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I am reading John Watrous' quantum information theory book. In the proof of Theorem 3.19 (practically the Alberti's theorem on the characterization of the fidelity function) he claims the following fact: given two Hermitian operators $Y_0, Y_1$ on a complex Euclidean space $\mathcal{X}$, the operator $$ \begin{pmatrix} Y_0 & -\mathbb{1} \\ -\mathbb{1} & Y_1 \end{pmatrix} $$ on $\mathcal{X} \oplus \mathcal{X}$ is positive semidefinite if and only if both $Y_0$ and $Y_1$ are positive definite and satisfy $Y_1 \geq Y_0^{-1}$. I do not immediately see how this can be proven. He says that Lemma 3.18 can be used for this purpose, but I still do not get how to link these two results. Someone is willing to help?


For convenience I write here the statement of Lemma 3.18. It says the following: given two positive semidefinite operators $P,Q$ and a linear operator $X$ on $\mathcal{X}$, the operator $$ \begin{pmatrix} P & X \\ X^* & Q \end{pmatrix} $$ on $\mathcal{X} \oplus \mathcal{X}$ is positive semidefinite if and only if there exists an operator $K$ such that $X = \sqrt{P} K \sqrt{Q}$ and $||K|| \leq 1$. Here $^*$ stands for the conjugate transposition and $|| \cdot ||$ is the spectral norm.

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    $\begingroup$ The initial statement follows from the Schur complement characterization of block PSD matrices, see the final part of the wiki page. $\endgroup$
    – Rammus
    Jun 15 at 18:22
  • $\begingroup$ Perfect, thank you, I didn't know about this result. I suppose it can be posted as an answer! However, I still wonder how this fact follows from the Lemma 3.18 cited above. $\endgroup$
    – adabb
    Jun 15 at 18:47
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Firstly, observe that if $Y_0$ were not positive semi-definite, the whole thing cannot be positive semi-definite, because if $|\lambda\rangle$ were an eigenvector of $Y_0$ with negative eigenvalue, then $\left(\begin{array}{c} |\lambda\rangle\\0\end{array}\right)$ would have a negative expectation on the overall operator, and hence there would be a negative eigenvalue, and the operator would not be positive semi-definite.

So, that means $P=Y_0$ and $Q=Y_1$ are both positive semi-definite with $X=-1$, so the Lemma applies. It requires $$ -I=\sqrt{Y_0}K\sqrt{Y_1}. $$ In other words, $$ -1\leq K=-\sqrt{Y_0}^{-1}\sqrt{Y_1}^{-1}\leq 1 $$ So, let's rearrange the first inequality $$ \sqrt{Y_1}\geq\sqrt{Y_0}^{-1}. $$ Since both $Y_0$ and $Y_1$ are positive semi-definite, I can square this, with no ambiguity on the direction of the inequality: $$ Y_1\geq Y_0^{-1}. $$ You probably want to go through this carefully to make sure that the "if and only if" conclusion is valid, but this should give a sense of how you could use stated lemma.

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  • $\begingroup$ Thank you for the insight in the first paragraph! I just don't get why you say that $K \geq -\mathbb{1}$. I suppose it comes from the requirement $||K|| \leq 1$, but how? $\endgroup$
    – adabb
    Jun 16 at 7:49
  • $\begingroup$ Alternatively, we can prove that $||K|| \leq 1$ is equivalent to $Y_0^{-1} \leq Y_1$ in the following way. If $Y_0^{-1} \leq Y_1$ then $||K|| = ||\sqrt{Y_0^{-1}}\sqrt{Y_1^{-1}}|| \leq ||\mathbb{1}|| = 1$. Vice versa, if $||K|| \leq 1$ then $K^* K \leq \mathbb{1}$, which is equivalent to $Y_0^{-1} \leq Y_1$. $\endgroup$
    – adabb
    Jun 16 at 7:52
  • $\begingroup$ I guess you've pretty much answered your own question there. The reasoning I was using was that $\|K\|\leq 1$ basically means that the absolute value of all eigenvalues is $\leq 1$, i.e. all eigenvalues are bounded between $\pm 1$. You can equivalently write this as $-1\leq K\leq 1$. $\endgroup$
    – DaftWullie
    Jun 16 at 8:18
  • $\begingroup$ Right, I was missing the fact that our $K$ is Hermitian. $\endgroup$
    – adabb
    Jun 16 at 8:55
  • $\begingroup$ This answer is on the right track, but there are two problems. First, $K$ may not be Hermitian, and second, squaring is not operator monotone (i.e., $P\leq Q$ does not necessarily imply $P^2\leq Q^2$ for $P,Q\geq 0$). These problems can be fixed at the same time, though: starting from $Y_0^{-1/2} = - K Y_1^{1/2}$, left multiply each side to its adjoint to obtain $Y_0^{-1} = Y_1^{1/2} K^{\ast} K Y_1^{1/2} \leq Y_1$ (using $K^{\ast} K\leq \mathbb{1}$, as adabb has noted in a comment). $\endgroup$ Jun 16 at 15:14
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Thanks to John Watrous itself, in the comments the following answer emerged. I will call $B$ the block operator defined in the question in terms of $Y_0, Y_1$.

  • Suppose $B \geq 0$. Then, as DaftWullie pointed out in another answer, it must be true that $Y_0 \geq 0$. In fact, if there exists a vector $u \in \mathcal{X}$ such that $u^* Y_0 u < 0$, then we can construct the vector $v = \begin{bmatrix} u \\ 0 \end{bmatrix} \in \mathcal{X} \oplus \mathcal{X}$ and find out that $v^* B v = u^* Y_0 u < 0$, in contradiction with the hypothesis $B \geq 0$. The same reasoning can be applied to say that $Y_1 \geq 0$ as well. By means of the Lemma, there must exist an operator $K$ such that $Y_0^{1/2} K Y_1^{1/2} = -\mathbb{1}$ and $\lVert K \rVert \leq 1$. Note that $\lVert K \rVert \leq 1$ implies $\lVert K^* K \rVert \leq 1$, and since $K^* K$ is Hermitian this means that its largest eigenvalue has modulus bounded by one, i.e. $K^* K \leq \mathbb{1}$. We can now see the following facts. First of all, $Y_0 > 0$ and $Y_1 > 0$, since if they were singular $Y_0^{1/2} K Y_1^{1/2}$ should have been singular too, which is false because it is equal to the nonsingular operator $-\mathbb{1}$. Moreover, if we write $Y_0^{-1/2} = -KY_1^{1/2}$ we can left multiply each side by its adjoint to obtain $Y_0^{-1} = Y_1^{1/2} K^* K Y_1^{1/2} \leq Y_1$.

  • Suppose $Y_0 > 0$, $Y_1 > 0$, and $Y_0^{-1} \leq Y_1$. Let us define $K = -Y_0^{-1/2} Y_1^{-1/2}$. By definition, we have that $Y_0^{1/2} K Y_1^{1/2} = -\mathbb{1}$. Moreover $$ K^* K = Y_1^{-1/2} Y_0^{-1} Y_1^{-1/2} \leq Y_1^{-1/2} Y_1 Y_1^{-1/2} = \mathbb{1} \, . $$ Since $K^* K$ is Hermitian, this implies that $\lVert K^* K \rVert \leq 1$, hence $\lVert K \rVert \leq 1$. By means of the Lemma, we conclude that $B \geq 0$. Alternatively, one can invoke the Löwner-Heinz theorem, according to which the square root is a monotone operation with respect to the Löwner order, that is $Y_0^{-1} \leq Y_1 \Rightarrow Y_0^{-1/2} \leq Y_1^{1/2}$. Since the spectral norm is monotone, we have then $$ \lVert K \rVert = \lVert Y_0^{-1/2} Y_1^{-1/2} \rVert \leq \lVert Y_1^{1/2} Y_1^{-1/2} \rVert = \lVert \mathbb{1} \rVert = 1 \, . $$

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