2
$\begingroup$

How can one show in detailed steps that Fermionic annihilation and creation operators, under Jordan-Wigner transformation, satisfy the Fermionic commutation relations?

The Fermionic commutation relations are: $$\{\hat{a}_i,\hat{a}_j\}= \{\hat{a}_i^\dagger,\hat{a}_j^\dagger\} =0 , \{\hat{a}_i,\hat{a}_j^\dagger\} = \delta_{ij}.$$

$\endgroup$
1
  • 1
    $\begingroup$ Do you have nearest-neighbor coupling again? $\endgroup$ Mar 14, 2021 at 23:43

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$ Mar 18, 2021 at 3:49
  • 1
    $\begingroup$ 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 :) $\endgroup$ Mar 18, 2021 at 4:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.