This question is strongly related to this other question regarding the definition of spinor.

Let's take for example the following exercise:

Given a particle with mass $m$ and spin $1/2$, described by the spinor: $$\psi(\vec{x})=Ae^{-br}\begin{pmatrix}z \\ \frac{x-iy}{\sqrt{2}}\end{pmatrix}$$ (with $b>0$ and $A$ normalization constant) what are the possible results of a measurement of $L^2,L_z,S_z$, and the respective probabilities?

To solve this I would think to shift everything to polar coordinates: $$\psi(r , \theta , \phi)=Ae^{-br} r\left( \cos{\theta} \chi _+ +\frac{\sin{\theta}}{\sqrt{2}}e^{-i\phi}\chi_-\right)$$ where $$\chi _+=\begin{pmatrix}1 \\ 0\end{pmatrix} \ \ \ ; \ \ \ \chi_-=\begin{pmatrix}0 \\ 1\end{pmatrix}$$ are the eigenstates of the spin.
Then we can notice that we can rewrite the spinor with the eigenfunctions of angular momentum, so the spherical harmonics $Y(\theta , \phi)$ (this should be useful since we want to find information about $L^2$ and $L_z$). Looking up the table of spherical harmonics we can see that: $$Y_{l=1,m=0}=\sqrt{\frac{3}{4\pi}}\cos{\theta} \ \ \ ; \ \ \ Y_{l=1,m=-1}=\sqrt{\frac{3}{8\pi}}\sin{\theta}e^{-i\phi}$$ so we can write the spinor as: $$\psi=F(r)\left(\frac{Y_{1,0}\chi _+}{\sqrt{2}}+\frac{Y_{1,-1}\chi _-}{\sqrt{2}}\right)$$ and from here we can deduce that $l$ must be one, so the result of the measurement of $L^2$ is for certain $\hbar^2l(l+1)=2\hbar^2$, on the other hand we can see from the $Y$ that $m$ can be $0$ or $1$, with probability one half, and so now we can easily understand that the possible result of a measurement of $L_z,S_z$ are: $S_z=\hbar/2 , L_z=0$ with probability one half and $S_z=-\hbar/2 , L_z=-\hbar$ with also probability one half.

I was pretty happy with this method of resolution and it is also the one featured in my lecture notes. But we can see from the comments of Cosmas Zachos on my previous question as well as from what written in page 194 of Modern Quantum Mechanics by Sakurai, that this procedure of expanding the spinor with the spherical harmonics is incorrect. Again this fact is strongly related with a potential answer to my previous question.

So what is going on here? Is my method of solving this exercise correct? And if so why since multiple sources state that writing the spin in terms of the spherical Harmonics is incorrect?


1 Answer 1


Your setup is correct, as you are tensoring/entangling eigenstates of $L_z$ and such of $S_z$, and your expectation values are sound. (This is not quite the notation suggested in the referent question. There, you start with a two-spinor, but, implicitly, and confusingly, you tensor it with infinite-dimensional kets you never write down, which you then dot on $\langle x|$. I left that question alone, because I believe it makes no sense: dotting an infinite dimensional bra to a two-dimensional ket?)

Sakurai/Napolitano certainly do not suggest you cannot do this one, here. They only prove you cannot define spherical harmonics for half integral S, as you do with L.

In the more mainstream tensor product notation $|l,~ m\rangle |s,~ s_z\rangle$ your wave function amounts to $$ Ae^{-br}(|1, 0\rangle |1/2, 1/2\rangle +|1, -1\rangle |1/2, -1/2 \rangle)/\sqrt{2} ~, $$ which you dotted by $\langle r|\langle \theta,\phi|$.

Observe this is not an eigenstate of $L_z+S_z$, so you are not quite adding S to L, but you may certainly write it down as a formal object wave function.

  • $\begingroup$ I have one more perplexity: why did you say that I am tensoring/entangling the eigenstates? I would describe the step I took as rewriting the spinor in terms of the eigenstates of $L_z$, so the spherical harmonics. What do you mean by tensoring/entangling? $\endgroup$
    – Noumeno
    Nov 30, 2020 at 14:56
  • $\begingroup$ Should be apparent in the way I wrote your state: it is a sum (entanglement) of two terms; each term is a tensor product of $L_z$ eigenstates with $S_z$ eigenstates, which otherwise don't know about each other. Their only connection is the way you grouped them together, somewhat counterintuitively, in your state which has no clear eigenoperators transcending the tensor factors. $\endgroup$ Nov 30, 2020 at 16:28

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