# Trying to understand spin in quantum mechanics

I'm trying to understand the concept of spin in Quantum Mechanics. I'm reading "Road to Reality" by Penrose, which despite not being a textbook, is reputed to give one a deep insight into physical processes.

1. Let us suppose that we have a spin $$\frac{1}{2}$$ particle. It has two eigenstates- $$|\uparrow\rangle$$ and $$|\downarrow\rangle$$. I would assume that spin $$S$$ is an operator such that when it acts on the wavefunction $$\alpha |\uparrow\rangle+\beta|\downarrow\rangle$$, it collapses it to one eigenstate, with an eigenvalue which would be the spin (so $$\frac{1}{2}$$ here). However, Penrose says that the spin can be thought of as the point $$[\alpha:\beta]$$ in $$\Bbb{C}P^1$$.

Why is this? Why do we not have a collapse to an eigenstate?

1. Let us suppose that we have a particle with spin $$j$$. Then $$N=2j$$. An angular momentum eigenstate of such a particle can be written as $$\psi_{AB\dots N}$$. I know that there are $$N+1$$ independent eigenstates, and that for even values of $$N$$, the eigenstates are spherical harmonic functions while for odd values of $$N$$ the eigenstates are spin-weighted spherical harmonic functions.

But how do you calculate the spin of this particle? One might say that the spin is just $$j$$. However, from the example of spin $$\frac{1}{2}$$, I would figure that the spin is suppose to be an element of a projective space. Is this not true?

• Why do you think the spin operator should collapse the wavefunction? Does the energy operator collapse the state $(|1\rangle+|2\rangle)/\sqrt{2}$ of the 1D infinite square well? – G. Smith Aug 19 '19 at 1:42
• @G.Smith- All other observables that I've come across, like momentum, position, etc, collapse the wavefunction to one eigenstate, and the corresponding eigenvalue is the momentum, position, respectively. I would expect that the energy operator (which I'm assuming is differentiation with respect to time) does the same? – fierydemon Aug 19 '19 at 1:45
• No, you have a very serious misunderstanding. None of these operators collapse a superposition of their eigenstates. For example, if you have two energy eigenstates, $\hat{H}\Psi_1=E_1\Psi_1$ and $\hat{H}\Psi_2=E_2\Psi_2$, then $\hat{H}(\Psi_1+\Psi_2)=E _1\Psi_1+E_2\Psi_2$. This is just simple linearity. You are confusing what an operator does to a superposition with what measurement does to a superposition. – G. Smith Aug 19 '19 at 2:00
• @G.Smith- I see. Thanks for clarifying the difference between the action of an operator and the process of measurement (which I guess does collapse the wavefunction to an eigenstate). However, say we take the momentum operator $P_x=ih\frac{\partial}{\partial x}$. Does this not collapse the wavefunction to an eigenstate? Is the action of this operator on the wavefunction not the same as the measurement of momentum? – fierydemon Aug 19 '19 at 2:04
• No, it is not. If you have two momentum eigenstates, $\hat{P}_x\Psi_1=p_{x1}\Psi_1$ and $\hat{P}_x\Psi_2=p_{x2}\Psi_2$, then $\hat{P}_x(\Psi_1+\Psi_2)=p_{x1}\Psi_1+p_{x2}\Psi_2$. No collapse from applying a linear operator! – G. Smith Aug 19 '19 at 2:09

You're confusing the measurement of an operator $$\hat{\cal O}$$ - which collapses the wave function to one of the eigenstates of $$\hat{\cal O}$$ - with a state, which can be a general linear combination of eigenstates of $$\hat{\cal O}$$.
In the case of spin (or more generally angular momentum) we speak of spin-$$s$$ when the largest possible eigenvalue is $$\hbar s$$. Thus a spin-1 particle can have eigenvalue $$\pm \hbar$$ or $$0\hbar$$.
Note that the direction of the angular quantization axis is irrelevant since any direction is equally good as any other. Thus, the possible eigenvalues of spin along $$\hat x$$, i.e. the eigenvalues of $$\hat S_x$$ are the same the eigenvalues of $$\hat S_z$$. This does NOT mean the eigenstates are the same: just the eigenvalues are the same. You can verify for yourself that the linear combinations $$\vert\psi_\pm\rangle=\vert \uparrow \rangle \pm \vert \downarrow\rangle$$ of eigenstates of $$\hat S_z$$ are actually eigenstates of $$\hat S_y$$.
A general spin-1/2 state will be a linear combination $$\vert \psi\rangle =\alpha \vert \uparrow\rangle+\beta\vert\downarrow\rangle.$$ can be represented as a point on $$\mathbb{CP}$$ as in the Bloch sphere.