According to REF 1 equation 3, a SWAP operation can be achieved via the Heisenberg Hamiltonian for spins $H=J(t)\mathbf{S}_1\mathbf{S_2}$


where $\int_0^td\tau J(\tau)=\pi/2$

I wrote out the expression of $U^{1/2}_{SWAP}$ and it indeed results into the matrix $U^{1/2}_{SWAP} =\left(\begin{array}{cccc}{1} & {0} & {0} & {0} \\ {0} & {\frac{1}{2}(1+i)} & {\frac{1}{2}(1-i)} & {0} \\ {0} & {\frac{1}{2}(1-i)} & {\frac{1}{2}(1+i)} & {0} \\ {0} & {0} & {0} & {1}\end{array}\right)$

The physical interpretation of the term $\exp\left(i\frac{\pi}{2}\mathbf{S}_1\mathbf{S}_2\right)$ is by coupling two electron spins in quantum dots via a tunnel coupling $t_0(t)$, $J=4t_0(t)^2/U$, where $U$ is the coulomb interaction between the electrons, for a time such that the SWAP operation is achieved and then turning off the coupling.

But how do I interpret the prefactor $e^{-i\pi/8}$ in physical terms?

REF 1: https://arxiv.org/abs/cond-mat/9905230


The prefactor is there to get the global phase of the SWAP right, likely for aesthetic reasons. It bears no physical relevance. You can interpret it as shifting the overall energy of the system, but this again has no physical relevance, as you can define energy offsets as will.

| cite | improve this answer | |
  • $\begingroup$ Is this similar to the freedom of adding a phase to a quantum state like $|\psi\rangle\rightarrow e^{i\phi}|\psi\rangle$ since this doesn't change anything about the physics $\langle\psi|A|\psi\rangle$. I mean without the prefactor, the matrix obviously doesn't look like the SWAP matrix and the SWAP matrix is a matrix with specific values, however physically somehow it is just a SWAP matrix with some energy shift? I dont understand this when obviously the matrix without prefactor has different values than a SWAP matrix. $\endgroup$ – PhysicsMan Jun 14 '19 at 11:11
  • $\begingroup$ Global phases are not relevant. This becomes evindent when you work with density operators, where $U$ acts as $\rho\mapsto U\rho U^\dagger$, where phases clearly cancel out. $\endgroup$ – Norbert Schuch Jun 14 '19 at 11:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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