For the Pauli spin matrix Ain Problem 6 , find the matricessin(kA) ,cos(kA) , ${\mathbf{e}}^{\mathbf{kA}}\mathbf{,}{\mathbf{e}}^{\mathbf{ikA}}\mathbf{,}$where $\mathbf{=}\sqrt{\mathbf{-}\mathbf{1}}$.

Trigonometry is dealing with specific functions of angles and their application in calculations. There are six angle functions commonly used in trigonometry. Their names and abbreviations are sine (sin), cosine (cos), tangent (tan), cotangent (cot), secant (sec), and cosecant (CSC).

In solutions of second-order systems of differential equations, the trigonometric functions (particularly sine and cosine) for real or complex square matrices appear. They are defined by the same Taylor series that governs real and complex trigonometric functions.

The given matrix is

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

Calculate a couple of powers of the matrix A .

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

Solve further calculate the matrix for ${\mathrm{A}}^3$.

$$\begin{gathered} {\mathrm{A}}^3=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right) \\ =\mathrm{A}=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right) \\ {\mathrm{A}}^3=\mathrm{A} \end{gathered}$$

Solve further calculate the matrix for ${\mathrm{A}}^4$.

$$\begin{gathered} {\mathrm{A}}^4=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right) \\ =\mathrm{A}=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right) \\ {\mathrm{A}}^4=\mathrm{I} \end{gathered}$$

And so on.

Find the value of sin kA.

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

Substitute the value of sin k for localid="1658990100959" $\mathrm{k}-\frac{{\mathrm{k}}^3}{3!}+\frac{{\mathrm{k}}^5}{5!}......$

$$\begin{gathered} \sin \mathrm{kA}=\mathrm{A}\sin \mathrm{k} \\ =\left(\begin{array}{cc}0& \sin \mathrm{k}\\ \sin \mathrm{k}& 0\end{array}\right) \end{gathered}$$

Solve for cos kA .

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

Substitute the value of 1 for $1-\frac{{\mathrm{k}}^2}{2!}+\frac{{\mathrm{k}}^4}{4!}......$

$$\begin{gathered} \cos \mathrm{kA}=\mathrm{Icos}\mathrm{k} \\ =\left(\begin{array}{cc}\cos \mathrm{k}& 0\\ 0& \cos \mathrm{k}\end{array}\right) \end{gathered}$$

Find the value of ${\mathrm{e}}^{\mathrm{kA}}$.

$$\begin{gathered} {\mathrm{e}}^{\mathrm{kA}}=\sum _{\mathrm{n}=0}^{\infty }\frac{{\mathrm{k}}^{\mathrm{n}}}{\mathrm{n}!}{\mathrm{A}}^{\mathrm{n}} \\ =\mathrm{I}+\mathrm{kA}+\frac{{\mathrm{k}}^2}{2!}{\mathrm{A}}^2+\frac{{\mathrm{k}}^3}{3!}{\mathrm{A}}^3 \\ {\mathrm{e}}^{\mathrm{kA}}=\mathrm{I}\left(1+\frac{{\mathrm{k}}^2}{2!}+\frac{{\mathrm{k}}^4}{4!}....\right)+\mathrm{A}\left(\mathrm{k}+\frac{{\mathrm{k}}^3}{3!}+\frac{{\mathrm{k}}^5}{5!}....\right) \end{gathered}$$

(1)

Follow the Taylor’s expansions of the hyperbolic functions.

role="math" localid="1658990421530" $$\begin{gathered} \sinh \mathrm{k}=\sum _{\mathrm{n}=1}^{\infty }\frac{{\mathrm{k}}^{2\mathrm{n}+1}}{\left(2\mathrm{n}+1\right)!} \\ \cosh \mathrm{k}=\sum _{\mathrm{n}=1}^{\infty }\frac{{\mathrm{k}}^{2\mathrm{n}}}{\left(2\mathrm{n}\right)!} \end{gathered}$$

Substitute $\left(1+\frac{{\mathrm{k}}^2}{2!}+\frac{{\mathrm{k}}^4}{4!}....\right)$ as cosh k and $\left(\mathrm{k}+\frac{{\mathrm{k}}^3}{3!}+\frac{{\mathrm{k}}^5}{5!}....\right)$as sinh k into equation (1).

$$\begin{gathered} {\mathrm{e}}^{\mathrm{kA}}=\mathrm{I}\cosh \mathrm{k}+\mathrm{A}\sinh \mathrm{k} \\ =\left(\begin{array}{cc}\cosh \mathrm{k}& 0\\ 0& \cosh \mathrm{k}\end{array}\right)+\left(\begin{array}{cc}0& \sinh \mathrm{k}\\ \sinh \mathrm{k}& 0\end{array}\right) \\ =\left(\begin{array}{cc}\cosh \mathrm{k}& \sinh \mathrm{k}\\ \sinh \mathrm{k}& \cosh \mathrm{k}\end{array}\right) \end{gathered}$$

Similarly, solve for ${\mathrm{e}}^{\mathrm{ikA}}$.

role="math" localid="1658990968005" $$\begin{gathered} {\mathrm{e}}^{\mathrm{ikA}}=\sum _{\mathrm{n}=0}^{\infty }\frac{\left(\mathrm{ik}\right)}{\mathrm{n}!}{\mathrm{A}}^{\mathrm{n}} \\ =\mathrm{I}+\mathrm{ikA}-\frac{{\mathrm{k}}^2}{2!}{\mathrm{A}}^2-\frac{{\mathrm{i}}^3{\mathrm{k}}^3}{3!}{\mathrm{A}}^3+\frac{{\mathrm{k}}^4}{4!}{\mathrm{A}}^4+\mathrm{i}\frac{{\mathrm{k}}^5}{5!}{\mathrm{A}}^5.... \\ {\mathrm{e}}^{\mathrm{ikA}}=\mathrm{I}\left(1-\frac{{\mathrm{k}}^2}{2!}+\frac{{\mathrm{k}}^4}{4!}....\right)+\mathrm{iA}\left(\mathrm{k}-\frac{{\mathrm{k}}^3}{3!}+\frac{{\mathrm{k}}^5}{5!}....\right) \end{gathered}$$

(2)

Substitute role="math" localid="1658990976808" $\left(1-\frac{{\mathrm{k}}^2}{2!}+\frac{{\mathrm{k}}^4}{4!}....\right)$as cos k and $\left(\mathrm{k}-\frac{{\mathrm{k}}^3}{3!}+\frac{{\mathrm{k}}^5}{5!}....\right)$as sin k into equation (2).

$$\begin{gathered} {\mathrm{e}}^{\mathrm{ikA}}=\mathrm{Icosk}+\mathrm{iA}\sin \mathrm{k} \\ =\cos \mathrm{k}\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)+\mathrm{i}\sin \mathrm{k}\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right) \\ =\left(\begin{array}{cc}\cos \mathrm{k}& \mathrm{i}\sin \mathrm{k}\\ \mathrm{i}\sin \mathrm{k}& \cos \mathrm{k}\end{array}\right) \end{gathered}$$

Hence, the values of the desired matrices have been calculated.