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.


1 Answer 1


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 $\sqrt{X}$.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.