Timeline for Is every unitary operator induced by a Hamiltonian?
Current License: CC BY-SA 4.0
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| Jan 9, 2020 at 10:10 | comment | added | yuggib | The representations that are not continuous or weakly but not strongly continuous cannot be represented in the exponential form. Another, different question is instead if it is possible to take the logarithm of any unitary operator. The answer has to come from the spectral theorem, but for "complex" operators (operators with complex spectrum) it is tricky to do a logarithm. | |
| Jan 9, 2020 at 9:49 | comment | added | yuggib | The answer to the above question is no, because the representations of the form $e^{-itH}$ are in one-to-one correspondence with strongly continuous unitary groups (and representation of the same form with $H$ self-adjoint and bounded are in one-to-one correspondence with uniformly continuous unitary groups). | |
| Jan 9, 2020 at 9:47 | comment | added | yuggib | A first observation is the following: a representation of the type $e^{-itH}$ for a unitary is meaningful only if the said unitary is parametrized by a real number $t$. Given that, since $e^{-itH}$ satisfies the properties of an abelian group when $t$ ranges over the real numbers (it is in fact a unitary representation of the abelian group of real numbers with sum as group operation), the natural question is in my opinion the following: are all unitary representations of the real numbers (seen as an abelian group) of the form $e^{-itH}$ for some self-adjoint $H$? | |
| Jan 8, 2020 at 18:59 | comment | added | Marsl | Thank you for the answer. This answer seems to be neither "no" nor "yes" if I understand correctly. You write "there are unitary operators that are not generated by a self-adjoint Hamiltonian in the above sense", could you reason why this is the case, or can you provide an example (I would not even know how to write these down)? And then you say that we can still identify a corresponding bounded self-adjoint but this identification is flawed in the sense that it does not lead to what we would usually say is the Hamiltonian in obvious cases. Is this correct? | |
| Jan 8, 2020 at 13:35 | history | answered | yuggib | CC BY-SA 4.0 |