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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$\begingroup$ related: quantumcomputing.stackexchange.com/q/28104/55, physics.stackexchange.com/q/460762/58382 $\endgroup$– glS ♦Oct 7, 2022 at 8:59
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1 Answer
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$^1$ Set $\rho$ to the zero operator to see that.
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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).
answered Oct 6, 2022 at 23:38
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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? $\endgroup$– IGYOct 7, 2022 at 1:03
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1$\begingroup$ 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 $p\ne 0$. $\endgroup$ Oct 7, 2022 at 2:04
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$\begingroup$ If $\text{Tr}\rho\neq0$, can I understand the action of this channel is first scaling the input state and then shifting the eigenvalues? $\endgroup$– IGYOct 7, 2022 at 9:11