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Let $\lVert\cdot\rVert$ be the matrix/sup norm on $M_n(\mathbb{C})$: $$\lVert A \rVert = \sup\{\lVert Av\rVert~;~\lVert v\rVert = 1\}$$ which for positive operators is equal to their highest eigenvalue.

Let $\rho$ and $\pi$ be two states in $M_n(\mathbb{C})$. That is: $\rho\geq 0$ and Tr$(\rho) = 1$ and so for $\pi$. And define $\rho^{-1}$ as the generalised inverse of $\rho$: the operator with the same eigenspaces and kernel, but every nonzero eigenvalue $\lambda$ changed to $\lambda^{-1}$, and define $\rho^{-\frac{1}{2}} = \sqrt{\rho^{-1}}$, where $\sqrt{}$ is the usual positive matrix square root.

Assuming that ker$(\rho)\subseteq $ker$(\pi)$, is it true that $$\lVert \rho^{-\frac{1}{2}}\pi\rho^{-\frac{1}{2}}\rVert\geq 1$$ for any such states $\rho$ and $\phi$?

I know this is true when $\rho$ and $\pi$ can be simultaneously diagonalised, because then this question reduces to a trivial statement about probability distributions. I'm looking for a proof (or counterexample) in the general case.

edit: Proof of this last statement: Suppose $\rho$ and $\pi$ are both diagonalised by some unitary $U$: $\rho=UPU^\dagger$, $\pi = UQU^\dagger$, for some diagonal $P$ and $Q$. Since the norm is unitary invariant, the expression we want to prove is:

$$\sup_i \frac{P_i}{Q_i} \geq 1$$

Suppose this is not the case, then $P_i\leq Q_i$ for all $i$, but we have $\sum_i P_i = \sum_i Q_i = 1$ and $P_i,Q_i\geq 0$, so this is only possible when $P_i=Q_i$ for all $i$.

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  • $\begingroup$ Positive means $z^* \rho z \ge 0$ for any complex vector $z$? $\endgroup$ – user251257 Oct 28 '16 at 13:06
  • $\begingroup$ Perhaps I have missed something but if $\rho=e_1 e_1^*$ (pure state) your definitions seem to set $\rho^{-1/2} = e_1 e_1^*$ as well? In which case $\rho^{-1/2}\pi\rho^{-1/2}$ could be zero? $\endgroup$ – H. H. Rugh Oct 28 '16 at 13:06
  • $\begingroup$ @H.H.Rugh hence $\ker \rho \subseteq \ker \pi$ $\endgroup$ – Omnomnomnom Oct 28 '16 at 13:08
  • $\begingroup$ You are right I did miss something. $\endgroup$ – H. H. Rugh Oct 28 '16 at 13:10
  • $\begingroup$ Yeah, that's why I require that extra property. @user251257: That is indeed what I mean by positive. Alternatively, that every eigenvalue is positive. $\endgroup$ – John Oct 28 '16 at 13:15
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You can prove the general case analogously. Assume wlog. $\rho$ be invertible after dividing out $\ker \rho$.

Notice that $\rho^{-1/2} \pi \rho^{-1/2}$ is positive again. Now, assume the converse $$ \| \rho^{-1/2} \pi \rho^{-1/2} \| = \lambda_\max(\rho^{-1/2} \pi \rho^{-1/2}) < 1. $$ Then, we have $$ tr(\pi) = \sum_{i=1}^n e_i^* \rho^{1/2} (\rho^{-1/2} \pi \rho^{-1/2}) \rho^{1/2} e_i < \sum_{i=1}^n e_i^* \rho^{1/2} \rho^{1/2} e_i = tr(\rho) = 1. $$

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Let $Z=\ker \rho$ and let $e_1,...,e_k$ be an orthonormal base for $Z^\perp$ which we supplement with $e_{k+1},...,e_n$ to get an orthonormal base for the full space. Then $$\rho=\sum_{i=1}^k \lambda_{i} e_i e_i^*, \ \ \ \ \rho^{-1/2} = \sum_{i=1}^k\lambda_i^{-1/2} e_i e_i^*$$ We then have for the trace norm $$ |\rho^{-1/2} \pi \rho^{-1/2}|_1 = \sum_{i=1}^k \frac{1}{\lambda_i} (e_i^* \pi e_i) \geq \sum_{i=1}^k (e_i^* \pi e_i) =\sum_{i=1}^n (e_i^* \pi e_i)= {\rm Tr} \; \pi =1$$ where we used $\ker \rho\subset \ker \pi$ in the third equality. And similarly (with $q=\rho^{-1/2} \pi \rho^{-1/2}$): $$ 1=|\pi|_1 = \sum_{i=1}^n (e_i^* \pi e_i) = \sum_{i=1}^k \lambda_i (e_i^* q e_i)\leq \|q\| \sum_{i=1}^k \lambda_i = \|q\| \; |\rho|_1 = \|q\|$$

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  • $\begingroup$ @H.H.Rugh it looks like you're using some kind of trace norm instead of the operator norm $\endgroup$ – Omnomnomnom Oct 28 '16 at 13:29

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