# Quantum circuit for the ZZ feature map

Havlicek et al. propose a feature map for embedding $$n$$-dimensional classical data on $$n$$ qubits: $$U_{\phi(x)}H^{\otimes n}$$, where $$U_{\phi(x)} = \exp (i \sum_{S \subseteq [n]} \phi_S(x) \prod_{i \in S} Z_i) \\ \phi_i(x) = x_i, \; \phi_{\{i, j\}}(x) = (\pi - x_0)(\pi - x_1)$$ and $$Z_i$$ is a $$Z$$-gate on the $$i$$-th qubit.

I'm currently considering the 2-dimensional, 2-qubit case, $$U_{\phi(x)} = \exp(i x_0 Z_0 + i x_1 Z_1 + i (\pi - x_0) (\pi - x_1) Z_0 Z_1),$$ and trying to find out, how do I get from the above definition to the circuit, as it is implemented in Qiskit:

     ┌───┐┌──────────────┐
q_0: ┤ H ├┤ U1(2.0*x) ├──■─────────────────────────────────────■──
├───┤├──────────────┤┌─┴─┐┌───────────────────────────────┐┌─┴─┐
q_1: ┤ H ├┤ U1(2.0*x) ├┤ X ├┤ U1(2.0*(π - x)*(π - x)) ├┤ X ├
└───┘└──────────────┘└───┘└───────────────────────────────┘└───┘


My reasoning for the first 2 $$U1$$ rotations is that since $$\exp(i \theta Z) = RZ(2 \theta)$$ up to global phase and $$RZ$$ is equivalent to $$U1$$ up to global phase, we might as well use $$U1(2 \phi_i(x))$$. Is that correct?

And if so, how do we then get $$CX \; U1(2 \phi_{\{0, 1\}}(x)) \; CX$$ from $$\exp(i \phi_{\{0, 1\}}(x) Z_0 Z_1)$$?

By the definition of $$U1(\lambda)$$ in qiskit, it is equivalent to $$RZ$$ up up a phase factor .

Now note that: $$Z_0 Z_1 = Z \otimes Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & 1\end{pmatrix}$$

since this is diagonal, we have that $$e^{-i\lambda Z_0Z_1} = \begin{pmatrix} e^{-i\lambda} & 0 & 0 & 0 \\ 0 & e^{i\lambda} & 0 & 0\\ 0 & 0 & e^{i\lambda} & 0\\ 0 & 0 & 0 & e^{-i\lambda}\end{pmatrix} = \begin{pmatrix} R_Z(2\lambda) & \boldsymbol{0}\\ \boldsymbol{0} & X R_Z(2\lambda) X\end{pmatrix}$$

since $$XR_Z(2\lambda) X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} e^{-i\lambda} & 0 \\ 0 & e^{i\lambda} \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} e^{i\lambda} & 0 \\ 0 & e^{-i\lambda} \end{pmatrix}$$

Thus, to implement $$e^{-i \lambda Z_0Z_1}$$ we would have a circuit like: Note that this circuit can be written as $$CX \cdot (I \otimes R_Z ) \cdot CX$$. Also by knowing this, you can check this explicitly as well, by first note that

$$I \otimes R_Z(2\lambda) = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \otimes \begin{pmatrix} e^{-i\lambda} & 0 \\ 0 & e^{i\lambda}\end{pmatrix} = \begin{pmatrix} e^{-i\lambda} & 0 & 0 & 0 \\ 0 & e^{i\lambda} & 0 & 0\\ 0 & 0 & e^{-i\lambda} & 0\\ 0 & 0 & 0 & e^{i\lambda}\end{pmatrix}$$ then $$CX \cdot \big(I \otimes RZ(2\lambda) \big) \cdot CX$$ is equal to $$\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\end{pmatrix} \begin{pmatrix} e^{-i\lambda} & 0 & 0 & 0 \\ 0 & e^{i\lambda} & 0 & 0\\ 0 & 0 & e^{-i\lambda} & 0\\ 0 & 0 & 0 & e^{i\lambda}\end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\end{pmatrix} = \begin{pmatrix} e^{-i\lambda} & 0 & 0 & 0 \\ 0 & e^{i\lambda} & 0 & 0\\ 0 & 0 & e^{i\lambda} & 0\\ 0 & 0 & 0 & e^{-i\lambda}\end{pmatrix}$$

• Thank you for the answer! Just one more question. Why can we go from $e^{i \lambda Z_0 Z_1}$ (as in $U_{\phi(x)}$) to $e^{-i \lambda Z_0 Z_1}$? Wouldn't that imply that $R_z(\lambda) = R_z(-\lambda)$, which seems to only be true for $\lambda = k \pi, \; k \in \mathbb{Z}$? Jun 9, 2021 at 0:10
• I just realize that you are considering $e^{i\lambda ZZ}$ instead of $e^{-i \lambda ZZ}$. In such cases, the $RZ$ rotation gate should sandwiched between the two CNOT gates should be $RZ(-2\lambda)$ instead of $RZ(2 \lambda)$. Jun 9, 2021 at 0:48
• That's the issue. The circuit still uses $U1(\lambda)$ instead of $U1(-\lambda)$. Found a relevant article, where they're using the same circuit Qiskit does, however, they define $U1(\lambda) = diag\{1, e^{-i \lambda}\}$, which would solve the sign issue. Could it be that the direction of rotation doesn't really matter? I'm using this feature map to estimate the kernel $K(x, y) = \Phi(x) \Phi(y)$, by measuring $(U_{\phi(y)} H^{\otimes n})^\dagger U_{\phi(x)} H^{\otimes n}$. Since $K(x, y) = K(y, x)$ and $U1(\lambda)^\dagger = U1(-\lambda)$, it seems so? Jun 9, 2021 at 12:10
• Or, more simply put (since I've run out of space), we're still going to measure the same overlap $\Phi(x) \Phi(y)$, as long as we're consistently rotating in one direction, be it clockwise or counter-clockwise. What do you think? Jun 9, 2021 at 12:20