What is the physical meaning of these commutation relations: $$[L_{z},L_{\pm}]=\pm\hbar L_{\pm}\tag{1}$$ and $$[L_{+},L_{-}]=2\hbar L_{z} ~?\tag{2}$$

  • $\begingroup$ I know that if they don't commute, the quantities can't be measured simultaneously. I was actually looking for some different answer to this question besides this since I have got ladder operators here. $\endgroup$ Aug 11, 2014 at 0:33
  • $\begingroup$ Yes, since $L_\pm$ aren't observables. My mistake. $\endgroup$
    – BMS
    Aug 11, 2014 at 1:40
  • 2
    $\begingroup$ Comment to the question (v3): Eqs. (1) and (2) are in one-to-one correspondence with the $so(3)$ Lie algebra $[L_i,L_j]=i\hbar\epsilon_{ijk} L_k$. Physically, the $so(3)$ Lie algebra governs how the system behaves under rotational transformations. $\endgroup$
    – Qmechanic
    Aug 11, 2014 at 9:10

1 Answer 1


Short intro to ladders

As you say, they're ladder operators. Let's get rid of the annoying $\hbar$ by setting it to one, and call them more systematically $L_{-1},L_0,L_1$ instead of $L_-,L_z,L_+$.

Then, the commutation relations take the uniform form

$$[L_n,L_m] = (n-m)L_{m+n}$$

If we had countably many of these, we'd have a Witt algebra, if there was a central charge, that would become a Virasoro algebra, but lets stay with these three for now.

Now, ladder operators should raise and lower stuff, right, just like going up and down a ladder. It all starts with eigenvectors $\lvert l \rangle$ of $L_0$, i.e. $L_0\lvert l \rangle = l \lvert l \rangle$. Now, through the commutation relations, we get that

$$ L_0(L_{-1})\lvert l \rangle = (l-1)(L_{-1})\lvert l \rangle \; \text{and} \; L_0(L_{1})\lvert l \rangle = (l+1)(L_{1})\lvert l \rangle$$

so $L_{1}$ raises the weight $l$ of the vector by $1$, while $L_{-1}$ lowers the weight of the vector by $1$.

The physical importance

Whenever you see an algebra like this, it means that the eigenvalue of $L_0$ is quantized, since the ladder operators raise/lower the weight in discrete steps. It means that, in terms of the natural operations on the vector space on which a representation of this algebra exists, spaces spanned by eigenvectors of $L_0$ while not differing by natural numbers do not overlap. In particular, if you know there should be a highest/lowest weight state from which all others arise by applying the ladder operators, you know the complete discrete set of eigenvalues of $L_0$ allowed for the system under consideration, and we can in fact find all allowed representations.

The algebra we are looking at is actually $\mathfrak{su}(2)$, which Lie integrates to the universal cover $\mathrm{SU}(2)$ of the rotation group $\mathrm{SO}(3)$, so what we are constructing are the spinor representations of non-relativistic QM.

The technique of the highest weight

We seek unitary, irreducible representations $V$ of the algebra. Unitarity means that $L_n^\dagger = L_{-n}$, irreduciblility that there is no subrepresentation $W \subset V$.

Let $\lvert l \rangle$ be a highest weight vector, i.e. $L_{1}\lvert l \rangle = 0$. Define the Verma module (don't try to understand the mathematician's definition of this if you are not prepared for serious math)

$$ \tilde{V}_l := \mathrm{span}\{L_{-1}^n\lvert l \rangle \vert n \in \mathbb{N}\}$$

Unitarity demands further that the rep we would like to obtain should possess a positive-definite inner product. Normalize $\langle l \vert l \rangle$ and examine the level 1 vectors $L_{-1}\lvert l \rangle$:

$$\langle l \rvert L_1 L_{-1} \lvert l \rangle = 2l \overset{!}{\ge} 0 $$

So, $l < 0$ is out of the game. If $l = 0$, then $L_1(L_{-1}\lvert l \rangle) = 2l\lvert l \rangle = 0$, so $L_{-1}\lvert l \rangle$ is a second highest weight vector and generates the subrep $\tilde{V}_{-1} \subset \tilde{V}_0$. We can obtain an irreducible, unitary rep by setting

$$ V_0 := \tilde{V}_{0} / \tilde{V}_{-1}$$

which is the trivial spin-$0$ rep.

A generalization of the above argument leads us to the statement that level n vectors $\lvert v \rangle = L_{-1}^n\lvert l \rangle$ have norm

$$ \langle v \vert v \rangle = \prod_{i = 0}^{n-1} (2l - i)$$

which is non-unitary for $l \not \in \frac{1}{2}\mathbb{N}$ and has null-vectors otherwise. The irreducible unitary reps are generally obtained by

$$ V_l := \tilde{V}_{l}/\tilde{V}_{-l-1} \; \text{with} \; l \in \frac{1}{2}\mathbb{N}$$

The conclusion

It were the commutation relations alone (together with ordinary unitarity conditions) that have shown us that spin/angular momentum is restricted to half-integers and integers. One can, by further thought, see that the half-integer reps do not induce reps of the $\mathrm{SO}(3)$, but only of the $\mathrm{SU}(2)$, and that therefore "true" angular momentum is quantized as an integer. Quantization of such generalized charges (in the Noetherian sense) is thus a natural consequence of the commutation relation of the algebra of the associated (Lie) symmetry group.  


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