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Suppose $\rho$ is a self-adjoint postive semidefinite operator in $\mathbb{C}^{n \otimes n}$ (a quantum state). $L$ and $M$ are complex $m \times n$ matrices. How would one prove that if $L^\dagger L \rho L^\dagger L = M^\dagger M \rho M^\dagger M$, then $L^\dagger L = M^\dagger M$? $L^\dagger$ refers to the conjugate transpose of $L$.

I came across this equality in a research paper I am studying and was not able to see a general proof of the statement.

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    $\begingroup$ just an observation: $A^\dagger A$ is self-adjoint for any matrix $A$. Then you want to show that $ABA=CBC$ where $A$ and $C$ are self-adjoint and $B$ is a positive operator imply that $A=C$ (what Idk if it is true) $\endgroup$
    – Masacroso
    Commented Jul 31, 2017 at 21:50
  • $\begingroup$ moreover: $A^\dagger A$ is also positive semidefinite, the same that $\rho$. $\endgroup$
    – Masacroso
    Commented Jul 31, 2017 at 21:58
  • $\begingroup$ Just so we can all agree on the starting point: $0$ is self adjoint positive semi-definite, but such conclusion cannot be made (in this case). Is it possible that the requirement is: $\rho$ is positive definite? $\endgroup$
    – Ranc
    Commented Jul 31, 2017 at 21:59
  • $\begingroup$ Is spectral theorem useful here? Since, B and C are self-adjoint, they have a spectral decomposition. $\endgroup$
    – user39617
    Commented Jul 31, 2017 at 22:05
  • $\begingroup$ I think I must make my statement precise. If for all $\rho \in \mathbb{C}^{n \otimes n}$ such that $\rho$ is self adjoint, positive semi-definite of trace 1, $L^\dagger L \rho L^\dagger L = M^\dagger M \rho M^\dagger M $, then $L^\dagger L = M^\dagger M$. $\endgroup$
    – user39617
    Commented Jul 31, 2017 at 22:08

2 Answers 2

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So, reading the comments, the intent was "for all $\rho$". Fix $\varepsilon>0$ and let $$\rho=(L^\dagger L+\varepsilon I)^{-2}.$$ Then your equality looks like $$ L^\dagger L(L^\dagger L+\varepsilon I)^{-2}L^\dagger L=M^\dagger M (L^\dagger L+\varepsilon I)^{-2} M^\dagger M. $$ As $\varepsilon\to0$, the left-hand-side converges to the range projecton of $L^\dagger L$. As the right-hand-side always maps into the range of $M^\dagger M$, we conclude that $\text{Ran}\,L^\dagger L\subset\text{Ran}\,M^\dagger M$. As the roles of $L$ and $M$ are reversible, we get equality.

If we restrict to the range of $L^\dagger L$, both $L^\dagger L$ and $M^\dagger M$ are invertible, since $$ \ker L^\dagger L=\text{Ran}\,(L^\dagger L)^\perp. $$ If $X$ is such inverse for $L^\dagger L$, and $Y$ for $M\dagger M$, we now have $$\tag{1} \rho=XM^\dagger M\rho M^\dagger MX. $$ By taking limits of positive definite matrices we obtain any positive semi-definite matrix, and by taking linear combinations we can obtain any matrix. So the equality $(1)$ holds for any arbitrary matrix $\rho$. We can take for instance $\rho=Y^2$, and we get $$\tag{2} Y^2=X^2. $$ Since $X,Y$ are positive semidefinite, it follows from $(2)$ that $X=Y$. By uniqueness of inverses, $L^\dagger L=M^\dagger M$.

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  • $\begingroup$ Sorry. I think I do not understand your point. If $L$ and $M$ are unitary matrices, then the proof is straightforward. $L^\dagger L = M^\dagger M$ as you mentioned irrespective of $\rho$ and thats what we need. $\endgroup$
    – user39617
    Commented Jul 31, 2017 at 22:04
  • $\begingroup$ You are right. I'll change the example. $\endgroup$ Commented Jul 31, 2017 at 22:11
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Following the last comment the question is easy to answer, that is: we want to prove that

If $$A^\dagger A M A^\dagger A=B^\dagger B M B^\dagger B$$

for all self-adjoint positive semidefinite matrices $M\in\Bbb C^{n\times n}$ with $\operatorname{trace}(M)=1$ and any $A,B\in\Bbb C^{m\times n}$ then $A^\dagger A=B^\dagger B$.

Observe that the operator defined by the matrix $n^{-1} I$ is self-adjoint positive semi-definite and clearly $\operatorname{trace}(n^{-1}I)=1$, hence the statement to be proven imply that

$$A^\dagger An^{-1} IA^\dagger A=B^\dagger Bn^{-1} IB^\dagger B\implies n^{-1}(A^\dagger A)^2=n^{-1} (B^\dagger B)^2\tag1$$

It can be easily seen that $A^\dagger A$ is positive semi-definite and that also it is $(A^\dagger A)^2$ (just observe that $A^\dagger A$ is also self-adjoint). Then we can apply the following theorem:

If $M$ is a (bounded) positive semi-definite operator it have a unique positive semi-definite square root.

Thus $(1)$ can be simplified to

$$A^\dagger A=B^\dagger B$$

after simplifying and taking the positive square root from both sides, because $A^\dagger A$ is positive semi-definite, and thus it is the unique positive semi-definite square root of $(A^\dagger A)^2$.

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  • $\begingroup$ Thank you! That makes sense. Since, $A^\dagger A M A^\dagger A=B^\dagger BMB^dagger B$ is true for all self-adjoint positive semi-definite $M$ of trace $1$, you fixed $M$ to be $\frac{1}{n} I_{n}$. $\endgroup$
    – user39617
    Commented Jul 31, 2017 at 23:31

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