Timeline for What does a unitary transformation mean in the context of an evolution equation?
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Jan 2, 2012 at 23:14 | comment | added | Ron Maimon | What handwaving? I just told you the exact discretization! | |
| Jan 2, 2012 at 18:53 | comment | added | joseph f. johnson | handwaving at this point is negative. | |
| Jan 2, 2012 at 10:11 | comment | added | Ron Maimon | This is the standard discretization of the Schrodinger Hamiltonian: make space a lattice, replace the p^2 operator with (T- -2I + T+) where T- and T+ are shifting operators by +- 1 spacing, the potential is still V(x), where x is restricted to lattice spots. If you make the lattice finite, this is now a finite dimensional Hilbert space, and the questions about the limit are the lattice spacing goes to zero. You always have to do this in the back of your head for any problem involving path integrals or infinite dimensional determinants, and disguising the limiting process is less than helpful. | |
| Jan 2, 2012 at 9:47 | comment | added | joseph f. johnson | It is time for you to put up the regularisation of the determinant in the original post so that we can see whether it is at all useful in this context, i.e., finding stationary energy levels.... | |
| Jan 2, 2012 at 8:28 | comment | added | Ron Maimon | This is the question of whether the zero eigenvalue is sensible in the limit. The other shift-operator zeros (like those in Fermion doubling) are real problems, and they jiggle ever-faster as the lattice gets finer, escaping from the long-range continuum limit. The other answer didn't discuss these issues, but it didn't pretend they weren't there, like yours does, by talking glibly about the infinite dimensional determinant as if there were no issues of the limit. The implicit idea in the first answer was that you would take the limit by hand, as you should. | |
| Jan 1, 2012 at 21:50 | comment | added | joseph f. johnson | I fear your argument is falsified by the shift operator. For every finite dimenstional « approximation », it has a kernel. But in the limit, there is no kernel. Not that I know this for a fact, but it means there is something to prove. You, if you wanted to make an answer along those lines, would hae to check that there was some other suitable notion of approximation which prevented this from happening. There was nothing in the other answer that even showed any awareness of this issue. | |
| Jan 1, 2012 at 17:38 | comment | added | Ron Maimon | All the arguments in physics about infinite dimensional determinants are from path integral considerations, and there is no point in being pedantically rigorous, because the mathematicians have no idea what the right concepts are for being rigorous about such things. Just imagine a space-time lattice, and take the lattice finer, and at each stage you can conclude that det=0 implies non-invertibility, and the only question is whether the limit of the zero eigenvalue is sensible in the limit. | |
| Dec 31, 2011 at 16:54 | comment | added | joseph f. johnson | Perhaps I am wrong, but when you extend the definition of det. that way, it is not clear that you can still conclude to the non-invertibility of the linear transformation of which it is the det, so the rest of the argument, which depends on their being a kernel, is left hanging. So using the power series is more direct and allows one to immediately conclude there is a non-trivial kernel. Furthermore, if one says « implications of the ....», well, that means « implications ». There was nothing in the other answer about the many things the homogeneity degree can lead to in regularisation..... | |
| Dec 31, 2011 at 10:17 | comment | added | Ron Maimon | His answer is superior to yours pedagogically, precisely because he doesn't shy away from the implication of the homogeneity degree! He explains that scaling an operator by a constant C introduces a factor of C^N where N is the dimension of the space. There is no point in separating the finite and infinite dimensional cases, because you need to define the determinants arising in quantum mechanics by a limiting procedure from the finite dimensional case anyway. This limiting process is best left explicit, and you check at the end if there is no dependence on the regulator. | |
| Dec 31, 2011 at 7:00 | history | edited | joseph f. johnson | CC BY-SA 3.0 |
eliminated advanced technical term, 'kernel'.
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| Dec 30, 2011 at 18:24 | history | edited | joseph f. johnson | CC BY-SA 3.0 |
included an implicit hypothesis
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| Dec 30, 2011 at 17:37 | comment | added | joseph f. johnson | There is a substantive difference with great pedagogical significance: Mr. Zalcman includes an unnecessary step and leaves out a necessary one: he decided it was important to explain the homogeneity degree of det, but unimportant to explain why the eigenvalues are the roots of det(tI-A). To me, this seemed to misjudge the level of the question. Anyway, it was an omission. | |
| Dec 30, 2011 at 11:42 | comment | added | Ron Maimon | This is a repeat of the fine answer by Adam Zalcman. There is no need for duplicate answers in general, but I won't downvote you, because you probably didn't know. | |
| Dec 30, 2011 at 3:00 | history | edited | joseph f. johnson | CC BY-SA 3.0 |
minus signs
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| Dec 30, 2011 at 2:55 | comment | added | joseph f. johnson | In infinite dimensions, I think one can still do something, and that is look for the poles of the power series (these are bounded operators so it makes sense) for $1\over I-U(t)e^{itE}$ regarded as a series in $E$, a real variable. A pole leads to a lack of invertibility here, too. | |
| Dec 30, 2011 at 2:45 | history | edited | joseph f. johnson | CC BY-SA 3.0 |
added 82 characters in body
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| Dec 30, 2011 at 1:47 | history | answered | joseph f. johnson | CC BY-SA 3.0 |