Find the eigenstates and eigenvalues of the Hamiltonian.

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Homework Help Overview

The discussion revolves around finding the eigenstates and eigenvalues of the Hamiltonian for a particle constrained to move on a circular path, as well as exploring the implications of a one-dimensional attractive potential that is negative everywhere. The participants are examining the variational principle and its application to bound states in quantum mechanics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of the variational principle and trial wave functions to demonstrate the existence of bound states. There is uncertainty about the appropriate potential for the circular motion problem, with some suggesting it may be zero while others propose a classical interpretation. Questions arise regarding the setup of the Schrödinger equation and the implications of different potential forms.

Discussion Status

The conversation is ongoing, with various interpretations of the potential being explored. Some participants have made attempts to solve the first question, while others are still clarifying the second question's requirements. There is no explicit consensus on the correct approach or potential, but several productive lines of reasoning have been presented.

Contextual Notes

Participants are working under the constraints of typical homework rules, which may limit the information provided in the problem statements. There is a discussion about the implications of assuming a potential when it is not explicitly stated, as well as the need to consider the quantum mechanical nature of the system.

  • #31
MathematicalPhysicist said:
The wave function is defined as exp(-ax^2), so for a=0 it's 1.

No it isn't, you are forgetting about the normalization constant...which depends on a:wink:

Now, &lt;H(0)&gt;=\int V(x)dx Now if I assume that this integral converges then it mustn't be bigger than zero, if it were zero then the integrand would equal almost everywhere to zero, but V(x)<0 for every x, so this integral is negative.

Again, \langle H(0) \rangle is undefined, but \lim_{a\to 0^{+}}\langle H(a)\rangle is well defined.
 
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  • #32
Ok, thanks.
 
  • #33
wait a minute, how do I calculate this limit?
&lt;H(a)&gt;=a\hbar^2/2m + \sqrt(2a/\pi)\int exp-2ax^2)dx
it looks to me it converges to zero.
 
  • #34
MathematicalPhysicist said:
wait a minute, how do I calculate this limit?
&lt;H(a)&gt;=a\hbar^2/2m + \sqrt(2a/\pi)\int exp-2ax^2)dx
it looks to me it converges to zero.

Shouldn't there be a V in the integrand?
 
  • #35
Yes, there should, and square root is over (2a/pi).

I believe this question is the easiest from the questions from my HW.
:-)
 

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