# Clarification on Watrous' proof of Alberti's theorem on the fidelity function

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.

• The initial statement follows from the Schur complement characterization of block PSD matrices, see the final part of the wiki page. Commented Jun 15, 2021 at 18:22
• 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. Commented Jun 15, 2021 at 18:47

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.

• 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? Commented Jun 16, 2021 at 7:49
• 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 $Y_0^{-1} \leq Y_1$. Commented Jun 16, 2021 at 7:52
• 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 $-1\leq K\leq 1$. Commented Jun 16, 2021 at 8:18
• Right, I was missing the fact that our $K$ is Hermitian. Commented Jun 16, 2021 at 8:55
• 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). Commented Jun 16, 2021 at 15:14

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 $$\lVert K \rVert = \lVert Y_0^{-1/2} Y_1^{-1/2} \rVert \leq \lVert Y_1^{1/2} Y_1^{-1/2} \rVert = \lVert \mathbb{1} \rVert = 1 \, .$$