Physics Asked on July 5, 2021
Given two density matrices $rho_1,rho_2$ (let them be finite dimensional). I try to show that $mathrm{Tr}((rho_1-rho_2)^2) = 0 implies rho_1 = rho_2$. What I tried is the following.
I realised that $(rho_1 – rho_2)^2$ is semi positive-definite since $rho_1,rho_2$ are Hermitian and semi positive-definite. Therefore the eigenvalues are real and non-negative. Since the sum of these eigenvalues is zero all the eigenvalues of $(rho_1 – rho_2)^2$ must be zero. However I don’t see how to connect this to the conclusion that $rho_1 = rho_2$?
It is true that we have $(rho_1 – rho_2)^2 v = 0$ for all $v in mathbb{C}^n$ (since you can express $v$ in terms of complete eigenbasis of $(rho_1 – rho_2)^2$) and hence $(rho_1 – rho_2)^2 = 0$. But this does not imply $rho_1 = rho_2$ right?
Note that $rhoequiv (rho_1-rho_2)$ is hermitian. If $rho^2 =0$, then for an element of the underlying Hilbert space $ psiin mathscr{H}$ it holds that: $$ (psi,rho^2 psi) = (rho psi,rho psi) =((rho_1 - rho_2)psi,(rho_1 - rho_2)psi) = ||(rho_1-rho_2)psi||^2 =0 quad ,$$
where $(cdot,cdot)$ denotes the scalar product and $|| cdot||$ the norm. The properties of the norm then imply that $$ (rho_1-rho_2) psi = 0 quad . $$ Since $psi in mathscr{H}$ was chosen arbitrarily and an operator is defined by its action on the elements of the Hilbert space, we conclude that $rho_1 = rho_2$.
Correct answer by Jakob on July 5, 2021
If $A$ is hermitian and $A^2=0$, then $A=0$. The proof is quite simple, suppose $Aneq 0$
$$ langle i|A^2|irangle=sum_j langle i|A|jranglelangle j|A|irangle=sum_j |langle i|A|jrangle|^2neq 0$$
at least for one of the $i$s, because I assumed $Aneq 0$. Hence if $Aneq 0$, $A^2neq 0$. This clearly implies that if $A^2=0$, then $A=0$.
Answered by user2723984 on July 5, 2021
Get help from others!
Recent Answers
Recent Questions
© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP