# Expressing “Square root of Swap” gate in terms of CNOT

How could a $\sqrt{SWAP}$ circuit be expressed in terms of CNOT gates & single qubit rotations?

• CNOT & $\sqrt{SWAP}$ Gates

Any quantum circuit can be simulated to an arbitrary degree of accuracy using a combination of CNOT gates and single qubit rotations.

Both CNOT and $\sqrt{SWAP}$ are universal two-qubit gates and can be transformed into each other.

Edit

The questions are not identical, and are likely to attract different audiences. -DaftWullie

Can the difference be regarded as applied vs theoretic?

• quantumcomputing.stackexchange.com/questions/2228/… – Nelimee Jun 26 '18 at 5:53
• Possible duplicate of How to implement the "Square root of Swap gate" on the IBM Q (composer)? – Sanchayan Dutta Jun 26 '18 at 6:55
• @Blue I think that while the answers are essentially the same, the questions are not identical, and are likely to attract different audiences. – DaftWullie Jun 26 '18 at 7:01
• May have been better to post edit as comment? Also, to be clear, the circuit that @DaftWullie posted at the end is what I was after (which I did not see in the other post). – meowzz Jun 26 '18 at 7:55
• @Blue It is indeed a candidate for closure. But given there are slight differences and willing answerers, I'll let it live. – James Wootton Jun 26 '18 at 8:42

As pointed out by @Nelimee, this question is essentially answered in this question, even if that question seems more specific. However, for the sake of completeness... (Note that I make no claims about minimality of construction with respect to, for example, number of controlled-not gates.)

Let's start with a unitary matrix for the square root of SWAP: $$\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & \frac{1}{2}+\frac{i}{2} & \frac{1}{2}-\frac{i}{2} & 0 \\ 0 & \frac{1}{2}-\frac{i}{2} & \frac{1}{2}+\frac{i}{2} & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right)$$ Note that if we pre- and post-multiply by a controlled NOT, controlled off qubit 2, this transforms into the form of a controlled-$U$ gate: $$\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & \frac{1}{2}+\frac{i}{2} & \frac{1}{2}-\frac{i}{2} \\ 0 & 0 & \frac{1}{2}-\frac{i}{2} & \frac{1}{2}+\frac{i}{2} \\ \end{array} \right)$$ There are standard constructions for this controlled-$U$ gate. So, overall, we have a circuit that looks like where $ABC=\mathbb{I}$ and $AXBXC=e^{i\pi/4}\sqrt{X}$. Again, there are standard routes towards finding $A$, $B$ and $C$. For instance, we can define $R_Y(\theta)$ to be a rotation about the $Y$ axis by an angle $\theta$, i.e. $R_Y(\theta)=e^{i\theta Y}$ (there may be a factor of $\frac12$ here compared to some definitions). Then, if $$A=R_Y(\theta)R_Z(\phi)\qquad B=R_Z(-2\phi)\qquad C=R_Z(\phi)R_Y(-\theta),$$ then it's easy to verify that $ABC=\mathbb{I}$. Furthermore, $AXBXC=R_Y(\theta)R_Z(4\phi)R_Y(-\theta)$ since $XR_Z(\phi)X=R_Z(-\phi)$. You can think of this as $R_Z(4\phi)$ specifying the eigenvalues that we want, and the $R_Y(\theta)$ is changing the basis from the computational basis into something else. In the present case, we have $\theta=\pi/2$ and $4\phi=\pi/4$.

The one little thing that we haven't got right yet is a phase factor. Our $AXBXC$ is creating the correct rotation up to an overall phase of $e^{i\pi/4}$: $$\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & \frac{1}{\sqrt{2}} & -\frac{i}{\sqrt{2}} \\ 0 & 0 & -\frac{i}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \end{array} \right)$$ The trick to get this right is to apply an $R_Z(\pi/8)$ on the first qubit and remove a global phase of $e^{i\pi/8}$. Thus, we have

I think that you want answers to explicitly use full cnots, rather than partial versions. But since you already have an answer for that, I'll contribute a different perspective.

A $$\mathrm{SWAP}$$ can be thought of as a cnot that has been conjugated by oppositely oriented cnots.

$$\mathrm{SWAP} = {\rm cx}(k,j) \,\, {\rm cx}(j,k) \,\, {\rm cx}(k,j)$$

To make a $$\sqrt{\mathrm{SWAP}}$$, we can instead use a square root of cnot conjugated by oppositely oriented cnots.

$$\sqrt{\mathrm{SWAP}} = {\rm cx}(k,j) \,\, \sqrt{{\rm cx}(j,k)} \,\, {\rm cx}(k,j)$$

To verify that this is indeed a $$\sqrt{\mathrm{SWAP}}$$, you can simply square it and verify that it ends up in the way we expect (using the fact that cnots square to identity).

• Awesome! Your $\mathrm{SWAP}$ approach makes me think of an unanswered question.. – meowzz Jun 26 '18 at 9:12