# Confusion about the Wigner-Eckart theorem

## 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?

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.