I am doing some self-study in between undergrad and grad school and I came across the beastly Wigner-Eckart theorem in Sakurai's Modern Quantum Mechanics. I was wondering if someone could tell me why it is useful and perhaps just help me understand a bit more about it. I have had two years of undergrad mechanics and I think I have a reasonably firm grasp of the earlier material out of Sakurai, so don't be afraid to get a little technical.

  • 3
    $\begingroup$ Great question :) My understanding of the W-E theorem is pretty flaky too so I'll be very interested to see what people come up with. $\endgroup$
    – David Z
    Feb 8, 2011 at 4:11

5 Answers 5


I will not get into theoretical details -- Luboš ad Marek did that better than I'm able to.
Let me give an example instead: suppose that we need to calculate this integral:

$\int d\Omega (Y_{3m_1})^*Y_{2m_2}Y_{1m_3}$

Here $Y_{lm}$ -- are spherical harmonics and we integrate over the sphere $d\Omega=\sin\theta d\theta d\phi$.

This kind of integrals appear over and over in, say, spectroscopy problems. Let us calculate it for $m_1=m_2=m_3=0$:

$\int d\Omega (Y_{30})^*Y_{20}Y_{10} = \frac{\sqrt{105}}{32\sqrt{\pi^3}}\int d\Omega \cos\theta\,(1-3\cos^2\theta)(3\cos\theta-5\cos^3\theta)=$

$ = \frac{\sqrt{105}}{32\sqrt{\pi^3}}\cdot 2\pi \int d\theta\,\left(3\cos^2\theta\sin\theta-14\cos^4\theta\sin\theta+15\cos^6\theta\sin\theta\right)=\frac{3}{2}\sqrt{\frac{3}{35\pi}}$

Hard work, huh? The problem is that we usually need to evaluate this for all values of $m_i$. That is 7*5*3 = 105 integrals. So instead of doing all of them we got to exploit their symmetry. And that's exactly where the Wigner-Eckart theorem is useful:

$\int d\Omega (Y_{3m_1})^*Y_{2m_2}Y_{1m_3} = \langle l=3,m_1| Y_{2m_2} | l=1,m_3\rangle = C_{m_1m_2m_3}^{3\,2\,1}(3||Y_2||1)$

$C_{m_1m_2m_3}^{j_1j_2j_3}$ -- are the Clebsch-Gordan coefficients

$(3||Y_2||1)$ -- is the reduced matrix element which we can derive from our expression for $m_1=m_2=m_3=0$:

$\frac{3}{2}\sqrt{\frac{3}{35\pi}} = C_{0\,0\,0}^{3\,2\,1}(3||Y_2||1)\quad \Rightarrow \quad (3||Y_2||1)=\frac{1}{2}\sqrt{\frac{3}{\pi}}$

So the final answer for our integral is:

$\int d\Omega(Y_{3m_1})^*Y_{2m_2}Y_{1m_3}=\sqrt{\frac{3}{4\pi}}C_{m_1m_2m_3}^{3\,2\,1}$

It is reduced to calculation of the Clebsch-Gordan coefficient and there are a lot of, tables, programs, reduction and summation formulae to work with them.

  • 2
    $\begingroup$ This is a really nice, practical example. $\endgroup$
    – Cogitator
    Feb 9, 2011 at 3:22
  • $\begingroup$ I think I have to give most helpful to this answer, because although Marek gave one that obviously had more theoretical detail, this one helped me to better understand where the W-E theorem falls within what I already know in quantum mechanics. $\endgroup$
    – Cogitator
    Feb 9, 2011 at 3:31
  • $\begingroup$ @Kostya, $7\times 5\times 3 = 105$, is because of $2(l+1)$ values. $\endgroup$
    – L.K.
    Jul 1, 2014 at 14:16
  • $\begingroup$ @lavkush, yep. $2\ell + 1$, to be more precise... $\endgroup$
    – Kostya
    Jul 1, 2014 at 21:08
  • $\begingroup$ @Kostya Just a last clarification. You said So instead of doing all of them we got to exploit their symmetry. What symmetry are we talking about? Sperical symmetry? $\endgroup$
    – L.K.
    Mar 8, 2017 at 10:11

First, W-E theorem is just a simple (just bear with me, I know that the theorem can appear formidable if not explained properly) statement about the decomposition of the tensor product of representations into its irreducible components.

Suppose we have a group $G$ and a tensor operator $A_r$ transforming under a(n irreducible) representation $\Gamma^1$ of it ($r$ counting the number of components of the operator, e.g. 3 for the usual angular momentum in 3 dimensions). Also suppose that you have additional (irreducible) representation $\Gamma^2$ with basis $\left\{ \left| \psi_n \right>\right\}$. It is easy to check that then vectors $\Psi_{rn} \equiv A_r \left|\psi_n\right>$ transform as a tensor product $\Gamma^1 \otimes \Gamma^2$.

In the following we use that the group $G$ is compact in order to be able to decompose its representations into irreducible components. This can be weakened but in general representations of non-compact groups don't have to be reducible to their irreducible components; there's pretty subtle mathematics behind representation theory of non-compact groups; even seemingly simple ones as $SL(2, \mathbb{R})$.

So, we'll decompose the representation as $$\Gamma^1 \otimes \Gamma^2 = \bigoplus_{\alpha} \Gamma^{\alpha}$$ where $\alpha$ runs over irreducible representations of $G$ (possibly repeated). This amounts to finding a more suitable basis for vectors $\Psi_{rn}$. We will write $\Phi_{\alpha m}$ for that basis (with $m$ indexing the components of representation $\Gamma^{\alpha}$). Then we can write $$\Psi_{rn} = \sum_{\alpha, m} U^{\alpha m}_{rn} \Phi_{\alpha m}$$.

Now, back to the problem: we are interested in computing some element such as $\left<\omega_k \right| A_r \left | \psi_n\right>$ with $\omega_k$ transforming in some $\Gamma^3$ representation. Thanks to Schur orthogonality relations and the fact that we decomposed $A_r \left | \psi_n\right>$ into its irreducible components we can see that for this element to be non-zero, there has to $\Gamma^3$ representation in the $\Gamma^{\alpha}$ decomposition. That already saves us a lot of computation. If there is no $\Gamma^3$ present there then we're finished, all those elements will be zero. And if it is present, we only have to carry out calculations corresponding to just the $\Gamma^3$ part of the decomposition (which will usually be a small part of the total).

Okay, I think the above might have been a little confusing, so let's try an example. The canonical one is with $SO(3)$ of course. Suppose we want to compute something like $\left<j m\right| \mathbf X \left|j' m'\right>$ with $\mathbf X$ being the position operator (which transforms under $SO(3)$ vector irrep). So we are interesting of doing a tensor product ${\mathbf 3} \otimes (\mathbf{2j +1})$ which can be shown to be equal to $(\mathbf{2j-1}) \oplus (\mathbf {2j+1}) \oplus (\mathbf {2j+3})$ (supposing $j$ is high enough for simplicity). So $j'$ has to be equal to one $j-1, j, j+1$ and moreover $m = m_X + m'$ (if $\mathbf X$ is written in a diagonal basis of the vectorial representations one can decompose it into operators having eigenvalues of m_X = $-1, 0, 1$) will need to hold (this is again a consequence of orthogonality relations).

Anyway, the most important point (at least for me) is about the decomposition of the tensor product into irreducible components. Supposing you can carry out this decomposition (which can be pretty hard at times), the calculation of matrix elements will simplify greatly and you can immediately decide stuff such as whether some molecule with a given symmetry will radiate in IR (which amounts to computing matrix elements of a dipole operator between eigenstates of that molecule). I am sure you can imagine lots of other similar stuff you may compute yourself. Applications of this theorem are pretty much limitless.

  • $\begingroup$ Excellent explanation @Marek. +1 $\endgroup$
    – user346
    Feb 8, 2011 at 17:44
  • $\begingroup$ I know what a group is and what an irreducible representation is. Though I am a little shaky on the rest of the group theory and the notation (the first part), this answer is helpful, particularly the last three paragraphs. $\endgroup$
    – Cogitator
    Feb 9, 2011 at 3:18
  • $\begingroup$ @Cogitator: glad to hear that. Of course, my answer is nowhere near complete (I am sure books can be written about W-E and its applications) and in particular it misses the usual formulation found in QM (which was nicely addressed by Kostya). I provided this answer because for me that classical formulation didn't work at all and I only understood W-E after learning some group theory :) $\endgroup$
    – Marek
    Feb 9, 2011 at 7:35
  • 1
    $\begingroup$ Wigner-Eckhart is a little more than just decomposing the tensor product representation. The special properties of SO(3) allow us to say something stronger: the matrix elements of a tensor operator in the basis of the irreducible components are equal up to a constant. This is false for most Lie groups $\endgroup$ Jun 6, 2021 at 1:15

The Wigner-Eckart Theorem


is a formula that tells us about all "simple constraints" that group theory - the mathematical incarnation of the wisdom about symmetries, especially in the $SO(3)\approx SU(2)$ case (rotations in a three-dimensional space) - implies about matrix elements of tensor operators - those that transform as some representation of the same symmetry.

The dependence on the indices labeling basis vectors of the representations - $m_1, m_2, m_3$ in the $SU(2)$ case - is totally determined. Instead of thinking that some matrix elements depend on many variables, physicists may realize that the symmetry guarantees that the matrix elements only depend on a few labels labeling the whole "multiplets" of the states and operators rather than on all the labels identifying the individual components. It's always critically important to know how much freedom or how much uncertainty there is about some observables - for example, experimenters don't want to repeat their experiment $(2J+1)^3$ times without a good reason - and we would get a totally wrong idea without this theorem.

To see that the theorem is used all the time, check e.g. these 5700 papers


many of which are highly cited ones. The topics of the papers include optics, nanotubes, X-rays, spectroscopy, condensed matter physics, mathematical physics involving integrable systems, quantum chemistry, nuclear physics, and virtually all other branches of physics that depend on quantum mechanics. In many other cases, the theorem is being used without mentioning its name, and its generalizations are being used all the time in advanced theoretical physics, e.g. in the contexts with groups that are much more complicated than $SO(3)\approx SU(2)$.

It's interesting to note that in some sense, quantum mechanics allows the symmetry to impose as many constraints as classical physics. In classical physics, we could start with an initial state labeled by numbers $I_i$, apply some operations depending on parameters $O_j$, and we would obtain a final state described by parameters $F_k$. Classical physics would tell us Yes/No - whether we can get from $I_i$ via $O_j$ to $F_k$: that's the counterpart of the quantum probability amplitude.

The rotational $SO(3)$ symmetry - which has 3 parameters (3 independent rotations - or the latitude; longitude of the axis, and the angle) would only tell us that if we rotate all objects $I_i,O_j,F_k$ by the same rotation, we obtain a valid proposition again (Yes goes to Yes, No goes to No). So the dependence on 3 parameters - corresponding to the 3 rotations - is eliminated. In quantum mechanics, we also eliminate the dependence on 3 parameters - in this case $m_1,m_2,m_3$, the projections $j_z$ for the two state vectors and for the operator sandwiched in between them. In some proper counting, this is true for any $d$-dimensional group of symmetries, I think.


And if you are interested in "where" it might come useful: Some of the selection rules for optical transitions can be obtained from it, and I faintly recall that it helps rewriting the Hamiltonian for Spin-Orbit coupling into a much more convenient form.


1) Remember that a tensor operator is a collection (i.e. a set of 2j+1) operators which transform amongst themselves under group transformation (let's stick to SU(2) or SO(3)) or commutation with the generators of su(2) or su(3) (if we use infinitesimal transformation).

2) The Wigner-Eckart theorem ultimately states that some operators (the components of the tensor operator) can be written in terms of the spherical coordinates as a common function of the radial variable r multiplied by a spherical harmonic specified by the component.

3) The simplest examples are the observables x,y and z, which can be written as r*(spherical harmonic $Y_{1m}$), v.g. $z=r \cos(\theta) \propto r Y_{10}(\theta,\phi)$.
The common function here is r itself. For more complicated operators, more complicated functions are encountered, but the central point is that the components of a tensor operator of angular momentum L are always of the form $f(r) Y_{LM}$, having a common f(r) and different components M.

4) Because the function f(r) is common to all the components, the radial integral between various basis states can be evaluated using any component of the tensor, i.e. any component M. This radial integral is basically the reduced matrix element (or at least very closely related to this reduced matrix elements as some definitions have extra factors in the numerator or denominator). The angular part is an integration of spherical harmonics so proportional to some Clebsh-Gordan coefficient involving the angular quantum numbers $(L,M)$ of the component of the tensor and the angular quantum numbers, $(L_i, M_i)$ and $(L_f,M_f)$ of the initial and final states, respectively. The selection rules enter because $M_i+M=M_f$ and $L_f$ must be in the decomposition of $L\otimes L_i$ (i.e. must be in the usual range $\vert L_i-L \vert \le L_f \le L_i+L$).

5) The use is very simple: a) ratios of matrix elements of tensor operators will not depend on the reduced matrix elements, so for instance ratios of decay rates (or cross sections) of components of a tensor operator can be compared to experiment without any explicit knowledge of the function f(r) or the (often complicated) radial integral needed to evaluate the reduced matrix element - the integration over the radial variable is the same for all components so drops out of any ratio involving those components. b) If the reduced matrix element is known then the matrix element of any component of the tensor can be calculated as the integration over the angles (the $Y_{LM}$ part) is easily done analytically.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.