# Swap operation with the Heisenberg Hamiltonian

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

$$U^{1/2}_{SWAP}=e^{-i\frac{\pi}{8}}\exp\left(i\frac{\pi}{2}\mathbf{S}_1\mathbf{S}_2\right)$$

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?

• 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. – PhysicsMan Jun 14 '19 at 11:11
• 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. – Norbert Schuch Jun 14 '19 at 11:19