The Wigner-Eckart theorem tells us that for any tensor operator, $\mathbf{T}^{(k)}$ that

\begin{align} \langle jm|T^{(k)}_q|j'm'\rangle = \langle j'm'kq|jm\rangle \langle j||\mathbf{T}^{(k)}||j'\rangle \end{align}

Consider a system in quantum state $$ |\psi\rangle = \sum_{j',m'} |j'm'\rangle \langle j'm'|\psi\rangle $$

we then have

\begin{align} \langle T^{(k)}_q\rangle &= \langle \psi|T^{(k)}_q|\psi\rangle = \sum_{j,j',m,m'} = \langle\psi|jm\rangle \langle jm|T^{(k)}_q|j'm'\rangle\langle j'm'|\psi\rangle\\ &= \sum_{j,j',m,m'}\langle \psi|jm\rangle\langle j'm'kq|jm\rangle\langle j||\mathbf{T}^{(k)}||j'\rangle\langle j'm'|\psi\rangle \end{align}

Suppose that the system is in a state of fixed total $\mathbf{J}$ so that $\langle j'm'|\psi\rangle \propto \delta_{j'j_0}$ so we get

\begin{align} \langle T^{(k)}_q\rangle &= \sum_{m,m'} \langle \psi|j_0m\rangle \langle j_0mkq|j_0m'\rangle \langle j_0||\mathbf{T}^{(k)}||j_0 \rangle \langle j_0m'|\psi\rangle\\ &=\langle j_0||\mathbf{T}^{(k)}||j_0\rangle\sum_{m,m'} \langle \psi|j_0m\rangle\langle j_0mkq|j_0m'\rangle\langle j_0m'|\psi\rangle\\ &=\langle j_0||\mathbf{T}^{(k)}||j_0\rangle A_{kq\psi} \end{align}

Similarly, for a different tensor operator (also of rank $k$) we have

$$ \langle W^{(k)}_q\rangle = \langle j_0||\mathbf{W}^{(k)}||j_0\rangle A_{kq\psi} $$

We then see that

\begin{align} \langle T^{(k)}_q\rangle &= \frac{\langle j_0||\mathbf{T}^{(k)}||j_0\rangle}{\langle j_0||\mathbf{W}^{(k)}||j_0\rangle} \langle W^{(k)}_q\rangle\\ &= B_{k\mathbf{T}\mathbf{W}j_0} \langle W^{(k)}_q \rangle \end{align}

This tells us more generally that

\begin{align} \langle \mathbf{T}^{(k)}\rangle = B_{k\mathbf{T}\mathbf{W}j_0} \langle \mathbf{W}^{(k)}\rangle \end{align}

This says that the expectation value of any two tensor operators are linearly proportional to eachother. Consider two vector operator $\mathbf{X} = \mathbf{X}^{(1)}$ and $\mathbf{Y} = \mathbf{Y}^{(1)}$. Then we have that

\begin{align} \langle \mathbf{X}\rangle = B_{1\mathbf{X}\mathbf{Y}j_0}\langle \mathbf{Y}\rangle \end{align}

Telling us that the expcation values of $\mathbf{X}$ and $\mathbf{Y}$ are linearly proportional. This is surprising to me because classically it is possible to have a system containing vectors pointing in multiple directions. For example, one conclusion of the above analysis is that $\langle \mathbf{L} \rangle$ is always proportional to $\langle \mathbf{S} \rangle$. I didn't think this was always the case..

Perhaps it is important that I have restricted the system to a state of well defined total $\mathbf{J}$. Perhaps when that is done it indicates the system has a special type of rotational symmetry (not generic to all states) which exactly satisfies the above surprising constraints I have demonstrated.

Can anyone please shed some light on the correctness/surprisingness of my conclusions?

Related: Why must the electron's electric dipole moment (EDM) always be aligned with the spin? Also, this question arose from considering this derivation/definition of the Lande-g factor which relies on the fact that $\langle \mathbf{\mu} \rangle \propto\langle \mathbf{J} \rangle$.


Your conclusion is perfectly correct, but it only applies when the state of the system is known to lie in a single irreducible representation of $SU(2)$, i.e. it has a definite total angular momentum $j$.

For atomic nuclei, this is often a good approximation, because higher spin states will be substantially higher in energy. At low energies, the nucleus can be regarded as having a definite spin. Then it follows by your argument that all vector operators must be parallel, e.g. $\langle \mu \rangle \propto \langle \mathbf{J} \rangle$, even though generally they would be independent. In fact, I'm not sure that very common, quantum mechanics 101 statement makes any sense at all without using the Wigner-Eckart theorem.

The conclusion only sounds strange because most systems do not satisfy the condition. Classical systems have relatively sharply defined angular orientations and hence must occupy many values of $j$ by the angular version of the uncertainty principle.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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