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I've been working through the great introduction to quantum computing on Quantum Country. One exercise there is to find a possible quantum gate matrix that is not the $X,I$ or $H$ matrix.

I thought of $B = \frac{1}{\sqrt2}\begin{bmatrix}i & 1\\1 & i\end{bmatrix}$. However, I cannot find any article / paper mentioning this quantum gate.

From what I understood, all quantum gates must be $2\times2$, unitary matrices. I have even checked whether $B$ is indeed unitary, and yes, it is.

Why does this gate $B$ not have any useful applications? Or is it just not a valid gate?

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    $\begingroup$ It's a fine question that deserves an answer longer than a comment, but not all quantum gates need to be $2\times 2$ - similarly, not all classical gates need to be two-input (think of an AOI - and/or/invert - gate). The Toffoli gate is $3\times 3$. They do have to be reversible, as in unitary, however. $\endgroup$ – Mark S May 20 at 1:38
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    $\begingroup$ @MarkS As a point of note with respect to the Toffoli gate, that's a three-qubit operation, such that its unitary representation is a $2^3 \times 2^3$ operator. $\endgroup$ – Chris Granade May 20 at 3:51
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    $\begingroup$ I must say that it's a weird exercise. There are infinitely many possible single qubit quantum gates that are not $X$, $I$, or $H$. We don't even need to delve into two-qubit or three-qubit operations. $\endgroup$ – Sanchayan Dutta May 20 at 7:59
  • $\begingroup$ @ChrisGranade thanks! Sorry for the blunder. $\endgroup$ – Mark S May 20 at 12:25
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That's not the right way to look at it. In quantum mechanics, time evolutions are considered to be unitary and any unitary evolution can be written as a sequence of unitary operators $U_1, U_2, U_3,\ldots$ acting on a quantum state $|\Psi\rangle$. Any single-qubit unitary operation is a $2\times 2$ matrix of the form:

$$U=\begin{pmatrix}a&b\\-e^{i\phi}b^*&e^{i\phi} a^*\end{pmatrix}, |a|^2+|b|^2=1.$$

In your case, $a=\frac{i}{\sqrt{2}}$, $b=\frac{1}{\sqrt{2}}$ and $\phi=\pi$. Now, $B$ isn't one of the "popular" quantum gates like the Pauli or the Hadamard, but that doesn't mean it can't be a valid evolution operator! The Hadamard and the Pauli are generally chosen because they have some neat properties; for instance, the Pauli matrices form a basis for all $2\times 2$ Hermitian matrices. For more, read the Wikipedia pages on Hadamard matrix and Pauli matrices. Moreover, note that some gates can be easier to engineer than others; I had written an answer about this before. Also see: Why do we use the standard gate set that we do?.

So for drawing quantum circuits we generally try to choose a universal gate set that's also relatively easy to physically implement. Now any unitary evolution can be replicated with an $\epsilon>0$ precision, with a sufficient number of elementary quantum gates which form a universal set. Niel de Beaudrap also wrote a nice answer on this topic. Perhaps, as an exercise, you can try to write the $B$ gate in terms of the elementary quantum gates listed here?

If you notice carefully, $\sqrt{\operatorname{NOT}}=\frac{1-i}{\sqrt{2}}B$. That is, $B$ differs from the $\sqrt{\operatorname{NOT}}$ (or $\sqrt{X}$) by a simple phase factor! Thus, $\sqrt{\operatorname{NOT}}$ and $B$ are essentially the same gate cf. Norbert Schuch's answer. For a physical implementation of $\sqrt{\operatorname{NOT}}$ check Jones, Hansen & Mosca (1998).

I suggest playing around with it a bit on Quirk. The Bloch sphere intuition should help:

                         Source:https://journals.aps.org/prb/abstract/10.1103/PhysRevB.67.094510

You see, both $\sqrt{X}$ and $B$ rotate a qubit state around $\hat{x}$ by $90^{\mathrm{o}}$, albeit with different phase factors.

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    $\begingroup$ Another way to notice that $B = e^{i\phi} \sqrt X$ is by noticing that $B^2=iX$. Same thing, just a different way of writing it that I found easier to spot $\endgroup$ – Mithrandir24601 May 20 at 12:38
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    $\begingroup$ @Mithrandir24601 you made a typo $B^2=iX$ not $iX^2$. $\endgroup$ – AHusain May 20 at 19:39
  • $\begingroup$ Whoops... I've fixed that now... $\endgroup$ – Mithrandir24601 May 20 at 21:37

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