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The CNOT gate is usually written as

$|0\rangle\langle0|\otimes I + 1\rangle\langle1|\otimes X$

(with $X,Y,Z$ being the Pauli Basis and $I$ the Identity).

I have yet to stumble across the representation Wikipedia gives when looking at books on the topic:

$e^{i\frac{\pi}{4}\left(I-Z\right)\otimes\left(I-X\right)}$

Why is it that none of the books on quantum mechanics / computing /information I have looked at uses this exponential representation? Does somebody know, which book does?


Cross-posted on physics.SE

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    $\begingroup$ People probably don't quote the exponential formula because it doesn't make it clear what the gate is doing to basis states $|z_1z_2\rangle$. The exponential formula does give you a hint as to what kind of interaction might allow you to build a CNOT gate, though. $\endgroup$ Commented May 18, 2022 at 21:58
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    $\begingroup$ I think this exponential representation is useful if you want to build a Hamiltonian that recreates a CNOT. $\endgroup$
    – Mauricio
    Commented May 19, 2022 at 8:10
  • $\begingroup$ Do you know of books that give the exponential representation? $\endgroup$
    – manuel459
    Commented May 19, 2022 at 10:40
  • $\begingroup$ @Mauricio In most cases, it would probably be more instructive to rewrite the CNOT as an exponential of $X\otimes X$ and correct with local Cliffords. I don't see the point of this representation. Usually, you can't choose your interactions. $\endgroup$ Commented May 19, 2022 at 17:09

2 Answers 2

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You can also do it by expanding the terms in the exponential:

$$ e^{i\frac{\pi}{4}(I−Z)\otimes(I−X)} = e^{i\frac{\pi}{4}(II - IX - ZI + ZX)} $$

here I omitted the tensor product, e.g $II = I \otimes I$. Then you can group the diagonal and no diagonal operators as:

$$ e^{i\frac{\pi}{4}([II - ZI] + [ZX - IX])} = e^{i\frac{\pi}{4} \bigl( \bigl[ \begin{smallmatrix}0 & 0\\ 0 & 2I\end{smallmatrix}\bigr] + \bigl[ \begin{smallmatrix}0 & 0\\ 0 & -2X\end{smallmatrix}\bigr] \bigr)} $$

The matrices $\bigl[ \begin{smallmatrix}0 & 0\\ 0 & 2I\end{smallmatrix}\bigr] $ and $ \bigl[ \begin{smallmatrix}0 & 0\\ 0 & -2X\end{smallmatrix}\bigr]$ commute, so we can write:

$$ e^{i\frac{\pi}{4} \bigl( \bigl[ \begin{smallmatrix}0 & 0\\ 0 & 2I\end{smallmatrix}\bigr] + \bigl[ \begin{smallmatrix}0 & 0\\ 0 & -2X\end{smallmatrix}\bigr] \bigr)} = e^{i\frac{\pi}{4} \bigl[ \begin{smallmatrix}0 & 0\\ 0 & 2I\end{smallmatrix}\bigr] } e^{i\frac{\pi}{4} \bigl[ \begin{smallmatrix}0 & 0\\ 0 & -2X\end{smallmatrix}\bigr] }$$

this becomes:

$$ \begin{pmatrix} I & 0 \\ 0 & iI\end{pmatrix}\begin{pmatrix} I & 0 \\ 0 & e^{-i\frac{\pi}{2}X}\end{pmatrix} $$

using the property $e^{iaX} = \cos(a)I + i X \sin(a) $ with $a=\frac{\pi}{2}$ (you can look it up in the section 'Exponential of a Pauli vector' in https://en.wikipedia.org/wiki/Pauli_matrices ) you get $e^{-i\frac{\pi}{2}X} = - i X $. So finally:

$$ \begin{pmatrix} I & 0 \\ 0 & iI\end{pmatrix}\begin{pmatrix} I & 0 \\ 0 & -iX\end{pmatrix} = \begin{pmatrix} I & 0 \\ 0 & X \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0\end{pmatrix}$$

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  • $\begingroup$ You carried out the calculation but that is not what is being asked here. $\endgroup$
    – Mauricio
    Commented May 19, 2022 at 8:08
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We can derive this equality by direct matrix exponentiation. First, note that $I-X$ satisfies the following: \begin{align} (I-X)^2 &= (I-X)(I-X) = 2(I-X)\\ (I-X)^3 &= (I-X) (I-X)^2 = (I-X)(2 (I-X)) = 2(I-X)^2 = 4(I-X) \\&\vdots \\(I-X)^k &= 2^{k-1} (I-X) \tag{1} \end{align}

for $k\geq 1$, and $(I-X)^0 = I$ by definition. Then, using the definition of matrix exponentiation we can exponentiate $I-X$ as follows:

\begin{align} \exp(i \theta (I-X)) &= \sum_{k=0}^\infty \frac{(i \theta)^k (I-X)^k}{k!} \tag{2} \\&= I + \sum_{k=1}^\infty \frac{(i \theta)^k (I-X)^k}{k!} \tag{3} \\&= I + (I-X)\sum_{k=1}^\infty \frac{(i \theta)^k 2^{k-1}}{k!} \tag{4} \\&= I + (I-X)\frac{1}{2}\left(\sum_{k=0}^\infty \frac{(i 2\theta)^k }{k!} - 1\right) \tag{5} \\&= I + (I-X)\frac{1}{2}(e^{i2\theta} - 1) \tag{6} \\&= I + (I-X)ie^{i\theta}\sin\theta \tag{7} \end{align}

Line (4) uses Eq. (1), line (6) uses the definition of (scalar) exponentiation, and line (7) uses $$ (e^{i\theta} - e^{-i\theta}) = 2i\sin\theta \tag{8} $$

We can rewrite $I-Z$ as: \begin{align} I - Z = \begin{pmatrix} 0 & 0 \\ 0 & 2 \tag{9} \end{pmatrix} \end{align}

and so we need to solve \begin{align} \exp\left[i \frac{\pi}{4} (I-Z) \otimes (I-X)\right] &= \exp\left[i \frac{\pi}{2} |1\rangle \langle 1| \otimes (I-X)\right] \tag{10} \\&= \exp\left[ \begin{pmatrix} 0 & 0 \\ 0 & i\frac{\pi}{2} (I-X) \end{pmatrix}\right] \tag{11} \\&= \begin{pmatrix} I & 0 \\ 0 & \exp\left[i\frac{\pi}{2} (I-X)\right] \end{pmatrix}\tag{12} \end{align} which is a block-diagonal matrix ("$0$" means a $2 \times 2$ matrix of zeros here). The last equality maybe isn't obvious, but basically since the matrix exponential is just matrix multiplication -- e.g. Eq. (2) -- and the matrix product of block-diagonal matrices is just a block diagonal of matrix products, we can exponentiate each block separately. Then, applying Eq. (7) for $\theta = \pi/2$ we get $$ \exp\left[i\frac{\pi}{2} (I-X)\right] = I + (I - X)(i^2) = X $$ And so plugging into Eq (12) we find \begin{equation} \exp\left[i \frac{\pi}{4} (I-Z) \otimes (I-X)\right] = \begin{pmatrix} I & 0 \\ 0 & X\end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0\end{pmatrix} \end{equation}

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  • $\begingroup$ You carried out the calculation but that is not what is being asked here. $\endgroup$
    – Mauricio
    Commented May 19, 2022 at 8:08
  • $\begingroup$ Check the edits. "Why can the CNOT be represented as $e^{i\frac{\pi}{4}\left(I-Z\right)\otimes\left(I-X\right)}$" was the original question title. $\endgroup$
    – forky40
    Commented May 19, 2022 at 16:15
  • $\begingroup$ well, the title at the time I answered $\endgroup$
    – forky40
    Commented May 19, 2022 at 16:26
  • $\begingroup$ Sorry for this. I saw my original title was edited by somebody slightly misleading so I corrected it - sadly to late. Nevertheless forky your comment is still very helpful for me! Thank you. $\endgroup$
    – manuel459
    Commented May 19, 2022 at 21:20

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