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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.

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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).

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  • $\begingroup$ 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? $\endgroup$
    – IGY
    Commented Oct 7, 2022 at 1:03
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    $\begingroup$ 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 $p\ne 0$. $\endgroup$ Commented Oct 7, 2022 at 2:04
  • $\begingroup$ If $\text{Tr}\rho\neq0$, can I understand the action of this channel is first scaling the input state and then shifting the eigenvalues? $\endgroup$
    – IGY
    Commented Oct 7, 2022 at 9:11

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