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These are answers submitted by jicri

Hi again, I progress only very little on this one. The way to define anticommuting operators in the Physics package is Setup(quantumop={c,C}, algebrarules={%AntiCommutator(c[i],c[j])=0, %AntiCommutator(C[i],C[j])=0}); Then c[i]^2; and C[i]^2; both return zero as expected. Troubles begin if I want to define the rules for mixing the c's and the C's like Setup(algebrarules={%AntiCommutator(c[i],C[j])=KroneckerDelta[i,j]}); This last rule is ignored and the 2 sets rather anticommute, that is simplify(c[i].C[j]+C[j].c[i]); returns 0!
this seem so simple! Yet I have no clue, and the help pages are mum on the subject... So thanks for your concern... Otherwise anyone, any idea?
Ok, thanks a lot. So I started with the same Physics:-Setup command as you. Then I define the following operators J[i,j] := I*(C[i].c[j]-C[j].c[i]): and I want to identify and factor bilinears such as C[i].c[i], in products of the type J[1,2]*J[2,1]; BTW, the above command returns terms such as C[1]^2*c[2]^2 that should be evaluated to zero but are seemingly not recognized.
Thanks, I guess my question was not properly formulated. After some trial and error here's where I'm at. I define anticommuting operators {c[i],C[i], i=1..N} that obey the following algebra AntiCommutator(c[i],C[j])=KroneckerDelta[i,j] and AntiCommutator(c[i],c[j])=AntiCommutator(C[i],C[j])=0 Then I define the product of a certain number of these operators, and I want to simplify it not only using the above algebra, which I believe one can do by simply applying simplify(productofoperators) but also by identifying some patterns. Namely I'd like to identify some bilinears, say Dagger(c[i]).c[i], collect and factor them out where possible, and eventually maybe rename them, like Dagger(c[i]).c[i]="n[i]" in the present case. The latter step may cause a problem since there is no "n" in the algebra... Hope what I want to achieve is clearer... Thanks in advance!
thanks a lot guys! I still have to wrap my mind a bit around the powerful first solution but heck! both work.
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