# what is square root of a density matrix power two?

I know that in algebra for a variable we have

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

What if $$x$$ is a density matrix?

• does this have any relation with quantum computing or quantum information? math.stackexchange.com might be a better place to ask this question
– glS
Jun 3, 2023 at 13:24
• Have you tried wikipedia? Jun 3, 2023 at 13:30
• of course it has relation. for a Herminian matrix p = p-dagger. then p.p-dagger = p^2 wikipedia has nothing about this question.
– reza
Jun 3, 2023 at 13:32
• @glS the important consideration is that density matrices are positive - otherwise it's just mathematics en.wikipedia.org/wiki/… Jun 3, 2023 at 13:34
• @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
– glS
Jun 3, 2023 at 14:13

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.