Central potential QM problem

In summary, the conversation discusses the last part of a long question regarding the time-independent Schrodinger equation for a central potential. It is suggested to use separation of variables to simplify the equation and then make a substitution to further simplify it. The question then asks if the spectrum of the solution is the same as for a one-dimensional problem. It is argued that the solutions will not be identical due to the additional term in the potential, and the concept of boundary conditions is also brought up. Finally, it is suggested that the question may actually be asking if the solutions to the 1D problem can be carried over to the 3D problem, which is deemed unlikely due to the different domains of the solutions.
  • #1
sachi
75
1
this is the last part of a long question.
We first have to serparate the TISE for a central potential into two parts using separation of variables (i.e into and R(r) and Y(theta, phi). Then we take the total differential equation in R(r) and substitute R(r) = f(r)/r and get:

-(hbar^2)/2m * (d^2)f/(dr^2) + (V + c/2m(r^2))f = Ef

where c =l(l+1) where l is the angular momentum quantum no. all the mathematical derivations I can do fine.
The question notes that the equation is exactly the same form as the 1-d TISE with V replaced by V +c/2m(r^2). It then asks if we can conclude that the spectrum is the same as for the 1-d problem.
I'm a bit stumped with this. It seems obvious that if you include another term in the potential of a different form, then the energy levels should change (except for l=0), but presumably they want something better than this. Any suggestions would be greatly appreciated.

Sachi
 
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  • #2
Hi sachi,

Clearly the differential equation is the same, but you know there is more to solving a differential equation than just the equation itself. In general you need to specify things like initial conditions or boundary conditions, domain of the solutions, and so forth. Think about some of these issues and see what progress you can make.
 
  • #3
I think I've essentially nailed the point you were hinting at (i.e we no longer have the boundary condition that psi = 0 outside the box, so we are not going to get the same set of stationary waves corresponding to psi, so the energy levels will be different). There's just one thing that you said that I'm a little bit confused about; surely the differential equation is no longer the same? What if V(r) is not proportional to 1/(r^2). then our modified potential has a completely different form and so does the differential equation?

Sachi
 
  • #4
sachi said:
this is the last part of a long question.
We first have to serparate the TISE for a central potential into two parts using separation of variables (i.e into and R(r) and Y(theta, phi). Then we take the total differential equation in R(r) and substitute R(r) = f(r)/r and get:

-(hbar^2)/2m * (d^2)f/(dr^2) + (V + c/2m(r^2))f = Ef

where c =l(l+1) where l is the angular momentum quantum no. all the mathematical derivations I can do fine.
The question notes that the equation is exactly the same form as the 1-d TISE with V replaced by V +c/2m(r^2). It then asks if we can conclude that the spectrum is the same as for the 1-d problem.
I'm a bit stumped with this. It seems obvious that if you include another term in the potential of a different form, then the energy levels should change (except for l=0), but presumably they want something better than this. Any suggestions would be greatly appreciated.

Sachi

You are right that it is obvious that the spectrum will not be the same! (except for the case l=0, as you noted). The extra term contains 1 over r^é so it chqnges co,pletely the solution (as you said). An example is the infinite 1-D square well vs the infinite spherical well. For l=0, the spectrum is identical with the wavefunctions also identical, i.e. sine functions (with the width of the 1-D well replaced by the radius of the spherical well) but for l not equal to zero, the solutions are the Bessel functions and the energies are given in terms of the zeroes of those functions.

Pat
 
  • #5
nrqed,

I think they aren't asking the rather dull question of whether two different differential equations have the same solutions. I believe the question is if you already knew the solutions to the 1d problem with potential [tex] V_{eff}(x) [/tex] (not just [tex] V(x) [/tex] ), can you carry those solutions over wholesale. The answer must in general be no. Just for starters, the variable r in the radial equation must always be positive, but you would probably have solved the 1d equation on the whole real line. It isn't hard to imagine situations where this can mess things up.
 
  • #6
Physics Monkey said:
nrqed,

I think they aren't asking the rather dull question of whether two different differential equations have the same solutions. I believe the question is if you already knew the solutions to the 1d problem with potential [tex] V_{eff}(x) [/tex] (not just [tex] V(x) [/tex] ), can you carry those solutions over wholesale. The answer must in general be no. Just for starters, the variable r in the radial equation must always be positive, but you would probably have solved the 1d equation on the whole real line. It isn't hard to imagine situations where this can mess things up.

Ah ok. Yes, this makes much more sense as a question! Sorry I did not get it at first.
And I agree with your point completely.

sorry if I was way too naive in my interpretation of the question. Thansk for your comment!

Patrick
 

1. What is a central potential quantum mechanics problem?

A central potential quantum mechanics problem is a type of problem in which a particle (such as an electron) is subjected to a central force, meaning the force acts towards the center of a given reference frame. This type of problem is commonly encountered in the study of atomic and molecular systems.

2. How is the Schrödinger equation used to solve central potential quantum mechanics problems?

The Schrödinger equation is a fundamental equation in quantum mechanics that describes the time evolution of a quantum system. It can be used to solve central potential problems by setting up the equation for the system and then solving it for the wave function of the particle under the influence of the central force.

3. What are some common examples of central potential quantum mechanics problems?

Some common examples of central potential quantum mechanics problems include the hydrogen atom, the helium atom, and the simple harmonic oscillator. These systems all involve a central force that acts on the particle and can be solved using the Schrödinger equation.

4. How does the angular momentum affect central potential quantum mechanics problems?

The angular momentum of a particle is an important factor in central potential problems. The angular momentum of the particle is quantized, meaning it can only take on certain discrete values. This quantization is due to the symmetry of the central force, which leads to conserved angular momentum in the system.

5. What is the significance of central potential quantum mechanics problems in the study of quantum mechanics?

Central potential quantum mechanics problems are important because they provide a framework for understanding the behavior of particles in many physical systems. They also allow for the application of mathematical tools and techniques used in quantum mechanics, such as the Schrödinger equation, to solve for the behavior of these systems. Additionally, the solutions to central potential problems can provide valuable insights into the behavior of more complex quantum systems.

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