Are spinless fermions just fermions with fixed spin?

Question

I was wondering if I could recycle code used for spinful fermions by simply specifying the total spin quantum number to be N/2, i.e. the maximum possible for N spinful fermions, in order to create a spinless fermion model.

Would this cause the hilbert space to be larger or different from the results I'd get using the pre-packaged spinless siteset operators?

Thanks!

Jon

Answer 1

Hi Jon,
I think this is correct and will work, however, I'd advise just translating your code to use the Spinless site set. All that should involve for the most part is switching the site set names and then the names of certain operators, but otherwise the structure and logic of the code could be left intact and you wouldn't have to worry about any weird surprises -

Best regards,
Miles

Comments

Ok thanks Miles. I scanned through the caveats about fermionic commutations, and am I correct to say that it is the same as for spinful fermions? That is, the commutators for spinless fermions are the same as for spinful fermions in my MPO operators.

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Hi Jon, as with many things in physics, the answer is “it depends”. Yes, if you only consider spinful fermions all of the same spin type, then the commutation relations among them are the same as for spinless fermions. If you bring in the other spin type then it’s different. It would be cleaner to just use the type you want to physically study, and not that much harder than reusing your spinful code - hope that helps