# Depolarizing channel for $n$ qubits: why is there a trace term?

The depolarizing channel for an n-qubit quantum circuit is defined as $$\mathcal{E}(\rho) = \frac{pI}{2^n}\text{Tr}(\rho)+(1-p)\rho,\quad\text{where} \quad\rho \equiv\sum_ip_i|\psi_i\rangle\langle\psi_i|.$$ My question is: is $$\text{Tr}(\rho)$$ necessary in the definition? Since the density matrices have a trace of $$1$$, I was wondering if the term here has something to do with generalizing the definition for different input states or circuits.

• – glS
Oct 7, 2022 at 8:59

There are situations where it is profitable or convenient to think about quantum channels as defined on the space of all linear operators on a given Hilbert space, not just density matrices. In this case the $$\mathrm{Tr}(\rho)$$ factor is necessary to ensure that $$\mathcal{E}$$ is linear$$^1$$. If the input is known to have unit trace, then the $$\mathrm{Tr}(\rho)$$ factor is not necessary.

## Notational convention

Sometimes people distinguish the two situations by writing the argument as $$\rho$$ when it is a density matrix and $$X$$ when it is a general linear operator.

## Example: Constructing the Choi matrix

As an example of a situation where it is helpful to think of $$\mathcal{E}$$ as acting on all suitable linear operators, consider the computation of the Choi matrix $$J(\mathcal{E})$$ which is defined as $$J(\mathcal{E})=\sum_{ij}|i\rangle\langle j|\otimes\mathcal{E}(|i\rangle\langle j|).\tag1$$ Clearly, in the process of computing $$J(\mathcal{E})$$ we'll find ourselves appying $$\mathcal{E}$$ to operators such as $$|0\rangle\langle 1|$$ which is not unit trace. Therefore, it is useful to have a formula for $$\mathcal{E}$$ that applies to all linear operators.

## Linear extension

Note that given a description of a channel's action on density matrices there is no ambiguity about its action on other linear operators. This follows from the fact that the set of density matrices contains a basis of the space of all linear operators$$^2$$.

$$^1$$ Set $$\rho$$ to the zero operator to see that.

$$^2$$ Curiously, this is not the case in quantum mechanics over the real numbers where distinguishable channels may agree on all real density matrices (defined as symmetric matrices with unit trace).

• Thanks so much for the answer! For this example, $\text{Tr}(|0\rangle\langle1|) = 0$, and $p = 0$, so this becomes the identity channel. Is that correct?
– IGY
Oct 7, 2022 at 1:03
• Yes, $\mathrm{tr}(|0\rangle\langle 1|)=0$, but $p$ doesn't have to be zero - it's just a parameter, i.e. a part of the specification of a given channel. In fact, if you wish to see how the omission of $\mathrm{tr}(\rho)$ in the definition of the depolarizing channel leads to a function that fails to be linear you should set $p\ne 0$. Oct 7, 2022 at 2:04
• If $\text{Tr}\rho\neq0$, can I understand the action of this channel is first scaling the input state and then shifting the eigenvalues?
– IGY
Oct 7, 2022 at 9:11