Suppose that $U_1,U_2$ are unitary matrices. A paper here (page 4) I am reading gives the following set of inequalities (using Dirac notation here for vectors representing pure quantum states, aka. tensor products of vectors in $\mathbb{C}^2$ with length $1$).

$$ \|U_1 - U_2\| \geq \|U_1 |\psi \rangle - U_2 |\psi \rangle\| \geq \frac {1}{2}\operatorname{Trace}\left |U_1 |\psi \rangle \langle \psi| U_1^{\dagger} - U_2 |\psi \rangle \langle \psi| U_2^{\dagger}\right| \\ = D (U_1 |\psi \rangle \langle \psi| U_1^{\dagger} - U_2 |\psi \rangle \langle \psi| U_2^{\dagger}),$$

where $D$ is the trace distance.

The norm used is not stated. Here are some things I know related to these inequalities:

1) Unitary matrices preserve the length of vectors

2) If $|| A||$ is a matrix norm induced by a vector norm, then for $||x||=1$, $||A|| \geq ||Ax||$, which would explain the first inequality. I'm not sure what is happening in the second inequality. Insights appreciated.

  • $\begingroup$ I've edited your question to make it clearer, is the current version accurate? $\endgroup$ – Omnomnomnom Jul 18 at 1:24
  • $\begingroup$ Also, do you have a link to the paper? $\endgroup$ – Omnomnomnom Jul 18 at 1:25
  • $\begingroup$ @Omnomnomnom yes it is. I will link it. $\endgroup$ – IntegrateThis Jul 18 at 1:27
  • $\begingroup$ I think that the norm is indeed the induced vector norm, as you've stated. I think that the inequality being used in the second step is $$ \|x - y\| \geq \frac 12 \operatorname{Trace}|xx^* - yy^*| $$ for vectors $x,y \in \Bbb C^n$. I don't immediately see why this would hold, though. $\endgroup$ – Omnomnomnom Jul 18 at 1:39
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    $\begingroup$ The trace in question would be $\sum |x_i^2 - y_i^2|$. Supposing that the norm we used was euclidian, i.e. $||x-y||^2 = \sum |x_i - y_i|^2$, we can say that $|x_i^2-y_i^2| \leq |x_i-y_i|( |x_i+y_i|)$. Next we use Cauchy Schwarz and the triangle inequality for $||.||_2$ norm to get the result. I've used the fact that the 2 norm of these vectors is 1, which is maybe bold haha $\endgroup$ – George Dewhirst Jul 18 at 1:52

The chain of inequalities makes sense if we add in the following.

Claim: For any unit vectors $x,y$, we have $$ \|x - y\| \geq \frac 12 \operatorname{Tr}|xx^* - yy^*|. $$

Proof: Noting that $x$ and $y$ are unit vectors, we compute $$ \|x - y\|^2 = (x-y)^*(x-y) = x^*x - x^*y - y^*x + y^*y \\ = 1 - \langle x, y\rangle - \langle y,x \rangle + 1 = 2(1 - \operatorname{Re}\langle x,y \rangle) $$ So, we have $\|x - y\| = \sqrt{2}\sqrt{1 - \operatorname{Re}\langle x,y \rangle}$.

On the other hand, $A = (xx^* - yy^*)$ is a rank-2 Hermitian matrix with trace zero, which means that it has non-zero eigenvalues $\pm \lambda$ for some $\lambda>0$. We compute $$ \begin{align*} 2\lambda^2 &= \lambda^2 + (-\lambda)^2 = \operatorname{Tr}(A^2) = \operatorname{Tr}[(xx^* - yy^*)^2] \\ & = \operatorname{Tr}[xx^*xx^* - xx^*yy^* - yy^*xx^* + yy^*yy^*] \\ & = \operatorname{Tr}[x^*xx^*x - x^*yy^*x - y^*xx^*y + y^*yy^*y] \\ &= \langle x,x\rangle^2 - 2|\langle x,y\rangle|^2 + \langle y,y \rangle^2 \\ & = 2 - 2|\langle x,y\rangle|^2 \end{align*} $$ We thereby conclude that $\lambda = \sqrt{1 - |\langle x,y \rangle|}$. Thus, we compute $$ \frac 12 \operatorname{Tr}|xx^* - yy^*| = \frac 12 (2 \lambda) = \sqrt{1 - |\langle x,y \rangle|}. $$

Thus, it suffices to show that $$ \sqrt{2}\sqrt{1 - \operatorname{Re}\langle x,y \rangle} \geq \sqrt{1 - |\langle x,y \rangle|}. $$ Indeed, we have $$ \sqrt{2}\sqrt{1 - \operatorname{Re}\langle x,y \rangle} \geq \sqrt{1 - |\langle x,y \rangle|} \iff\\ 2(1 - \operatorname{Re}\langle x,y \rangle) \geq 1 - |\langle x, y \rangle| \iff \\ 1 + |\langle x, y \rangle| \geq 2 \operatorname{Re}\langle x,y \rangle \iff\\ \frac{1 + |\langle x, y \rangle|}{2} \geq \operatorname{Re}\langle x,y \rangle. $$ The last inequality can be shown to hold as follows: by Cauchy Schwarz, $|\langle x, y \rangle| < 1$. So, $$ \frac{1 + |\langle x, y \rangle|}{2} \geq |\langle x,y \rangle| = \sqrt{(\operatorname{Re}\langle x,y \rangle)^2 + (\operatorname{Im} \langle x,y\rangle)^2} \\ \qquad \qquad \quad\geq \sqrt{(\operatorname{Re}\langle x,y \rangle)^2} = |\operatorname{Re}\langle x,y \rangle| \geq \operatorname{Re}\langle x,y \rangle. $$

  • $\begingroup$ Why does $A$ have rank $2$, and why is it's trace $0$? Thanks! $\endgroup$ – IntegrateThis Jul 18 at 2:11
  • $\begingroup$ Sorry for the rushed answer. Assuming that $x \neq y$ so that $A \neq 0$, $A$ has trace $0$ since $$ \operatorname{Tr}(A) = \operatorname{Tr}(xx^* - yy^*) = x^*x - y^*y = 1 - 1 = 0 $$ As for the rank: $A$ is the sum of $2$ rank-1 operators, so its rank is at most $2$. Since $A$ is a non-zero matrix with rank at most 2 and trace zero, it must have eigenvalues $\pm \lambda$ (and therefore rank exactly $2$). $\endgroup$ – Omnomnomnom Jul 18 at 2:14
  • $\begingroup$ Note that for a Hermitian matrix, the rank of $A$ is the total number of non-zero eigenvalues of $A$. $\endgroup$ – Omnomnomnom Jul 18 at 2:17
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    $\begingroup$ If you want a reference for the inequalities, I think that this is one of the Fuchs van der Graaf inequalities. Watrous's text "theory of quantum information" should have this in there somewhere. $\endgroup$ – Omnomnomnom Jul 18 at 2:20
  • $\begingroup$ Thanks so much for the help, I need to study it all before I can say much more but this is exactly what I was looking for. $\endgroup$ – IntegrateThis Jul 18 at 2:21

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