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How to show in detailed steps that Fermionic annihilation and creation operators under Jordan-Wigner transformation satisfy the Fermionic commutation relation

$\{\hat{a}_i,\hat{a}_j\}= \{\hat{a}_i^\dagger,\hat{a}_j^\dagger\} =0 , \{\hat{a}_i,\hat{a}_j^\dagger\} = \delta_{ij}.$

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    $\begingroup$ Do you have nearest-neighbor coupling again? $\endgroup$
    – user1271772
    Mar 14, 2021 at 23:43

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Based on my answer to this: Fermionic occupation operator and nearest neighbor Fermionic hopping interaction as a qubit operator, you can see that we have:

\begin{align} \hat{a}_i &= \frac{1}{2} Z^{\otimes (i-1)} (X - iY),\\ \hat{a}_i^\dagger &=\frac{1}{2} Z^{\otimes (i-1)} (X + iY).\\ \end{align}

If $i=j$ we have:

\begin{align} \{\hat{a}_i,\hat{a}_i^\dagger\} &\propto \frac{1}{4} ((X - iY)(X + iY) + (X + iY)(X - iY)),\\ &= \frac{1}{4} (X^2 + iXY + - iXY + Y^2 + X^2 -iXY + iXY + Y^2) \\ &=\frac{1}{4}(2X^2 + 2Y^2)\\ & =\frac{1}{4}(4I) \\ & = I. \end{align}

All $Z$ operators are replaced by $I$ operators since $Z \times Z = I$, and this is also what was done in the penultimate step, where $X^2 = I$ and $Y^2 = I$ were used.

For the other anti-commutators we have:

\begin{align} \{\hat{a}_i,\hat{a}_j\} &= \frac{1}{4} ((X - iY)(X - iY) + (X - iY)(X - iY)),\\ \{\hat{a}^\dagger_i,\hat{a}^\dagger_j\} &= \frac{1}{4} ((X + iY)(X + iY) + (X + iY)(X + iY)). \end{align}

You can then do the same type of arithmetic that I went through in detail for the first anti-commutator.

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  • $\begingroup$ Shouldn't the last two equations be zero? When $i=j$ this follows from the fact that $\hat{a}_i$ is nilpotent with degree 2. When $i \ne j$, this follows from the fact that $Z$ and $X\pm iY$ anticommute (and all other operator pairs in tensor products resulting from JW commute). Am I missing something? $\endgroup$
    – Adam Zalcman
    Mar 18, 2021 at 3:49
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    $\begingroup$ The question asks why they are zero. I gave the full work for the $\{ a_i , a_j^\dagger \}$ case when i=j, then showed the first line for the other two cases (still with i=j). The user asked several questions of this same nature within a couple days, and I wanted to let them do some of the work after giving some starting steps :) $\endgroup$
    – user1271772
    Mar 18, 2021 at 4:16

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