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Nov 13, 2020 at 17:00 vote accept tmph
Nov 13, 2020 at 2:02 history edited Phil Hasnip CC BY-SA 4.0
Expanded detail on how to optimise with respect to both C and alpha
Nov 13, 2020 at 2:01 comment added Phil Hasnip @tmph OK, I've reworked it to be more explicit. The short answer is that in steepest descent you do a single 1D optimisation which changes both $C$ and $\alpha$ simultaneously - there are better methods! I've also tried to briefly explain what co- and contravariance mean here, but in essence $(H-ES)\phi$ transforms like $S\phi$ if we change the basis, and if we're going to use this as a direction to add to $\phi$ it needs to transform in the same way as $\phi$.
Nov 12, 2020 at 21:11 comment added tmph Also, could you explain the introduction of $S^{-1}$ into $D$ some more? I do not know how to read co/contravariance or why we care.
Nov 12, 2020 at 21:08 comment added tmph Okay, that does help but I am still not really sure how $\alpha$ is updated. Are $C$ and $\alpha$ updated by separate 1D optimisations or do they somehow share the same $\lambda$ from the functional derivative w.r.t. $\phi$ (sorry, my variational calculus is rusty)? I do not see how to get $D_\alpha$, $D_C$.
Nov 12, 2020 at 17:33 comment added Phil Hasnip Also updated now to include explicitly how you would treat the C and alpha changes. Is that clearer?
Nov 12, 2020 at 17:33 history edited Phil Hasnip CC BY-SA 4.0
Added detail on how to handle variation of different parameters (\alpha and C)
Nov 12, 2020 at 16:02 comment added Phil Hasnip I've rewritten it in terms of phi, to try to clarify it. Is that better? Would you like d/dC and d/dalpha explicitly?
Nov 12, 2020 at 16:01 history edited Phil Hasnip CC BY-SA 4.0
Recast in terms of variation with respect to the full wavefunction phi, rather than simply C.
Nov 12, 2020 at 15:48 comment added Phil Hasnip Sorry that wasn't clear, nothing about this procedure is unique to C -- you simply compute dE/dalpha as well and change alpha in the same way as C. In general you need a functional derivative, for which the conventional symbol is delta.
Nov 12, 2020 at 7:10 comment added tmph Thanks for the detailed response. One thing I am confused about, how does any of this update $\alpha$? It looks like this only works by modifying $C$. Also, as a side note, it looks like your derivative is a gradient, is that correct? I have not seen such notation using $\delta$ before.
Nov 12, 2020 at 1:32 comment added Nike Dattani - No Free Time +100! Such an excellent and thorough answer!!
Nov 12, 2020 at 0:27 history answered Phil Hasnip CC BY-SA 4.0