(a) Show that the Pauli spin matrices (Problem ) are Hermitian.

(b) Show that the Pauli spin matrices satisfy the Jacobi identity$\mathbf{[}\mathbf{\text{A}}\mathbf{,}\mathbf{[}\mathbf{\text{B}}\mathbf{,}\mathbf{\text{C}}\mathbf{\left]}\mathbf{\right]}\mathbf{+}\mathbf{[}\mathbf{\text{B}}\mathbf{,}\mathbf{[}\mathbf{\text{C}}\mathbf{,}\mathbf{\text{A}}\mathbf{\left]}\mathbf{\right]}\mathbf{+}\mathbf{[}\mathbf{\text{C}}\mathbf{,}\mathbf{[}\mathbf{\text{A}}\mathbf{,}\mathbf{\text{B}}\mathbf{\left]}\mathbf{\right]}\mathbf{=}\mathbf{0}$where$\mathbf{[}\mathbf{\text{A}}\mathbf{,}\mathbf{\text{B}}\mathbf{]}$is the commutator of A, B,[see (6.3)].

(c) Generalize (b) to prove the Jacobi identity for any (conformable) matrices A, B, C. Also see Chapter 6, Problem (3.14).

Pauli spin matrices in problem 6.6$A=\left[\begin{array}{cc}3& -1\\ -4& 2\end{array}\right]andB=\left[\begin{array}{cc}5& 2\\ -7& 3\end{array}\right]$are given below.

Jacobi Identity$[A,[B,C\left]\right]+[B,[C,A\left]\right]+[C,[A,B\left]\right]=0$

A square matrix Ais said to be Hermitian if ${\mathit{A}}^{\mathbf{*}}\mathbf{=}{\mathit{A}}^{\mathbf{T}}$, where A*is the complex conjugate and A^Tis the transpose of A. The commutator of A, Bis mathematically presented as$\mathbf{[}\mathit{A}\mathbf{,}\mathit{B}\mathbf{]}\mathbf{}\mathbf{=}\mathbf{}\mathit{A}\mathit{B}\mathbf{}\mathbf{-}\mathbf{}\mathit{B}\mathit{A}$.

a)
Consider $A=\left[\begin{array}{cc}3& -1\\ -4& 2\end{array}\right]andB=\left[\begin{array}{cc}5& 2\\ -7& 3\end{array}\right]$

These matrices do not have any complex numbers; therefore, the complex conjugate of these will be the same as before. Also, the inspections the transpose of these matrices are not equal to the complex conjugate matrices. Thus, the matrices are not Hermitian.

b)

Consider $[A,[B,C\left]\right]+[B,[C,A\left]\right]+[C,[A,B\left]\right]$.

Apply the commutative law $[A,B]=AB-BA$, such that $AB-BA=0$in the above.

Then,

role="math" localid="1664251882463" $$\begin{gathered} =[A,[B,C\left]\right]+[B,[C,A\left]\right]+[C,[A,B\left]\right] \\ =[A,(BC-CB\left)\right]+[B,(CA-AC\left)\right]+[C,(AB-BA\left)\right] \\ =[A,0]+[B,0]+[C,0] \\ =A.0-0.A+B.0-B.0+C.0-0.C \\ =0 \end{gathered}$$

Thus,$[A,[B,C\left]\right]+[B,[C,A\left]\right]+[C,[A,B\left]\right]=0$

c)

Now, take three conformable matrices A, B, and C with the same dimension, such that, $AB=BA,BC=CB,AC=CA$.

Consider $[A,[B,C\left]\right]+[B,[C,A\left]\right]+[C,[A,B\left]\right]$.

Since, $AB=BA,BC=CB,AC=CA$,

$$\begin{gathered} [B,C]=BC-CB=0 \\ [C,A]=CA-AC=0 \\ [A,B]=AB-BA=0 \end{gathered}$$

Thus, the generalization of the matrices can be mathematically presented as $[A,[B,C\left]\right]+[B,[C,A\left]\right]+[C,[A,B\left]\right]=0$.