Graphical representations of the vector model of quantum angular momentum

Question

This question is in reference to the book "Introduction to Modern Physics" by Richtmyer and Kennard, particularly in their discussion of the graphical representation of quantized angular momentum.

They describe that for a given state $|J,m_j\rangle$, one "pictorial" alternative is to draw vectors with a length of $J$, and to pick the angle such that it projection towards the chosen axis is $m_J$, and they provide the following example (in their notation they write $m_J$ as $M$):

So far so good. Then they suggest that a more accurate representation would be to draw the vector with a length of $\sqrt{J(J+1)}$. They mention that in this case, the vector is necessarily inclined even when $m_J = \pm J$, so it represents better that in general $\vec{J}$ has no definite direction. I believe this agrees with this kind of representation:

But right after that they make this rather cryptic statement:

The square of its component perpendicular to the axis then represents correctly the value of the square of the corresponding component of the angular momentum

First, shouldn't it be the component parallel to the chosen axis? And secondly, I'm not sure why are they mentioning the square of the component $m_J^2$. Are they suggesting a different graphical representation or is this just a simple mistake on their end? The only thing I could think of is that in the particular case $m_J=\pm J$, then the component of the vector perpendicular to the axis is $\sqrt{J}$, so its square does equal $m_J$, but it is definitely not $m_J^2$ (so the statement is still wrong) and this only occurs at the extreme values.

Answer 1

Your text should indicate this is a graphical shared metaphor in the community to summarize the quantum picture geometrically, of help in rough estimates. Skipping the real McCoy, the cartoon metaphor cannot avoid appearing confusing.

The real Mc Coy is a 3-vector of operators $\hat J_𝑥,\hat J _𝑦,\hat J_𝑧$, all (2𝐽+1)×(2𝐽+1) matrices, $$ \vec{\hat J}=\begin{pmatrix}\hat J_𝑥\\ \hat J_𝑦\\ \hat J_𝑧\end{pmatrix}. $$

Because of the Lie algebra underlying these components, they cannot be diagonalized simultaneously, so their eigenvalues that physics cares about cannot be displayed in the same picture. There is a salutary fact, however: both $\hat J_z$ and $\vec{\hat J}⋅\vec{\hat J} ≡ \hat J_z^2+\hat J_y^2+\hat J_x^2≡\hat J_z^2+ \vec{\hat J}_⊥⋅\vec{\hat J}_⊥$ do have simultaneous eigenvectors, for their respective eigenvalues $J,J−1,...,−J$ and $𝐽(𝐽+1)$, and hence so does $\vec{\hat J}_⊥⋅\vec{\hat J}_⊥$, namely $𝐽(𝐽+1)−𝑀^2$, for all $|𝑀|≤𝐽$.

You may then summarize your J=2 cartoon plotting vectors of length $\sqrt{𝐽(𝐽+1)};𝑀;$ and $\sqrt{𝐽(𝐽+1)−𝑀^2}$ on your cones: the topmost vertical vector of length 2 in your picture is M=2; the radius of the topmost circle at its tip is $\sqrt{2⋅3−4}=\sqrt{2}$; the length of the "hypotenuse" vector from the origin to the circle is $\sqrt 6$; also, the radius of the central circle on the $𝐽_𝑥,𝐽_𝑦$ plane, a degenerate cone, is also $\sqrt 6$; the remaining three cones represent $𝑀=1;−1;−2$, respectively.

Indeed, as the authors reassure you, for $J=M=2$, there is still a cone, the topmost, ie, the x and y components don't vanish!

Ignore the axial/azimuthal direction: it's not really a precession. It reminds you that, at some level, the x and y components are symmetric/equivalent.

As indicated in the comments, without the real McQM, the picture is not very meaningful.