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The diamond norm distance between two operations is the maximum trace distance between their outputs for any input (including inputs entangled with qubits not being operated on).

Is it the case that the Diamond norm is subadditive under composition?

$$\text{Diamond}(U_1 U_2, V_1 V_2) \stackrel{?}{\leq} \text{Diamond}(U_1, V_1) + \text{Diamond}(U_2, V_2)$$

Niel de Beaudrap
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Craig Gidney
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  • Your notation is cumbersome/unclear in more than one way: Are U,V etc. TPCP maps? What set is A from? What is the definition of "Trace"? (Also, in the example a dagger seems to be missing?) [Note that without you saying what your definition of "Trace" you use, it will be hard to say if you are right or wrong!!] – Norbert Schuch Aug 17 '19 at 22:40
  • Could it be that you are not using the correct definition of the diamond norm either -- see, e.g. https://cstheory.stackexchange.com/a/4920/4047? There, it is defined over all extensions of the map to larger systems! – Norbert Schuch Aug 18 '19 at 13:20
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    @NorbertSchuch It absolutely could be a slightly wrong definition. Feel free to edit the question to have the correct definition. – Craig Gidney Aug 18 '19 at 17:32
  • What do you mean by "slightly" wrong?? -- The definition is in the linked question, feel free to copy it. After all, you have to know what is your question - maybe you are after the quantity you defined above rather than the diamond norm distance?? – Norbert Schuch Aug 18 '19 at 17:47
  • @NorbertSchuch I mean slightly wrong in the sense that the edit distance is small. I only included the definition as a clarification of what "diamond distance" means, and if I got that wrong then I want it corrected. – Craig Gidney Aug 18 '19 at 20:55
  • In that sense, forgetting a "not" in a statement is also only slightly wrong ;) – Norbert Schuch Aug 18 '19 at 21:25
  • @NorbertSchuch Yes, exactly. – Craig Gidney Aug 18 '19 at 22:36
  • You don't seem to bother much to put the correct definition in your question ... – Norbert Schuch Aug 19 '19 at 09:36

1 Answers1

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For arbitrary linear super-operators $U_j$ and $V_j\def\D{\mathrm{Diamond}} \def\Dn#1{\lVert #1 \rVert_\diamond}\def\le{\leqslant} $, we have $$\def\D{\mathrm{Diamond}} \def\Dn#1{\lVert #1 \rVert_\diamond}\def\le{\leqslant} \begin{aligned} \D(U_1 U_2, V_1 V_2) &= \Dn{U_1 U_2 - V_1 V_2} \\&\le \Dn{U_1 U_2 - V_1 U_2} + \Dn{V_1 U_2 - V_1 V_2} \\&= \Dn{(U_1 - V_1) U_2} + \Dn{V_1(U_2-V_2)} \\&\le \Dn{U_1 - V_1} \Dn{U_2} + \Dn{V_1} \Dn{U_2 - V_2}, \end{aligned}$$ writing compositions by juxtaposition for brevity. In the case that $U_j$ and $V_j$ are CPTP maps, we have $\Dn{U_j} = \Dn{V_j} = 1$, so that $$\begin{aligned} \D(U_1 U_2, V_1 V_2) &\le \Dn{U_1 - V_1} \Dn{U_2} + \Dn{V_1} \Dn{U_2 - V_2} \\&= \Dn{U_1 - V_1} + \Dn{U_2 - V_2} \\&= \D(U_1,V_1) + \D(U_2,V_2).\end{aligned}$$

Niel de Beaudrap
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