For the Pauli spin matrix B in Problem 6, find${\mathbf{e}}^{\mathbf{i\theta B}}$ and show that your result is a rotation matrix. Repeat the calculation for${\mathbf{e}}^{\mathbf{-}\mathbf{i\theta B}}$ .

A rotation matrix can be defined as a transformation matrix that works with a vector to produce a rotated vector such that the coordinate axes always remain fixed. These matrices rotate the vector counter-clockwise by an angle θ. A rotation matrix is always a square matrix with real entities.

A rotation matrix is a matrix used to rotate in Euclidean space in linear algebra. For example, the matrix rotates points in the xy -Cartesian plane in a counter-clockwise direction around the Cartesian coordinate system's origin.

The given Pauli spin matrix is,

$\mathrm{B}=\left(\begin{array}{cc}0& -\mathrm{i}\\ 1& 0\end{array}\right)$ ….. (1)

Find the exponential of the matrix ${\mathrm{e}}^{-\mathrm{i\theta B}}$.

The exponential of the matrix ${\mathrm{e}}^{\mathrm{i\theta B}}$is defined by its power series as follows:

$$\begin{gathered} {\mathrm{e}}^{\mathrm{i\theta B}}=\sum _{\mathrm{n}=0}^{\infty }\frac{{\left(-\mathrm{i\theta B}\right)}^{\mathrm{n}}}{\mathrm{n}!} \\ =\mathrm{l}-\mathrm{i\theta B}+\frac{{\left(-\mathrm{i\theta B}\right)}^2}{2!}+\frac{{\left(-\mathrm{i\theta B}\right)}^3}{3!}+\frac{{\left(-\mathrm{i\theta B}\right)}^4}{4!}+\frac{{\left(-\mathrm{i\theta B}\right)}^5}{5!}+...+\frac{{\left(-\mathrm{i\theta B}\right)}^{\mathrm{n}}}{\mathrm{n}!}+... \\ =\mathrm{l}-\mathrm{i\theta B}-\frac{{\mathrm{\theta }}^2{\mathrm{B}}^2}{2!}+\mathrm{i}\frac{{\mathrm{\theta }}^3{\mathrm{B}}^3}{3!}+\frac{{\mathrm{\theta }}^4{\mathrm{B}}^4}{4!}-\mathrm{i}\frac{{\mathrm{\theta }}^5{\mathrm{B}}^5}{5!}+...+{\mathrm{i}}^{\mathrm{n}}\frac{(-1){\mathrm{\theta }}^{\mathrm{n}}{\mathrm{B}}^{\mathrm{n}}}{\mathrm{n}!}+... \\ =\left[\mathrm{l}-\frac{{\mathrm{\theta }}^2{\mathrm{B}}^2}{2!}+\frac{{\mathrm{\theta }}^4{\mathrm{B}}^4}{4!}+...\right]+\left[-\mathrm{i\theta B}+\mathrm{i}\frac{{\mathrm{\theta }}^3{\mathrm{B}}^3}{3!}-\mathrm{i}\frac{{\mathrm{\theta }}^5{\mathrm{B}}^5}{5!}+...\right] \end{gathered}$$

The matrix ${\mathrm{B}}^2$with even power is the product between square $2\times 2$matrix B and itself.

$\begin{array}{l}{\mathrm{B}}^3={\mathrm{B}}^2.\mathrm{B}\\ ={\left(\begin{array}{cc}1& 0)\\ 0& 1\end{array}\right)}_{2\times 2}.{\left(\begin{array}{cc}1& -\mathrm{i}\\ 0& 1\end{array}\right)}_{2\times 2}\\ =\mathrm{l}.\mathrm{B}\\ =\mathrm{B}\end{array}$

${\mathrm{B}}^3=\left(\begin{array}{cc}0& -\mathrm{i}\\ 1& 0\end{array}\right)$

Where, l is the square $2\times 2$identity unit matrix.

The matrix ${\mathrm{B}}^3$ with odd power is the product between square $2\times 2$ matrix ${\mathrm{B}}^3$ and ${\mathrm{B}}^{2\mathrm{n}+1}$ with even and odd powers respectively.

$\begin{array}{l}{\mathrm{B}}^{\mathrm{s}}={\mathrm{B}}^2.\mathrm{B}\\ ={\left(\begin{array}{cc}1& 0)\\ 0& 1\end{array}\right)}_{2\times 2}.{\left(\begin{array}{cc}1& 0)\\ 0& 1\end{array}\right)}_{2\times 2}\\ =\mathrm{l}.\mathrm{B}\\ =\mathrm{B}\end{array}$

${\mathrm{B}}^3=\left(\begin{array}{cc}0& -\mathrm{i}\\ 1& 0\end{array}\right)$

Find the matrices ${\mathrm{B}}^{2\mathrm{n}}$ and ${\mathrm{B}}^{2\mathrm{n}+1}$.

${\mathrm{B}}^{2\mathrm{n}}=\mathrm{l}$

Evaluate the general value of ${\mathrm{B}}^{2\mathrm{n}+1}$ for any non-negative integer.

$\begin{array}{l}{\mathrm{B}}^{2\mathrm{n}+1}={\mathrm{B}}^{2\mathrm{n}}{\mathrm{B}}^1\\ =\mathrm{l}.\mathrm{B}\\ =\mathrm{B}.\mathrm{l}\\ =\mathrm{B}\end{array}$

The exponential of the matrix ${\mathrm{e}}^{\mathrm{i\theta B}}$.

role="math" localid="1658990769412" $$\begin{gathered} {\mathrm{e}}^{\mathrm{i\theta B}}=\left[\mathrm{l}-\frac{{\mathrm{\theta }}^2{\mathrm{B}}^2}{2!}+\frac{{\mathrm{\theta }}^4{\mathrm{B}}^4}{4!}+...\right]+\left[\mathrm{i\theta B}-i\frac{{\mathrm{\theta }}^3{\mathrm{B}}^3}{3!}+\mathrm{i}\frac{{\mathrm{\theta }}^5{\mathrm{B}}^5}{5!}+...\right] \\ =\left[\mathrm{l}-\frac{{\mathrm{\theta }}^2\mathrm{l}}{2!}+\frac{{\mathrm{\theta }}^4\mathrm{l}}{4!}+...\right]+\left[\mathrm{i\theta B}-i\frac{{\mathrm{\theta }}^3\mathrm{B}}{3!}+\mathrm{i}\frac{{\mathrm{\theta }}^5\mathrm{B}}{5!}+...\right] \\ =\left[1-\frac{{\mathrm{\theta }}^2}{2!}+\frac{{\mathrm{\theta }}^4}{4!}+...\right]+\mathrm{iB}\left[\mathrm{\theta }-\mathrm{i}\frac{{\mathrm{\theta }}^3}{3!}+\mathrm{i}\frac{{\mathrm{\theta }}^5}{5!}+...\right] \\ {\mathrm{e}}^{\mathrm{i\theta B}}=\mathrm{l}\sum _{\mathrm{n}=0}^{\infty }\frac{{\left(-1\right)}^{\mathrm{n}}{\mathrm{\theta }}^{2\mathrm{n}}}{\left(2\mathrm{n}\right)!}+\mathrm{iB}\sum _{\mathrm{n}=0}^{\infty }\frac{{\left(-1\right)}^{\mathrm{n}}{\mathrm{\theta }}^{2\mathrm{n}+1}}{\left(2\mathrm{n}+1\right)!}....\left(2\right) \end{gathered}$$

As

${\mathrm{e}}^{\mathrm{i\theta B}}=\mathrm{l}\mathrm{cos\theta }+\mathrm{iB}\mathrm{sin\theta }$ ….. (3)

Compare equation (2) and (3), and you have,

$$\begin{gathered} \cos \theta =\sum _{\mathrm{n}=0}^{\infty }\frac{{\left(-1\right)}^{\mathrm{n}}{\mathrm{\theta }}^{2\mathrm{n}}}{\left(2\mathrm{n}\right)!} \\ =1-\frac{{\mathrm{\theta }}^2}{2!}+\frac{{\mathrm{\theta }}^4}{4!}+...+\frac{{\left(-1\right)}^{\mathrm{n}}{\mathrm{\theta }}^{2\mathrm{n}+1}}{\left(2\mathrm{n}+1\right)!}+... \end{gathered}$$

And

$$\begin{gathered} \mathrm{sin\theta }=\sum _{\mathrm{n}=0}^{\infty }\frac{{\left(-1\right)}^{\mathrm{n}}{\mathrm{\theta }}^{2\mathrm{n}+1}}{\left(2\mathrm{n}+1\right)!} \\ =\mathrm{\theta }-\frac{{\mathrm{\theta }}^3}{3!}+\frac{{\mathrm{\theta }}^5}{5!}+...+\frac{{\left(-1\right)}^{\mathrm{n}}{\mathrm{\theta }}^{2\mathrm{n}+1}}{\left(2\mathrm{n}+1\right)!}+... \end{gathered}$$

As known that,

$\mathrm{l}=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$

Therefore,

$$\begin{gathered} \mathrm{iB}=\mathrm{i}\left(\begin{array}{cc}0& -\mathrm{i}\\ \mathrm{i}& 0\end{array}\right) \\ =\left(\begin{array}{cc}0& -{\mathrm{i}}^2\\ {\mathrm{i}}^2& 0\end{array}\right) \\ =\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right) \end{gathered}$$

Rewrite equation (3) and substitute $\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$for l , and you obtain

$$\begin{gathered} {\mathrm{e}}^{\mathrm{i\theta B}}=\mathrm{Icos\theta }-\mathrm{iBsin\theta } \\ =\mathrm{cos\theta }\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)-\mathrm{sin\theta }\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right) \\ =\left(\begin{array}{cc}\mathrm{cos\theta }& 0\\ 0& \mathrm{cos\theta }\end{array}\right)+\left(\begin{array}{cc}0& -\mathrm{sin\theta }\\ \mathrm{sin\theta }& 0\end{array}\right) \\ =\left(\begin{array}{cc}\mathrm{cos\theta }& -\mathrm{sin\theta }\\ \mathrm{sin\theta }& \mathrm{cos\theta }\end{array}\right) \end{gathered}$$

Find the exponential of the matrix ${\mathrm{e}}^{\mathrm{i\theta B}}$.

$$\begin{gathered} {\mathrm{e}}^{\mathrm{i\theta B}}=\mathrm{Icos\theta }+\mathrm{iBsin\theta } \\ =\mathrm{cos\theta }\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)+\sin \left(\mathrm{\theta }\right)\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right) \\ =\left(\begin{array}{cc}\mathrm{cos\theta }& 0\\ 0& \mathrm{cos\theta }\end{array}\right)+\left(\begin{array}{cc}0& \mathrm{sin\theta }\\ -\mathrm{sin\theta }& 0\end{array}\right) \\ =\left(\begin{array}{cc}\mathrm{cos\theta }& \mathrm{sin\theta }\\ -\mathrm{sin\theta }& \mathrm{cos\theta }\end{array}\right) \end{gathered}$$

The above matrix represents a rotation matrix.

The computation of the exponential of the matrix ${\mathrm{e}}^{\mathrm{i\theta B}}$is defined by its power.

localid="1658991881445" $$\begin{gathered} {\mathrm{e}}^{\mathrm{i\theta B}}=\sum _{\mathrm{n}=0}^{\infty }\frac{\left(-\mathrm{i\theta B}\right)}{\mathrm{n}!} \\ =\mathrm{l}-\mathrm{i\theta B}+\frac{{\left(-\mathrm{i\theta B}\right)}^2}{2!}+\frac{{\left(-\mathrm{i\theta B}\right)}^3}{3!}+\frac{{\left(-\mathrm{i\theta B}\right)}^4}{4!}+\frac{{\left(-\mathrm{i\theta B}\right)}^5}{5!}+...+\frac{{\left(-\mathrm{i\theta B}\right)}^{\mathrm{n}}}{\mathrm{n}!}+... \\ =\mathrm{l}-\mathrm{i\theta B}-\frac{{\mathrm{\theta }}^2{\mathrm{B}}^2}{2!}+\mathrm{i}\frac{{\mathrm{\theta }}^3{\mathrm{B}}^3}{3!}+\frac{{\mathrm{\theta }}^4{\mathrm{B}}^4}{4!}-\mathrm{i}\frac{{\mathrm{\theta }}^5{\mathrm{B}}^5}{5!}+...+{\mathrm{i}}^{\mathrm{n}}\frac{(-1){\mathrm{\theta }}^{\mathrm{n}}{\mathrm{B}}^{\mathrm{n}}}{\mathrm{n}!}+... \\ =\left[\mathrm{l}-\frac{{\mathrm{\theta }}^2{\mathrm{B}}^2}{2!}+\frac{{\mathrm{\theta }}^4{\mathrm{B}}^4}{4!}+...\right]+\left[-\mathrm{i\theta B}+\mathrm{i}\frac{{\mathrm{\theta }}^3{\mathrm{B}}^3}{3!}-\mathrm{i}\frac{{\mathrm{\theta }}^5{\mathrm{B}}^5}{5!}+...\right] \end{gathered}$$

Find the exponential of the matrix ${\mathrm{e}}^{-\mathrm{i\theta B}}$.

$$\begin{gathered} {\mathrm{e}}^{-\mathrm{i\theta B}}=\left[\mathrm{l}-\frac{{\mathrm{\theta }}^2{\mathrm{B}}^2}{2!}+\frac{{\mathrm{\theta }}^4{\mathrm{B}}^4}{4!}+...\right]+\left[-\mathrm{i\theta B}+\mathrm{i}\frac{{\mathrm{\theta }}^3{\mathrm{B}}^3}{3!}-\mathrm{i}\frac{{\mathrm{\theta }}^5{\mathrm{B}}^5}{5!}+...\right] \\ =\left[\mathrm{l}-\frac{{\mathrm{\theta }}^2\mathrm{l}}{2!}+\frac{{\mathrm{\theta }}^4\mathrm{l}}{4!}+...\right]+\left[-\mathrm{i\theta B}+\mathrm{i}\frac{{\mathrm{\theta }}^3\mathrm{B}}{3!}-\mathrm{i}\frac{{\mathrm{\theta }}^5\mathrm{B}}{5!}+...\right] \\ =\left[1-\frac{{\mathrm{\theta }}^2}{2!}+\frac{{\mathrm{\theta }}^4}{4!}+...\right]-\mathrm{iB}\left[\mathrm{\theta }-\mathrm{i}\frac{{\mathrm{\theta }}^3}{3!}+\mathrm{i}\frac{{\mathrm{\theta }}^5}{5!}+...\right] \\ =\mathrm{l}\sum _{\mathrm{n}=0}^{\infty }\frac{{\left(-1\right)}^{\mathrm{n}}{\mathrm{\theta }}^{2\mathrm{n}}}{\left(2\mathrm{n}\right)!}-\mathrm{iB}\sum _{\mathrm{n}=0}^{\infty }\frac{{\left(-1\right)}^{\mathrm{n}}{\mathrm{\theta }}^{2\mathrm{n}+1}}{\left(2\mathrm{n}+1\right)!} \end{gathered}$$

Solve further.

${\mathrm{e}}^{-\mathrm{i\theta B}}=\mathrm{lcos\theta }-\mathrm{iBsin\theta }$

Deduce that the exponential of the matrix is rotation matrix.

$$\begin{gathered} {\mathrm{e}}^{\mathrm{i\theta B}}=\mathrm{Icos\theta }-\mathrm{iBsin\theta } \\ =\mathrm{cos\theta }\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)-\mathrm{sin\theta }\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right) \\ =\left(\begin{array}{cc}\mathrm{cos\theta }& 0\\ 0& \mathrm{cos\theta }\end{array}\right)+\left(\begin{array}{cc}0& -\mathrm{sin\theta }\\ \mathrm{sin\theta }& 0\end{array}\right) \\ =\left(\begin{array}{cc}\mathrm{cos\theta }& -\mathrm{sin\theta }\\ \mathrm{sin\theta }& \mathrm{cos\theta }\end{array}\right) \end{gathered}$$

Hence, proved that the exponential of the matrix is localid="1658991824922" ${\mathrm{e}}^{-\mathrm{i\theta B}}=\left(\begin{array}{cc}\mathrm{cos\theta }& -\mathrm{sin\theta }\\ \mathrm{sin\theta }& \mathrm{cos\theta }\end{array}\right)$which represent a rotation matrix.