Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Is there any case of potential $V$, such that the continuity of the operator $H=c\ \Delta+V$ is not spoiled?

And I don't know any non-differnetial operator examples for continous spectra. I guess I don't know the full spectrum of operators (ba dum tss!).

My question comes about as I wonder about the justification of the term "bounded state", especially in relatoion to the fairly straight forward classical concept.

share|improve this question
add comment

3 Answers

Examples of operators with continuous spectrum that are not differential operators in the position representation are the components of the position operator.

Adding to the free Hamiltonian a repulsive potential doesn't change the continuous spectrum, except for shifting the ground state energy. It also introduces no resonances.

In order to have a bound state, the minimal potential at infinity must be larger than the minimal finite potential. (But because of quantum effects, this is only a necessary condition.)

The reason a bound state is called bound is that the probablility of moving far apart decays exponentially with the distance.

share|improve this answer
add comment

There are many potentials V such that there are no bounded states : For instance all positive potentials which converge to zero at infinity (at least if they are not to pathological).

In this case there might be resonances with finite lifetime, but eventually the particle will allways escape to infinity.

share|improve this answer
    
Interesting, although it stands in contrast to the other answer I think. How does this come about. Like do you maybe have some simple reference (a Sturm Luiville theory theorem or something)? –  NiftyKitty95 Aug 6 '12 at 19:25
    
Consider for instance the simple one-dimensional case with non-negative V with compact support. Since V is non-negative all eigenvalues of H must be non-negative. But then each eigenfunction of H must be a superposition of two plane waves outside the support of V and can't therefore be square integrable. So there are no bounded states. –  jjcale Aug 7 '12 at 17:53
    
Mhm, I don't see that so easily. –  NiftyKitty95 Aug 7 '12 at 18:36
add comment

The trivial case is where V=const. Any potentials that offer boundaries, I believe will always create non-continuous spectra due to the nature of differential equations and boundary conditions.

share|improve this answer
    
I can't seem to add comments to an answer that is not my own, but I would like to comment on jcale's answer and Nick's response. Non-continuity in the energy arises from the boundary conditions set up by the potential V. In any case where the system is not bounded, you have a continuum of possible energy levels. The easiest way to see this is to simply plot a graph of V and a flat line that represents the energy of the system. The intersection between V and the energy E represents a boundary condition. For two intersection points, you have a bounded state with quantized energy. –  mcFreid Aug 7 '12 at 20:35
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.