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The fully depolarizing quantum channel in a $d$ dimensional Hilbert space is defined by $$ \mathcal N^D (\rho) = \text{Tr}[\rho]\frac I d $$ I've seen that definition in several places but I don't understand what the trace is doing in there. The trace of a density operator is by definition one. So why can it be a different value here and how is this important?

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Quantum channels are linear maps, that is, $\mathcal E(\alpha \rho+\beta\sigma)=\alpha\mathcal E(\rho)+\beta\mathcal E(\sigma)$. This is only satisfied if you include the normalization $\mathrm{tr}(\rho)/d$ in the definition of the depolarizing channel.

(Of course, as said by @glS, if you know that you will only every apply $\mathcal E$ to normalized states, you don't need to take this factor along. However, it is still advisable to do so -- otherwise, you might end up forgetting this normalization assumption at some point and get a wrong result.)

To illustrate a case where this is necessary, imagine computing the Choi state of $\mathcal E$, this is, applying $\mathcal E$ to a maximally entangled state: This is for instance done by computing

$$(\mathcal E\otimes 1\!\!1)(\sum|i\rangle\langle j|\otimes |i\rangle\langle j|) = \sum\mathcal E(|i\rangle\langle j|)\otimes |i\rangle\langle j|\ ,$$ and $\mathcal E(|i\rangle\langle j|)$ requires this normalization.

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  • $\begingroup$ This makes a lot of sense and I should have remembered that normalization is something you actually have to decide to do. I was asking the question specifically because I have encountered a case were that trace was still there and important - clearly because it wasn't automatically normalized. $\endgroup$ – S. Move Feb 14 at 17:53
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Indeed, it's superfluous if you consider the action of that map of quantum states.

For example, in Nielsen and Chuang (page 379 of the tenth-anniversary edition) the definition of depolarizing channel is given as $$\mathcal E(\rho)=\frac{p I}{d} + (1-p)\rho,$$ which becomes $\mathcal E(\rho)=I/d$ for $p=1$.

Note that it's not strictly necessary to define states as having trace 1, just as it's not strictly needed to require ket states to be normalized. If you don't impose normalization on the states, however, you have to change everything else accordingly.

You can also decide to use a convention in which for whatever reason states are defined as having a fixed trace that is different than one. Again, this will have consequences in the way all the other quantities are computed, and you have here an example of how the formula for the depolarizing channel would change.

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