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I was reading Square root of Pauli operators: is there a common convention to define them uniquely? and it got me thinking about square roots.

Recall the Pauli $ X $ gate $$ X=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} $$

How many unitary matrices $ U $ are there such that $$ U^2=X $$ In other words, how many unitary square roots does $ X $ have? I would expect that there are $ 4 $ square roots because each eigenvalue has 2 square roots.

However this naïve approach fails for some matrices, for example the identity $ I $. Not only does $ \pm I $ square to $ I $ but also every matrix of the form $ VZV^{-1} $ for $ V $ unitary is a unitary square root of $ I $. For example $ \pm I, \pm X, \pm Z , \pm Y $ are all unitary matrices that square to $ I $. As are all matrices of the form $$ \begin{bmatrix} 0 & e^{i \theta} \\ e^{-i \theta} & 0 \end{bmatrix} $$

Edit: The relevant distinction here between $ X $ having finitely many roots and $ I $ having infinitely many roots is that $ X $ has all distinct eigenvalues while $ I $ has repeated eigenvalues.

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I want to answer this on two levels.

What are the $X$ and $\sqrt{X}$ gates?

A good way of seeing this is on the Bloch Sphere. The generic rotation around the $x$-axis is defined as $$ R_x(\theta) = \cos(\theta/2)I - i\sin(\theta/2)X. $$ If $\theta=\pi$, then we get the $X$ gate, up to an overall phase. In simple terms, the $X$ gate is just a rotation of 180 degrees around the $x$-axis on the Bloch sphere.

Now $\sqrt{X}$ is just breaking up this rotation into two equal rotations. It should be apparent, there are only two ways of doing this. Either $\sqrt{X}$ is a 90 degrees clockwise rotation around the $x$-axis, or it is a 90 degrees anti-clockwise rotation around the $x$-axis.

Do we have more freedom algebraically?

In your linked question, this answer tells us that there can be exactly four square roots $$ \sqrt{X} = \pm 1 |+\rangle \langle +| \pm i |-\rangle \langle -|, $$ which yield $X$ without an overall phase. Two of these correspond to a 90 degree clockwise rotation with a phase inbuilt, and other two are a 90 degrees anti-clockwise rotation. To see this, note that with the choice of $1,i$, in the above, we get $$ \sqrt{X} = \frac{1}{2} \begin{pmatrix} 1+i & 1-i \\ 1-i & 1+i \end{pmatrix} = \frac{1}{\sqrt{2}}e^{i\pi/4}(I-iX). $$ And if we make the choice $-1,-i$, we get $$ \sqrt{X} = \frac{1}{2} \begin{pmatrix} -1-i & -1+i \\ -1+i & -1-i \end{pmatrix} = -\frac{1}{\sqrt{2}}e^{i\pi/4}(I-iX). $$ These differ by a minus sign and square to $X$.

Finally, if you allow an overall phase in $X$, i.e. if $A = \sqrt{X}$ and $A^2 = e^{i\phi}X$, then, you can define infinitely many $\sqrt{X}$.

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  • $\begingroup$ Just to clarify are you saying that the 4 unitary square roots of $ X $ are $ H diag(1,\pm i) H $ and $ H diag(\pm i,1) H $? What about $ H diag(-1,\pm i) H $ and $ H diag(\pm i,-1) H $? Here $ H $ is the hadamard. $\endgroup$ Commented Feb 14, 2023 at 22:30
  • $\begingroup$ I think you should compute these by hand. Only $H \text{diag}(1,\pm i)H$ and $H \text{diag}(-1,\pm i)H$ square to exactly $X$. These are the four matrices I (and the other answer) have indicated. The other four choices that you have given square to $-X$. $\endgroup$ Commented Feb 14, 2023 at 22:54
  • $\begingroup$ Ah you're right I computed them by hand. In retrospect it should have been obvious that flipping the elements on the diagonal gives the negative because $ HZH=X $ but flipping the entries of $ Z $ gives $ -Z $ so $ H(-Z)H=-X $. So any $ 2\times 2 $ unitary matrix with distinct eigenvalues has exactly 4 square roots $\endgroup$ Commented Feb 15, 2023 at 0:32
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    $\begingroup$ Thanks again for this answer. I think the normalizations are off. It should be $$ \sqrt{X} = \frac{1}{2} \begin{pmatrix} 1+i & 1-i \\ 1-i & 1+i \end{pmatrix} = \frac{1}{\sqrt{2}}e^{i\pi/4}(I-iX). $$ and $$ \sqrt{X} = \frac{1}{2} \begin{pmatrix} -1-i & -1+i \\ -1+i & -1-i \end{pmatrix} = -\frac{1}{\sqrt{2}}e^{i\pi/4}(I-iX). $$ $\endgroup$ Commented Apr 2, 2023 at 22:22
  • $\begingroup$ Thanks, fixed the error. $\endgroup$ Commented Apr 2, 2023 at 22:35

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