Find the eigenstates and eigenvalues of the Hamiltonian.

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SUMMARY

The discussion focuses on finding the eigenstates and eigenvalues of the Hamiltonian for a particle confined to a circular path, specifically addressing the variational principle and the implications of a negative potential. The participants clarify that the potential is zero everywhere for the given problem, which simplifies the analysis. They emphasize the necessity of showing that the expectation value of the Hamiltonian is negative to confirm the existence of at least one bound state. The final conclusion is that a bound state exists primarily due to the negative potential condition.

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  • Familiarity with variational methods in quantum mechanics.
  • Knowledge of expectation values and their significance in determining bound states.
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  • #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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