Timeline for Curves on $SU(4)$ whose adjoint action on $\mathfrak{su}(4)$ integrates to $0$
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Sep 3, 2015 at 18:14 | comment | added | Benjamin | Also, what about when $T$ is close to the minimum $T$ which makes the system fixed time controllable? Would this make solutions no longer exist? | |
| Sep 2, 2015 at 0:33 | comment | added | Benjamin | I am familiar with the concept of fixed time controlability, however, if you know a particularly clear mathematical ref that would be great as there seems to be quite a lot of vagueness surrounding this point in the quantum control community. It is the equidistributed bit I'm not seeing, what ensures that such a curve exists even given fixed time controlability? | |
| Sep 1, 2015 at 21:59 | comment | added | Robert Bryant | If $T$ is large enough and $a$ and $b$ are generic, then, yes, you will be able to reach every point in the group and in an equidistributed sense. The proof is not hard, but it uses ideas from control theory. If $T$ is small (how small depends on $a$), then, no, you may not be able to reach every point, even ignoring the equidistribution condition. | |
| Sep 1, 2015 at 21:42 | comment | added | Benjamin | I see, I also had a similar thought. But I also thought that this would not work as not every route to every end-point can be implemented. Wouldn't we need to show that ever point can be reached by a curve which is 'equidistributed' in the sense you mention. | |
| Sep 1, 2015 at 21:34 | comment | added | Robert Bryant | It would not be easy to explain in a short note, but the basic point is that the system you are describing above is a right-invariant control system with drift on the compact manifold $\mathrm{SU}(4)$, and, for generic choice of $a$ and $b$, it is controllable via the single control ($w(s)$), as is not difficult to show using standard control theory. Thus, as long as $T$ is sufficiently large, you have enough time to steer all over $\mathrm{SU}(4)$ and to equidistribute the resulting curve with respect to the position map defined by $\xi$. | |
| Sep 1, 2015 at 19:33 | comment | added | Benjamin | How can one see that there are many? Should this be obvious to me! | |
| Sep 1, 2015 at 16:39 | comment | added | Robert Bryant | I believe that, if $a$, $b$, and $\xi$ are generically chosen in ${\frak{su}}(4)$, then there will be such curves and, in fact, an infinite dimensional family of them, though they might not be easy to construct. | |
| Sep 1, 2015 at 16:15 | comment | added | Benjamin | I'm somewhat banking on there being none, or the examples being somehow trivial. However, the commuting case is not important. I've not been able to cook up any example. | |
| Sep 1, 2015 at 15:42 | comment | added | Robert Bryant | Yes, there will be far fewer such curves. Indeed, there could be none. For example, if $a$ and $b$ happen to commute with $\xi\not=0$, then there will be none since, for all such curves $U_s\xi U_s^\dagger$ will be constant and equal to $\xi$. | |
| Sep 1, 2015 at 13:40 | comment | added | Benjamin | So, to avoid asking another similar question, what if we now insist that $\frac{d U_s}{ds}= (a+w(s)b)U_s$ for some $a,b \in \mathfrak{su}(n)$ and some smooth, real $w$. This seems that there should be far fewer curves with both properties. | |
| Sep 1, 2015 at 13:31 | comment | added | Benjamin | This was pretty much what I meant when I said "find all Xs", I really just meant, "please can someone tell me about the nature of all Xs". As there are an infinite dim family (which seems obvious now you say it), I see that I need to add more conditions for my application to work at all, so there's no point in "finding them all" in any strong sense than your answer. Thanks for your help. | |
| Sep 1, 2015 at 11:18 | history | answered | Robert Bryant | CC BY-SA 3.0 |