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Studying quantum angular momentum from my lectures and also from this useful collections of lectures (from Leonard Susskind) I have of course stumbled upon rising and lowering operators (a.k.a ladder operators).

I am completely fine with the definition of those operators and with the fact that applying them allows us to discover new eigenstates of the angular momentum, with different eigenvalues associated to them.

But from this my lectures go on to state that the existence of this operators alone allows us to prove that:

  1. Angular momentum is quantized.

  2. We can find all eigenstates of angular momentum by applying the ladder operators to a known eigenstate of it.

In both my lectures and the Leonard Susskind's ones this two statements are dropped like the most obvious thing in the world to prove, so I tried to find an easy way to show that (1.) and (2.) must be true given the definition of the ladder operators and their main proprieties. After a bit I gave up and went here in search for answers and I finded this related question.

Problem is: from the answers to the linked question seems that to understand why (1.) and (2.) follows we must first be familiar with representation theory of the Lie algebra $\mathfrak{su}(2)$. I am not at all familiar with this topic and I am very much afraid of getting sidetracked if I try to dig into this topic right now.

Question is: Is there a more direct, maybe less formal, way of showing that (1.) and (2.) must be true or am I doomed to not understand this topic until I study representation theory of lie algebra?

The fact that we must understand representation theory to get (1.) and (2.) seems really strange to me since all my resources on the topic seem to suggest that (1.) and (2.) are an obvious consequence of the existence of the ladder operators. Furthermore a similar situation is present in the analysis of the quantum harmonic oscillator, with creation and destruction operators in place of the ladder ones, so a clear understanding of this topic seems to me essential. (At least is also useful to understand why the eigenstates of the harmonic oscillators are also quantized.)

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You can proof this using three facts

  1. Using the ladder operators we can always create a state with higher $L_z$ eigenvalue
  2. The spectrum of $L_z$ is bounded. (There is a lowest and highest eigenvalue)
  3. The action of $L_{\pm}$ on a state $|l,m\rangle$ is $L_\pm|l,m\rangle=\hbar\sqrt{l(l+1)-m(m+1)}|l,m\pm1\rangle$ where $l$ is the quantum number corresponding to $L^2$.

I will use $L_+$ here but for $L_-$ the reasoning is the same. The first fact follows from the commutator $[L_z,L_+]=\hbar L_+$. Suppose we have an eigenstate $|m\rangle$ such that $L_z|m\rangle=\hbar m|m\rangle$. Then $L_+|m\rangle$ is again an eigenstate of $L_z$ but with eigenvalue $\hbar (m+1)$. By applying $L_z$ to this state we see that is indeed an eigenstate of $L_z$. \begin{align}L_zL_+|m\rangle&=(L_+L_z+[L_z,L_+])|m\rangle\\ &=L_+L_z|m\rangle+[L_z,L_+]|m\rangle\\ &=L_+(\hbar m)|m\rangle+\hbar L_+|m\rangle\\ &=\hbar(m+1)L_+|m\rangle \end{align} The proof of the second fact is bit more involved but there is a nice proof on https://en.wikipedia.org/wiki/Angular_momentum_operator#Derivation_using_ladder_operators.

So if I start with a state $|l,m\rangle$ and keep applying $L_+$ then at some point $m$ will exceed the bounds mentioned in fact 2. The only solution is that $L_+$ gives the zero vector for some $m$ meaning that applying $L_+$ after that keeps on giving the zero vector. Using fact 3 this will give us a condition on $m$. To quote the article I mentioned above:

If this is zero, then ${l(l+1)=m_{\text{max}}\left(m_{\text{max}}+1\right)}$, so ${\displaystyle l=m}$ or $l=-m-1$. However, because $L^{2}-L_{z}^{2}$ is positive-semidefinite, ${\displaystyle \hbar ^{2}l(l+1)\geq (\hbar m)^{2}}$, which means that the only possibility is ${\displaystyle m_{\text{max}}=l}$.

Here positive-semidefinite means that all of its eigenvalues are larger or equal to zero. A similar reasoning gives $m_{\text{min}}=-l$. So because we get from $m=-l$ to $m=l$ using a number of raising operators we can conclude $2l$ must be an integer (so $l$ a half-integer and $m$ must be an integer (or half-integer). Finally note that I could have used $J$ or $S$ just as well in this derivation. If $J$ is defined as the usual $J=L+S$ then $J$ is an integer (not half-integer) so this last paragraph does not apply.

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  • $\begingroup$ But why can we be sure that there isn't an eigenstate of the angular momentum hidden somewhere? To say it in another way: how can we be sure that the ladder operators step on all the possible eigenstates? We can always create a new eigenstate by applying the ladder operator, but this does not guaranties that the ladder operators allow us to create all possible eigenstates, right? $\endgroup$
    – Noumeno
    Nov 4 '20 at 15:25
  • $\begingroup$ @Noumeno Good point. What is certain is that the eigenvalues are (half) integers exactly as described above because the reasoning is general enough. So if we had some hidden quantum number $\alpha$ labelling the hidden states then the spectrum would be degenerate i.e. $L_z|m,\alpha\rangle=m|m,\alpha\rangle$ for all $\alpha$. If this reasoning is enough to rule out any hidden eigenstates I do not know to be honest. $\endgroup$ Nov 4 '20 at 21:29
  • $\begingroup$ I edited my answer because I switched $J$ and $L$ in my last paragraph, my bad. $\endgroup$ Nov 4 '20 at 21:32
  • $\begingroup$ I am not sure about the explanation you just provided in the comments.. Can you expand on it by editing your answer? The problem I brought up in the comments, the one about the possibility of hidden eigenstates, was the central problem in my question. I simply cannot understand why the ladder operators allow us to be certain to find all the eigenstates. $\endgroup$
    – Noumeno
    Nov 5 '20 at 15:34
  • $\begingroup$ @Noumeno My point was that the ladder operators are definitely able to find all the eigenvalues but this still leaves room for multiple states with the same eigenvalue. I don't know how one would proof you have found all the eigenstates. I'm sorry $\endgroup$ Nov 5 '20 at 18:03
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It is, of course, not neccesary to be well versed in Representation theory of Lie algebras to understand the results of the spectrum of $\mathbf{J}^{2}$ and $J_{z}$, it's a matter of operator algebra and the properties of Hilbert spaces. I'll give you an outline, but for a complete exposition I recommend you to take a look at the classic book "quantum mechanics" by Albert Messiah.

(1) $\mathbf{J}^{2}$ and $J_{z}$ commute so they have a set of common eigenvectors. For simplicity, let suppose that $\mathbf{J}^{2}$ and $J_{z}$ form a complete set of commuting observable.

(2) $\mathbf{J}^{2}$ is a definite semi-positive operator, i.e., it's eigenvalues are positive numbers or zero (for any Hermitian operator $A$, $A^{2}$ is definite positive since, for any vector, $\left\langle u\right|A^{2}\left|u\right\rangle$ is the norm of $A\left|u\right\rangle$ and norms are positive by definition).

(3) Call the common set of eigenvectors $\left|jm\right\rangle$ , where

$$\mathbf{J}^{2}\left|jm\right\rangle =j(j+1)\left|jm\right\rangle ,$$ $$J_{z}\left|jm\right\rangle =m\left|jm\right\rangle .$$

We write the eigenvalues of $\mathbf{J}^{2}$ as $j(j+1)$ because why not? there is no restriction so far on $j$, so at this stage $j(j+1)$ can be any non-negative real number.

(4) Here is where the mathematics starts, I'll only give the conclusions. Some identities of the ladder operator are used to show that, for a given $j$, we necessarily have $-j\leq m\leq j$. This result is, again, based on the fact that norms on Hilbert spaces cannot be negative.

It can also be proven that $J_{+}\left|jm\right\rangle =0$ if an only if $m=j$, and $J_{-}\left|jm\right\rangle =0$ if an only if $m=-j$.

(5) You can prove that if $m\neq j$ then then $J_{+}\left|jm\right\rangle$ is proportional to $\left|j,m+1\right\rangle$ . A similar result holds for n $J_{-}\left|jm\right\rangle$.

(6) By repeating operation by $J_{+}$, we can keep increasing $m$ of any give $\left|jm\right\rangle$. So we can create a series of vectors proportional to $\left|j,m+1\right\rangle , \left|j,m+2\right\rangle , ...,\left|j,m+k\right\rangle ,...,$ but this series have to stop somewere since $m\leq j$. Hence, there have to exist a integer number $q$ such that $m+q=j$, and, from point (4) above, $J_{+}\left|j,m+q\right\rangle =0$. The same happens for $J_{-}$ and the series of vectors $\left|j,m-1\right\rangle , \left|j,m-2\right\rangle , ...,\left|j,m-k\right\rangle ,...,$ there exist a number $p$ such that $m-p=-j$ so $J_{+}\left|j,m-p\right\rangle =0$.

Since $q$ and $p$ are non-negative integers ttheir addition $q+p=2j$ is also a non-negative integer. Thus, $j$ can only take the values $$j=0,\frac{1}{2},1,\frac{3}{2},2,\ldots$$.

From (4) and (6), we can only have $$m=-j,-j+1,\ldots,j-1,j.$$

All of the above is based solely on the commutation properties of the components of the angular momentum.

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