4
$\begingroup$

In the John Watrous TQI notes and he states that for a given square operator $A \in \mathbf{L}(X)$,

$$ \Vert A \Vert_1 = \max \{\vert \langle U,A\rangle \vert: U \in \mathbf{U}(x) \}$$ He also states that tho achieve this the identity is useful $$\Vert A \Vert_p = \Vert \mathbf{s}(A) \Vert_p$$ where $\mathbf{s}(A)$ are the singular values of $A$.

In my attempt to prove this I used that given $A \in \mathbf{L}(X)$ and $ U,V \in \mathbf{U}(x)$, taking the SVD $A = U \Sigma V^*$ and $W = UV^*$ $$\max_{T \in \mathbf{U}(X)} \vert \langle T,A\rangle \vert \geq \vert\langle W,A\rangle \vert = \vert {\rm Tr}(W^* A) \vert = \vert {\rm Tr}(VU^*U\Sigma V^*) \vert= \vert \sum_i s_i \vert = \sum_i \vert s_i \vert$$

To prove now that $\max_{T \in \mathbf{U}(X)} \vert \langle T,A\rangle \vert \leq \sum_i \vert s_i \vert $ i've tried, and is where i couldn't reach any satisfactory result

Given $Z \in \mathbf{U}(x) $ such that $\Vert Z \Vert \geq \Vert T \Vert$, then \begin{align*}\max_{T \in \mathbf{U}(X)} \vert \langle T,A\rangle \vert &\leq \vert\langle ZW,ZA\rangle \vert \\ &= \vert {\rm Tr}(W^*Z^*Z A) \vert \\ &= \vert {\rm Tr}(VU^*U\Sigma V^*) \vert= \Big\vert \sum_i s_i \Big\vert = \sum_i \vert s_i \vert \end{align*}

Thus we conclude that $$ \max_{T \in \mathbf{U}(X)} \vert \langle T,A\rangle \vert = \sum_i \vert s_i \vert = \Vert A \Vert_1$$ Does someone have a better way to state that $\max_{T \in \mathbf{U}(X)} \vert \langle T,A\rangle \vert \leq \sum_i \vert s_i \vert $ ?

$\endgroup$
1
  • $\begingroup$ I'm not quite sure what the operator $Z$ is doing in your calculation: because $Z$ is unitary $\langle ZW,ZA\rangle=\langle W,A\rangle$ so your bound reads $\max_T|\langle T,A\rangle|\leq|\langle W,A\rangle|$ (i.e. the same as in your first step but this time the inequality flipped) which can in general not be true. $\endgroup$ Commented Jun 24 at 6:59

2 Answers 2

3
$\begingroup$

The converse also follows from the singular value decomposition: for all $T\in\mathbf U(X)$ \begin{align*} |\langle T,A\rangle|&=|{\rm tr}(T^*U\Sigma V^*)|=|{\rm tr}(V^*T^*U\Sigma)|\\ &=\Big|\sum_i\langle i|V^*T^*U\Sigma|i\rangle\Big|\\ &=\Big|\sum_is_i\langle i|V^*T^*U|i\rangle\Big|\\ &\leq\sum_i s_i|\langle i|V^*T^*U|i\rangle|\\ &\leq\sum_i s_i\|\,|i\rangle\|\|V^*T^*U |i\rangle\|=\sum_i s_i=\|A\|_1\,. \end{align*} In the fourth line we used the triangle inequality, in the fifth line we used Cauchy-Schwarz, and finally $\|V^*T^*U|i\rangle\|=1$ because $\|\,|i\rangle\|=1$ and $V,T,U$ are all unitary (i.e. length-preserving).

More generally, the above is a special case of the following standard inequality for the trace: $|\langle A,B\rangle|\leq\|AB\|_1\leq\|A\|_1\|B\|$ for all $A,B$

$\endgroup$
2
$\begingroup$

Write the SVD of $A$ as $$A = \sum_k s_k |u_k\rangle\!\langle v_k|,$$ for some pair of orthonormal bases $\{|u_k\rangle\}$ and $\{|v_k\rangle\}$. If $T$ is unitary, then it can always be written as $T=\sum_k |u_k\rangle\!\langle w_k|$ for some third basis $\{|w_k\rangle\}$. It follows that $$\langle T,A\rangle = \sum_k s_k \langle v_k|w_k\rangle.$$ The conclusion then follows immediately from $$ \lvert \sum_k s_k \langle v_k|w_k\rangle\rvert \le \sum_k s_k \lvert \langle v_k|w_k\rangle \rvert \le \sum_k s_k.$$ To show that the maximum can be achieved you can simply use $T=\Pi_+-\Pi_-$ with $\Pi_\pm$ projections onto positive and negative eigenspaces of $A$ (see also this related answer for more details). And also the answer to Proving the inequality $|\mathrm{tr}(AU)|\le \mathrm{tr}|A|$ in Uhlmann's theorem, which I've just realised is essentially the same question as this one. You'll also find very similar calculations in Intuitive role of the polar decomposition in proof of Uhlmann's theorem for fidelity.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.