I want to know if the commutation relation for angular momentum, $$\left[ J_\alpha,J_\beta\right] = i\hbar\epsilon_{\alpha\beta\gamma}J_\gamma$$ is enough for defining its unique eigenstate of spin, $\lvert s,m\rangle$.

As far as I know, the relation is enough for defining ladder operator $J_\pm$.
In the case of orbital angular momentum L, ode $L_+\psi_t=0$ and $L_-\psi_b=0$ defines unique state, thus uniquely defining the whole ladder $\lvert l,m\rangle$.

However, in the case of spin, the state vector lies in abstract space, so the equation $S_+\psi_t=0$ and $S_-\psi_b=0$ can't guarantee the uniqueness of $\psi_t$ and $\psi_b$. Moreover, the relation $S_z(S_+\lvert s,m\rangle)=(m+\hbar)(S_+\lvert s,m\rangle)$ shows that $S_+\lvert s,m\rangle$ is one of eigenstate for $S_z$ with eigenvalue $m+\hbar$, but this may not be unique.
Isn't there a possibility that $S^2$ and $S_z$ have some degeneracy? If then, the quantum ladder may not cover the whole basis for $S^2$ and $S_z$.

I think ignoring these possibilities, and regarding these procedure as unique construction of $\lvert j,m_j\rangle$ will be enough, but I want a strict demonstration. Does this have something to do with generator-nature of angular momentum?

Edit) Considering other index k, such as position $\vecr$, will give a state vector $\psi\left( r \right) \lvert s,m\rangle$ which will generate eigenvectors with same eigenvalue as well as different spacial component $\psi\left( r \right)$ is given.

What I'm wondering precisely is :
The standard theory for angular momentum, starting from the commutation relation, will in the end argue the existence of eigenstate $\lvert j,m_j\rangle$, satisfying $$J^2\lvert j,m_j\rangle = j\left( j+1 \right)\hbar\lvert j,m_j\rangle\\J_z\lvert j,m_j\rangle = m_j\hbar\lvert j,m_j\rangle$$ I'm wondering if it is possible to find $\lvert j,m_j\rangle'$ without introducing any other index.
In the case of angular momentum L, it is possible to directly solve the eigenvalue equation to find the unique solution(spherical harmonics), so the problem I'm wondering happens on the case of spin S, where algebraic relation is the only restriction.


1 Answer 1


Yes it's possible there is some degeneracy. If you consider say, a three dimensional harmonic oscillator, or an electron in a Coulomb potential, or any other spherically symmetric problem (except a quantum rotor), you can certainly define angular momentum operators $S_\alpha$ obeying those commutation relations, but there are multiple states associated to any given $s,m$ due to the radial motion.

  • $\begingroup$ I'm sorry, but aren't those states are just multiple of Radial component R(r) and spherical harmonics |l m>? In that case, since |l m> is not a function of r, these states could be considered as constant multiple of |l m>, I think... I'm wondering if there is a nontrivial degenerate case, so I'll edit my question. $\endgroup$
    – user291938
    Mar 13, 2021 at 17:30
  • $\begingroup$ @user291938, They can't be considered a 'multiple' of |l,m>, they are different states, but they have the same values of $l,m$. Yes it's trivial, but that is the answer to your question. Just looking at the angular momentum operators you can not determine if there is a degeneracy or not. $\endgroup$
    – octonion
    Mar 13, 2021 at 17:52

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