Timeline for Mathematical equivalent to ladder operators?
Current License: CC BY-SA 3.0
9 events
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| May 24, 2014 at 21:31 | comment | added | user37929 | Sure, every comment is highly useful. Especially, I would appreciate it if you could comment on the application in the edit I tried.(particularly interesting would : where my error is or why this method failed) | |
| May 24, 2014 at 21:17 | comment | added | john mangual | @Lipschitz I do know a way to generalize the ladder operators... and I would love to write about them. As your question discusses periodic potentials I had left it out... may I put a second answer here? | |
| May 24, 2014 at 21:12 | comment | added | user37929 | right, apparently classical ladder operators only exist for 'shape invariant' potentials. but apparently, there are also some steps done in the direction of more general potentials, as you can for example see in the paper given by Carlo. And this is actually, why I am asking. Because I was wondering, when exactly it is possible to extract the full spectrum from the ground state? I mean I am asking about periodic operators because I have to deal with them at the moment, but my question is more general. | |
| May 24, 2014 at 21:09 | comment | added | john mangual | @Lipschitz I am suggesting a ladder structure may not exist... the Harmonic oscillator in phase space is a circle $H = p^2 + x^2$ which has a rotational symmetry, but your hamiltonain is $H = p^2 + \sin(x)$ which does not have the same type of symmetry. We have $[p,x] = i$ but $[p, \sin x] = i \cos x$. | |
| May 24, 2014 at 21:07 | comment | added | user37929 | so you say: $\psi(x+2 \pi) = \psi(x) e^{i 2\pi x}$? I mean, this is clear, from $\psi(0) = \psi(2\pi)$. I agree, Floquet-theory and Bloch theorem were the first classical results in some sense to the theory of periodic Schrödinger operators, but I don't see how all this helps me with my actual problem: finding the ladder operators for higher eigenvalues if the ground state is known. | |
| May 24, 2014 at 20:46 | history | edited | john mangual | CC BY-SA 3.0 |
added 961 characters in body
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| May 24, 2014 at 20:36 | comment | added | john mangual | @Lipschitz Your Hamiltonian is invariant under translations $x \mapsto x + 2\pi$. Therefore, the translation group $\mathbb{Z}^2$ induces a representation on the space of wave-functions $\psi_k(x)$. | |
| May 24, 2014 at 20:22 | comment | added | user37929 | mhmm... I don't really see through your answer. My first concern is, that your theorem applies only to translations, which is slightly different from my problem which is more like rotation. Even if it applies to rotation (as I restrict myself to $[0,2\pi]$ with $\psi(0)=\psi(2\pi)$ and the same for the derivative), then I don't see how it helps me getting all the energy eigenvalues if I just know the ground state ? Actually, your theorem does not give much more insight into the problem, as I already assumed my wavefunction to be periodic. So sorry, I don't see how this helps. | |
| May 24, 2014 at 19:48 | history | answered | john mangual | CC BY-SA 3.0 |