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?

  • $\begingroup$ 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? $\endgroup$
    – G. Smith
    Aug 19, 2019 at 1:42
  • $\begingroup$ @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? $\endgroup$ Aug 19, 2019 at 1:45
  • 4
    $\begingroup$ 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. $\endgroup$
    – G. Smith
    Aug 19, 2019 at 2:00
  • $\begingroup$ @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? $\endgroup$ Aug 19, 2019 at 2:04
  • $\begingroup$ 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! $\endgroup$
    – G. Smith
    Aug 19, 2019 at 2:09

1 Answer 1


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.


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