+1 vote
asked by (490 points)

Hi,

On the page of http://itensor.org/docs.cgi?page=tutorials/fermions
, it shows the anticommutation relation of spinless case can be derived by treating it for i < j.
However, I think the derivation doesn't work for i = j.
Is there anything I missed?
Thank you very much!

Victor.

1 Answer

+1 vote
answered by (770 points)
edited by

Hello Victor,

For i = j,
$$
ci ci^\dagger + ci^\dagger ci = ai ai^\dagger + ai^\dagger ai = 1.
$$
Maybe you missed the fact that for Hard Core bosons, it's
$$
\{a, a^\dagger\} = 1,
$$
not
$$
[a, a^\dagger] = 1.
$$

Hope this helps. Thanks.

commented by (490 points)
Hi,

You are right. Thanks for your help.
commented by (24.7k points)
Thanks for the answer hermit0308. I thought this was probably the correct answer but wanted to do some reading to make sure. Certainly it's consistent to define the hard core boson operators to anticommute on site (and commute when the sites are different of course).

A possibly clearer definition of the Jordan-Wigner transformation is to transform the fermion operators into spin operators attached to string instead of hard-core bosons attached to string. As you may know, in the spin interpretation, the string operators are the Pauli Z operators. Furthermore, the Pauli matrices X,Y,Z anticommute with each other so the on-site behavior is like fermions already.
commented by (770 points)
Hi Miles, thanks for the comment.
commented by (490 points)
Thanks for your help.
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