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I have been studying quantum mechanics, specifically angular momentum, but I have a question that concerns raising and lowering operators as a whole. For total angular momentum, you can define: $$J_\pm=J_x\pm iJ_y $$ Anyone who is familiar with angular momentum will recognize these as the raising and lowering operators, but I will continue on with the problem to better explain my question.

An analysis of this problem shows that: $$ [J_z, J_\pm]=\pm \hbar J_\pm$$ $$ [J^2, J_\pm]=0 $$ From here it is easy to see that if $J_z|\alpha\beta\rangle= \beta|\alpha\beta\rangle, $ and $J^2|\alpha\beta\rangle= \alpha|\alpha\beta\rangle$, $$ J_z(J_+|\alpha\beta\rangle)=(J_+J_z+\hbar J_+)|\alpha\beta\rangle= (J_+\beta+\hbar J_+)|\alpha\beta\rangle=(\beta +\hbar)J_+|\alpha\beta\rangle $$ And thus we can say $J_+|\alpha\beta\rangle=C|\alpha,\beta + \hbar\rangle $.

However, while this approach is very clean cut, in my mind it doesn't exactly show that the eigenvalues of $J_z$ exist only in increments of $\hbar$. For instance, if I were able to find some arbitrary set of operators $W_\pm$, such that $[J_z, W_\pm]=\pm (\hbar /4)W_\pm$, then I could easily show by the logic above that the eigenvalues of $J_z$ exist in increments of $\hbar /4$. So then, what guarantees that I cannot find such operators? More specifically, what part of the "raising and lowering operator" method guarantees that there are not more possible eigenvalues of $J_z$ (or any operator), than those found using raising and lowering operators?

Qmechanic
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Phantom101
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2 Answers2

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The formal answer lies in representation theory, in this case, the representation theory of the Lie algebra $\mathfrak{su}(2)$, which is spanned by the three operators $J_z,J_+,J_-$. That there are no more eigenvalues of $J_z$ than those found by the ladder operator method follows from two facts:

  1. Every representation of $\mathfrak{su}(2)$ is completely decomposable, i.e. the direct sum of irreducible representations.

  2. The irreducible representations of $\mathfrak{su}(2)$ are precisely the "spin representations" of physics, labeled by the half-integer largest eigenvalue ("highest weight") $s$ of $J_z$, which have dimension $2s+1$, consisting of the states with eigenvalues $-s,-s+1,\dots,s-1,s$.

$s$ has to be half-integer because one can directly show that if $s$ is the highest weight, then the lowest eigenvalue is $-s$, and if the difference between the highest and the lowest weight were not an integer, we would be able to reach an even lower weight by appling the lowering operator to the highest weight state.

ACuriousMind
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  1. There is no combination of angular momentum operator that satisfy a condition like $[J_z,W_{\pm}]=\pm (\hbar /4)W_\pm$. The only possible ladder operators constructed out of $J_x$ and $J_y$ are $J_\pm$, and their commutation relations are $[J_z,J_{\pm}]=\pm \hbar J_\pm$, which does imply that neighbouring $m$ values differ by $1$. (Since we only have $J_x,J_y$ and $J_z$ to play with, it's not hard to show that $[J_z,J_{\pm}]=\pm \hbar J_\pm$: just start with a generic $J_+=a L_x+bL_y$ and you will find that $b=\pm i a$. The actual value of $a$ is irrelevant for computing the shift in $m$.)
  2. It is possible for an operator $\hat A$ to satisfy (for instance) $[J_z,\hat A]=2 \hbar \hat A$. An example is any operator proportional to $(x+iy)^2$. The action of this operator changes $m$ by $+2\hbar$ but $\hat A$ is NOT an angular momentum operator.
  3. Angular momentum operators have a Lie algebraic structure, and from the representation theory of Lie algebras we know that the set $\{\vert jm\rangle\}$ must contain $2j+1$ elements and must contain $m=j$ and $m=-j$. Thus the ladder by angular momentum ladder operators can only change $m$ by one unit of $\hbar$.
ZeroTheHero
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