0
$\begingroup$

I am studying the physical implementation of the CNOT gate from "Introduction to Quantum Computing" (Ray LaPierre) and I am confused about the order of operators in the tensor product of the calculation below - why is it swapped for the two different qubits? Hamiltonian calculation

$\endgroup$

1 Answer 1

3
$\begingroup$

The total Hamiltonian $\hat{H}$ is written as the sum of the following three terms: $$ \hat{H}_1 = \frac{\hbar \omega_1}{2} \hat{\sigma}_{z1} \otimes \hat{I} $$ $$ \hat{H}_2 = \hat{I} \otimes \frac{\hbar \omega}{2} \hat{\sigma}_{z2} $$ $$ \hat{H}_{\mathrm{int}} = J \hat{\sigma}_{z1} \otimes \hat{\sigma}_{z2} $$

Each of these is a $4 \times 4$ operator since it acts on a 2-spins system and, as such, it is written as a tensor product of two $2 \times 2$ operators ($\hat{\sigma}_{z1}$, $\hat{\sigma}_{z2}$, and the identity $\hat{I}$). In each tensor product left $\otimes$ right, the left operator is acting on the first spin while the right operator is acting on the second spin. This is why for $\hat{H}_1$ and $\hat{H}_2$, the left and right parts are swapped with $\hat{I}$ whereas for $\hat{H}_{\mathrm{int}}$ both of them are different from $\hat{I}$ (representing the interaction Hamiltonian).

$\endgroup$

Your Answer

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

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