# Why does zero $x$ angular momentum imply nonzero $z$ angular momentum?

Assume we have a state $$\psi =|n, L =1, L_x =0\rangle$$. If we compute it's reprentation in $$L_z$$ basis we get: $$\psi = \frac{1}{\sqrt{2}}|n, L =1, L_z =1\rangle - \frac{1}{\sqrt{2}}|n, L =1, L_z =-1\rangle.$$

I am perfectly comfortable with the fact that I get a superposition of states as $$L_x$$ and $$L_z$$ don't commute. Furthermore, the expected value of $$L_z$$ is $$0$$ just as expected.

However the eigenstate $$|n, L =1, L_z =0\rangle$$ doesn't contribute. What is the physical interpretation of this fact?

There seems to be a logical mismatch between the title of your question and the final coda question, unless you are conflicted about some internal connection between the two?

The triplet representation of angular momentum in the spherical basis is $$L_z=\operatorname{diag}(1,0-1)$$, $$L_+=\sqrt{2} \begin{bmatrix} 0& 1 &0 \\ 0& 0& 1\\ 0&0&0\end{bmatrix}, \qquad J_-=J_+^T,$$ hence $$\sqrt{2} ~ L_x= \begin{bmatrix} 0& 1 &0 \\ 1& 0& 1\\ 0&1&0\end{bmatrix},$$ whose evident null vector $$\psi=1/\sqrt{2} \begin{bmatrix} 1 \\ 0\\ -1\end{bmatrix},$$ lacks a middle component. You see this is unavoidable from the sparse first row of $$L_x$$, which, in turn, follows the sparse first row of the raising operator. Likewise, you may further observe the null vector of $$L_y$$ also lacks a middle component, for essentially the same reason.

What connection are you conjecturing?

• Bonus point: Now repeat for L=2, instead. Do you see the failure of the null vectors now? They do have a nonvanishing middle component.
• I am trying to get intuition as to why the eigenvector L_z =0 isn't part of the superposition. May 23, 2019 at 15:44
• The argument of @CosmasZachos shows this is not related to the choice of quantization axis but rather generic. May 23, 2019 at 19:14
• I understand the math involved. I am trying to understand the physical meaning behind it. May 24, 2019 at 8:51
• Context? Any spurious "meaning" will have to survive the unforgiving obviation of higher spins. May 24, 2019 at 14:39

# No classical reason

I don't think that there is an obvious “deeper reason” here.

So let's suppose you have $$L_x, L_y, L_z$$ satisfying standard spin relations that e.g. $$[L_x,L_y]=iL_z$$ with eigenkets e.g. $$\def\ket#1{~\big|{#1}\big\rangle}\def\bra#1{\big\langle{#1}\big|~}\def\braket#1#2{\big\langle{#1}\big|{#2}\big\rangle}\def\lket#1{|{#1}\rangle} L_x\ket{-1_x}=-\ket{-1_x},\\ L_x\ket{0_x}=0,\\ L_x\ket{1_x}=\ket{1_x}.$$I believe your question is why $$\braket{0_x}{0_z}=\braket{0_y}{0_z}=\braket{0_x}{0_z}=0.$$

The problem is that this is not true, at least not given the facts that I have just given you. Like, those facts are certainly enough to establish, say, $$e^{-i\phi L_z}$$ as a unitary operator performing a rotation about the $$z$$-axis. But for example for a spin-2 particle we should have $$L_x = \begin{bmatrix}0&1&0&0&0\\ 1&0&1&0&0\\ 0&1&0&1&0\\ 0&0&1&0&1\\ 0&0&0&1&0\end{bmatrix},$$with $$\ket{0_x}=\sqrt{\frac13}\Big[\ket{-2_z} - \ket{0_z} + \ket{2_z}\Big].$$ That doesn't have $$\braket{0_z}{0_x}=0$$.

What is clear is that this alternating series plus a "gap by two units of spin" pattern is likely to recur, and so $$\braket{0_x}{0_z}$$ is distinctive of spin-odd systems (spin-1, spin-3, ...) as opposed to spin-even systems (spin-2, spin-4, ...) where it is likely to be $$\pm 1.$$ And it doesn't seem like there's going to be any easy classical physical intuition which is going to give you some insight, “oh, so that’s how spin-even and spin-odd systems are different and therefore they must have orthogonal spin-zero states in the odd case, but not in the even case.”

# One possibility: spin compositions?

The only thing I can think might be a bit of a way to generate nice intuition would perhaps be to use spin-composition rules. So for example it's well-known that given two spin-$$1/2$$ particles they have four spin states which they can be in, but a Hamiltonian measuring total angular momentum splits them into a singlet state with zero angular momentum,$$\sqrt{1/2}\ket{\uparrow\downarrow}-\sqrt{1/2}\ket{\downarrow\uparrow}$$, plus a spin-1 particle, \begin{align} \ket{1} &= \ket{\uparrow\uparrow}\\ \ket{0} &= \sqrt{\frac12}\ket{\uparrow\downarrow} + \sqrt{\frac12} \ket{\downarrow\uparrow}\\ \ket{-1} &=\ket{\downarrow\downarrow} \end{align}Identifying $$\lket\rightarrow\propto\lket\uparrow+\lket\downarrow,$$ and $$\lket\leftarrow\propto \lket\uparrow-\lket\downarrow$$, the $$x$$-triplet state also has angular momentum 1 but is \begin{align} \ket{1_x} &\propto ~\uparrow\uparrow + \uparrow\downarrow + \downarrow\uparrow+\downarrow\downarrow\\ \ket{0_x} &\propto~ \uparrow\uparrow - \downarrow\downarrow\\ \ket{-1_x} &\propto~ \uparrow\uparrow - \uparrow\downarrow - \downarrow\uparrow+\downarrow\downarrow, \end{align}showing the basic $$\langle 0_x|0_z\rangle = 0$$ behavior. And now there are more places to hang your physicist’s intuition on; the most conspicuous intuition here is that you get an $$\uparrow\uparrow + \downarrow\downarrow$$ term for both $$\rightarrow\rightarrow$$ and $$\leftarrow\leftarrow$$, since $$(-1)^2 = 1,$$ so that in some sense the spin-zero state must be entirely devoted to this $$\lket{\uparrow\uparrow}-\lket{\downarrow\downarrow}$$ term to be orthogonal to them.

There is a lot of math available in spin composition, connection to Young diagrams for example. So if you can repeat this argument for the spin-2 case, whether it's as a subset of two spin-1 particles or as a subset of four spin-1/2 particles, you might have some chances to create interesting intuitions.

Here you might want to look at spherical harmonics (at $$r=1$$):

$$Y_1^{\pm 1} \propto x \pm iy$$

$$Y_1^0 \propto z$$

so those are by default in the $$z$$-basis. It is pretty clear that $$m=0$$ harmonic in the $$x$$ basis (which I'll call...$$X$$? not that great of a notation, but it's all I have):

$$X_1^0 \propto x$$

and that is completely orthogonal to $$z$$. It's immediately clear that:

$$x \propto Y_1^{+1} + Y_1^{-1}$$

I know that's not a "physics" reason, it's just math, but if you consider waves on a spherical surface...then it's physics: counter rotating waves going around the equator (w.r.t to $$z$$) add up to be an $$x$$-dipole, and that contains zero $$z$$-dipole.