MHB Problem of the Week # 198 - January 12, 2016

[Ackbach]

Here is this week's POTW:

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Let
\begin{align*}
\sigma_1&=\begin{bmatrix}0&1\\1&0\end{bmatrix} \\
\sigma_2&=\begin{bmatrix}0&-i\\i&0\end{bmatrix} \\
\sigma_3&=\begin{bmatrix}1&0\\0&-1\end{bmatrix}
\end{align*}
be the three Pauli spin matrices. Let $\vec{v}$ be a real, three-dimensional unit vector, and let $\theta$ be a real number. Compute $\exp(i\theta \, \vec{v}\cdot\vec{\sigma}),$ where
$$\vec{v}\cdot\vec{\sigma}=\sum_{j=1}^3 v_j \, \sigma_j.$$

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[Ackbach]

Re: Problem Of The Week # 198 - January 12, 2016

This is Exercise 2.35 on page 75 of Quantum Computation and Quantum Information, by Nielsen and Chuang.

Congratulations to kiwi and Opalg for their correct answers. kiwi's solution follows:

Spoiler

Let \(\vec{v}=(a,b,c)\) where \(a^2+b^2+c^2=1\) because v is a unit vector.

It is straightforward to verify:
1. \(\sigma_1^2=\sigma_2^2=\sigma_3^2=I\)
2. \(\sigma_1\sigma_2=\sigma_2\sigma_3=\sigma_3\sigma_1=\sigma_2\sigma_1=\sigma_2\sigma_3=\sigma_3\sigma_1=0.I\)

Now \((\vec{v}\cdot\vec{\sigma})^2=(a\sigma_1+b\sigma_2+c\sigma_3)^2\)
\(\therefore(\vec{v}\cdot\vec{\sigma})^2=a^2\sigma_1^2+b^2\sigma_2^2+c^2\sigma_3^2+ab\sigma_1\sigma_2+ba\sigma_2\sigma_1+bc\sigma_2\sigma_3+cb\sigma_3\sigma_2+ca\sigma_3\sigma_1+ac\sigma_1\sigma_3\)
using (1) and (2)
\(\therefore(\vec{v}\cdot\vec{\sigma})^2=(a^2+b^2+c^2)I=I\)
So
3. \((\vec{v}\cdot\vec{\sigma})^{2n}=I\) for any n

Finally using a Taylor expansion:

\(e^{i\theta(\vec{v}\cdot\vec{\sigma})}=\sum_{n=0}^{\inf} \frac{(i)^n[\theta(\vec{v}\cdot\vec{\sigma})]^n}{n!}\)

\(\therefore =\sum_{n=0}^{\inf} \frac{(i)^{2n}[\theta(\vec{v}\cdot\vec{\sigma})]^{2n}}{(2n)!}+\sum_{n=0}^{\inf} \frac{(i)^{2n+1}[\theta(\vec{v}\cdot\vec{\sigma})]^{2n+1}}{(2n+1)!}\)

Using (3) we get:

\(\therefore =\sum_{n=0}^{\inf} I\frac{(-1)^{n}[\theta]^{2n}}{(2n)!}+\sum_{n=0}^{\inf} \frac{i(-1)^{n}[\theta]^{2n+1}(\vec{v}\cdot\vec{\sigma})}{(2n+1)!}\)

\(\therefore =I\sum_{n=0}^{\inf} \frac{(-1)^{n}[\theta]^{2n}}{(2n)!}+i(\vec{v}\cdot\vec{\sigma})\sum_{n=0}^{\inf} \frac{(-1)^{n}[\theta]^{2n+1}}{(2n+1)!}\)

Now recognising the Taylor expansions of cos and sin gives:

\( e^{i\theta(\vec{v}\cdot\vec{\sigma})}=I\cos(\theta)+i(\vec{v}\cdot\vec{\sigma})\sin(\theta)\)