# solutions to 2.7

Exercise: find the product of rotations by $$\frac{\pi}{2}$$ around $$x, y$$ and then $$z$$ axes.
Solution: $$\newcommand{Rot}[1]{\overset{R_#1}{\longrightarrow}}$$ Let $$R_x,R_y$$ and $$R_z$$ represent the rotation by $$\frac{\pi}{2}$$ matrices about the $$x,y$$ and $$z$$ axes respectively, as defined in the lectures notes: $$R_x = \left[\begin{array}{c|cc} 1 & 0 & 0\\ \hline 0 & 0 & 1 \\ 0 & -1 & 0 \end{array}\right],~~~ R_y = \left[\begin{array}{c|c|c} 0 & 0 & -1 \\ \hline 0 & 1 & 0 \\ \hline 1 & 0& 0 \end{array}\right],~~~ R_z = \left[\begin{array}{cc|c} 0 & 1 & 0 \\ -1 & 0 & 0 \\ \hline 0 & 0 & 1 \end{array}\right]$$ Our goal is to compute $$R_zR_yR_x$$, and although this is a very simple product, we can also deduce the overall rotation by pure geometrical means. The idea is to learn how each axis behaves upon rotation – this can be done by observing the rotation matrices, giving the relation: $$\hat x = R_x \hat x = – R_y \hat z = R_z \hat y,~~~~~~ \hat y = R_y \hat y = R_x \hat z = – R_z \hat x,~~~~~~ \hat z = R_z \hat z = – R_x \hat y = R_y \hat x$$ Apply the above rules to to compute the rotation $$R_zR_yR_x$$ to each of the standard unit axes $$\hat x, \hat y, \hat z$$, we get: $$\begin{array} {cccccccc} \hat x & \Rot{x} &\hat x &\Rot{y} &\hat z &\Rot{z} & \hat z \\ \hat y & \Rot{x} &\hat z &\Rot{y} &-\hat x & \Rot{z} & -\hat y \\ \hat z & \Rot{x} &-\hat y & \Rot{y} &-\hat y &\Rot{z} &\hat x \end{array}$$ Hence the overall rotation is given by $$R = \begin{bmatrix} 0 & 0 & 1 \\ 0 & -1 & 0 \\ 1 & 0 & 0 \end{bmatrix}$$ which can be verified by taking the product as well. The product $$R=R_zR_yR_x$$ is a rotation matrix. Its axis is any vector $$v$$ such that $$Rv = v$$, which in this case is $$v= \hat x + \hat z$$. Furthermore, we have $$R\hat y = -\hat y$$, so this is a rotation by $$\pi$$, namely the rotation vector, as given in part 5 of the lecture notes is $$(a,b,c) = \frac{\pi}{\sqrt{2}}(\hat x+ \hat z))$$.
Exercise: let $$A$$ is the diagonal matrix with elements $$1,2,\ldots,d$$ on the diagonal. Find all matrices commuting with $$A$$.
$$\newcommand{bra}[1]{\left\langle #1\right|} \newcommand{ket}[1]{\left| #1\right\rangle} \newcommand{braket}[1]{\left\langle #1\right\rangle}$$
Solution: the commutator is the set of all matrices $$B$$ satisfying $$[A,B] = AB-BA = 0$$ which is a linear equation in the entries of $$B$$. Denote by $$\ket{k},~~k=1,\ldots,d$$ the set of all standard vectors, and identify the matrix $$A$$ with a operator on the standard basis. Note that we can write $$A$$ as the sum $$A = \sum_{k=1}^d k \ket{k}\bra{k}$$, hence $$BA = B\sum_{k=1}^d k \ket{k}\bra{k} = \sum_{k=1}^d k B\ket{k}\bra{k}.$$ Thus, the $$(i,j)$$ entry of the matrix $$BA$$ is given by $$\braket{i|BA|j} = \sum_{k=1}^d k \braket{i|B|k}\underbrace{\braket{k|j}}_{\delta_{k,j}} = j\braket{i|B|j}$$ Similarly, the $$(i,j)$$ entry of $$AB$$ is given by $$\braket{i|BA|j} = i\braket{i|B|j}$$. In conclusion if $$AB-BA=0$$ then the the entries of $$B$$ must satisfy the condition $$(j – i)B_{i,j} =0 ,~~~ i,j = 1,\ldots,d,$$
or, equivalently, the matrix $$B$$ is diagonal itself.

Exercise: do there exist matrices $$A,B$$ such that $$AB-BA=E$$?
Solution: let $$f$$ be some central function, say, the trace $$f(A) = \text{tr}(A)$$. If $$BA-AB = E$$, then $$\underbrace{\text{tr}(BA-AB)}_{=0} = \underbrace{\text{tr}(E)}_{=N}$$ However, the left hand side vanish (as the trace is a central function), whereas the right hand is $$N$$ – the dimension of the square (!) matrix $$A$$, hence no such