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

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SUMMARY

The discussion revolves around the analysis of a particle's wavefunction, psi(r,theta,phi) = A*Exp[-r/a0], in the context of Schrödinger's equation. The key findings indicate that the angular momentum quantum number l is zero due to the absence of theta and phi dependence in the wavefunction. The calculated energy E was initially misinterpreted, with the correct value being E = -h_bar^2/(2*mu*a0^2) when considering the potential V(r) approaching zero as r approaches infinity. The solution emphasizes the importance of carefully reading the problem statement to avoid miscalculations.

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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 Schrödinger's equation)
(3) Having found E, consider finite r and find V(r)

Homework Equations


Schrödinger'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 Schrödinger'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
 
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jsc314159 said:

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 Schrödinger's equation)
(3) Having found E, consider finite r and find V(r)

Homework Equations


Schrödinger'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 Schrödinger'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.
 
kdv,

Thanks, that is the solution.

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

jsc
 
jsc314159 said:
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.
 
I will keep that in mind.

Thanks again.
 

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