PSet 2 Discussion
Conventions used below: we keep \(\hbar\) explicit, and use \(\hat J_\pm=\hat J_x\pm i\hat J_y\). For coupled angular-momentum states we write \(\ket{j,m}\), and for spherical tensors we use components \(\hat T_q^{(k)}\) with \(q=-k,-k+1,\dots,k\).
Problem 2-1: Spherical tensor operator basics
(a)
We are given the defining commutator for a spherical tensor component:
\[
[\hat J_z,\hat T_q^{(k)}]=\hbar q\,\hat T_q^{(k)}.
\]
Take a matrix element between \(\bra{j',m'}\) and \(\ket{j,m}\):
workout
\[
\bra{j',m'}[\hat J_z,\hat T_q^{(k)}]\ket{j,m}
=\bra{j',m'}\hat J_z\hat T_q^{(k)}\ket{j,m}-\bra{j',m'}\hat T_q^{(k)}\hat J_z\ket{j,m}.
\]
Using \(\hat J_z\ket{j,m}=\hbar m\ket{j,m}\) and \(\bra{j',m'}\hat J_z=\hbar m'\bra{j',m'}\) gives
\[
\bra{j',m'}[\hat J_z,\hat T_q^{(k)}]\ket{j,m}=\hbar(m'-m)\bra{j',m'}\hat T_q^{(k)}\ket{j,m}.
\]
But the defining commutator also implies
\[
\bra{j',m'}[\hat J_z,\hat T_q^{(k)}]\ket{j,m}=\hbar q\,\bra{j',m'}\hat T_q^{(k)}\ket{j,m}.
\]
Therefore, if the matrix element is nonzero, we must have
\[
\boxed{m'=m+q.}
\]
(b)
For a vector operator \(\hat{\mathbf V}=(\hat V_x,\hat V_y,\hat V_z)\), the defining commutators are
\[
[\hat J_i,\hat V_j]=i\hbar\sum_{k}\epsilon_{ijk}\hat V_k.
\]
Define spherical components (standard convention)
\[
\hat V_0=\hat V_z,\qquad
\hat V_{\pm 1}=\mp\frac{\hat V_x\pm i\hat V_y}{\sqrt{2}}.
\]
First check the \(\hat J_z\) commutator. From the vector-operator relations,
\[
[\hat J_z,\hat V_x]=i\hbar \hat V_y,\qquad
[\hat J_z,\hat V_y]=-i\hbar \hat V_x,\qquad
[\hat J_z,\hat V_z]=0,
\]
so
\[
[\hat J_z,\hat V_x\pm i\hat V_y]=\pm \hbar(\hat V_x\pm i\hat V_y).
\]
Therefore \([\hat J_z,\hat V_{\pm 1}]=\pm\hbar \hat V_{\pm 1}\) and \([\hat J_z,\hat V_0]=0\), i.e.
\[
[\hat J_z,\hat V_q]=\hbar q\,\hat V_q,\qquad q\in\{-1,0,1\}.
\]
Next, with \(\hat J_\pm=\hat J_x\pm i\hat J_y\) one finds the ladder structure (equivalently, one can evaluate \([\hat J_\pm,\hat V_q]\) using the same Cartesian commutators):
From \([\hat J_i,\hat V_j]=i\hbar\sum_k \epsilon_{ijk}\hat V_k\), we will use
\[
[\hat J_x,\hat V_x]=0,\qquad
[\hat J_x,\hat V_y]=i\hbar \hat V_z,\qquad
[\hat J_x,\hat V_z]=-i\hbar \hat V_y,
\]
\[
[\hat J_y,\hat V_x]=-i\hbar \hat V_z,\qquad
[\hat J_y,\hat V_y]=0,\qquad
[\hat J_y,\hat V_z]=i\hbar \hat V_x.
\]
We now compute the commutators explicitly and rewrite the answers in terms of the spherical components \(\hat V_0=\hat V_z\) and \(\hat V_{\pm 1}=\mp(\hat V_x\pm i\hat V_y)/\sqrt{2}\).
First, for \(\hat J_+\):
workout
\[
\begin{aligned}
\left[\hat J_+, \hat V_{+1}\right]
&=-\frac{1}{\sqrt{2}}\,[\hat J_x+i\hat J_y,\hat V_x+i\hat V_y] \\
&=-\frac{1}{\sqrt{2}}\Big([\hat J_x,\hat V_x]+i[\hat J_x,\hat V_y]+i[\hat J_y,\hat V_x]+i^2[\hat J_y,\hat V_y]\Big) \\
&=-\frac{1}{\sqrt{2}}\Big(0+i(i\hbar \hat V_z)+i(-i\hbar \hat V_z)-0\Big)=0,
\end{aligned}
\]
\[
\begin{aligned}
\left[\hat J_+,\hat V_{0}\right]
&=[\hat J_x+i\hat J_y,\hat V_z]=[\hat J_x,\hat V_z]+i[\hat J_y,\hat V_z] \\
&=(-i\hbar \hat V_y)+i(i\hbar \hat V_x)=-\hbar(\hat V_x+i\hat V_y)
=\hbar\sqrt{2}\,\hat V_{+1},
\end{aligned}
\]
\[
\begin{aligned}
\left[\hat J_+,\hat V_{-1}\right]
&=\frac{1}{\sqrt{2}}\,[\hat J_x+i\hat J_y,\hat V_x-i\hat V_y] \\
&=\frac{1}{\sqrt{2}}\Big([\hat J_x,\hat V_x]-i[\hat J_x,\hat V_y]+i[\hat J_y,\hat V_x]-i^2[\hat J_y,\hat V_y]\Big) \\
&=\frac{1}{\sqrt{2}}\Big(0-i(i\hbar \hat V_z)+i(-i\hbar \hat V_z)+0\Big)
=\frac{2\hbar}{\sqrt{2}}\hat V_z
=\hbar\sqrt{2}\,\hat V_0.
\end{aligned}
\]
So we have
\[
[\hat J_+,\hat V_{+1}]=0,\qquad
[\hat J_+,\hat V_{0}]=\hbar\sqrt{2}\,\hat V_{+1},\qquad
[\hat J_+,\hat V_{-1}]=\hbar\sqrt{2}\,\hat V_{0}.
\]
Similarly, for \(\hat J_-=\hat J_x-i\hat J_y\):
\[
\begin{aligned}
\left[\hat J_-,\hat V_{-1}\right]
&=\frac{1}{\sqrt{2}}\,[\hat J_x-i\hat J_y,\hat V_x-i\hat V_y] \\
&=\frac{1}{\sqrt{2}}\Big([\hat J_x,\hat V_x]-i[\hat J_x,\hat V_y]-i[\hat J_y,\hat V_x]+(-i)(-i)[\hat J_y,\hat V_y]\Big) \\
&=\frac{1}{\sqrt{2}}\Big(0-i(i\hbar \hat V_z)-i(-i\hbar \hat V_z)+0\Big)=0,
\end{aligned}
\]
\[
\begin{aligned}
\left[\hat J_-,\hat V_{0}\right]
&=[\hat J_x-i\hat J_y,\hat V_z]=[\hat J_x,\hat V_z]-i[\hat J_y,\hat V_z] \\
&=(-i\hbar \hat V_y)-i(i\hbar \hat V_x)=\hbar(\hat V_x-i\hat V_y)
=\hbar\sqrt{2}\,\hat V_{-1},
\end{aligned}
\]
\[
\begin{aligned}
\left[\hat J_-,\hat V_{+1}\right]
&=-\frac{1}{\sqrt{2}}\,[\hat J_x-i\hat J_y,\hat V_x+i\hat V_y] \\
&=-\frac{1}{\sqrt{2}}\Big([\hat J_x,\hat V_x]+i[\hat J_x,\hat V_y]-i[\hat J_y,\hat V_x]+(-i)(i)[\hat J_y,\hat V_y]\Big) \\
&=-\frac{1}{\sqrt{2}}\Big(0+i(i\hbar \hat V_z)-i(-i\hbar \hat V_z)+0\Big)
=\hbar\sqrt{2}\,\hat V_0.
\end{aligned}
\]
These cases can be summarized compactly as
\[
[\hat J_\pm,\hat V_q]=\hbar\sqrt{(1\mp q)(1\pm q+1)}\,\hat V_{q\pm 1}.
\]
These are precisely the rank-\(k=1\) spherical-tensor commutators, so the \(\hat V_q\) form a \(k=1\) spherical tensor.
(c)
Consider the operator-space “Casimir” built from double commutators:
\[
\sum_{i=x,y,z}[\hat J_i,[\hat J_i,\hat T_q^{(k)}]].
\]
workout
It is convenient to rewrite the Cartesian sum in terms of \((z,\pm)\):
\[
\sum_{i=x,y,z}[\hat J_i,[\hat J_i,\hat T]]
=[\hat J_z,[\hat J_z,\hat T]]+\frac12[\hat J_+,[\hat J_-,\hat T]]+\frac12[\hat J_-,[\hat J_+,\hat T]].
\]
Using the defining spherical-tensor commutators
\[
\left[\hat J_z,\hat T_q^{(k)}\right]=\hbar q\,\hat T_q^{(k)},\qquad
\left[\hat J_\pm,\hat T_q^{(k)}\right]=\hbar\,C_\pm(k,q)\,\hat T_{q\pm 1}^{(k)},
\]
where
\[
C_+(k,q)=\sqrt{(k-q)(k+q+1)},\qquad
C_-(k,q)=\sqrt{(k+q)(k-q+1)}.
\]
we now compute the three pieces explicitly for \(\hat T=\hat T_q^{(k)}\).
First,
\[
\begin{aligned}
\left[\hat J_z,\left[\hat J_z,\hat T_q^{(k)}\right]\right]
&=\left[\hat J_z,\hbar q\,\hat T_q^{(k)}\right] \\
&=\hbar q\,\left[\hat J_z,\hat T_q^{(k)}\right] \\
&=\hbar^2 q^2\,\hat T_q^{(k)}.
\end{aligned}
\]
Next,
\[
\begin{aligned}
\left[\hat J_+,\left[\hat J_-,\hat T_q^{(k)}\right]\right]
&=\left[\hat J_+,\hbar\,C_-(k,q)\,\hat T_{q-1}^{(k)}\right] \\
&=\hbar\,C_-(k,q)\,\left[\hat J_+,\hat T_{q-1}^{(k)}\right] \\
&=\hbar^2\,C_-(k,q)\,C_+(k,q-1)\,\hat T_q^{(k)} \\
&=\hbar^2\,(k+q)(k-q+1)\,\hat T_q^{(k)},
\end{aligned}
\]
where in the last step we used
\[
C_-(k,q)\,C_+(k,q-1)
=\sqrt{(k+q)(k-q+1)}\,\sqrt{(k-(q-1))(k+(q-1)+1)}
=(k+q)(k-q+1).
\]
Similarly,
\[
\begin{aligned}
\left[\hat J_-,\left[\hat J_+,\hat T_q^{(k)}\right]\right]
&=\left[\hat J_-,\hbar\,C_+(k,q)\,\hat T_{q+1}^{(k)}\right] \\
&=\hbar\,C_+(k,q)\,\left[\hat J_-,\hat T_{q+1}^{(k)}\right] \\
&=\hbar^2\,C_+(k,q)\,C_-(k,q+1)\,\hat T_q^{(k)} \\
&=\hbar^2\,(k-q)(k+q+1)\,\hat T_q^{(k)}.
\end{aligned}
\]
Putting these into the Cartesian-to-\((z,\pm)\) decomposition (with the \(\tfrac12\) factors) gives
\[
\begin{aligned}
\sum_{i=x,y,z}\left[\hat J_i,\left[\hat J_i,\hat T_q^{(k)}\right]\right]
&=\left[\hat J_z,\left[\hat J_z,\hat T_q^{(k)}\right]\right]
+\frac12\left[\hat J_+,\left[\hat J_-,\hat T_q^{(k)}\right]\right]
+\frac12\left[\hat J_-,\left[\hat J_+,\hat T_q^{(k)}\right]\right] \\
&=\hbar^2\left(q^2+\frac12\Big((k+q)(k-q+1)+(k-q)(k+q+1)\Big)\right)\hat T_q^{(k)} \\
&=\hbar^2\left(q^2+\frac12\Big((k(k+1)+q-q^2)+(k(k+1)-q-q^2)\Big)\right)\hat T_q^{(k)} \\
&=\hbar^2 k(k+1)\,\hat T_q^{(k)}.
\end{aligned}
\]
workout
So the proportionality constant is \(\hbar^2 k(k+1)\), and we can summarize the result as
\[
\boxed{\sum_{i=x,y,z}[\hat J_i,[\hat J_i,\hat T_q^{(k)}]]=\hbar^2 k(k+1)\,\hat T_q^{(k)}.}
\]
This mirrors \(\hat{\mathbf J}^2\ket{j,m}=\hbar^2 j(j+1)\ket{j,m}\): the commutator action of \(\hat{\mathbf J}\) defines an SU(2) representation on operator space, and \(\hat T_q^{(k)}\) transforms as an irreducible rank-\(k\) multiplet under that action.
Problem 2-2: Rotations and spherical tensors
(a)
Work in the \(\hat J_z\) eigenbasis \(\{\ket{1,1},\ket{1,0},\ket{1,-1}\}\). The rotation about the \(x\) axis by angle \(\phi\) is
\[
\hat D[R_x(\phi)]=e^{-\,\frac{i}{\hbar}\phi \hat J_x}.
\]
workout
Using \(\hat J_x=(\hat J_+ + \hat J_-)/2\) and \(\hat J_\pm\ket{1,m}=\hbar\sqrt{2-m(m\pm 1)}\,\ket{1,m\pm 1}\), one finds
\[
\frac{\hat J_x}{\hbar}=
\frac{1}{\sqrt{2}}
\begin{pmatrix}
0&1&0\\
1&0&1\\
0&1&0
\end{pmatrix}
\equiv A,
\qquad
A^2=\frac12
\begin{pmatrix}
1&0&1\\
0&2&0\\
1&0&1
\end{pmatrix},
\qquad
A^3=A.
\]
Exponentiating (e.g. by using that powers of this matrix reduce to a quadratic polynomial, \(e^{-i\phi A}=I - i(\sin\phi)A + (\cos\phi-1)A^2\) ), the Wigner \(D\)-matrix in this basis is
\[
D^{1}(R_x(\phi))=
\begin{pmatrix}
\frac{1+\cos\phi}{2} & -\frac{i}{\sqrt{2}}\sin\phi & \frac{\cos\phi-1}{2} \\[4pt]
-\frac{i}{\sqrt{2}}\sin\phi & \cos\phi & -\frac{i}{\sqrt{2}}\sin\phi \\[4pt]
\frac{\cos\phi-1}{2} & -\frac{i}{\sqrt{2}}\sin\phi & \frac{1+\cos\phi}{2}
\end{pmatrix},
\]
with rows/columns ordered by \(m=1,0,-1\).
(b)
Given a vector operator \(\hat{\mathbf V}\), form the rank-1 spherical tensor \(\hat T_q^{(1)}\) via
\[
\hat T_0^{(1)}=\hat V_z,\qquad
\hat T_{\pm 1}^{(1)}=\mp\frac{\hat V_x\pm i\hat V_y}{\sqrt{2}}.
\]
A spherical tensor transforms under rotations as
\[
\hat D(R)^\dagger \hat T_q^{(1)} \hat D(R)=\sum_{q'=-1}^{1}\hat T_{q'}^{(1)}\,D^{1*}_{q',q}(R).
\]
For \(R=R_x(\phi)\), inserting the matrix from (a) (and complex conjugating it) gives
\[
\hat D^\dagger \hat T_{+1}^{(1)}\hat D
=\frac{1+\cos\phi}{2}\hat T_{+1}^{(1)}
+\frac{i}{\sqrt{2}}\sin\phi\,\hat T_{0}^{(1)}
+\frac{\cos\phi-1}{2}\hat T_{-1}^{(1)},
\]
\[
\hat D^\dagger \hat T_{0}^{(1)}\hat D
=\frac{i}{\sqrt{2}}\sin\phi\,\hat T_{+1}^{(1)}
+\cos\phi\,\hat T_{0}^{(1)}
+\frac{i}{\sqrt{2}}\sin\phi\,\hat T_{-1}^{(1)},
\]
\[
\hat D^\dagger \hat T_{-1}^{(1)}\hat D
=\frac{\cos\phi-1}{2}\hat T_{+1}^{(1)}
+\frac{i}{\sqrt{2}}\sin\phi\,\hat T_{0}^{(1)}
+\frac{1+\cos\phi}{2}\hat T_{-1}^{(1)}.
\]
(c)
Convert back to Cartesian components using
\[
\hat V_x=\frac{\hat T_{-1}^{(1)}-\hat T_{+1}^{(1)}}{\sqrt{2}},\qquad
\hat V_y=\frac{i(\hat T_{+1}^{(1)}+\hat T_{-1}^{(1)})}{\sqrt{2}},\qquad
\hat V_z=\hat T_0^{(1)}.
\]
Substituting the transformed \(\hat T_q^{(1)}\) from (b) gives the expected vector rotation about the \(x\) axis:
workout
\[
\hat D^\dagger \hat V_x \hat D=\hat V_x,
\]
\[
\hat D^\dagger \hat V_y \hat D=\cos\phi\,\hat V_y-\sin\phi\,\hat V_z,\qquad
\hat D^\dagger \hat V_z \hat D=\sin\phi\,\hat V_y+\cos\phi\,\hat V_z.
\]
This matches the standard rotation of a (vector) operator under \((R_x(\phi))\) written in operator form as
\[
\hat D[R]^\dagger \hat V_i \hat D[R] = \sum_j R_{ij}\hat V_j,
\]
with \((R_{x}(\phi))\) acting on \((y,z)\) by the usual \(2\times2\) rotation.
Problem 2-3: Spherical tensors as vectors in an abstract space
(a)
Define the rotation superoperator
\[
\tilde D[R]\cdot \hat O \equiv \hat D[R]\hat O \hat D^{-1}[R].
\]
This defines a map from operators to operators by conjugation. We now check it is a representation of rotations on the operator vector space.
Linearity. For scalars \(a,b\) and operators \(\hat O_1,\hat O_2\),
\[
\tilde D[R]\cdot (a\hat O_1+b\hat O_2)=a(\tilde D[R]\cdot \hat O_1)+b(\tilde D[R]\cdot \hat O_2),
\]
because conjugation distributes over sums and pulls out scalars.
Identity. For the identity rotation \(I\), we have \(\hat D[I]=\mathbb I\), hence
\[
\tilde D[I]\cdot \hat O=\mathbb I\,\hat O\,\mathbb I=\hat O.
\]
Composition. Using \(\tilde D[R_1]\tilde D[R_2]\cdot \hat O\equiv \tilde D[R_1]\cdot(\tilde D[R_2]\cdot \hat O)\), we compute
\[
\begin{aligned}
(\tilde D[R_1]\tilde D[R_2])\cdot \hat O
&=\tilde D[R_1]\cdot\left(\hat D[R_2]\hat O\hat D^{-1}[R_2]\right) \\
&=\hat D[R_1]\left(\hat D[R_2]\hat O\hat D^{-1}[R_2]\right)\hat D^{-1}[R_1] \\
&=\hat D[R_1]\hat D[R_2]\;\hat O\;\hat D^{-1}[R_2]\hat D^{-1}[R_1].
\end{aligned}
\]
Now use the group property \(\hat D[R_1]\hat D[R_2]=\hat D[R_1R_2]\) together with \((AB)^{-1}=B^{-1}A^{-1}\) to identify
\(\hat D^{-1}[R_1R_2]=\hat D^{-1}[R_2]\hat D^{-1}[R_1]\), giving
\[
(\tilde D[R_1]\tilde D[R_2])\cdot \hat O
=\hat D[R_1R_2]\hat O\hat D^{-1}[R_1R_2]
=\tilde D[R_1R_2]\cdot \hat O.
\]
Inverse. Because \(\hat D[R]\) is unitary for rotations, \(\hat D^{-1}[R]=\hat D^\dagger[R]\), and
\[
\tilde D[R^{-1}]\cdot(\tilde D[R]\cdot \hat O)
=\hat D[R^{-1}]\left(\hat D[R]\hat O\hat D^{-1}[R]\right)\hat D^{-1}[R^{-1}]
=\hat O,
\]
so \(\tilde D[R^{-1}]\) is the inverse of \(\tilde D[R]\).
(b)
For a rotation by angle \(\phi\) about axis \(\hat{\mathbf n}\),
\[
\hat D[R]=\exp\!\left(-\frac{i}{\hbar}\phi\,\hat{\mathbf n}\cdot \hat{\mathbf J}\right).
\]
For \(\phi\to 0\),
\[
\hat D[R]=\mathbb I-\frac{i}{\hbar}\phi\,\hat{\mathbf n}\cdot\hat{\mathbf J}+O(\phi^2),\qquad
\hat D^{-1}[R]=\mathbb I+\frac{i}{\hbar}\phi\,\hat{\mathbf n}\cdot\hat{\mathbf J}+O(\phi^2).
\]
Thus, for any operator \(\hat O\), we expand the conjugation to first order:
\[
\begin{aligned}
\tilde D[R]\cdot \hat O
&=\hat D[R]\hat O\hat D^{-1}[R] \\
&=\left(\mathbb I-\frac{i}{\hbar}\phi\,\hat{\mathbf n}\cdot\hat{\mathbf J}\right)\hat O
\left(\mathbb I+\frac{i}{\hbar}\phi\,\hat{\mathbf n}\cdot\hat{\mathbf J}\right)+O(\phi^2) \\
&=\hat O-\frac{i}{\hbar}\phi\,(\hat{\mathbf n}\cdot\hat{\mathbf J})\hat O
+\frac{i}{\hbar}\phi\,\hat O(\hat{\mathbf n}\cdot\hat{\mathbf J})+O(\phi^2) \\
&=\hat O-\frac{i}{\hbar}\phi\,\left[(\hat{\mathbf n}\cdot\hat{\mathbf J}),\hat O\right]+O(\phi^2).
\end{aligned}
\]
Writing \(\hat{\mathbf n}\cdot\hat{\mathbf J}=\sum_{j=x,y,z} n_j\hat J_j\), we can rewrite the commutator as
\[
\left[(\hat{\mathbf n}\cdot\hat{\mathbf J}),\hat O\right]
=\sum_{j=x,y,z} n_j\left[\hat J_j,\hat O\right],
\]
so
\[
\tilde D[R]\cdot \hat O
=\hat O-i\phi\sum_{j=x,y,z}n_j\underbrace{\left(\frac{1}{\hbar}\left[\hat J_j,\hat O\right]\right)}_{\equiv\ \mathcal J_j\cdot \hat O}
+O(\phi^2).
\]
workout
Comparing with \(\tilde D[R]=\tilde{\mathbb 1}-i\phi\sum_j n_j\mathcal J_j+O(\phi^2)\) identifies the angular-momentum superoperators as
\[
\boxed{\mathcal J_j\cdot \hat O\equiv \frac{1}{\hbar}[\hat J_j,\hat O],\qquad j=x,y,z.}
\]
Equivalently, \(\mathcal J_\pm\cdot \hat O=\frac{1}{\hbar}[\hat J_\pm,\hat O]\).
(c)
A rank-\(k\) spherical tensor is defined by
\[
\left[\hat J_z,\hat T_q^{(k)}\right]=\hbar q\,\hat T_q^{(k)},\qquad
\left[\hat J_\pm,\hat T_q^{(k)}\right]=\hbar\sqrt{k(k+1)-q(q\pm 1)}\,\hat T_{q\pm 1}^{(k)}.
\]
workout
Using \(\mathcal J_j\cdot \hat O=\frac{1}{\hbar}[\hat J_j,\hat O]\) from part (b), we get immediately
\[
\mathcal J_z\cdot \hat T_q^{(k)}
=\frac{1}{\hbar}\left[\hat J_z,\hat T_q^{(k)}\right]
=\frac{1}{\hbar}\left(\hbar q\,\hat T_q^{(k)}\right)
=q\,\hat T_q^{(k)},
\]
\[
\boxed{\mathcal J_z\cdot \hat T_q^{(k)}=q\,\hat T_q^{(k)}},
\]
and similarly
workout
\[
\mathcal J_\pm\cdot \hat T_q^{(k)}
=\frac{1}{\hbar}\left[\hat J_\pm,\hat T_q^{(k)}\right]
=\sqrt{k(k+1)-q(q\pm 1)}\,\hat T_{q\pm 1}^{(k)}.
\]
\[
\boxed{\mathcal J_\pm\cdot \hat T_q^{(k)}=\sqrt{k(k+1)-q(q\pm 1)}\,\hat T_{q\pm 1}^{(k)}}.
\]
This is the same ladder structure as for kets \(|j,m\rangle\):
\[
\hat J_z|j,m\rangle=\hbar m\,|j,m\rangle,\qquad
\hat J_\pm|j,m\rangle=\hbar\sqrt{(j\mp m)(j\pm m+1)}\,|j,m\pm 1\rangle.
\]
So the correspondence is:
the “vector” in operator space is \(\hat T_q^{(k)}\), the “vector” in Hilbert space is \(|j,m\rangle\), and \((k,q)\) play the roles of \((j,m)\) under the commutator-generated action \(\mathcal J_i\cdot \hat O=\frac{1}{\hbar}[\hat J_i,\hat O]\).
Problem 2-4: Two spin ensembles
(a)
Let \(\{\hat A_q^{(k)}\}\) and \(\{\hat B_q^{(k)}\}\) be spherical tensors of the same rank \(k\). By definition, under a rotation \(R\) implemented by \(\hat U(R)\),
\[
\hat U(R)\hat A_q^{(k)}\hat U(R)^\dagger=\sum_{q'=-k}^{k} D^{(k)}_{q',q}(R)\,\hat A_{q'}^{(k)},\qquad
\hat U(R)\hat B_q^{(k)}\hat U(R)^\dagger=\sum_{q'=-k}^{k} D^{(k)}_{q',q}(R)\,\hat B_{q'}^{(k)}.
\]
Define \(\hat C_q^{(k)}=\alpha\hat A_q^{(k)}+\beta\hat B_q^{(k)}\) with \(\alpha,\beta\) complex numbers. Then, using linearity of conjugation,
\[
\begin{aligned}
\hat U(R)\hat C_q^{(k)}\hat U(R)^\dagger
&=\hat U(R)\left(\alpha\hat A_q^{(k)}+\beta\hat B_q^{(k)}\right)\hat U(R)^\dagger \\
&=\alpha\,\hat U(R)\hat A_q^{(k)}\hat U(R)^\dagger+\beta\,\hat U(R)\hat B_q^{(k)}\hat U(R)^\dagger \\
&=\alpha\sum_{q'=-k}^{k} D^{(k)}_{q',q}(R)\,\hat A_{q'}^{(k)}+\beta\sum_{q'=-k}^{k} D^{(k)}_{q',q}(R)\,\hat B_{q'}^{(k)} \\
&=\sum_{q'=-k}^{k} D^{(k)}_{q',q}(R)\left(\alpha\hat A_{q'}^{(k)}+\beta\hat B_{q'}^{(k)}\right) \\
&=\sum_{q'=-k}^{k} D^{(k)}_{q',q}(R)\,\hat C_{q'}^{(k)}.
\end{aligned}
\]
so \(\{\hat C_q^{(k)}\}\) is also a rank-\(k\) spherical tensor.
(b)
Define total angular momentum operators
\[
\hat J_i^{\mathrm{tot}}=\hat J_i^{(1)}+\hat J_i^{(2)},\qquad
\hat J_\pm^{\mathrm{tot}}=\hat J_\pm^{(1)}+\hat J_\pm^{(2)},
\]
and assume the two subsystems commute (standard on a tensor-product Hilbert space):
\[
[\hat J_i^{(1)},\hat J_j^{(2)}]=0\qquad \text{for all }i,j\in\{x,y,z\}.
\]
Introduce the “difference” vector operator
\[
\hat{\mathbf D}\equiv \hat{\mathbf J}^{(1)}-\hat{\mathbf J}^{(2)},
\]
so in particular \(\hat D_z=\hat J_z^{(1)}-\hat J_z^{(2)}\).
To show that \(\hat{\mathbf D}\) transforms as a vector under the total angular momentum \(\hat{\mathbf J}^{\mathrm{tot}}\), compute
workout
\[
\begin{aligned}
\left[\hat J_i^{\mathrm{tot}},\hat D_j\right]
&=\left[\hat J_i^{(1)}+\hat J_i^{(2)},\hat J_j^{(1)}-\hat J_j^{(2)}\right] \\
&=\left[\hat J_i^{(1)},\hat J_j^{(1)}\right]-\left[\hat J_i^{(2)},\hat J_j^{(2)}\right] \\
&=i\hbar\sum_{k}\epsilon_{ijk}\hat J_k^{(1)}-i\hbar\sum_{k}\epsilon_{ijk}\hat J_k^{(2)} \\
&=i\hbar\sum_{k}\epsilon_{ijk}\left(\hat J_k^{(1)}-\hat J_k^{(2)}\right)
=i\hbar\sum_{k}\epsilon_{ijk}\hat D_k.
\end{aligned}
\]
This is exactly the vector-operator commutation relation, so \(\hat{\mathbf D}\) is a rank-\(k=1\) spherical tensor under rotations generated by \(\hat{\mathbf J}^{\mathrm{tot}}\). In particular, its \(q=0\) spherical component is \(\hat D_0=\hat D_z\).
For completeness, we can also see directly that \(\hat D_z\) has \(q=0\) by checking the \(\hat J_z^{\mathrm{tot}}\) commutator:
\[
\begin{aligned}
\left[\hat J_z^{\mathrm{tot}},\hat D_z\right]
&=\left[\hat J_z^{(1)}+\hat J_z^{(2)},\hat J_z^{(1)}-\hat J_z^{(2)}\right] \\
&=\left[\hat J_z^{(1)},\hat J_z^{(1)}\right]-\left[\hat J_z^{(2)},\hat J_z^{(2)}\right]=0.
\end{aligned}
\]
and that it ladders correctly under \(\hat J_\pm^{\mathrm{tot}}\):
\[
\begin{aligned}
\left[\hat J_\pm^{\mathrm{tot}},\hat D_z\right]
&=\left[\hat J_\pm^{(1)}+\hat J_\pm^{(2)},\hat J_z^{(1)}-\hat J_z^{(2)}\right] \\
&=\left[\hat J_\pm^{(1)},\hat J_z^{(1)}\right]-\left[\hat J_\pm^{(2)},\hat J_z^{(2)}\right] \\
&=\mp\hbar\,\hat J_\pm^{(1)}\pm\hbar\,\hat J_\pm^{(2)}
=\mp\hbar\left(\hat J_\pm^{(1)}-\hat J_\pm^{(2)}\right).
\end{aligned}
\]
This is consistent with \(\hat D_z\) being the \(q=0\) component of a rank-1 tensor.
(c)
Define
\[
\hat D_{\mathrm{int}}\equiv \hat J_+^{(1)}\hat J_-^{(2)}-\hat J_+^{(2)}\hat J_-^{(1)}.
\]
Since operators on different subsystems commute, we can also rewrite \(\hat J_+^{(2)}\hat J_-^{(1)}=\hat J_-^{(1)}\hat J_+^{(2)}\), so
\[
\hat D_{\mathrm{int}}=\hat J_+^{(1)}\hat J_-^{(2)}-\hat J_-^{(1)}\hat J_+^{(2)}.
\]
First compute the \(q=0\) condition by commuting with \(\hat J_z^{\mathrm{tot}}=\hat J_z^{(1)}+\hat J_z^{(2)}\). Using \([\hat J_z,\hat J_\pm]=\pm\hbar\hat J_\pm\) and the product rule \([A,BC]=[A,B]C+B[A,C]\),
workout
\[
\begin{aligned}
\left[\hat J_z^{\mathrm{tot}},\hat J_+^{(1)}\hat J_-^{(2)}\right]
&=\left[\hat J_z^{(1)},\hat J_+^{(1)}\right]\hat J_-^{(2)}+\hat J_+^{(1)}\left[\hat J_z^{(2)},\hat J_-^{(2)}\right] \\
&=(+\hbar \hat J_+^{(1)})\hat J_-^{(2)}+\hat J_+^{(1)}(-\hbar \hat J_-^{(2)})=0,
\end{aligned}
\]
and similarly
\[
\begin{aligned}
\left[\hat J_z^{\mathrm{tot}},\hat J_-^{(1)}\hat J_+^{(2)}\right]
&=\left[\hat J_z^{(1)},\hat J_-^{(1)}\right]\hat J_+^{(2)}+\hat J_-^{(1)}\left[\hat J_z^{(2)},\hat J_+^{(2)}\right] \\
&=(-\hbar \hat J_-^{(1)})\hat J_+^{(2)}+\hat J_-^{(1)}(+\hbar \hat J_+^{(2)})=0.
\end{aligned}
\]
so
\[
\left[\hat J_z^{\mathrm{tot}},\hat D_{\mathrm{int}}\right]=0,
\]
so \(\hat D_{\mathrm{int}}\) has \(q=0\).
To see it is rank \(k=1\), it is helpful to relate it to a cross product. Using \(\hat J_\pm=\hat J_x\pm i\hat J_y\) and commuting operators across subsystems,
\[
\begin{aligned}
\hat D_{\mathrm{int}}
&=\left(\hat J_x^{(1)}+i\hat J_y^{(1)}\right)\left(\hat J_x^{(2)}-i\hat J_y^{(2)}\right)
-\left(\hat J_x^{(1)}-i\hat J_y^{(1)}\right)\left(\hat J_x^{(2)}+i\hat J_y^{(2)}\right) \\
&=2i\left(\hat J_y^{(1)}\hat J_x^{(2)}-\hat J_x^{(1)}\hat J_y^{(2)}\right)
=-2i\left(\hat J_x^{(1)}\hat J_y^{(2)}-\hat J_y^{(1)}\hat J_x^{(2)}\right).
\end{aligned}
\]
Define the operator-valued cross product
\[
\hat{\mathbf W}\equiv \hat{\mathbf J}^{(1)}\times \hat{\mathbf J}^{(2)},\qquad
\hat W_i=\sum_{j,k}\epsilon_{ijk}\hat J_j^{(1)}\hat J_k^{(2)}.
\]
Then \(\hat W_z=\hat J_x^{(1)}\hat J_y^{(2)}-\hat J_y^{(1)}\hat J_x^{(2)}\), and the above shows
\[
\hat D_{\mathrm{int}}=-2i\,\hat W_z.
\]
Finally, \(\hat{\mathbf W}\) transforms as a vector under joint rotations of the two ensembles. Concretely, for a proper rotation \(R\) with matrix elements \(R_{ij}\), each angular-momentum vector transforms as
\[
\hat U(R)\hat J_i^{(\alpha)}\hat U(R)^\dagger=\sum_{j}R_{ij}\hat J_j^{(\alpha)},\qquad \alpha\in\{1,2\}.
\]
workout
Therefore,
\[
\begin{aligned}
\hat U(R)\hat W_i\hat U(R)^\dagger
&=\sum_{j,k}\epsilon_{ijk}\,\hat U(R)\hat J_j^{(1)}\hat U(R)^\dagger\;\hat U(R)\hat J_k^{(2)}\hat U(R)^\dagger \\
&=\sum_{j,k}\epsilon_{ijk}\left(\sum_{j'}R_{jj'}\hat J_{j'}^{(1)}\right)\left(\sum_{k'}R_{kk'}\hat J_{k'}^{(2)}\right) \\
&=\sum_{i'}R_{ii'}\sum_{j',k'}\epsilon_{i'j'k'}\,\hat J_{j'}^{(1)}\hat J_{k'}^{(2)}
=\sum_{i'}R_{ii'}\hat W_{i'},
\end{aligned}
\]
where we used the standard identity for proper rotations \(R\):
\(\sum_{j,k}\epsilon_{ijk}R_{jj'}R_{kk'}=\sum_{i'}R_{ii'}\epsilon_{i'j'k'}\).
Thus \(\hat{\mathbf W}\) is a vector operator (rank \(k=1\)), so \(\hat W_z\) (and hence \(\hat D_{\mathrm{int}}\)) is its \(q=0\) component under \(\hat{\mathbf J}^{\mathrm{tot}}\).
(d)
For a Hamiltonian of the form
\[
\hat H=\lambda_z\hat D_z+\lambda_{\mathrm{int}}\hat D_{\mathrm{int}},
\]
workout
the term \(\lambda_z(\hat J_z^{(1)}-\hat J_z^{(2)})=\lambda_z\hat D_z\) is a differential Zeeman (detuning) term: in the product basis of \(\hat J_z^{(1)}\) and \(\hat J_z^{(2)}\) eigenstates, it assigns energy proportional to the imbalance \(m_1-m_2\) between the two ensembles.
The term \(\hat D_{\mathrm{int}}\) is an inter-ensemble “flip-flop” generator. For example, \(\hat J_+^{(1)}\hat J_-^{(2)}\) takes one unit of \(m\) from ensemble 2 and gives it to ensemble 1 (schematically \((m_1,m_2)\mapsto (m_1+1,m_2-1)\)), while \(\hat J_-^{(1)}\hat J_+^{(2)}\) does the opposite; \(\hat D_{\mathrm{int}}\) selects an antisymmetric combination of these transfers. As written, \(\hat D_{\mathrm{int}}^\dagger=-\hat D_{\mathrm{int}}\), so a Hermitian interaction term would use \(i\hat D_{\mathrm{int}}\) (or take \(\lambda_{\mathrm{int}}\) purely imaginary).
If the interaction term is present (and the overall Hamiltonian is Hermitian), eigenstates are generically not simple tensor products of the two ensembles, so one expects entangled ground states except in limits where the interaction is perturbatively small compared to \(|\lambda_z|\).
(e)
From (b,c), both \(\hat D_z\) and \(\hat D_{\mathrm{int}}\) transform as rank-\(k=1\) tensors with \(q=0\) under \(\hat{\mathbf J}^{\mathrm{tot}}\). Define
\[
F^{(\alpha)}(J,m)\equiv \bra{J+1,m}\hat D_\alpha\ket{J,m},\qquad \alpha\in\{z,\mathrm{int}\}.
\]
Because \(q=0\), the selection rule is \(m\)-conserving (so \(m\) is the same on the bra and ket), and because \(k=1\) the triangle rule allows \(J\to J+1\).
By the Wigner-Eckart theorem, for any rank-\(k\) spherical tensor \(\hat T_q^{(k)}\),
\[
\bra{j_2,m_2}\hat T_q^{(k)}\ket{j_1,m_1}
=\langle j_1,m_1;k,q|j_2,m_2\rangle\,\langle j_2 \| \hat T^{(k)} \| j_1\rangle,
\]
where the reduced matrix element \(\langle j_2\|\hat T^{(k)}\|j_1\rangle\) is independent of \(m_1,m_2,q\).
workout
Apply this with \(j_1=J\), \(j_2=J+1\), \(k=1\), \(q=0\), and \(\hat T^{(1)}_0=\hat D_\alpha\):
\[
F^{(\alpha)}(J,m)=\langle J,m;1,0|J+1,m\rangle\,\langle J+1 \| \hat D_\alpha \| J\rangle.
\]
Taking the ratio cancels the Clebsch-Gordan coefficient:
\[
\begin{aligned}
\frac{F^{(\mathrm{int})}(J,m)}{F^{(z)}(J,m)}
&=\frac{\langle J,m;1,0|J+1,m\rangle\,\langle J+1 \| \hat D_{\mathrm{int}} \| J\rangle}{\langle J,m;1,0|J+1,m\rangle\,\langle J+1 \| \hat D_{z} \| J\rangle} \\
&=\frac{\langle J+1 \| \hat D_{\mathrm{int}} \| J\rangle}{\langle J+1 \| \hat D_{z} \| J\rangle},
\end{aligned}
\]
so
\[
\boxed{\frac{F^{(\mathrm{int})}(J,m)}{F^{(z)}(J,m)}=\frac{\langle J+1 \| \hat D_{\mathrm{int}} \| J\rangle}{\langle J+1 \| \hat D_{z} \| J\rangle}\ \text{is independent of }m.}
\]