Speed up H^k by another k! factor

[lukas-weber]

The product of Pauli strings commutes up to a minus sign, right?

Probably this can be used to speed up moments quite a bit: For H^k, enumerate just the ordered k-tuples. For each ordered tuple, compute the sum of the signs of all permutations. Maybe this can be done efficiently using some group theory and Young tableaux or something, but even if not, it should be largely more efficient than doing the equivalent amount of dict lookups.

I thought if it can also be used to speed up the normal product A*B, but probably it’s not worth it unless you know that the two operators are the same.

[lkdvos]

That's a really cool idea! I'm also not sure if this could work for the normal product, although we could consider checking for the intersection of A and B first, i.e. A = C + A2 and B = C + B2 and then compute the cross terms separately. The main issue here is that typically we'd have one large factor and one much smaller one, so this might not actually save that much since C could be quite small

[nicolasloizeau]

ahah yeah thanks for seeing this! that seems crazy there is still a k! factor to gain !
maybe that's what we need to unlock moments expansions for low temperature stuff

[nicolasloizeau]

actually I think that for computing the moments tr(H^k), this comes for free, because only the tuples that contract to identity contribute to the trace, so there is no need to compute all the signs

[lukas-weber]

There are cancellations though, no?

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-------- Original Message --------

On 8/7/25 08:31, Nicolas Loizeau wrote: actually I think that for computing the moments tr(H^k), this comes for free, because only the tuples that contract to identity contribute to the trace, so there is no need to compute all the signs — Reply to this email directly, [view it on GitHub](#51 (comment)), or [unsubscribe](https://github.com/notifications/unsubscribe-auth/ALX63H52VJE4KPZ4MWU6HYD3MNBJBAVCNFSM6AAAAACC5IPOFSVHI2DSMVQWIX3LMV43URDJONRXK43TNFXW4Q3PNVWWK3TUHMYTIMBTGQ2DMOI). You are receiving this because you authored the thread.Message ID: ***@***.***>

[nicolasloizeau]

There are cancellations though, no?
…
-------- Original Message --------
On 8/7/25 08:31, Nicolas Loizeau wrote: actually I think that for computing the moments tr(H^k), this comes for free, because only the tuples that contract to identity contribute to the trace, so there is no need to compute all the signs — Reply to this email directly, [view it on GitHub](#51 (comment)), or unsubscribe. You are receiving this because you authored the thread.Message ID: @.***>

ah my bad, yes, XYZ=-YXZ