## Quick question about Two-qubit SWAP gate from the Exchange interaction

0

I am reading the following paper: Optimal two-qubit quantum circuits using exchange interactions.

I have a problem with the calculation of the unitary evolution operator $$U$$ (Maybe it is stupid):

I have figure out the matrix of $$H$$:

$$$$H = J \begin{bmatrix}1 & 0 & 0 & 0\\ 0 & -1 & 2 & 0\\ 0 & 2 & -1 & 0\\ 0 & 0 & 0 & 1\\ \end{bmatrix}$$$$

But I cannot write the matrix of Operator $$U$$ and get the result of $$(SWAP)^α$$.

Could you please help me to calculate it? I really want to know how to get the matrix of U.

Thank you so much.

The figure is shown as below:

1

Generally speaking, it is frowned upon to include pictures with text in questions, as it becomes hard to copy etc. Also, if imgur fails, the question becomes more or less unreadable. Furthermore, check https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference for a quick tutorial on how to perform proper markup for math.

– JSdJ – 2020-11-17T13:20:34.593

Thank you, it is my first time to use this forum. I will, thanks again! – None – 2020-11-17T14:09:53.110

1

You need to calculate $$U=e^{-iHt}$$. The trick to doing this is working out the eigenvectors of $$H$$: there's $$|00\rangle$$ and $$|11\rangle$$ with eigenvalues J, and $$|\Psi^{\pm}\rangle=(|01\rangle\pm|10\rangle)/\sqrt{2}$$ with eigenvalues $$(-1\pm 2)J$$. In particular, notice that this means 3 of the eigenvalues are $$J$$. Hence, there are two eigenspaces of $$H$$, $$|\Psi^-\rangle\langle\Psi^-|$$ and $$I-|\Psi^-\rangle\langle\Psi^-|$$. Hence, we can find $$U=e^{-iJt}(I-|\Psi^-\rangle\langle\Psi^-|)+e^{3iJt}|\Psi^-\rangle\langle\Psi^-|.$$ If you remove an irrelevant global phase, this is just the same as $$U=(I-|\Psi^-\rangle\langle\Psi^-|)+e^{4iJt}|\Psi^-\rangle\langle\Psi^-|.$$ This is exactly what you were after, with $$4Jt=\pi\alpha$$.

Thank you for your reply! I have got the third equation as you write, but I am confused about when should we ignore the global phase and shouldn't. It is tricky. – None – 2020-11-17T14:07:56.603

You can always ignore a global phase as it makes no observable difference. The only time you have to be careful is when moving from $U$ to controlled-$U$: a global phase on $U$ is a relative phase in controlled-$U$ and cannot be ignored. – DaftWullie – 2020-11-17T14:40:48.783

Cool! That makes sense! By the way, how could this operator become a swap gate by controlling the coupling coefficient $J$, in the exchange Hamiltonian? I thought $J$ is a constant as usual but it can be changed with time-varying. Am I right? – None – 2020-11-17T15:12:41.863

You could make it time varying. Then all you'd have instead of $Jt$ is $\int J(t)dt$. – DaftWullie – 2020-11-17T15:57:32.787

Thanks, Bro! That really makes sense. – None – 2020-11-17T22:45:28.913