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

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