# Quantization of angular momentum: is Dirac's proof wrong?

I'm trying to understand the physicist's proof of the theorem on the spectral structure of angular momentum operators (I'm being told that this proof is due to Dirac). I will refer to Ballentine's book, chap. 3, section 1.

Angular momentum operators $J_x, J_y, J_z$, whatever definition we adopt, are bound to satisfy the commutation relations

$$\tag{AMCR} [J_x, J_y]=iJ_z,\ [J_y, J_z]=iJ_x,\ [J_z, J_x]=iJ_y,$$ which imply $$[J^2, J_z]=0.$$ Following Ballentine's book, we claim that from those relations alone we can infer:

1. $J^2$ and $J_z$ have pure point spectrum;
2. the possibile eigenvalues for $J^2$ are of the form $j(j+1)$ for a half-integer $j$;
3. for every eigenvalue $j(j+1)$ of $J^2$, $J_z$ has the corresponding eigenvalues $m=-j, -j+1 \ldots j$. Those eigenvalues are simultaneous, meaning that for each one of them there exists a ket $\lvert j, m\rangle$ such that $J^2\lvert j, m\rangle=j(j+1)\lvert j, m\rangle$ and $J_z\lvert j, m\rangle=m\lvert j, m\rangle$. $J_z$ hasn't got any more simultaneous eigenvalues with $J^2$: in particular the spectrum of both operators is discrete.

I'll give an outline of the proof, stressing the troublesome parts in italic. If necessary, you can follow the complete proof at the link above.

Let $\lvert \beta, m\rangle$ be a simultaneous eigenket of $J^2, J_z$ respectively.

Step 1) We prove that $\beta \ge m^2$.

Step 2) Let $j$ be the maximum allowable value of $m$ for fixed $\beta$. Then, applying a ladder operator $J_+$ to $\lvert \beta, j\rangle$, we see that $J_+\lvert \beta, j\rangle=0$. (*)

Step 3) Turns out that $\beta=j(j+1)$ and that $-j=\min(m)$.

Step 4) Because of the existence of ladder operators, eigenvalues for $J_z$ are integer spaced (!). In particular, the difference between $j$ and $-j$ must be integer and so $j$ must be half integer. $\square$

Step 2 and Step 4 are troublesome. Step 4 seems to assume implicitly that if $\lvert j, m'\rangle$ is an eigenket then it can be reached by successive applications of ladder operators. For example, if $m'>m$ then it is assumed implicitly that $\lvert j, m'\rangle=J_+J_+\ldots J_+\lvert j,m\rangle$. I don't understand why this should be the case.

I would gladly accept any hint or reference on the question. Thank you.

(*) EDIT There used to be a question here, which is now solved. See Peter Taylor's comment and AlbertH answer below. Question was: "To do so the author uses implicitly the fact that $j$ is an eigenvalue of $J^2$ and so that an eigenket $\lvert \beta, j\rangle$ exists. This doesn't look obvious to me". end of edit

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With step 2 as written (and BTW it would be easier to follow without overloading $j$ - it was the first element of the ket earlier, and now it's the second), isn't the existence of the eigenket $\lvert \beta, j\rangle$ true by definition of $j$ as "the maximum allowable value of $m$ for fixed $\beta$"? – Peter Taylor Jul 5 '12 at 15:17
@PeterTaylor: I'm sorry for the confusing notation, it is the one the book adopts here end of page. if I have the time I'll try to fix it.  Your suggestion is interesting. Indeed, the spectrum of any operator is closed, and so if it is also bounded then it must have maximum and minimum. I think that this might clarify Step 2. – Giuseppe Negro Jul 5 '12 at 16:32
Could you define $J^2$? Google Books will not show me the page where it is defined. – Nate Eldredge Jul 6 '12 at 14:41
@NateEldredge: Of course. $J^2=\mathbf{J}\cdot\mathbf{J}=J_x^2+J_y^2+J_z^2$. – Giuseppe Negro Jul 6 '12 at 17:08
@GiuseppeNegro Hi. By looking at your profile, I guess you may have an interest in mathematical physics. It would be great if you have time to look at my question here : math.stackexchange.com/questions/470666/… . Sorry for disturbing you if you are not interested. – Amr Aug 18 '13 at 20:50

I don't know the mentioned book.
However, about Step 2, $J_x, J_y, J_z$ are generators of an unitary representation of $SO(3)$ that is a compact Lie group. By Peter-Weyl theorem all irreducible unitary representation of a compact Lie group are finite-dimentional. So $J_z, J^2$ are made up of self-adjoint finite-dimensional operators and as such they admits at least an eigenvector.

Edit

Let $J_x, J_y, J_z$ be an irreducible representation of the above commutation rules.
By Schur's lemma, $[J^2, J_i] = 0$ for each $i\in\{x, y, z\}$, implies $$J^2 = \beta \mathbb I$$ $J_z$ is a finite-dimensional self adjoint operator, hence diagonalizable.
Let's take an eigenvector, $\lvert m \rangle$, of $J_z$, with repeated applications of $J_-$ and $J_+$ we can construct a sequence of eigenvectors $$s := \{\lvert m - h \rangle, \dotsc, \lvert m - 1\rangle, \lvert m \rangle, \lvert m + 1 \rangle, \dotsc,\lvert m + k \rangle \}$$ Above sequence is finite because its elements are independent vectors of the finite-dimensional representation space $V$.
Let's call $S$ the subspace spanned by vectors in $s$. The equalities $$J_x = \frac 1 2 (J_+ + J_-)\\ J_y = \frac 1 {2i} (J_+ - J_-)$$ show that $S$ is an invariant space for $J_x, J_y, J_z$. Since the representation is irreducible $S$ must coincide with $V$ and no other (indipendent) eigenvectors of $J_z$ can exist.

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Thank you, this is enlightening. I think that this observation can be useful in understanding the other problem, too. – Giuseppe Negro Jul 5 '12 at 16:25
@Giuseppe: Yes, I think it can. IIRC it is easy to see that the vectors $J_-J_-\cdots J_-|j(j+1),j\rangle$ span a space that is stable under all the operators $J_x,J_y,J_z$. IOW a subrepresentation $V$. So the claims of Steps 1 thru 3 are valid for this subrepresentation. But clearly here all the weights differ from the maximal one by an integer, because EVERYTHING in this subspace was gotten by applying the descending ladder to the maximal weight vector. – Jyrki Lahtonen Jul 5 '12 at 18:35
@JyrkiLahtonen Hello everyone. It would be great if you have time to look at my question here: math.stackexchange.com/questions/470666/… . Sorry for disturbing you if you are not interested. Thank you – Amr Aug 18 '13 at 20:54