Timeline for Unitarity of PMNS matrix
Current License: CC BY-SA 3.0
13 events
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| Jul 10, 2017 at 9:34 | history | edited | SRS | CC BY-SA 3.0 |
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| Aug 10, 2014 at 19:16 | comment | added | user22180 | Let us continue this discussion in chat. | |
| Aug 10, 2014 at 19:05 | history | edited | dmckee --- ex-moderator kitten | CC BY-SA 3.0 |
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| Aug 10, 2014 at 19:02 | comment | added | dmckee --- ex-moderator kitten | @user22180 At last I get it and you are right, I was conflating the requirement that the time evolution operator must be unitary (which follows from the Hermitivity of the Hamiltonian) with a condition on the Hamiltonian itself. Edits to follow. | |
| Aug 10, 2014 at 18:38 | comment | added | user22180 | I don't know whether you are getting my question or not. I want to ask whether Hamiltonian is Unitary always. DO you mean $H^{\dagger}H=I$? Then again $H^{\dagger}=H$ ,so that $H^2=I$? And you know the consequences of this. All the eigenvectors are degenerate, moreover all the vectors are eigenvector of $H^2$. | |
| Aug 10, 2014 at 18:22 | comment | added | dmckee --- ex-moderator kitten | @user22180 Perhaps physics.stackexchange.com/questions/15858/… would help. | |
| Aug 10, 2014 at 17:08 | comment | added | user22180 | I knew the Hamiltonian must have to be Hermitian. But should it be Unitary also? | |
| Aug 10, 2014 at 15:59 | comment | added | dmckee --- ex-moderator kitten | @user22180 It must be the complete Hamiltonian, with all corrections and the bits we would normally neglect, and I should probably add at least one more weasel word line "isolated" to imply that if we hold some interaction out as "external" this rule no longer applies; but yes or the reason explained. | |
| Aug 10, 2014 at 13:49 | comment | added | user22180 | Do you really mean this: "The Hamiltonian for any system must be unitary"? | |
| Apr 6, 2014 at 10:52 | comment | added | innisfree | oh i see what it means, in the see-saw model, the $3\times3$ (flavor) PMNS mixing matrix might not be unitary, but the full mixing matrix with all flavors and LH and RH neutrinos must be unitary. a bit misleading, but i suppose it;s true that $3\times3$ PMNS need not be unitary. | |
| Apr 6, 2014 at 10:49 | comment | added | innisfree | What do you make of this note in the wiki article? en.wikipedia.org/wiki/PMNS_matrix#cite_note-1 | |
| Apr 6, 2014 at 10:38 | comment | added | agemO | The PMNS matrix is not an Hamiltonian, but you're right, it's more general, any change of observable basis (here from mass egeinstate to flavor eigenstate) must be done with a unitary matrix so that probability are conserved | |
| Apr 5, 2014 at 17:37 | history | answered | dmckee --- ex-moderator kitten | CC BY-SA 3.0 |