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I was reading up on the quantum depolarizing channel (Preskill's Notes) (stack exchaange explanation), and I'm failing to see how the form

\begin{align} \sigma &= (\mathcal E \otimes \mathbb I)(|\Omega\rangle\langle \Omega|) = \sum_{ij} \mathcal E(|i\rangle\langle j|)\otimes |i\rangle\langle j|\ \end{align}

gets simplified to:

$$ \sigma = \frac{p}{2D}\mathbb I\otimes \sum_{i}|i\rangle\langle i| + (1-p)\frac1D \sum_{ij}|i\rangle\langle j|\otimes |i\rangle\langle j|\ . $$

where then the first term becomes the identity. When I solve the same channel representation, I find the matrix looks like:

$$ \sigma = \frac{p}{2D}\mathbb I\otimes \sum_{ij}|i\rangle\langle j| + (1-p)\frac1D \sum_{ij}|i\rangle\langle j|\otimes |i\rangle\langle j|\ . $$

Where the first term isn't able to simplify into something as simple as the identity, but is of the form $$I_{2} + \sigma^{x}_{2}$$ where the subscript denotes that the operation is on the second qubit.

I can't seem to pinpoint the mistake in my derivation here, so any pointers are much appreciated.

Apple Meson
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1 Answers1

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For $\mathcal E(X)=p\mathrm{tr}(X)\,\frac{\mathbb I}{2}+(1-p)X$, then

$$\begin{align} \sigma &= (\mathcal E \otimes \mathbb I)(|\Omega\rangle\langle \Omega|)\\ &= \sum_{ij} \mathcal E(|i\rangle\langle j|)\otimes |i\rangle\langle j|\ \\ &=\sum_{ij}\left[p\mathrm{tr}(|i\rangle\langle j|)\,\frac{\mathbb I}{2}+(1-p)|i\rangle\langle j|\right] \otimes |i\rangle\langle j| \\ &= \sum_{ij}\left[p\langle j|i\rangle\,\frac{\mathbb I}{2}+(1-p)|i\rangle\langle j|\right] \otimes |i\rangle\langle j| \\ &= \sum_{ij}p\delta_{ij}\frac{\mathbb I}{2} \otimes |i\rangle\langle j| +(1-p)|i\rangle\langle j| \otimes |i\rangle\langle j| \\ &=p\sum_{i}\frac{\mathbb I}{2} \otimes |i\rangle\langle i| + \sum_{ij}(1-p)|i\rangle\langle j| \otimes |i\rangle\langle j| \end{align}$$

The term $\mathrm{tr}(|i\rangle\langle j|)$ is missing in your calculation I think.

HerrWarum
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