Deriving optimal QPD for the CZ gate#
Here we will briefly go over where the decomposition for the CZ gate used comes from and how that is transfromed to operations implementable on a quantum computer. The derivation is based on [1].
Background#
The goal is to decompose a quantum channel implementing the CZ gate into a set of local channels implementable as single qubits operations. This decomposed channel can then be sampled to approximate the expectation values of the original channel.
Factor the CZ gate#
The CZ gate can (upto global phase) be decomposed as:
We can immediately see that the first two channels are just local single qubit Z rotations, S gates in this case. The only non-local part is then the third term \(\exp(-i\pi Z \otimes Z/4)\) and so has the form \(\exp(i\theta A_1 \otimes A_2)\) with \(A_1 = A_2 = Z, \theta = -\pi/4\)
Apply Lemma 1 from [1]#
The lemma states that we can decompose the super operator \(S(\exp(i\theta A_1 \otimes A_2))\) as:
if \(A_1^2 = A_2^2 = I\), which is the case for the pauli operators.
Transform to gates#
In less mathematical terms we can decompose any gate of form \(S(\exp(i\theta A_1 \otimes A_2))\) into a set of 6 single qubit operations. Looking at the decomposition we can see that term 1 is just applying identity on both qubits aka doing nothing. The second term is applying \(A_1\) on qubit 1 and \(A_2\) on qubit 2, remember that A are pauli operators and the total operations of term 2 is then the Z gate on both qubits. The third term is a bit more complex so let us look at it in detail.
Term 3#
Let us look at the structure that repeats twice in the sum: \(S((I+\alpha_1A_1) \otimes S(I+i\alpha_2A_2))\)
We can see that the superoperator acts separately on each qubit so we can look at in in two parts. By calculating out the matrices Appendix 1 we can see that the imaginary term is just the S gate (for \(\alpha=-1\)) and the \(S^\dagger\) gate for \(\alpha=+1\). Similarly we can see that the real term is the \(\ket{0}\bra{0}\) projector for \(\alpha=+1\) and the \(\ket{1}\bra{1}\) for \(\alpha=-1\). The total operation from the third term then becomes either measure + apply S/\(S^\dagger\) or apply S/\(S^\dagger\) + measure.
Now we can get the full decomposition by applying the extra terms from equation 1, which are also just S gates and calculating the coefficients Appendix 2. The full decomposition is then:
Qubit 1 |
Qubit 2 |
Coef |
|---|---|---|
S |
S |
1/2 |
Z S |
Z S |
1/2 |
Meas |
Z |
-1/2 |
Meas |
I |
1/2 |
Z |
Meas |
-1/2 |
I |
Meas |
1/2 |
Here Meas is the computational basis measurement and Z S means first applying the Z gate and the the S gate. I is the identity gate. Z S could also be simplified to the \(S^\dagger\) gate here since they are equal.
Summing over the absolute coefficients we get \(\gamma=3\) which is optimal (without communication).
References#
K. Mitarai and K. Fujii, “Constructing a virtual two-qubit gate by sampling single-qubit operations”, New J. Phys., vol. 23, no. 2, p. 023021, Feb. 2021, doi: 10.1088/1367-2630/abd7bc.
Appendix#
1. Explicitly calculate term 3 matrices#
\( \begin{aligned} Z =& \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \\ \end{aligned} \), \( \begin{aligned} I =& \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \\ \end{aligned} \)
Imagnary term#
\(I+i\alpha A\), with
A = Z, \(\alpha = \pm 1\)
Expanding gives: \( \begin{aligned} & \begin{pmatrix} i\pm 1 & 0 \\ 0 & i\mp 1 \end{pmatrix} \\ &= (1 \pm i)\begin{pmatrix} 1 & 0 \\ 0 & \pm i \end{pmatrix} \\ \end{aligned} \)
This is exactly equal to the S/\(S^\dagger\) gate (upto global phase which gets absorber by the coefficient).
Real term#
\(I+\alpha A\), with
A = Z, \(\alpha = \pm 1\)
Expanding gives: For \(\alpha = +1\)
\( \begin{aligned} & 2\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \\ &= 2\ket{0}\bra{0} \end{aligned} \)
For \(\alpha = -1\)
\( \begin{aligned} & 2\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \\ &= 2\ket{1}\bra{1} \end{aligned} \)
These are exactly the 0 and 1 projectors (the factor of 2 gets absorbed by the coefficient)
2. Calculating coefficients#
The coefficients follow immediately from the coefficients of equation 2. Let us go term wise:
Term 1#
\(\cos^2(\pi /4)=\frac{1}{2}\)
Term 2#
\(\sin^2(\pi /4)=\frac{1}{2}\)
Term 3#
\(\frac{1}{8}\cos(\pi/4)\sin(\pi/4) = \frac{1}{16}\)
Nos this gets multiplied by the extra factors from the terms in Appendix A (since they can be moved out of the sum). This then gives us:
\(\frac{1}{16}*2^2*|1 + i|^2 = \frac{1}{16}*4*2=\frac{1}{2}\)
And finally this is multiplied by the \(\alpha_1 \alpha_2\) factor from the sum yielding \(\pm \frac{1}{2}\) as the coeffcient for term 3 depending on the \(\alpha's\).