# What applications does the quantum gate [(i,1),(1,i)] have?

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?

• 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. May 20, 2019 at 1:38
• @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. May 20, 2019 at 3:51
• 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. May 20, 2019 at 7:59
• @ChrisGranade thanks! Sorry for the blunder. May 20, 2019 at 12:25

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:

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

• 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 May 20, 2019 at 12:38
• @Mithrandir24601 you made a typo $B^2=iX$ not $iX^2$. May 20, 2019 at 19:39
• Whoops... I've fixed that now... May 20, 2019 at 21:37