# Inverting the depolarizing channel

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 $$(1)$$?

• 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?
– glS
Jul 30 at 23:05
• 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. Jul 31 at 1:47

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.

• 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. Jul 31 at 20:00

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 $$p=2^n + 1$$.