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It is a well-known fact that the values for the square of the orbital angular momentum of a particle $L^2$ and it's projection in the $z$-direction $L_z$ are $m\hbar$ and $l(l+1)\hbar$ and that $l$ and $m$ can take integer values ($...,-2,-1,0,1,2,...$) And they are the values you obtain from solving the wave equation for the observables.

Despite this, if you were to solve algebraically for the eigenvalues of $$\hat L_z|l,m\rangle$$ and $$\hat L^2|l,m\rangle$$ using ladder operators or outright solving for the individual matrix components (Like in Born's, Heisenberg's and Jordan's On Quantum Mechanics II) we obtain that the quantum number can have half-integer value ($...,1,-\frac{1}{2},0,\frac{1}{2},1,...$) and we can only reduce them to only integers by appealing to rotation operators or, once again, the wave equation.

I find the fact that we get an extra set of results extremely bizarre, since the wave equation can be derived from the matrices (like in Heisenberg's The Physical Principles of the Quantum Theory). I also find bizarre that, if one didn't know about rotation operators or the wave equation one could live their life thinking either orbital angular momentum can have half-integer values, or that Quantum Theory is wrong, as it disagrees with experiment, since those wrong values are eigenvalues of $L_z$ and $L^2$ The only explanation I can think of is that orbital angular momentum is actually not truly observable, only total angular momentum is, but this reason seems neither right,nor satisfactory.

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  • $\begingroup$ I guess the result also includes spin, so half integer is ok. $\endgroup$ – K_inverse Nov 26 '18 at 5:57
  • $\begingroup$ So would that mean that both orbital angular momentum and spin cannot be directly measured, only their combination? And then again, why can you get the proper values from the wave equation but not algebraically? $\endgroup$ – user140323 Nov 26 '18 at 6:05
  • $\begingroup$ Basically, this is because the algebraic method can only "see" the local properties of the symmetry group at the identity (i.e. its Lie algebra) and therefore it cannot distinguish between SO(3) and its double cover SU(2). How familiar are you with that language? $\endgroup$ – Emilio Pisanty Nov 26 '18 at 7:57
  • $\begingroup$ I am almost completely unfamiliar with it, sadly. I have heard of it, but I haven't quite grasped what it means. Is there anywhere I could find an intuitive explanation of it? And if it isn't much trouble, could you please explain without resorting to such language? $\endgroup$ – user140323 Nov 26 '18 at 14:28
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If I understand your question correctly, then the answer is that not all properties of an operator are encoded in its commutation relations.

If you are given three operators called $L_1, L_2$, and $L_3$ and told only that they obey the commutation relations $[L_i,L_j]=(i\hbar) \epsilon_{ijk} L_k$, one might ask what (if anything) they can conclude about the operators or the states on which they act. It is a priori possible that you can't conclude anything about the operators without providing a concrete implementation of them on some Hilbert space, but of course this is not true; this information is sufficient to establish the existence of the Casimir operator $L^2$, as well as to derive all possible eigenvalues of $L^2$ and $L_3$ which would be consistent with the commutators.

But this isn't the whole story. A concrete realization of these operators contains more information than the commutation relations alone. When we define $L_i := (-i\hbar)\epsilon_{ijk}x_j \partial_k$ which acts on some subset of $L^2(\mathbb R^3)$, then not only do we recover the commutation relations we started with, we also find that only integer values of $l$ are allowed.

We should not be surprised by this. The spectra which are derived "algebraically" - which admit the possibility of half-integer values of $l$ - are based on the commutators and the commutators alone. In other words, we used the commutators to constrain (note: not exactly specify) the possible spectra of the operators. A concrete implementation of the operators contains the commutation relations, but also more detailed information about how the operator acts on the Hilbert space. Therefore, it's entirely possible that such an implementation would not exhaust the algebraically derived spectra, and indeed this turns out to be the case.


You don't need to resort to a wave equation to see this, by the way. The angular momentum operators which act on $\mathbb C^2$ (spin 1/2) and those which act on $\mathbb C^3$ (spin 1) obey precisely the same commutation relations but have entirely different spectra; it's therefore obvious that commutation relations are not sufficient to tell you everything you need to know about the associated operators.

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  • $\begingroup$ Oh, I'd never quite considered that, thank you very much! If you don't mind, how could you derive the proper eigenvalues using $\hat L_i=\epsilon _{ijk}\hat x \hat p$ algebraically? Hopefully as elementary a derivation as possible please? $\endgroup$ – user140323 Nov 26 '18 at 14:34
  • $\begingroup$ The simplest derivation I know is something like this. As far as directly solving the differential equations, using a power series solution is pretty straightforward. $\endgroup$ – J. Murray Nov 26 '18 at 15:29
  • $\begingroup$ Oh I know how to solve the differential equations. I just want to know how to obtain the correct values algebraically. $\endgroup$ – user140323 Nov 26 '18 at 16:26
  • $\begingroup$ I don't quite understand your question. You seem to be asking for a way to calculate the exact spectrum of an operator without using the actual definition of the operator in terms of its action on a Hilbert space, and in general I do not think you can do that. The concrete implementation you're referring to is a differential operator, and solving the resulting eigenvalue equation yields the spectrum. There are "algebraic" tricks you can do to constrain the spectrum, but I don't think you can state it exactly without actually solving the problem. $\endgroup$ – J. Murray Nov 26 '18 at 17:30
  • $\begingroup$ My bad, what I mean is, is it possible to solve for the correct eigenvalues of the orbital angular momentum operator without having to limit ourselves to a specific eigenbasis and solving differential equations? I've recently beem reading up on the creation of Quantum Mechanics by Heisenberg, and I suppose if his and Schrödinger's QM are equivalent, one should be able to obtain the same results in Hesenberg's formulation, shouldn't we? $\endgroup$ – user140323 Nov 26 '18 at 17:43
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It is well-known that the finite dimensional irreps $V_{\ell}$ of the Lie algebra $so(3)$ are classified by spin $\ell\in\frac{1}{2}\mathbb{N}_0$. To rule out half-integer representations for the orbital angular momentum (OAM) Lie algebra $${\rm span}_{\mathbb{R}}(L_1,L_2,L_3)~\cong~ so(3),$$ one should use the fact the OAM operators $$ L_j~=~\sum_{k,\ell=1}^3\epsilon_{jk\ell}x^kp^{\ell} $$ are realized in terms of position and momentum operators, which in turn satisfy CCR. For details, see e.g. my Phys.SE answer here.

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$\def\cH{\cal H} \def\vL{\vec L} \def\vn{\vec n}$ I would like to answer from another viewpoint. You found eigenvalues of angular momentum components to be ${1 \over 2}\,n\hbar\ (n \in \Bbb Z)$. Let's assume that your physical system do possess states with all these eigenvalues. It's convenient to separate integer eigenvalues from odd half-integer ones, by introducing two Hilbert spaces, $\cH_+$ and $\cH_-$, the former containing all eigenvectors with integer eigenvalues (and linear combinations thereof), the latter eigenvectors with odd half-integer eigenvalues. The full Hilbert space $\cH$ will be the direct sum $$\cH = \cH_+ \oplus \cH_-.$$ Note that in general eigenspaces belonging to a given eigenvalue - say of $L_z$ - will be many-dimensional (even infinite-dimensional) since your physical system will possess other observables commuting with $L_z$. But this will not harm my argument.

The reason behind my notation is the following. Define $$R(\vn,\phi) = \exp\left(\!-{i \over \hbar}\,\phi\,\vn\cdot\vL\right)$$ where $\vn$ is a unit vector, $\phi\in[0,2\pi]$. It's natural to interpret $R$ (which is a unitary operator on $\cH$) as a rotation of angle $\phi$ around an oriented axis with unit vector $\vn$.

Let's study the action of $R(\vn,2\pi)$. When acting on an eigenvector of $L_z$ with integer eigenvalue $R(\vn,2\pi)$ leaves it invariant, whereas for odd half-integer eigenvalue the eigenvector gets multiplied by $-1$. In both cases the state (vector up to a phase factor) is unaltered, which is satisfactory. From definition of $\cH_+$, $\cH_-$, the same happens for any vector in those subspaces, and this explains subscripts $_+$, $_-$.

But if we had to consider a vector which is a linear superposition of one of $\cH_+$ and one of $\cH_-$ we would get an unpleasant result: the final vector would represent a different state from the initial one. We are thus led to postulate that such superpositions do not occur. As far as $\vL$ is concerned we are on safe ground, as these operators leave invariant $\cH_+$ and $\cH_-$. But our assumption is not trivial when other observables are taken into account: we are requiring that all observables of our system leave $\cH_+$ and $\cH_-$ invariant. This is known as a superselection rule, after Wick, Wightman, Wigner.

In simpler words, a quantum system may be born in one of two possible kinds of states, differentiated by angular momentum eigenvalues: either integer valued or odd half-integer valued. No internal event in igs evolution can change this mark. (It's necessary to say "internal" because an interaction may have that effect: think e.g. of an atom losing an electron. Although it could be thought of as another system.)

A final note. Thinking of atoms it's usual to introduce many angular momenta. Orbital and spin, of a single electron or sum on several electrons. All of them obey the same commutation relations and are acceptable observables for the whole system, that they be constants of motion or not. But the $+$ or $-$ mark for the atom only depends on number of electrons, as each one contributes with a 1/2 spin. (I'm neglecting nuclear spin, which isn't always allowed.)

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