I will start by giving a brief explanation to the Clebsch-Gordan coef. It's because how I perceive this coef. that I don't understand the Wigner-Eckart theorem.

The Clebsch-Gordan coefficient related the kets of the coupled basis (eigenstates of the total angular momentum) to kets of the uncoupled Tensor product basis (which are expressed as tensor product among the eigenstates of the angular momentums in consideration, that the system displays). (I hope my wording is correct, and if not, I would like a proper way of saying what I just wrote above). So:

$|J,M\rangle=\sum_{m_1=-j_1}^{j_1}\sum_{m_2=-j_2}^{j_2} |j_1,j_2;m_1,m_2\rangle\langle j_1,j_2;m_1,m_2|J,M\rangle$.

So in the Clebsch-Gordan coef. we have quantum numbers from 3 types of angular momentums, the two the system displays and the total angular momentum (the result of their coupling).

The Wigner-Eckard theorem is the following:

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

It's hard for me to put into words what I don't understand, but I will try my best:

The total angular momentum has, in the formula, the quantum numbers: $j,m$ and $j',m'$ which are used to represent different eigenstates of it.

Now, in the right side, we have the Clebsch-gordan coeficients, which as I described above should be an inner product between the tensor product of eigenstates of angular momentums present in the system, with eigenstates of the total angular momentum. So while $|k,q\rangle$ is an eigenstate of one angular momentum, represented via it's angular and magnetic quantum numbers, the other angular momentum has as it's quantum numbers $j',m'$ which are used in the left side as quantum numbers of the total angular momentum. This means than, that the other operator that we are considering is the total angular momentum itself, so in other words the Clebsch-Gordan coefficients in the theorem, are the inner product between the tensor product of eigenstates of some arbitrary angular momentum and total angular momentum with eigenstates of the total angular momentum. How is this possible? What's the point here, or the main idea? As I said above, the coef. are used when we want to transition (if I can say that) between the coupled basis, which uses eigenstates of the total hamiltonian and of the uncoupled basis which uses eigenstates, that are the result of the tensor product between eigenstates of the angular momentums involved in the coupling. But here is not the case.So what is going on? And how does it make sense?

  • 1
    $\begingroup$ I feel I already answered this question here a year back (second-to-last paragraph of the second point) $\endgroup$
    – ACuriousMind
    Aug 9, 2023 at 16:41

2 Answers 2


This is confused but the key point is that $T^{k}_q$ is constructed or defined so it transforms in the same way as $\vert k,q\rangle$: basically \begin{align} R(\Omega)T^k_q R^{-1}(\Omega)&=\sum_{q'} T^{k}_{q'} D^{k}_{q'q}(\Omega) \tag{1}\\ [L_z,T^{k}_{q}]&=k T^{k}_{q} \quad\hbox{etc} \tag{2} \end{align} whereas of course \begin{align} R(\Omega)\vert kq\rangle=\sum_{q'} \vert kq'\rangle D^{k}_{q'q}(\Omega)\, ,\\ L_z\vert kq\rangle=q\vert kq\rangle \quad\hbox{etc} \end{align}

Now, consider $T^{k}_q\vert j'm'\rangle$ and look at $$ L_z T^{k}_q\vert j'm'\rangle = [L_z,T^k_q]\vert j'm'\rangle + T^k_q L_z\vert j'm'\rangle = (m'+q)T^{k}_q\vert j'm'\rangle\, . $$ You can do the same with $L_\pm T^{k}_q\vert j'm'\rangle$. The point is that the state $T^{k}_q\vert j'm'\rangle$ behaves under rotation "just as" the uncoupled state $\vert k,q;j'm'\rangle$ except that whereas $\vert k,q\rangle$ is normalized in the sense that $\langle k,q\vert k,q\rangle=1$, the tensor operator does not necessarily satisfy an obvious normalization condition.

Now, if you know that $\langle j,m\vert k,q;j'm'\rangle$ is just a CG, then surely $\langle j,m \vert T^{k}_q\vert j'm'\rangle$ will also be proportional to the CG $\langle j,m\vert k,q;j'm'\rangle$, and the proportionality constant $\langle j\Vert T^k\Vert j'\rangle$ accounts for the fact the operator $T^{k}_q$ isn't necessarily normalized.

Indeed the reduced matrix element $\langle j\Vert T^k\Vert j'\rangle$ is in fact this "norm". It clearly depends on the specifics of $T^k$ for if $T^k_q$ is the q'th component of a tensor operator, so is $\tilde{T}^k_q=\alpha T^k_q$ where $\alpha\in \mathbb{C}$ is an constant: this constant does not affect (1) or (2).

BTW there is usually a dimensionality factor $1/\sqrt{2j'+1}$ that is inserted so that $\langle jm\vert T^k_{-q}\vert j'm'\rangle$ basically gives the same reduced matrix element $\langle j\Vert T^k\Vert j'\rangle = \langle j' \Vert T^k\Vert j\rangle$

  • $\begingroup$ I am fine with what your wrote, my main issue is with the fact that the ket $|j',m'\rangle$ is present in both sides. In the left side as an eigenstate of the total angular momentum, while in the right size, in the tensor product eigenstate, projected into an eigenstate of the total angular momentum. This is the part that confuses me. If we had L (l,m_l) S (s,m_s) and J=L+S (j,m_j), than how does the theorm looks like with these quantum numbers ? $\endgroup$
    – imbAF
    Aug 9, 2023 at 17:56
  • $\begingroup$ Not sure I follow with your statement that $\vert j'm'\rangle$ appears on both sides: of course it must since $\langle j'm'\vert T^k_q\vert jm\rangle$ must depend on $j',m'$ on the left, the right hand side must also depend on $j'$. Now, in your notation $\langle J M_J\vert T^S_q\vert LM_L\rangle$ would just be $\langle J\Vert T^S\Vert L\rangle\langle J M_J\vert S,q; LM_L\rangle$ again divided by some extra dimension factor. $\endgroup$ Aug 9, 2023 at 18:10
  • $\begingroup$ The physical interpretation here is that $T^k$ "carries" angular momentum $k$ so that, when combined to a state of angular momentum $j'$, the system has possible angular momenta in the range $\vert j'-k\vert \le j\le \vert j'+k\vert$. $\endgroup$ Aug 9, 2023 at 18:12
  • $\begingroup$ oups $\uparrow\uparrow$ my $j'\leftrightarrow j$ w/r to my initial answer. $\endgroup$ Aug 9, 2023 at 18:13
  • $\begingroup$ And why here $JM_J|T_s^q|LM_L$ is the orbital and not spin ? I was under the impression that the ket and bra in the left side where eigenstates of the total angular momentum, and we were "inspecting" matrix elements of a component of the tensor operator in a sub Hilbert space corresponding to a degenerate eigenvalue of the total angular momentum $\endgroup$
    – imbAF
    Aug 9, 2023 at 18:21

I think the key is that using the notation $\langle J M|j_1 m_1, j_2 m_2\rangle$ to denote the Clebsch-Gordan coefficient in the case of the Wigner-Eckart theorem is technically and abuse of notation. Typically the Clebsch-Gordan coefficients arise from considering the change of basis coefficients between two bases for the tensor product space $\mathcal{H}_1 \otimes \mathcal{H}_2$. Within the context of this problem, it is ok to write the Clebsch-Gordan coefficients in that way

However, in the case of the Wigner-Eckart theorem, if we write the coefficients that way we're opening up room for confusion. In this case we have a SINGLE Hilbert space $\mathcal{H}_1$ We have

$$ \langle j_1 m_1|T^{(k)}_q|j_2m_2\rangle = \langle j_1||T^{(k)}||j_2\rangle \langle j_1m_1|j_2m_2,k q\rangle $$ But what is the the thing on the right? Is $|j_2 m_2, kq\rangle$ a ket? If so, then what Hilbert space does it live in? In this problem we just have ONE Hilbert space. The way out of the confusion is to realize that, in the expression on the right hand side, $|j_2 m_2, kq\rangle$ is not a ket, and $\langle j_1 m_1|$ is not a bra. Rather $\langle j_1 m_1|j_2m_2, k q\rangle$ is just a number. A number that happens to equal the same number that you get in the OTHER problem that involved two Hilbert spaces.

So a less confusing way to write things would be \begin{align} C_{j_1m_1,j_2m_2}^{JM} =& \langle JM |j_1m_1, j_2m_2\rangle\\ \langle j_1 m_1|T^{(k)}_q|j_2m_2\rangle =& \langle j_1||T^{(k)}||j_2\rangle C_{j_2m_2, k_q}^{j_1m_1} \end{align}

The problem is, the former, confusing notation, makes some identities look very natural that would be hard to remember in the other notation. For example \begin{align} \sum_{m_1, m_2} \langle JM|j_1m_1, j_2m_2\rangle \langle j_1m_1, j_2m_2|J'M'\rangle = \langle JM|J'M'\rangle \delta_{JJ'}\delta_{MM'} \end{align} is just a resolution of the identity in the two Hilbert space problem. But expressing it the other way \begin{align} \sum_{m_1, m_2} C_{j_1m_1, j_2m_2}^{JM} C_{j_1, m_1, j_2m_2}^{J'M'*} = \delta_{JJ'}\delta_{MM'} \end{align} Is just ugly. You also have to remember which coefficient goes in each of the six slots and I don't think there's standardized format for the subscripts and super scripts. It's just not as nice of a notation as the dirac notation.

Maybe I would suggest notations like \begin{align} &\langle JM|j_1m_1, j_2m_2\rangle_{CG}\\ &\langle JM||j_1m_1, j_2m_2\rangle\\ &\overline{\langle JM|j_1m_1, j_2m_2\rangle}\\ \end{align} Just some way to notate that "this thing that looks like an inner product of two kets is not actually that, it is actually just a number. Just like $\langle j_1||T^{(k)}||j_2\rangle$ looks like the matrix element for a certain operator, but it isn't quite that.

Ah, an obvious alternative notation is to use the Wigner 3j symbol in the expression for the Wigner-Eckart theorem instead of the Clebsch-Gordan coefficient $$\begin{pmatrix}j_1&&j_2&&J\\m_1&&m_2&&-M\end{pmatrix}$$


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