We have a scalar operator $A$, being invariant under rotations which commutes with the angular momentum, i.e. $$[A,J_i]=0 \text{ where } i=x,y,z$$ $$[A,J^2]=0 $$

So eigenfunctions of $A$ can be chosen such that they are eigenfunctions of $J^2$ and $J_z$. I know the corresponding eigenvalues to these operators are $\hbar^2j(j+1)$ and $\hbar m$ respectively.

I shall show, that the eigenvalues of $A$ do not depend on the magnetic quantum number $m$. I feel this should go along the lines of the commutator ones again but I don't get anywhere.

Something like this:

Suppose $\psi$ are eigenfunctions of all three operators and $$A\psi = \lambda \psi$$ where $\lambda=\lambda(m)$ and lead this to a contradiction. Any help appreciated!

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    $\begingroup$ Schur's lemma... $\endgroup$ – Valter Moretti Oct 28 '17 at 12:31
  • $\begingroup$ Sorry but I am not yet familiar with group theory / representation theory and I have never studied this lemma. Is there no easier way to get there? $\endgroup$ – Marsl Oct 28 '17 at 13:01
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    $\begingroup$ The crucial remark is that the eigenspace at fixed $j$ whose basis is labeled by the eigenvalues of $L_z$ is irreducible. By Schur's lemma every matrix commuting with the $L_k$ is a multiple of the identity matrix... $\endgroup$ – Valter Moretti Oct 28 '17 at 13:06
  • $\begingroup$ Another proof, more direct, can be obtained by means of the ladder operators: $A$ commutes with them... $\endgroup$ – Valter Moretti Oct 28 '17 at 13:08
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    $\begingroup$ @ValterMoretti Your comments would be very useful to a great many undergraduates if you could expand them into a crisp and well-written answer. $\endgroup$ – ZeroTheHero Oct 28 '17 at 14:00

First of all the question has to be stated into a more precise form.

The Hilbert space $H$ is an orthogonal direct sum of the eigenspaces $H_j$ of $J^2$: $$H = \oplus_{j}H_j \tag{1}$$ where, obviously, $$J|_{H_j}= j(j+1)I\:.\tag{2}$$ Since $A$ and $J_k$ commute with $J^2$, every $H_j$ is invariant under the action of these operators: $$A(H_j) \subset H_j\:,\quad J_k(H_j) \subset H_j\:.\tag{3}$$

It is cleat that (1) and (3) imply that $A$ is known provided it is known in each subspace $H_j$. I therefore henceforth consider the restrictions $A|_{H_j}$ and $J_z|_{H_j}$ to a generic $H_j$, considering $H_j$ as the Hilbert space of the theory, though I will use the simpler notation $A$ and $J_j$ in place of $A|_{H_j}$ and $J_z|_{H_j}$.

Since $A$ and $J_z$ commute, it may happen that, $A=f(J_z)$ for some non-constant function $f$. In other words, an eigenvector $|j,m\rangle$ of $J_z$ with eigenvalue $m$ is also an eigenvector of $A$ with eigenvalue $f(m)$ for some non-constant function $f$.

Let us prove that the function $f$ is actually constant. This is the thesis written into a more precise form.

In other words, $A$ restricted to $H_j$ is of the form $cI$.

As $L_k$ commutes with $A$, $A$ commutes with $J_\pm$ which are linear combinations of $J_x$ and $J_y$ and therefore, from $$A|j,-j\rangle = f(-j)|j,-j\rangle$$ we have $$J_+A|j,-j\rangle = f(-j)J_+|j,-j\rangle$$ that is $$AJ_+|j,-j\rangle =C_j f(-j)|j,-j+1\rangle$$ namely $$C_j A|j,-j+1\rangle = C_jf(-j)|j,-j+1\rangle\:.$$ For some non vanishing $C_j$, so that $$A|j,-j+1\rangle = f(-j)|j,-j+1\rangle\:.$$ Repeating the operation (finding constants $C_m\neq 0$) we get $$A|j,m\rangle = f(-j)|j,m\rangle \quad \mbox{if $m=-j,-j+1,\ldots, j$.}$$ In other words, explicitly writing the restriction to $H_j$, since the vectors $|j,m\rangle$ forms a basis of $H_j$, $$A|_{H_j} = \sum_{m} f(-j) |j,m\rangle \langle j,m| = f(-j)I\:.$$ In every subspace $H_j$, $A$ is the constant $aI$ operator and this constant may depend on $j$. In the entire Hilbert space: $$A = \sum_{j,m} a_j |j,m\rangle \langle j,m| = \oplus_j a_jI_j\:.$$

As a matter of fact we have established the most elementary version of Wigner-Eckart theorem.

  • $\begingroup$ Nicely and simply done. As a technical detail, since in general the action of $J_+$ on $\vert jm\rangle$ depends on $m$, how to you "eliminate" this $m$ dependence? i.e. $AJ_+\vert jm\rangle =f(-j) \sqrt{(j-m)(j+m+1)}\,\vert j,m+1\rangle$ $\endgroup$ – ZeroTheHero Oct 28 '17 at 15:49
  • $\begingroup$ Yes, sure, however these coefficients cancel since they appear on both sidesl. I corrected my text taking them into account. $\endgroup$ – Valter Moretti Oct 28 '17 at 15:58
  • $\begingroup$ Wow, thanks a lot. Actually the first part of stating the question in a precise form might have helped me even more than the second one. I will think through it a few times, now and see whether there is something not quite clear to me. Cheers! $\endgroup$ – Marsl Oct 28 '17 at 16:09
  • $\begingroup$ @ValterMoretti Right... I was thinking you could have used (unnormalized) states defined as $J^k_+\vert j,-j\rangle$ without explicit reference to $\vert j,-j+k\rangle$ without affecting your argument. $\endgroup$ – ZeroTheHero Oct 28 '17 at 16:41
  • $\begingroup$ No, I prefer to use normalized eigenvectors in order to obtain the standard spectral decomposition... $\endgroup$ – Valter Moretti Oct 28 '17 at 16:47

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