Confusion about the Wigner-Eckart theorem

Question

Background

This will be a lengthy thread, but I made sure that all 3 questions are related to each other and related to the same topic. I currently encountered the W.E-theorem and while I do understand some things, when we consider tensor operators of order 1 (vector operators), I am confused about the several other things, when I try to understand the general formula. I will start with what I understand and then list some things for which I am not sure. First I will write down the formula and try explain what I understand and what I find confusing:

$$\langle k,j,m|T^{(r)}_q|k',j',m'\rangle=\langle j',r; m',q|j,m\rangle \frac{\langle k,j||T^{(r)}||k',j'\rangle}{\sqrt{2j +1}}$$

Because the Clebsch–Gordan coefficients are present: $\langle j',r; m',q|j,m\rangle$, this means that we have an angular momentum coupling, in this case of $j'$ and $r$ into $j$. The setup can be two particles with 2 different angular momenta, or 1 particle where we consider orbital and intrinsic angular momentum. For the Clebsch–Gordan coefficients in my lecture, the following notation was used: $\langle j_1,j_2; m_1,m_2|j,m\rangle$ when considering 2 particles each with angular momentum $\vec J_1$/$J_{1_z}$ and $\vec J_2$/$J_{2_z}$ and main angular quantum numbers $j_1$/$j_2$ and secondary angular quantum numbers $m_1$/$m_2$. Drawing conclusion from this simple case, in the W.E-Theorem I have:

$j_1=j'$, $j_2=r$, $m_1=m'$, $m_2=q$, $J=j$ (I wrote $J$ but I could have very well left it as $j$), $M=m$.

First question

One can have $j=\frac 1 2$ and $j_2=\frac 3 2$. I just said that in the W.E -Theorem we consider angular momentum coupling (because of the C.G Coef.) and also that $j_2=r$. Then for the 2 given values of $j_1$ and $j_2$, I get that $r= \frac 3 2$. Are there tensors operators of order $\frac 3 2$ or other similar values ? (I don't think there are though). If not, then does it mean that the W.E-Theorem is valid for certain values of $j_2$ ?

$|j,m\rangle$ is a joint eigenstate of the squared total angular momentum $\vec J^2$ and its z-component $J_z$, where:

$j=j_1 +j_2=j' + r$ the main total angular momentum quantum number.

$-j\le m \le j$ the secondary total angular momentum quantum number.

Second question

What exactly is the meaning of this: $\langle k,j,m|T^{(r)}_q|k',j',m'\rangle$?

I know it's a matrix component, but what confuses me are the bra and ket vectors over here. The bra is an eigenstate of $\vec J^2$ and $J_z$, a basis element of the total hilber space $H = H_1 \otimes H_2$, while the ket, as it can be seen is an eigenstate of $\vec J_1^2$ and $J_{1_z}$ or a basis element of $H_1$. The questions here are many: 1)What does it mean to have bra and ket of different hilbert spaces? Why is it considering an eigenstate of $\vec J_1^2$ and $J_{1_z}$ and not of $\vec J_2^2$ and $J_{2_z}$ instead. How does it make sense to find a matrix component using basis kets/bras of different hilbert spaces? As one can see I am very confused about this part.

I want to draw an analogy to when one considers a vector operator and we also have the total angular momentum $\vec J = \vec J_1 + \vec J_2$. In a sub-Hilbert space $H(k,j)$ according to the W.E-Theorem all vector operators are proportional to the angular momentum. For vector operators the theorem is:

$$\langle k,j,m|\vec V|k,j,m'\rangle= \alpha(k,j)\langle k,j,m|\vec J|k,j,m'\rangle$$ ($\alpha$ is some proportionality constant).

I will put both expressions close to each other in order to emphasize the difference there is, which confuses me:

$$\langle k,j,m|T^{(r)}_q|k',j',m'\rangle$$

$$\langle k,j,m|\vec V|k,j,m'\rangle$$

In the first we have a bra that is eigenstate of $\vec J^2$ and $J_z$ and ket that is eigenstate of $\vec J_1^2$ and $J_{1_z}$.

In the second we have a bra and a ket both eigenstates of $\vec J^2$ and $J_z$.

How can the eigenstate of one of the two angular momentum change into an eigenstate of the total angular momentum? Or said in general terms, how can a basis ket of one Hilbert space change so that it is a basis ket of another Hilbert space?

Third question

In our lecture it was said without proof or an explanation that

$\langle k,j,m|T^{(r)}_q|k',j',m'\rangle \neq 0$ only if:

$$q=m-m'$$ $$|j-j'|\le r \le j+j'$$

How did we end up with these 2 inequalities?

Answer 1

1. In principle there exist tensor operators of arbitrary angular momentum/spin - if you consider that the angular momentum generators $L_i$ themselves are a vector operator, then e.g. $L_i$ in a spin-3/2 representation is a "tensor operator of order 3/2".

2. Since you are confused about how the states here work, let's set the scene for the Wigner-Eckart theorem again:

   We have a space of states $H$ upon which we have a reducible representation of the rotation group $\mathrm{SO}(3)$, so that $H$ decomposes as $H = \bigoplus_l H_l^{c_l}$, where $H_l$ is the irreducible representation with spin $l$ and $c_l$ its multiplicity (i.e. how many copies of it there are). By the usual argument, this means that any state $\lvert \psi\rangle \in H$ can be decomposed as $\lvert \psi\rangle = \sum_{k,l,m} \psi_{klm}\lvert k,l,m\rangle$ where the $\lvert k,l,m\rangle \in H_l \subset H$ are eigenstates of $L^2,L_z$ and the $k$ is some label that distinguishes states with the same $l,m$ in different subrepresentations with the same $l$ (i.e. you don't need any $k$ if $c_l = 1$ for all $l$).

   So when you want to look at an operator $T$ on $H$, then since $\lvert k,l,m\rangle$ is a basis of $H$, the matrix elements $\langle k,l,m\vert T\vert k',l',m'\rangle$ tell you everything you need to know about the operator.

   The Wigner-Eckart theorem now says that when $T$ is a "nice" operator that itself transforms under rotations as having $r,q$ as total angular momentum and $q$ as the eigenvalue of $L_z$, then there these matrix elements have a particular (perhaps "simpler") form, namely $$\langle k,l,m\vert T\vert k',l',m'\rangle = C_{lmrql'm'} T_R(k,l,k',l'),$$ where $ C_{lmrql'm'}$ is the Clebsch-Gordan coefficient for these angular momentum values and $T_R$ is some term that does not depend on $m,m',q$ - the "reduced matrix element". Note that the only states that occur here are the $\lvert k,l,m\rangle$ that are eigenstates of the angular momentum $L$ on the full Hilbert space $H$. The usual notation with some strange $\lvert k,l\rangle$ or whatever is intended to be mnemonic but there isn't actually any single state like $\lvert k,l\rangle$ or $\lvert l,m\rangle$ inside $H$.

   If you have only looked at Clebsch-Gordan coefficients as "coupling different angular momentum operators", then a) this is a bit of a silly imposition of a particular physical interpretation on a purely mathematical way of decomposing the tensor product of $\mathrm{SO}(3)$ representations into the direct sum of irreducible $\mathrm{SO}(3)$ representations and b) the "different" angular momentum operators here are simply the restrictions of $L$ to the individual $H_l$ the states and the operator exist in. The Wigner-Eckart theorem is nothing more than decomposing $H_l\otimes H_r$ - the space in which $T\lvert k,l,m\rangle$ lives - and looking for a $H_{l'}$ in there because the matrix element needs to be a scalar under rotation and the only non-zero part of an inner product of $T\lvert k,l,m\rangle$ and $\lvert k',l',m'\rangle$ is the part where both sides have the same angular momentum.

   (By the way, this view in terms of representations shows that the Wigner-Eckart theorem is really just a special case of how representations in general work, and lets us generalize the theorem to operators and states in arbitrary representations, not just of $\mathrm{SO}(3)$, see this question and its answer)

3. Lastly, your lecture did state the proof that this matrix element is non-vanishing only for certain values of the angular momenta involved: The proof is the Wigner-Eckart theorem - the Clebsch-Gordan coefficient on its right-hand side is zero for values outside these certain values.