# Show that quantum channels act as affine transformations in the Bloch sphere

I am referring to Equation (8.89) to (8.92) in Chapter 8 of "Quantum Computing and Information 10th Anniversary Edition" by Nielsen and Chuang. This section deals with the geometric picture of single qubit quantum operations on a Bloch sphere. I am trying to arrive at (8.91) and (8.92) which the authors claimed that "...it is not difficult to check..."

We have the state of a single qubit in Bloch representation:

$$\rho=\frac{I+\vec{r}\cdot\vec{\sigma}}{2}\hspace{5em}(8.87)$$

The affine map $$\varepsilon$$ is given as:

$$\vec{r}\longrightarrow\vec{r'}=M\space\vec{r}+\vec{c}\hspace{5em}(8.89)$$

The operators that generate the operator-sum-representation for $$\varepsilon$$ are given in the form:

$$E_{i}=\alpha_{i}I+\sum^{3}_{k=1}a_{ik}\sigma_{k}\hspace{5em}(8.90)$$

The text went on to say it is not difficult to check that:

$$M_{jk}=\sum_{l}\bigg[a_{lj}a^{*}_{lk}+a^{*}_{lj}a_{lk}+\big(|a_l|^2-\sum_pa_{lp}a^*_{lp}\big)+i\sum_p\epsilon_{jkp}(\alpha_la^*_{lp}-\alpha^*_la_{lp})\bigg]\hspace{1em}$$

$$c_k=2i\sum_l\sum_{jp}\epsilon_{jkp}a_{lj}a^*_{lp}\hspace{5em}\text{(8.91) and (8.92)}$$

and they claim $$c_k$$ was written as such because $$\sum_{i}E^{\dagger}_{i}E_{i}=I$$ is used to simplify it.

So I tried to see how I can somehow arrive at (8.91) and (8.92) by doing these:

$$\varepsilon(\rho)=\frac{I+\sum_{l}E_{l}\vec{r}\cdot\vec{\sigma}E^{\dagger}_{l}}{2}=\frac{I+(M\space\vec{r}+\vec{c})\cdot\vec{\sigma} }{2}$$

The second term in the above equality is what I tried to work out,

\begin{align} \mathcal{E}&(\rho)=\sum_l E_l\rho E_l^\dagger=\frac{I+\sum_l E_l(\vec{r}.\vec{\sigma})E_l^\dagger}{2}\\ &=\frac{1}{2}\bigg[I+\sum_l\Big[ \Big(\alpha_lI+\sum_{j=1}^3a_{lj}\sigma_j\Big)\Big(\sum_{k=1}^3r_k\sigma_k\Big)\Big(\alpha_l^*I+\sum_{p=1}^3a^*_{lp}\sigma_p\Big) \Big]\bigg]\\ &=\frac{1}{2}\bigg[I+\sum_l\Big[ |\alpha_l|^2\sum_{k=1}^3r_k\sigma_k+\alpha_l\sum_{k=1}^3r_k\sigma_k\sum_{p=1}^3a_{lp}^*\sigma_p+\alpha_l^*\sum_{j=1}^3a_{lj}\sigma_j \sum_{k=1}^3r_k\sigma_k+\sum_{j=1}^3a_{lj}\sigma_j\sum_{k=1}^3r_k\sigma_k\sum_{p=1}^3a_{lp}^*\sigma_p \Big]\bigg]\\ &=\frac{1}{2}\bigg[I+\sum_l\Big[ |\alpha_l|^2\sum_{k=1}^3r_k\sigma_k+\alpha_l\sum_{k=1}^3\sum_{p=1}^3r_ka_{lp}^*\sigma_k\sigma_p+\alpha_l^*\sum_{j=1}^3\sum_{k=1}^3r_ka_{lj}\sigma_j \sigma_k+\sum_{j=1}^3\sum_{k=1}^3\sum_{p=1}^3r_ka_{lj}a_{lp}^*\sigma_j\sigma_k\sigma_p \Big]\bigg]\\ \end{align}

Before I attempt to write out the subsequent expansion in latex, I'll like to know am I on the right track? Because after expansion I get the terms with $$r_k$$ and we know $$c_{k}$$ is a constant. Can someone guide me on how to understand (8.91) and (8.92)? Or even better, show me on how to arrive at these two equations through any means. Help will be very much appreciated. Thanks.

There is already mistake in the very first line of your computation: When you write $$\mathcal{E}(\rho)=[\ldots]=\frac12(I+\sum_l E_l(\vec{r}.\vec{\sigma})E_l^\dagger)$$ you forgot to apply the channel to the first summand, as well: while the completeness relation $$\sum_l E_l^\dagger E_l=I$$ holds for every trace-preserving map, $$\sum_l E_lE_l^\dagger=I$$ only holds for unital maps, i.e. those channels which map the identity to itself. In the Bloch picture the latter would correspond to the special case $$\vec c=0$$ meaning if you proceeded with your approach then you would only get the expression for $$M$$.
With this in mind let me give a few hints on how the computation works and what the main steps are. First, as you did as well, one expands the operator sum form using (8.87) as well as (8.90) \begin{align*} \sum_lE_l\rho E_l^\dagger=\sum_l\Big( \alpha_l I+\sum_k a_{lk}\sigma_k \Big)\Big( \frac12I+\sum_p \frac{r_p}2\sigma_p \Big)\Big( \alpha_l^* I+\sum_j a_{lj}^*\sigma_j \Big) \end{align*} This gives rise to two types of terms: some which feature $$r_p$$ and some without $$r_p$$ appearing. Comparing this to the desired form $$M\vec r+\vec c$$ shows that all the information about $$M$$ is contained in the expression $$\sum_l( \alpha_l I+\sum_k a_{lk}\sigma_k )(\sum_p \frac{r_p}2\sigma_p )( \alpha_l^* I+\sum_j a_{lj}^*\sigma_j )$$ and everything about $$\vec c$$ is contained in the remaning term $$\frac12\sum_l( \alpha_l I+\sum_k a_{lk}\sigma_k )( \alpha_l^* I+\sum_j a_{lj}^*\sigma_j )$$. The final insight here is that---because $$M$$ describes how Pauli components are mapped to Pauli components---the entries of $$M$$ are obtained by starting from $$\rho$$ being a Pauli matrix and computing the overlap (trace) with "the corresponding other Pauli". More precisely, $$M_{rs}=\frac12{\rm tr}\Big(\sigma_r \sum_l\Big( \alpha_l I+\sum_k a_{lk}\sigma_k \Big) \sigma_s\Big( \alpha_l^* I+\sum_j a_{lj}^*\sigma_j \Big)\Big)$$ from which it does not take much effort to get to (8.91) (this time for real!). Similarly, the components of $$c$$ are obtained via $$c_r=\frac12{\rm tr}\Big(\sigma_r \sum_l\Big( \alpha_l I+\sum_k a_{lk}\sigma_k \Big)\Big( \alpha_l^* I+\sum_j a_{lj}^*\sigma_j \Big)\Big)\,;$$ two of the four emerging terms $$\sum_{j,l}\frac12{\rm tr}(\sigma_r \alpha_l a_{lj}^*\sigma_j )=\sum_{l}\alpha_l a_{lr}^*$$ and $$\sum_{k,l}\frac12{\rm tr}(\sigma_r a_{lk}\sigma_k \alpha_l^* )=\sum_{l}a_{lk}\alpha_l^*$$ can be taken care of via the completeness relation $$\sum_iE_i^\dagger E_i$$. With this one, eventually, gets out (8.92) as desired.
• Without computing the matrix elements for $M$ as you suggest and instead making the direct expansion, among the terms there are two which read $\alpha_i (\vec{r}\cdot\vec{\sigma})(\vec{a}_i^*\cdot\vec{\sigma})$ and $\alpha_i^* (\vec{a}_i \cdot\vec{\sigma})(\vec{r}\cdot\vec{\sigma})$. Expanding for instance the first term will yield $\alpha_i (\vec{r} \cdot \vec{a}_i^* \mathbb{1} + i (\vec{r} \times \vec{a}_i^*)\cdot\vec{\sigma})$. However, there are no terms like $a_k^*$ in the book's expansion for $M$ (ok there are some, but they involve the $i$ of the cross-product). Apr 17 at 16:28