Intuition for Total Angular Momentum Quantum Number

Question

I am trying to build intuition about angular momentum states in quantum mechanics. I'll use $\vec{\boldsymbol{V}}$ to represent a quantum angular momentum operator. This could be orbital or spin angular momentum for any sort of particle or system. I'll take the $z$ axis to be the quantization axis for reference.

My general question is suppose I have two states with differing TOTAL spin quantum number, $V$ but the same spin projection quantum number $m_V$. What is an intuitive picture for the difference between two such states? Or perhaps what general statement can be made about the differences between two such states? I'll focus on states with $m_V=0$ since I think that should be the simplest case to think about but hopefully the intuitions can extend to higher angular momenta projections.

For one example, what is the difference between the $\vert n=3,l=2,m=0\rangle$ and $\vert n=3, l=1,m=0\rangle$ states for the hydrogen electron? I can look up or simulate how these states look and of course they look different because they involve spherical harmonics of different orders. However, I'm not able to verbalize what generically makes them different. Since $m=0$ for both states they are both azimuthally symmetric. If I look at the spherical harmonics $Y_2^0$ and $Y_1^0$ I can see that $Y_1^0$ is basically an upper and a lower hemisphere whereas $Y_2^0$ has 3 (rather than 2) azimuthal bands so that sort of helps..

Another example which throws some wrenches into the intuition I might build from the previous example is the case of a spin singlet vs. a spin triplet. We know that for two spin $\frac{1}{2}$ particles we have

$$ \vert S=0,m_s=0\rangle = \frac{1}{2}\left(\vert \uparrow \downarrow \rangle - \vert\downarrow \uparrow\rangle \right) $$

Whereas

$$ \vert S=1,m_s=0\rangle = \frac{1}{2}\left(\vert \uparrow \downarrow \rangle + \vert\downarrow \uparrow\rangle \right) $$

The symmetric state has total spin $1$ whereas the anti-symmetric state has total spin $0$. In this case we can sort of see again that the total spin 1 state is just "different" than the total spin 0 case but again i can't articulate exactly in what way it is different. A more general case of looking at this example is what is the difference between $\vert 0,0 \rangle$ AND $\vert 1,0 \rangle$ intrinsic spin states regardless of that state arose as the composition of multiple smaller spin systems or not.

Here are some incomplete propositions for what difference total spin makes for a quantum state.

-$\vert n=3,l=2,m=0 \rangle$ and $\vert 1,0\rangle$ are different from $\vert n=3,l=1,m=0\rangle$ and $\vert 0,0\rangle$ in that they belong to different multiplets of angular momentum states. For example, $\vert 1,0\rangle$ has something in common with $\vert 1,-1 \rangle$ and $\vert1,+1 \rangle$ which $\vert 0,0\rangle$ does not have in common with those states.

-All of these states have $0$ projection of angular momentum along the $z$ axis. So something that they have in common is that I should NOT imagine any rotation happening for such states (whereas for $m_V \neq 0$ I can imagine the particle or intrinsic spin rotating about the $z$ axis in some way). However, if we consider what the states look like in terms of basis states of $x$ or $y$ (rather than $z$) there will be differences. For example, the singlet state is total spin 0 so that means it is more like a scalar then a vector. So whether you look in the $x$, $y$, or $z$ basis it will look the same. That is $m_x$ and $m_y$ and $m_z$ would all be zero for a singlet state. However, for the triplet state if $m_z=0$ then one would find the expression in $x$ and $y$ bases to be a superposition of $m_x=\pm 1$ or $m_y = \pm 1$. So somehow multiple dimensions of rotation need to be taken into account. Similar results would be found for the spherical harmonic case where you are looking at orbital angular momentum.

-Irreducible representation etc. etc. I've taken a few stabs at understanding representation theory for quantum angular momentum but it has never stuck. I'm sure the answer is probably explicated pretty well in that formalism so if someone can help me put together the right words from this perspective that would be great.

-Along with the last point I wonder if there is a simple group theory answer to my questions.

Short statement of question: The short statement of my question is what in particular makes states of different total angular momentum different. Or, perhaps, what do states of the same total angular momentum have in common?

Answer 1

The total angular momentum quantum number j tells you to use a (2j+1) x (2j+1) matrix to rotate the (2j+1) component vector that stands for the particle. It is said that j labels the different representations of the rotation group.

For example, a pion transforms like 1-vector (j=0) under rotations. All rotations just take the m=0 state into itself with a phase factor.

For example, an electron transforms like a 2-vector (j=1/2) under rotations. Rotations (2 x 2) matrices can turn the m=-1/2 state into linear combinations of all the states m=-1/2,+1/2.

For example, a $\rho$ meson transforms like a 3-vector (j=1) under rotations. Rotations (3 x 3) matrices can turn the m=0 state into linear combinations of all the states m=-1,0,+1.

For example, a $\Delta$ baryon transforms like a 4-vector (j=3/2) under rotations. Rotations (4 x 4) matrices can turn the m=-3/2 state into linear combinations of all the states m=-3/2,-1/2,+1/2,+3/2.

For example, a graviton transforms like a 5-vector (j=2) under rotations. Rotations (5 x 5) matrices can turn the m=0 state into linear combinations of all the states m=-2,-1,0,+1,+2.

While true that j$\hbar$ can be related to mechanical angular momentum, j really just tells you the representation of the rotation group to use to rotate an object.

Answer 2

I realized the answer to this question is actually pretty straightforward and was staring me in the face all along. My second comment about a possible answer has the key. I was focusing to hard on what was happening in the $z$ axis and not thinking about the $x$ and $y$ axes.

I said that since $m_z=0$ I should not imagine any rotation happening. Well, that is only true around the $z$ axis. If the $\vert 1,0\rangle$ state is written in the $x$ basis we have

$$ \vert 1,0\rangle_z = \frac{1}{\sqrt{2}}\left(\vert 1,+1\rangle_x + \vert1,-1\rangle_x\right) $$

That is, the state $\vert 1,0\rangle$ has no angular momentum in the $z$ direction BUT it has non-zero total angular momentum so it must have angular momentum somewhere! Well this shows up (symmetrically) in the $x$ and $y$ directions in a superposition sense.

In short I was too quick to think that $m_z=0$ means there is no rotation going on.

The answer to my question is that the difference between states with different values for $V$ but the same value for $m_V$ is that the state with a larger $V$ does really have larger angular momentum in the $x$ and $y$ directions than the state with the smaller $V$.