# Error in repeated applications of a quantum channel?

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?

• What do you mean by "in terms of the difference between these unitaries"? Will you only accept functions of $U-V$? Commented Sep 18, 2023 at 18:47
• Or something like $(U^\dagger\rho U - V^\dagger\rho V)$ (i.e. a single application of the channel). Or potentially something similar. Commented Sep 18, 2023 at 19:03
• 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 Commented Sep 18, 2023 at 20:23
• 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. Commented Sep 18, 2023 at 21:25
• @QuantumMechanic yeah, if $U$ and $V$ are simultaneously diagonalizable, then it would make things a lot easier. Commented Sep 18, 2023 at 23:01

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*}