I know that in algebra for a variable we have

$ \sqrt {x^2} = |x|$

What if $x$ is a density matrix?

Please share resource for your answer.

  • $\begingroup$ does this have any relation with quantum computing or quantum information? math.stackexchange.com might be a better place to ask this question $\endgroup$
    – glS
    Commented Jun 3, 2023 at 13:24
  • $\begingroup$ Have you tried wikipedia? $\endgroup$ Commented Jun 3, 2023 at 13:30
  • $\begingroup$ of course it has relation. for a Herminian matrix p = p-dagger. then p.p-dagger = p^2 wikipedia has nothing about this question. $\endgroup$
    – reza
    Commented Jun 3, 2023 at 13:32
  • $\begingroup$ @glS the important consideration is that density matrices are positive - otherwise it's just mathematics en.wikipedia.org/wiki/… $\endgroup$ Commented Jun 3, 2023 at 13:34
  • 1
    $\begingroup$ @QuantumMechanic I mean, sure. My point is that without any context this is a bit off-topic here. Adding some modicum of context for where the question arose would be desirable $\endgroup$
    – glS
    Commented Jun 3, 2023 at 14:13

2 Answers 2


If $\rho$ is a density matrix, then $\sqrt{\rho^2} = \rho$.

To see why this is, let's start with the definition of the square root of a matrix. Ordinarily, if $A$ is a square matrix, there may be multiple choices of a square matrix $B$ such that $B^2 = A$. However, if $P$ is a positive semidefinite matrix, then there is a unique choice of a positive semidefinite matrix $Q$ such that $Q^2 = P$, and when people write $\sqrt{P}$ for a positive semidefinite matrix $P$, this is what is most typically meant. You can find a proof of this claim (that there exists a unique positive semidefinite matrix $Q$ such that $Q^2 = P$) by taking $k=2$ in Theorem 7.2.6 of Horn and Johnson, Matrix Analysis.

Once we have the definition of $\sqrt{\rho^2}$ in place, it's pretty trivial: $\rho$ is positive semidefinite and $\rho^2 = \rho^2$, so we have our unique positive semidefinite square root: $\sqrt{\rho^2} = \rho$.


Take any spectral decomposition of a density operator $$\rho=\sum_n \rho_n |n\rangle\langle n|.$$ The square is defined unambiguously as $$\rho^2=\sum_n \rho_n^2 |n\rangle\langle n|.$$ By inspection, any operator of the form $$T(k_1,k_2,\cdots)=\sum_n \rho_n (-1)^{k_n} |n\rangle\langle n|,\quad k_n\in(0,1)$$ will square to $\rho^2$, so we find the answer to not be unique. Of course, this is just an application of spectral theorem to the function $\sqrt{(\cdot)^2}$, where we have applied this function to the eigenvalues $\rho_n$ of the operator $\rho$. Only one of these square roots is its positive semidefinite because positive semidefinite matrices have unique positive semidefinite square roots. This really is all on the Wikipedia page.


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