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Suppose I have two quantum channels. Assume they they consist of $r\in \mathbb{Z}$ applications of unitaries, $U$ and $V$ respectively. Let the error between the channels acting on some state $\rho$ be: $$\varepsilon(U,V) = U^{\dagger r} \rho U^r - V^{\dagger r} \rho V^r. $$

Ideally, I want an expression for $\epsilon(U,V)$ in terms of the difference between these unitaries and $r$. Ideally I want an expression for $\varepsilon(U,V)$ as defined above, not the norm of the expression.

My first attempt was to write that $U^r - V^r = \sum_{k=1}^{r-1} U^{r-k}(U-V)V^k$ and substitute into the expression for $\varepsilon(U,V)$. However, this introduces a lot of cross terms which make things messy, and one ends up with $O(r^2)$ many terms when it feels like there should only be $O(r)$. Is there a better way to do this?

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  • $\begingroup$ What do you mean by "in terms of the difference between these unitaries"? Will you only accept functions of $U-V$? $\endgroup$ Commented Sep 18, 2023 at 18:47
  • $\begingroup$ Or something like $(U^\dagger\rho U - V^\dagger\rho V)$ (i.e. a single application of the channel). Or potentially something similar. $\endgroup$ Commented Sep 18, 2023 at 19:03
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    $\begingroup$ Hm. I was thinking along the lines of $\epsilon(U,V)=U^{\dagger r}[\rho - W \rho W^\dagger ]U^r$ for $W=U^r V^{\dagger r}$, and finding $W$ in terms of $U V^\dagger$, but even that requires lots and lots of arbitrary commutators $\endgroup$ Commented Sep 18, 2023 at 20:23
  • $\begingroup$ To be honest, I think I'd be happy with anything which is linear in $r$ and some characterisation of the error between a single application of the separate unitaries. $\endgroup$ Commented Sep 18, 2023 at 21:25
  • $\begingroup$ @QuantumMechanic yeah, if $U$ and $V$ are simultaneously diagonalizable, then it would make things a lot easier. $\endgroup$
    – FDGod
    Commented Sep 18, 2023 at 23:01

1 Answer 1

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The following should meet OP's expectations for a suitable expansion as specificed in their comment: Defining the difference channel $\Phi_{U,V}(\rho):=U^*\rho U-V^*\rho V$ one can indeed express $\varepsilon(U,V)$ as a sum of $r$ terms involving just $\Phi_{U,V}$ (and, of course, the original unitaries $U,V$): $$ \boxed{\varepsilon(U,V)=\sum_{j=0}^{r-1}(U^*)^{r-j-1} \Phi_{U,V}\big( (V^*)^{j}\rho V^{j} \big) U^{r-j-1}}\tag1 $$ While this holds for arbitrary $U,V$ and all $\rho$, in the special case where $[U,V]=0$ Eq.(1) simplifies to $$ \varepsilon(U,V)=\sum_{j=0}^{r-1}(U^*)^{r-j-1} (V^*)^{j}\Phi_{U,V}( \rho ) V^{j}U^{r-j-1}\tag2 $$ so there the only "error information" one needs is $\Phi_{U,V}( \rho )=U^*\rho U-V^*\rho V$; for the general case, however, I don't see a way around evaluating the error not only on $\rho$ but also on $(V^*)^{j}\rho V^{j}$ for all $j$. In other words I think this is the best one can do given the desired constraints on the expansion.

Proof of Eq.(1): \begin{align*} \sum_{j=0}^{r-1}&(U^*)^{r-j-1} \Phi_{U,V}\big( (V^*)^{j}\rho V^{j} \big) U^{r-j-1}\\ &=\sum_{j=0}^{r-1}(U^*)^{r-j-1} U^* (V^*)^{j}\rho V^{j} U U^{r-j-1}-\sum_{j=0}^{r-1}(U^*)^{r-j-1} V^*(V^*)^{j}\rho V^{j} V U^{r-j-1}\\ &=\sum_{j=0}^{r-1}(U^*)^{r-j} (V^*)^{j}\rho V^{j} U^{r-j}-\sum_{j=0}^{r-1}(U^*)^{r-j-1} (V^*)^{j+1}\rho V^{j+1}U^{r-j-1}\\ &=\sum_{j=0}^{r-1}(U^*)^{r-j} (V^*)^{j}\rho V^{j} U^{r-j}-\sum_{j=1}^{r}(U^*)^{r-j} (V^*)^{j}\rho V^{j}U^{r-j}\\ &=(U^*)^{r-j} (V^*)^{j}\rho V^{j} U^{r-j}\Big|_{j=0}-(U^*)^{r-j} (V^*)^{j}\rho V^{j}U^{r-j}\Big|_{j=r}\\ &=(U^*)^r\rho U^r-(V^*)^r\rho V^r=\varepsilon(U,V)\tag*{$\square$} \end{align*}

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