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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:

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

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

$$\frac{I+\sum_{l}E^{\dagger}_{l}\vec{r}\cdot\vec{\sigma}E_{l}}{2}=\frac{1}{2}\bigg[I+\sum_{l}\big(\alpha^*_lI+\sum^3_{j=1}a^*_{lj}\sigma_j\big)\big(\sum^3_{k=1}r_k\sigma_k\big)\big(\alpha_lI+\sum^3_{p=1}a_{lp}\sigma_p\big) \bigg]$$

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.

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