if I am not wrong, I think Energy is that of an excited state, NOT ground state, except I haven't figured it out (for I have not gotten b constant), and have no idea which energy level it is....

Any hints or notices to my mistakes would be appreciated!

2. Oct 20, 2014

### stevendaryl

Staff Emeritus
$\hat{H} g(r) = E g(r)$

gives you an equation of the form:

$\frac{\alpha}{r} + \beta + \gamma r = 0$

where $\alpha$, $\beta$ and $\gamma$ are combinations of your constants: $A, b, E$.

That equation can only be true for all values of $r$ if each coefficient is equal to zero, separately. So it must be that
$\alpha = 0$
$\beta = 0$
$\gamma = 0$

So if you figure out what $\alpha$, $\beta$ and $\gamma$ are in terms of your constants, you can get three equations for your two unknowns: $E$ and $b$. That's too many, but it will turn out that the equations are redundant--the third one is a linear combination of the other two. So you can solve for $E$ and $b$.

$D$ cancels out, so you can't solve for it using the Schrodinger equation, but instead you have to choose $D$ so that the integral of $|g(r)|^2$ over all space is equal to 1.

3. Oct 20, 2014

### terp.asessed

because, according to what you've said:

α = 4ħ2/(2mb) - A = 0 --> b = 2ħ2/(Am)
β = -5ħ2/(2mb2) + A/b = 0
γ = ħ2/(2mb3) + E/b = 0

...but I couldn't derive b from β and γ....does it mean I solved everything from the beginning wrong?

And

For the last one, do you mean I need to normalize?

Last edited by a moderator: Oct 20, 2014
4. Oct 20, 2014

### stevendaryl

Staff Emeritus
On the right-hand side of Schrodinger's equation for this case, you have
$E g(r) = E D e^{-r/b} - (E D r/b) e^{-r/b}$

So the equation for $\beta$ should have $E$ on the right-hand side of the equals.

But you can certainly solve for $E$ using the equation for $\gamma$. You already know $b$ from your equation for $\alpha$. Plug that into the equation for $\gamma$

Yes. $D$ is a normalization constant.

Last edited: Oct 20, 2014
5. Oct 20, 2014

### terp.asessed

Thank you--I guess I figured out b and E, but may I ask what I should do with β? Can I use it to check b and E I found out?

6. Oct 21, 2014

### stevendaryl

Staff Emeritus
When I looked at it, it seemed that the equation for $\beta$ gave the same information as the equation for $\gamma$.