Finding a constant from eigenfunction

[terp.asessed]

Homework Statement

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.

Homework 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

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!

 

[stevendaryl]

Homework Statement

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.

Homework 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

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!

\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.

 

[terp.asessed]

Thank you for your reply! Btw, could you please clarify:

the third one is a linear combination of the other two..

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

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.

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

 

[stevendaryl]

Thank you for your reply! Btw, could you pls clarify:
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?

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

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

Yes. D is a normalization constant.

 

[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?

 

[stevendaryl]

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?

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