Timeline for Characterizing the harmonic oscillator creation and annihilation operators in a rotationally invariant way
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 10, 2010 at 9:43 | comment | added | José Figueroa-O'Farrill | Yes, on hindsight, I should have perhaps focussed on the harmonic oscillator. Anyway, there are many hamiltonians you can write down using creation and annihilation operators. Each one defines a different quantum system and in most cases the grading of $V_k$ defined here will have nothing to do with the energy. The symmetry group of the harmonic oscillator is in fact the unitary group preserving the hermitian inner product on (the complexification of) $E$. In other words, the symmetry group is $\operatorname{U}(n)$ and not $\operatorname{O}(n)$. | |
| Aug 10, 2010 at 7:59 | comment | added | Otis Chodosh | Thank you! I think this pretty much answers my question. Can you explain what you mean about the "harmonic oscillator being a red herring?" I don't really understand why, for example, this describes the harmonic oscillator while being $GL(n)$ invariant... | |
| Aug 10, 2010 at 7:46 | vote | accept | Otis Chodosh | ||
| Aug 10, 2010 at 2:28 | comment | added | José Figueroa-O'Farrill |
In the case of the harmonic oscillator, the hamiltonian takes a very simple form: namely, choosing a basis $(e_i)$ for $E$, with canonical dual basis $(e^i)$ for $E^*$, the hamiltonian is given by $H = \sum_i C(e^i) A(e_i)$ up to a constant. In other words, this is the grading operator on $V_k$.
|
|
| Aug 10, 2010 at 2:15 | comment | added | David E Speyer | Maybe this is a dumb question, but what does this have to do with the eigenvectors of $H$? | |
| Aug 10, 2010 at 2:08 | comment | added | José Figueroa-O'Farrill | Thanks, Victor -- I have edited my answer. Without realising it, I found myself talking about the Clifford algebra, which is my usual arena. | |
| Aug 10, 2010 at 2:06 | history | undeleted | José Figueroa-O'Farrill | ||
| Aug 10, 2010 at 2:06 | history | edited | José Figueroa-O'Farrill | CC BY-SA 2.5 |
corrected some signs, added more disclaimer and deleted final paragraph
|
| Aug 10, 2010 at 1:56 | history | deleted | José Figueroa-O'Farrill | ||
| Aug 10, 2010 at 1:09 | comment | added | Victor Protsak | José, the automorphism group of the Heisenberg Lie algebra is the symplectic group (the isometry group of the nondegenerate skew-symmetric form defining the commutation relations). In your second displayed formula, the sign should have been minus instead of plus. | |
| Aug 10, 2010 at 0:53 | history | answered | José Figueroa-O'Farrill | CC BY-SA 2.5 |