if $HH = I$ and $XX =I$, then is $H=X$?

$HH = I = XX$ or, $HH = XX$ then, taking under root, is $H = X$?

This is absurd but how to disprove it?

  • 5
    $\begingroup$ The argument is obviously flawed since it fails even in the good ol' integers: $1\cdot 1=1$ and $(-1)\cdot(-1)=1$ but $1\ne-1$. One way to disprove it for the Hadamard and Pauli $X$ gate is to carry out the calculations. $\endgroup$ Nov 26, 2021 at 8:20

1 Answer 1


Obviously $H\neq X$, as they are simply different matrices/operators.

The reason for the apparent "contradiction" is that the square root of a matrix (or of any number, for that matter) is not unique. What you are seeing here is that both $H$ and $X$ are valid square roots of the identity matrix $I$.

In fact, you can derive infinitely many possible square roots of $I$. Start observing that you can write $I=\mathbb P(u)+\mathbb P(v)$ for any pair of orthonormal vectors, $\langle u,v\rangle=0$ and $\langle u,u\rangle=\langle v,v\rangle=1$, where $\mathbb P(u)\equiv uu^\dagger$ denotes the projector onto $u$. Taking the square root via the eigendecomposition in the usual way then gives you $$\sqrt{I} = s \mathbb P(u) + t\mathbb P(v),$$ where $s,t\in\mathbb C$ are square roots of $1$, and thus $s,t\in\{+1,-1\}$. You thus get four possible solutions for every choice of two orthonormal vectors.

You get $H$ and $X$ choosing $s=-t=1$ and $u=(1,1)^T/\sqrt2$ and $u=(1,0)$, respectively.

It is interesting to note that much of the freedom that allowed the above solutions is due to the degeneracy of the matrix. A non-degenerate matrix will instead have only four solutions, as there is no freedom in the choice of the eigenvectors. For more information see e.g. the relevant Wikipedia page.


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