I am trying to understand why the angular momentum quantum number $l$ can either be an integer or an half-integer. At least this is stated in the book that I am learning from. It is the book by Schwabl. In Chapter 5 the eigenvalues of $L^2$ and $L_z$ are derived using the algebraic approach with commutator relations.

The result is that the eigenvalues of $L^2$ are given by: $$\lambda_l = \hbar^2l(l+1)\tag{1}$$ In the last sentence in the derivation Schwabl states that

The angular momentum eigenvalues $l$ are either integral or half integral, and the eigenvalues $m$ range in integer steps from $l$ to $-l$.

I understand the proof, i.e. that we have the step operators $L_+$ and $L_-$ and we use them to show that it takes $k\in \mathbb{N}_0$ steps to step down (using $L_-$) from the highest possible ladder rung to the lowest one. So we need $k$ steps to go from $l$ (highest rung) to $-l$ (lowest rung), such that $l-k=-l$, from which it follows that $l=\frac{k}{2}.$

From $l=\frac{k}{2}$ with $k\in\mathbb{N}_0$ I would conclude that the spectrum of $L^2$ is given by:

$$\sigma_{L^2}=\left\{\hbar^2l(l+1)\;\middle|\; l=0,\frac{1}{2},1,\frac{3}{2},2,\frac{5}{2},3,...\right\}\tag{2}$$

However what Schwabl states is that the spectrum is either $$\sigma_{L^2}=\left\{\hbar^2l(l+1)\;\middle|\; l=0,1,2,3,...\right\}\tag{3}$$ or $$\sigma_{L^2}=\left\{\hbar^2l(l+1)\;\middle|\; l=\frac{1}{2},\frac{3}{2},\frac{5}{2},...\right\}\tag{4}$$

I also have other books (like Griffiths) where the word either is not used in the explanation. So my question is if using either is correct or not. And if yes, how can I see this?

  • $\begingroup$ A natural number k is either even or odd. In the first case, l is integer, and in the second, half integer. The possible eigenvalues are a set containing integers and half integers. What is this weirdo "sequence" stuff? $\endgroup$ Feb 19, 2023 at 22:06
  • $\begingroup$ Different l s characterize different representations and don't "talk" to each other, unless you have cooked up spectrum generating operators, highly unlikely in an introductory context. $\endgroup$ Feb 19, 2023 at 22:14
  • $\begingroup$ Thank you Cosmas Zachos, I updated the question using spectra of $L^2$. Regarding your second comment, does this mean that without additional knowledge the spectrum should be the one given in equation (2) in my question? $\endgroup$
    – pbit24
    Feb 19, 2023 at 23:05
  • 1
    $\begingroup$ Yes, of course the spectrum is in (2). It is a set textbooks ponderously illustrate by (2). You might treat all spins in subsets (3) and (4) differently, when you learn about bosons and fermions, but, as far as these arguments go, I have no clue why you are insisting on that "either". Of course l can be either an integer, or a half integer: it can't be both! You seem to suspect there is something deeper or systematic in the conclusion. $\endgroup$ Feb 19, 2023 at 23:36

1 Answer 1


$K$ can be an odd or an even number. In the first case the sequence is of half integers. For example: if $k = 1$, the values for $m$ are $-\frac{1}{2}$ and $\frac{1}{2}$.

In the second case, there is a sequence of integers. For example: if $k = 2$, the values for $m$ are $-1, 0, 1$.

There is no $k$ that leads to a complete sequence of half integers and integers.

  • $\begingroup$ Sorry for the confusion, I think my question is not precise enough. I think you confused the quantum number $l$ with the quantum number $m$. The $l$ is non negative and $m$ can take values in $m = -l, -l+1,-l+2,...,l-2,l-1,l$. So for $m$ your answer indeed is correct. However the question is why for all $l$'s there should only be either even or odd $k$'s. Because in all derivations I found the proof starts with fixing $\lambda_l$. Then for this paricular $\lambda_l$ a $k$ value of $k_l$ is found. To my understanding for different $l$'s we have different $k_l$'s. I will update my question later $\endgroup$
    – pbit24
    Feb 19, 2023 at 10:39
  • $\begingroup$ You are right. I corrected my answer. Anyway, $l$ can be theoretically any integer or half integer. But the sequence of $m$ in each case is: or only integers or only half integers. $\endgroup$ Feb 19, 2023 at 11:16
  • $\begingroup$ So would you say that the books claim of $l$ being either integer or half-integer is wrong? (the emphasis here is on the word either) $\endgroup$
    – pbit24
    Feb 19, 2023 at 11:21
  • $\begingroup$ Could be a translation issue? The author is not native English speaker. $\endgroup$ Feb 19, 2023 at 15:35
  • $\begingroup$ Unfortunately not, he is german (as me) and in the german version of the book also the word either is there. $\endgroup$
    – pbit24
    Feb 19, 2023 at 16:39

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