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Commutators, Lie groups, and quantum systems

by Einstein Mcfly
Tags: commutators, groups, quantum, systems
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Einstein Mcfly
#1
Oct18-06, 11:45 AM
P: 162
Hi folks. I've come across a method to determine the controllability of a quantum system that depends on the Lie group generated by the commutator of the skew-Hermetian versions of the field free and interaction (dipole) Hamiltonians. If, for an N dimensional system the dimension of the group generated is N^2, then the dynamical Lie group is U(N) (is it correct to say “is” or should I say “is isomorphic to”?) and “every unitary operator can be dynamically generated”. My questions are: What symmetry are we exploring here with this group? What does the commutator tell us about the symmetry of the operators that determines if they’re fully controllable? What does this mean physically?

I have little experience with this sort of argument (that is, deriving physical information from the structure of a group with the same symmetry), so any insight is much appreciated. Thanks.
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selfAdjoint
#2
Oct18-06, 02:59 PM
Emeritus
PF Gold
P: 8,147
I'll just answer this little part of your question:

Quote Quote by EinsteinMcfly
is it correct to say “is” or should I say “is isomorphic to”?
"Is" is fine. If two groups are isomorphic then the difference between them is just notation. Every element matches up and all products and inverses precisely match too.
Einstein Mcfly
#3
Oct19-06, 12:35 PM
P: 162
Quote Quote by selfAdjoint
I'll just answer this little part of your question:



"Is" is fine. If two groups are isomorphic then the difference between them is just notation. Every element matches up and all products and inverses precisely match too.
Thanks. That's what I thought.


Anyone else? I know there are a lot of very smart folks on here with a lot more experience than me. Is anyone good at interpreting these symmetries physically?

Epicurus
#4
Oct20-06, 01:31 AM
P: 76
Commutators, Lie groups, and quantum systems

Do you mean to say Lie algebra generated as opposed to Lie group. The Lie bracket defines the operator algebra.
Einstein Mcfly
#5
Oct20-06, 04:11 PM
P: 162
Quote Quote by Epicurus
Do you mean to say Lie algebra generated as opposed to Lie group. The Lie bracket defines the operator algebra.

Yes, indeed I did. Thanks.
Einstein Mcfly
#6
Feb5-07, 08:19 PM
P: 162
I just thought I'd bump this and see if anyone had any ideas.


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