Seeking a quality plain-language description of the Wigner-Eckart theorem

Question

I'm a third year physics undergrad with a very cursory knowledge of quantum mechanics and the formalism involved. For instance, I understand roughly how tensors work and what it means for a tensor to be irreducible, though it would take me a lot of work to apply this knowledge to a problem/extend it past what I've already seen.

As part of a project, I'm studying atomic nuclei in electric and magnetic fields. I'm trying to understand the energy of a nuclear quadrupole's interaction with an electric field gradient. The equation for this is

$E_Q = \sum_{\alpha,\beta} V_{\alpha\beta} Q_{\alpha\beta}$

where $\alpha$ and $\beta$ each iterate over $x, y, z$.

$Q$, the electric quadrupole moment, is given by

$Q_{\alpha\beta} = [\frac{3}{2}(I_{\alpha} I_{\beta} + I_{\beta} I_{\alpha}) - \delta_{\alpha\beta}I^2] * constant$

(taken from this powerpoint.) An electric quadrupole moment should have nothing to do with nuclear spin... or so I thought, until I ran across the idea of "spin coordinates" and the Wigner-Eckart theorem. This is roughly all I know about the theorem -- that it exists and that it can somehow convert between Cartesian and spin coordinates in quantum systems -- and I'd like to understand it better.

THE SHORT VERSION: I do not need a detailed mathematical understanding of the Wigner-Eckart theorem, but I'm very curious as to the general idea of it. Can anyone think of a plain-English (or rather, minimal-math) explanation of the theorem that would make sense to a beginning quantum physics student?

Answer 1

This question inspired me to try to write a conceptual introduction at the wikipedia article. To save you the trouble of clicking, I copied it below. (It's slightly inspired by what @Kostia wrote here)

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Motivating example: Position operator matrix elements for 4d→2s transition

Let's say we want to calculate transition dipole moments for an electron to transition from a 4d to a 2p orbital of a hydrogen atom, i.e. the matrix elements of the form $\langle 2p,m_1 | r_i | 4d,m_2 \rangle$, where r_i is either the x, y, or z component of the position operator, and m₁, m₂ are the magnetic quantum numbers that distinguish different orbitals within the 2p or 4d subshell. If we do this directly, it involves calculating 45 different integrals: There are three possibilities for m₁ (-1, 0, 1), five possibilities for m₂ (-2, -1, 0, 1, 2), and three possibilities for i, so the total is 3×5×3=45.

The Wigner–Eckart theorem allows one to obtain the same information after evaluating just one of those 45 integrals (any of them can be used, as long as it is nonzero). Then the other 44 integrals can be inferred just using algebra, with the help of Clebsch–Gordan coefficients, which can be easily looked up in a table, or computed by hand or computer.

Qualitative summary of proof

The Wigner–Eckart theorem works because all 45 of these different calculations are related to each other by rotations. If an electron is in one of the 2p orbitals, rotating the system will generally move it into a different 2p orbital (usually it will wind up in a quantum superposition of all three basis states, m=+1,0,-1). Similarly, if an electron is in one of the 4d orbitals, rotating the system will move it into a different 4d orbital. Finally, an analogous statement is true for the position operator: When the system is rotated, the three different components of the position operator are effectively interchanged or mixed.

If we start by knowing just one of the 45 values—say we know that $\langle 2p,m_1 | r_i | 4d,m_2 \rangle = K$—and then we rotate the system, we can infer that K is also the matrix element between the rotated version of $\langle 2p,m_1 |$, the rotated version of $r_i$, and the rotated version of $| 4d,m_2 \rangle$. This gives an algebraic relation involving K and some or all of the 44 unknown matrix elements. Different rotations of the system lead to different algebraic relations, and it turns out that there is enough information to figure out all of the matrix elements in this way.

(In practice, when working through this math, we usually apply angular momentum operators to the states, rather than rotating the states. But this is fundamentally the same thing, because of the close mathematical relation between rotations and angular momentum operators.)

In terms of representation theory

To state these observations more precisely, and to prove them, it helps to invoke the mathematics of representation theory. For example, the set of all possible 4d orbitals (i.e., the five states m=-2,-1,0,1,2 and their quantum superpositions) form a 5-dimensional abstract vector space. Rotating the system transforms these states into each other, so this is an example of a "group representation"—in this case, the 5-dimensional irreducible representation ("irrep") of the rotation group SU(2) or SO(3), also called the "spin-2 representation". Similarly, the 2p quantum states form a 3-dimensional irrep (called "spin-1"), and the components of the position operator also form the 3-dimensional "spin-1" irrep.

Now consider the matrix elements $\langle 2p,m_1 | r_i | 4d,m_2 \rangle$. It turns out that these are transformed by rotations according to the direct product of those three representations, i.e. the spin-1 representation of the 2p orbitals, the spin-1 representation of the components of r, and the spin-2 representation of the 4d orbitals. This direct product, a 45-dimensional representation of SU(2), is not an irreducible representation—instead it is the direct sum of a spin-4 representation, two spin-3 representations, three spin-2 representations, two spin-1 representations, and a spin-0 (i.e. trivial) representation. The nonzero matrix elements can only come from the spin-0 subspace. The Wigner–Eckart theorem works because the direct product decomposition contains one and only one spin-0 subspace, which implies that all the matrix elements are determined by a single scale factor.

Apart from the overall scale factor, calculating the matrix element $\langle 2p,m_1 | r_i | 4d,m_2 \rangle$ is equivalent to calculating the projection of the corresponding abstract vector (in 45-dimensional space) onto the spin-0 subspace. The results of this calculation are the Clebsch–Gordan coefficients.

Answer 2

Possibly (for this purpose) the simplest expression of the Wigner-Eckhart theorem in plain language is "what else could it be?" The angular motion of the nucleus is described by the spin. The spin operator is a vector. We need a second rank tensor for the quadrapole interaction. From the spin operator, you can only make one second rank traceless symmetric tensor, and so you use that. This is a simple example, of course, and (as usual) the simple examples in group theory allow you to get the correct answer without (really) knowing what you are doing. However, this is the "nub" of Wigner-Eckhart - there are only a finite number of possibilities (which can be computed using group theory) to express tensors in terms of operators that describe the states. You need to verify that you all such possibilities, represented at least once. To be sure, you really need non-plain language Wigner Eckhart / Group representation theory

Answer 3

So this explanation (my first post on stackexchange!) is based on H. Georgi's "Lie algebras in particle physics", chapter 4. Since $Q_{ij}$ is symmetric, real and traceless, it has 5 independent degrees of freedom. So it's possible to express $Q_{ij}$ into a 'spherical' basis $Q^s_l$, where $s=2$ in this case and $l$ takes on values -2, -1, 0, 1, 2. (For completness, a spherical tensor operator that transforms under the spin s representation of SU(2) is a set of operators such that: $[J_a,Q^s_l]=Q^s_m[J^s_a]_{ml}$).

The crux of the Wigner-Eckart theorem is that something like $Q^s_l |j,m,\alpha>$, which would physically correspond to electric effects of an atom which has an angular momentum, actually mathematically behaves like two angular momentum kets tensored together (recall adding angular momentum of particles in quantum mechanics). In terms of notation, this your quadrupole tensor operator acting on some state with total angular momentum squared j(j+1), and z angular momentum m. Here $\alpha$ parametrizes whatever other physics is going on in the system, for example non-perturbative quantum chromodynamic effects in the nucleus, which are super difficult to calculate.

The statement of the Wigner-Eckart theorem is that $<J,m',\beta|Q^s_l|j,m,\alpha>=\delta_{m',l+m}<J,l+m|s,j,l,m> * <J,\beta|Q^s|j,\alpha>$.

On the LHS, you have the probability amplitude of measuring $Q^s_l|j,m,\alpha>$ to have total angular momentum J(J+1), z angular momentum $m'$ and with 'new' physics $\beta$ (which can be quite complicated). On the RHS, the Wigner-Eckart theorem says that if you know the probability amplitude $<J,\beta|Q^s|j,\alpha>$, which can be calculated for ANY of your $Q^s_l$ (calculated means an experimentalist has made a measurement or some hard working graduate student did the calculation), then using only group theory you know $<J,m',\beta|Q^s_l|j,m,\alpha>$ for all the other $l$. All you need to do is calculate (or look up in tables) $<J,l+m|s,j,l,m>$. This term originates from the crux of the theorem that the tensor operator times the ket behaves like two kets, which can be described in the basis of individual angular momentums |s,j,l,m> or in the basis of combined angular momentums $|J,l+m>$.

All this arises because the $Q^s_l$ operators form an irreducible representation, namely from some highest weight vector $Q^2_2$ in your example, you can get all the other $Q^2_l$ just by applying the lowering operator. So they are all related. Thus, the term $<J,\beta|Q^s|j,\alpha>$ which contains all the nasty physics is universal constant within a given irreducible representation. So we don't need to calculate it for every $l$, only one of them (the easiest one).

Hope this helps!

Answer 4

Will attempt an answer. The theorem as you may know is based on representation theory.

Representation theory for Lie groups plays an important part because it states that observables can be constructed from an algebra of generators of the group.

The angular momentum operators are the generators of the spherical group (if i may say)

So each angular momentum operator transforms a state/observale across the sphere which is the underlying Lie group/manifold of a sphericaly symmetric system. And a compound transformation is a linear combination of simpler transformations.

Much like the momentum operator generates space translations (of the system) and the Hamiltonian operator generates time translations (of the system)

Answer 5

I believe that the simplest explanation, in the plainest language, is that the Wigner-Eckhart theorem is a quantum-mechanical expression of conservation of angular momentum.

This may not be self evident, but it's hard to get simpler.