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I have a depolarizing channel acting on $2^n \times 2^n$ Hermitian matrices, defined as

$$\tag{1} \mathcal{D}_p (X) = p X + (1-p) \frac{\text{Tr}(X)}{2^n} \mathbb{I}_{2^n} $$

where $\mathbb{I}_{d}$ is the identity operator on $d$ dimensions. I am wondering how to derive the inverse of this map, $\mathcal{D}_p^{-1}$. I found the following formula in (Huang, 2020, eqn S37) which is easily verifiable:

$$\tag{2} \mathcal{D}_{1/(2^n+1)} (X) = (2^n + 1) X - \text{Tr}(X)\mathbb{I}_{2^n} $$

but the authors never explicitly required $\text{Tr}(X)=1$ (nor is it implied from their notation). In this case, how does one derive the inverse to $(1)$?

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    $\begingroup$ I don't quite understand why ${\rm Tr}(X)=1$ should be required for this? Do you mean because if that's the case then ${\cal D}_p(X)$ is linear in $X$ and thus the inversion trivial? $\endgroup$
    – glS
    Jul 30 at 23:05
  • $\begingroup$ yes i now realize its not required, but i was able to initially assume pure states to make the inversion very simple. When I originally came up with the question I was getting stuck on the issue that trace isn't generally invertible but I see that's not an issue now. $\endgroup$
    – forky40
    Jul 31 at 1:47
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The existence of the inverse of a linear map is independent of the way the map affects the trace. Moreover, if an invertible map preserves a property then its inverse necessarily also preserves the property. Since depolarizing channel preserves the trace, so does its inverse.

Inverse of depolarizing channel

We can derive the formula for the inverse $\mathcal{D}_p^{-1}$ of a depolarizing channel $\mathcal{D}_p$ from the observation that the depolarizing parameter $p$ is multiplicative over channel composition, i.e. $\mathcal{D}_p\circ\mathcal{D}_q = \mathcal{D}_{pq}$. This implies that $\mathcal{D}^{-1}_p = \mathcal{D}_{1/p}$.

More explicitly, depolarizing channel $\mathcal{D}_p: L(\mathcal{H}) \to L(\mathcal{H})$ has the property that for any operator $X\in L(\mathcal{H})$

$$ \begin{align} D_p(D_q(X)) &= p\left(qX + (1-q)\frac{\mathbb{I}}{d}\mathrm{tr}(X)\right) + (1-p)\frac{\mathbb{I}}{d}\mathrm{tr}(X) \\ &= pqX + (1-pq)\frac{\mathbb{I}}{d}\mathrm{tr}(X) \\ &= D_{pq}(X) \end{align} $$

where $d=\dim\mathcal{H}$. Therefore, if $p\ne 0$ then the inverse of $\mathcal{D}_p$ is $\mathcal{D}_{1/p}$ since then we have that

$$ \mathcal{D}_p \circ \mathcal{D}_{1/p} = \mathcal{D}_{1/p} \circ \mathcal{D}_p = \mathcal{D}_{p \cdot 1/p} = \mathcal{D}_1 = \mathcal{I}. $$

If $p=0$, then $\mathcal{D}_0$ is constant on density operators and therefore does not have an inverse.

CP constraint

A linear map describes a physical process without post-selection, if it is completely positive (CP) and trace preserving (TP). Depolarizing channel $\mathcal{D}_p$ is completely positive if and only if $p\in\left[-\frac{1}{d^2-1}, 1\right]$. Moreover, if $p\in\left[-\frac{1}{d^2-1}, 1\right]$ and $\frac1p\in\left[-\frac{1}{d^2-1}, 1\right]$ then $p=1$. Therefore, the only completely positive depolarizing channel with completely positive inverse is the identity channel.

TP constraint

On the other hand, $\mathcal{D}_p$ preserves the trace for every $p$ and so the inverse $\mathcal{D}_{1/p}$ also preserves the trace. This is a special case of a general fact that if an invertible function $f$ preserves a property then its inverse $f^{-1}$ necessarily also preserves the property.

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    $\begingroup$ The multiplicative property is a neat observation. This makes me believe that other channels having this property would also have a corresponding shadow tomography protocol thats simple to represent. $\endgroup$
    – forky40
    Jul 31 at 20:00
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Ah, the channel is trace preserving so its straightforward to invert in this case. Let $Y = \mathcal{D}_p (X)$ so that \begin{align} \text{Tr}(Y) &= p\text{Tr}(X) + (1-p) \frac{\text{Tr}(X)}{2^n} \text{Tr}\left(\mathbb{I}_{2^n}\right) \\&= \text{Tr}(X) \end{align}

So that \begin{align} Y &= p X + (1-p) \frac{\text{Tr}(Y)}{2^n} \mathbb{I}_{2^n} \\\Rightarrow X &= \frac{1}{p}Y - (\frac{1}{p}-1) \frac{\text{Tr}(Y)}{2^n} \mathbb{I}_{2^n} \\&= \mathcal{D}_p^{-1}(Y) \end{align} and so \begin{align} \mathcal{D}_{1/p}^{-1}(X) = pX - (p-1) \frac{\text{Tr}(X)}{2^n} \mathbb{I}_{2^n} \end{align} which recovers $(2)$ with $p=2^n + 1$.

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