Premise: this is exercise 2.60 of Quantum Computation and Quantum Information, by Nielsen and Chuang, where I'm currently stuck.
Suppose $\vec{v}$ is any real three-dimensional unit vector, and $\sigma_{i}$ with $i \in \{1,2,3\}$ represents the Pauli matrices. Then we can define an Hermitian operator:
$$
\begin{aligned}
\vec{v}\cdot\vec{\sigma} &\equiv \sum_{i=1}^{3}v_{i}\sigma_{i} \\
&= v_{1} \sigma_{1} + v_{2} \sigma_{2} + v_{3} \sigma_{3}.
\end{aligned}
$$
Show that:
- $\vec{v}\cdot\vec{\sigma}$ has eigenvalues $\lambda_{\pm 1}=\pm 1$.
- the projectors onto the corresponding eigenspaces are given by $P_{\pm}=(I \pm \vec{v}\cdot\vec{\sigma}) / 2$.
My straightforward idea was to develop the observable:
$$
\begin{aligned}
\vec{v}\cdot\vec{\sigma} &\equiv v_{1} X + v_{2} Y + v_{3} Z \\
&= v_{1} \left[\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}\right] + v_{2} \left[\begin{matrix} 0 & -i \\ i & 0 \end{matrix}\right] + v_{3} \left[\begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix}\right] \\
&= \left[\begin{matrix} v_{3} & v_{1}-i v_{2} \\ v_{1}+i v_{2} & -v_{3} \end{matrix}\right]
\end{aligned}
$$
From which it is very easy to see that the eigenvalues are $\lambda_{\pm 1}=\pm 1$ and the corresponding eigenvectors are:
$$
\lvert \lambda_{-1} \rangle = \left[ \begin{array}{c} -\frac{v_{1}-i v_{2}}{1+v_{3}} \\ 1 \end{array}\right]; \quad \lvert \lambda_{+1} \rangle = \left[ \begin{array}{c} \frac{v_{1}-i v_{2}}{1-v_{3}} \\ 1 \end{array}\right]
$$
And then I would have constructed the projectors.
My question is: from the eigenvectors I can see that the first component of $\lvert \lambda_{-1} \rangle$ is going to diverge when $v_{3}=-1$, and also the first component of $\lvert \lambda_{+1} \rangle$ is going to diverge when $v_{3}=1$. I'm losing generality in the definition of the vector $\vec{v}$. This suggests my approach is wrong, and it is where I am stuck. Is there any other approach I could try to prove the given form of the projectors?
Best Answer
I'm not sure you're necessarily wrong. Because $\|\vec{v}\|=1,$ if $v_3=\pm 1,$ that forces $v_1=v_2=0,$ which is a special case: the case where $\vec{v}\cdot\vec{\sigma}=\pm\sigma_3.$ Why not treat this as a special case and do the math separately?