QM: Angular Momentum-Wavefunction "psi" & Schrodinger's Eq.

[jsc314159]

Homework Statement

Particle described by wavefunction psi(r,theta,phi) = A*Exp[-r/a0] (a0 = constant)

(1) What is the angular momentum content of the state
(2) Assuming psi is an eigenstate in a potential that vanishes as r -> infinity, find E (match leading terms in Schrodinger's equation)
(3) Having found E, consider finite r and find V(r)

Homework Equations

Schrodinger's equation in spherical coordinates.

The Attempt at a Solution

(1) The term A, in the wavefunction, is not given to be a function of theta or phi. I am thinking it is a normalization constant. Therefore, apparently there is no theta or phi dependence, l = 0. Does this seem reasonable.

(2) If l = 0, then Schrodinger's equation becomes:

(-h_bar^2/(2*mu) *(1/r^2 * partial/partial_r * r^2 partial/partial_r) + V(r))psi = E*psi

Let V(r) = 0 and plug in the given psi.

The answer if get is E = -h_bar^2/(2*mu*ao^2*r^2) * (r^2 - 2*r*a0).
The book's answer is E = -h_bar^2/(2*mu*ao^2)

The additional terms in my solution come from carring out the differentiation operations on psi. The (1/r^2 * partial/partial_r * r^2 partial/partial_r) on psi gets me 1/r^2)* (r^2 - 2*r*a0). Somehow the book solution eliminates the 2*r*ao term but I do not see how.

If I can figure out where I am going off track on part 2, I think I can manage part 3. Can you help?

jsc

 

[kdv]

Homework Statement

Particle described by wavefunction psi(r,theta,phi) = A*Exp[-r/a0] (a0 = constant)

(1) What is the angular momentum content of the state
(2) Assuming psi is an eigenstate in a potential that vanishes as r -> infinity, find E (match leading terms in Schrodinger's equation)
(3) Having found E, consider finite r and find V(r)

Homework Equations

Schrodinger's equation in spherical coordinates.

The Attempt at a Solution

(1) The term A, in the wavefunction, is not given to be a function of theta or phi. I am thinking it is a normalization constant. Therefore, apparently there is no theta or phi dependence, l = 0. Does this seem reasonable.

(2) If l = 0, then Schrodinger's equation becomes:

(-h_bar^2/(2*mu) *(1/r^2 * partial/partial_r * r^2 partial/partial_r) + V(r))psi = E*psi

Let V(r) = 0 and plug in the given psi.

The answer if get is E = -h_bar^2/(2*mu*ao^2*r^2) * (r^2 - 2*r*a0).
The book's answer is E = -h_bar^2/(2*mu*ao^2)

The additional terms in my solution come from carring out the differentiation operations on psi. The (1/r^2 * partial/partial_r * r^2 partial/partial_r) on psi gets me 1/r^2)* (r^2 - 2*r*a0). Somehow the book solution eliminates the 2*r*ao term but I do not see how.

If I can figure out where I am going off track on part 2, I think I can manage part 3. Can you help?

jsc

The potential is NOT zero. they just say that it goes to zero as r goes to infinity. So your solution is only valid when r goes to infinity. Take this limit in your answer and you will agree with the book.

 

[jsc314159]

kdv,

Thanks, that is the solution.

How can I learn to see these types of things more effectively?

jsc

 

[kdv]

kdv,

Thanks, that is the solution.

How can I learn to see these types of things more effectively?

jsc

In this case, you simply had to read very carefully the question. That was the key: that they say thatV goes to zero at infinity.

 

[jsc314159]

I will keep that in mind.

Thanks again.