1
$\begingroup$

I'm currently reading Classification with Quantum Neural Networks on Near Term Processors and I'm having trouble with one of the calculations.

The system is composed of $n+1$ qubits, $n$ of those are used as inputs and the last one as an output qubit. Given a label function $l(z)$ that returns +1 or -1 depending on the state $|z\rangle$, the unitary $U_l$ is defined as $$U_l|z,z_{n+1}\rangle = exp(i\frac{\pi}{4}l(z)X_{n+1})|z,z_{n+1}\rangle$$ It is mentionned in the paper that this operation is equivalent to rotating the output qubit about its x-axis by $\frac{\pi}{4}$ times the label of the string $z$. The next statement is the one that confuses me and I cannot derive it properly: $$U_l^\dagger Y_{n+1}U_l = cos(\frac{\pi}{2}l(Z))Y_{n+1} + sin(\frac{\pi}{2}l(Z))Z_{n+1}$$ where $l(Z)$ is interpreted as an operator diagonal in the computational basis.

I managed to derive a few lines but I'm having trouble matching the statement above, here is where I'm stuck: \begin{aligned} U_l^\dagger Y_{n+1}U_l &= U_l^\dagger [Y_{n+1},U_l] + U_l^\dagger U_l Y_{n+1}\\ &= U_l^\dagger [Y_{n+1},U_l] + Y_{n+1} \end{aligned} \begin{aligned}[] [Y_{n+1},U_l] &= [Y_{n+1}, \sum_{k=0}^\infty \frac{(i\frac{\pi}{4}l(z))^k}{k!}{X_{n+1}}^k]\\ &= \sum_{l=0}^{\infty}\frac{(-1)^l}{(2l)!}\left(\frac{\pi}{4}l(z)\right)^{2l}[Y_{n+1}, I_{n+1}] + i\sum_{m=0}^{\infty}\frac{(-1)^m}{(2m+1)!}\left(\frac{\pi}{4}l(z)\right)^{2m+1}[Y_{n+1}, X_{n+1}]\\ &= 0 + 2\sin(\frac{\pi}{4}l(z))Z_{n+1} \end{aligned} where we split the sum into even and odd terms. This gives in the end: \begin{aligned} U_l^\dagger Y_{n+1}U_l = Y_{n+1} + 2\sin(\frac{\pi}{4}l(z))Z_{n+1} \end{aligned} This is where I'm stuck. Did I make a mistake along the way?

$\endgroup$
2
  • $\begingroup$ could the general relation $e^A B e^{-A}=\exp(\operatorname{ad}(A))B\equiv B+[A,B]+\frac12[A,[A,B]]+...$ help here? $\endgroup$
    – glS
    Oct 19, 2022 at 23:47
  • $\begingroup$ Actually it does, I had not noticed that by choosing $A = -i\frac{\pi}{4}l(z)X_{n+1}$, we have $A^{\dagger} = -A$ and that satisfies the conditions of the Baker-Campbell-Hausdorf relation $\endgroup$ Oct 20, 2022 at 15:11

1 Answer 1

1
$\begingroup$

Following the remark of @glS, we can reformulate the problem using the Baker-Campbell-Hausdorf formula and we manage to derive the right results. Here are some details for the calculations: \begin{aligned} U_l^{\dagger}Y_{n+1}U_l &= e^{A}Y_{n+1}e^{-A} \qquad\qquad\qquad\qquad,\quad A = -i\frac{\pi}{4}l(z)X_{n+1}\\ &= \sum_{k=0}^{\infty}\frac{[(A)^k,Y_{n+1}]}{k!}\\ &= Y_{n+1} + (-i\frac{\pi}{4}l(z))[X_{n+1},Y_{n+1}] + \frac{1}{2!}(i\frac{\pi}{4}l(z))^2[X_{n+1},[X_{n+1},Y_{n+1}]]+...\\ &= Y_{n+1} + (-i\frac{\pi}{4}l(z))(2iZ_{n+1})+\frac{1}{2!}(i\frac{\pi}{4}l(z))^2(-(2i)^2Y_{n+1})+...\\ &= \sum_{k=0}^{\infty}\frac{(-i\frac{\pi}{4}l(z))^{2k}(2i)^{2k}(-1)^k}{(2k)!}Y_{n+1} + \sum_{k=0}^{\infty}\frac{(-i\frac{\pi}{4}l(z))^{2k+1}(2i)^{2k+1}(-1)^k}{(2k+1)!}Z_{n+1}\\ &= \cos(\frac{\pi}{2}l(z))Y_{n+1}+\sin(\frac{\pi}{2}l(z))Z_{n+1} \end{aligned} where we used the Pauli matrices commutation relations to simplify the sum of commutators, i.e. $[X,Y]=2iZ$ and $[X,Z]=-2iY$.

Another approach that was also successful is to compute the matrix product $U_l^{\dagger}Y_{n+1}U_l$ by writing $U_l$ explicitly: \begin{aligned} U_l = e^{i\frac{\pi}{4}l(z)X_{n+1}} &= \sum_{k=0}^{\infty}\frac{(i\frac{\pi}{4}l(z))^{2k}}{(2k)!}(X_{n+1})^{2k} + \sum_{k=0}^{\infty}\frac{(i\frac{\pi}{4}l(z))^{2k+1}}{(2k+1)!}(X_{n+1})^{2k+1}\\ &= \sum_{k=0}^{\infty}(-1)^k\frac{(\frac{\pi}{4}l(z))^{2k}}{(2k)!}I_{n+1} + \sum_{k=0}^{\infty}(-1)^k\frac{i(\frac{\pi}{4}l(z))^{2k+1}}{(2k+1)!}X_{n+1}\\ &= \cos(\frac{\pi}{4}l(z))I_{n+1} + i\sin(\frac{\pi}{4}l(z))X_{n+1} \end{aligned} we can then rewrite the product in matrix form: \begin{aligned} U_l^{\dagger}Y_{n+1}U_l &= \begin{pmatrix} \cos(\frac{\pi}{4}l(z)) & -i\sin(\frac{\pi}{4}l(z))\\ -i\sin(\frac{\pi}{4}l(z)) & \cos(\frac{\pi}{4}l(z)) \end{pmatrix} \begin{pmatrix} 0 & -i\\ i & 0 \end{pmatrix} \begin{pmatrix} \cos(\frac{\pi}{4}l(z)) & i\sin(\frac{\pi}{4}l(z))\\ i\sin(\frac{\pi}{4}l(z)) & \cos(\frac{\pi}{4}l(z)) \end{pmatrix}\\ &= \begin{pmatrix} \sin(\frac{\pi}{2}l(z)) & -i\cos(\frac{\pi}{2}l(z))\\ i\cos(\frac{\pi}{2}l(z)) & -\sin(\frac{\pi}{2}l(z)) \end{pmatrix}\\ &= \cos(\frac{\pi}{2}l(z))Y_{n+1} + \cos(\frac{\pi}{2}l(z))Z_{n+1} \end{aligned} and we finally get the expected result.

$\endgroup$
2
  • $\begingroup$ Is this the expected result though? The equation you quoted contained operator-valued terms $l(Z)$, and their notation suggests that $U_l^\dagger Y_{n+1} U_l$ is represented with a $2^{n+1}$-dimensional matrix, not a $2 \times 2$ matrix. $\endgroup$
    – forky40
    Oct 20, 2022 at 18:44
  • $\begingroup$ You're right I should replace $l(z)$ by $l(Z)$ to be exact. Since $l(Z)$ is diagonal in the Z-basis you have an operator $\cos(\frac{\pi}{2}l(Z))$ (or $\sin(\frac{\pi}{2}l(Z))$) acting on the first $n$ qubits and the operator $Y_{n+1}$ or ($X_{n+1}$ ) acting on the last qubit that we want to measure $\endgroup$ Oct 21, 2022 at 7:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.