How are commutation (of observables) and quantization related? Reading about the Stone-Von Neumann Theorem, it seems that commutativity is the classical limit of quantum mechanics, and hence non-quantization, but I don't understand the intuition behind the fact that commutativity of operators should imply any kind of quantization.

In a theory involving `quantized' quantities, such as quantum mechanics, why does commutation of operators (observables) suddenly become an important topic- why does quantization come hand in hand with the uncertainty principle?


1 Answer 1


I believe that you must study (not simply reading but trying to prove) that beautiful story of angular momentum quantization starting from the non-commutativity of the coordinate and momentum of a particle along any axis. The following is a starting point (found in any introductory book on Quantum Mechanics).

In Classical Mechanics the angular momentum of a particle $\mathbf{J}=\left(J_1,J_2,J_3\right)$ is defined by \begin{equation} \mathbf{J}\ \equiv \ \mathbf{r}\times\mathbf{p}\ = \begin{vmatrix} &\mathbf{e}_1&\mathbf{e}_2&\mathbf{e}_3&\\ &x_1&x_2&x_3&\\ &p_1&p_2&p_3& \end{vmatrix} \tag{01} \end{equation} where $ \mathbf{r}=\left(x_1,x_2,x_3\right)$ the position, $\mathbf{p}=\left(p_1,p_2,p_3\right)$ the momentum of the particle and $\{\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3 \}$ the basic vectors of an orthonormal Cartesian coordinate system. The definition (01) is valid in Quantum Mechanics too, but the $x_j,p_k$ are hermitian operators. So in coordinates \begin{equation} \mathbf{J}= \begin{bmatrix} &J_1&\\ &J_2&\\ &J_3& \end{bmatrix} \equiv \begin{bmatrix} &x_2p_3-x_3p_2&\\ &x_3p_1-x_1p_3&\\ &x_1p_2-x_2p_1& \end{bmatrix} =\ \mathbf{r}\times\mathbf{p} \tag{02} \end{equation} Now, we'll try to find if the components $J_1,\ J_2, \ J_3$ commute between each other or not, that is, if we can measure two components precisely at the same time. In order to find the commutators $[J_j,J_k],\ j\neq k$, we remind the commutation relations of $x_j,p_k$
\begin{eqnarray} &\left[x_j,p_k\right] &=\ x_jp_k-p_kx_j\ = \ i \hbar \delta_{jk} \tag{03-1}\\ &\left[x_j,x_k\right] &=\ x_jx_k-x_kx_j\ =\ 0 \tag{03-2}\\ &\left[p_j,p_k\right] &=\ p_jp_k-p_kp_j\ =\ 0 \tag{03-3} \end{eqnarray} So \begin{eqnarray} \left[J_1,J_2\right] &=& J_1J_2-J_2J_1 \nonumber\\ &=&\left(x_2p_3-x_3p_2\right)\left(x_3p_1-x_1p_3\right)-\left(x_3p_1-x_1p_3\right)\left(x_2p_3-x_3p_2\right) \nonumber \\ &=&\left(x_1p_2-x_2p_1\right)\left(x_3p_3-p_3x_3\right)\nonumber \end{eqnarray} or \begin{equation} \left[J_1,J_2\right]\ =\ J_1J_2-J_2J_1 \ =\ i \hbar J_3\\ \tag{04-1} \end{equation} and in a similar way for the other two commutators \begin{eqnarray} \left[J_2,J_3\right]&=& J_2J_3-J_3J_2 \ =\ i \hbar J_1 \tag{04-2}\\ \left[J_3,J_1\right]&=& J_3J_1-J_1J_3 \ =\ i \hbar J_2 \tag{04-3} \end{eqnarray} Equations (04) are expressed together under the symbolic and more simple relation \begin{equation} \mathbf{J}\times \mathbf{J}\ = \ i\;\hbar \;\mathbf{J}\\ \tag{05} \end{equation} or using the dimensionless operator \begin{equation} \mathbf{L}=\left(L_1,L_2,L_3\right)\equiv \left(\frac{J_1}{\hbar},\frac{J_2}{\hbar},\frac{J_3}{\hbar}\right)=\frac{\mathbf{J}}{\hbar}\\ \tag{06} \end{equation} \begin{equation} \mathbf{L}\times \mathbf{L}\ = \ i\;\mathbf{L}\\ \tag{07} \end{equation} It's impossible to measure precisely at the same time any two components of the angular momentum since they don't commute. The story continues by the fact that the operator \begin{equation} L^2\equiv L_1^2+L_2^2+L_3^2 \tag{08} \end{equation} which represents the square of the absolute value of angular momentum $\mathbf{L}$, commutes with its component along any axis, say $\:L_3 \:$ \begin{equation} \left[L^2,L_3\right]=L^2L_3-L_3L^2=0 \tag{09} \end{equation} and in general \begin{equation} \qquad \qquad \qquad \quad \left[L^2,L_k\right]=0 ,\qquad k=1,2,3 \tag{10} \end{equation} We can measure precisely at the same time the absolute value of angular momentum $L^2$ and its component $L_k$ along any arbitrary axis $x_k$. The finale is


The components of the dimensionless orbital angular momentum of a particle $\:\mathbf{L}=\mathbf{J}/\hbar =\left(\mathbf{r}\times\mathbf{p}\right)/ \hbar $ satisfy the commutation relations $\mathbf{L}\times \mathbf{L}= i\mathbf{L}$, by which

a) The allowable eigenvalues of the absolute value operator \begin{equation} L^2\equiv L_1^2+L_2^2+L_3^2 \nonumber\\ \end{equation} are
\begin{equation} \bbox[#FFFF88,5px]{j\left(j+1\right)\;, \qquad j=0\:,\:\tfrac{1}{2}\:,\:1\:,\:\tfrac{3}{2}\:,\:2\:,\:\tfrac{5}{2}\:,\:\ldots} \nonumber\\ \end{equation} b) The eigenvalue $\ j\left(j+1\right)\ $ has $\ (2j+1)$-multiplicity to which correspond the $\ (2j+1)\ $ possible eigenvalues of the component $L_3$ across an arbitrary axis $x_3$ \begin{equation} \bbox[#E6E6E6,5px]{m\;, \qquad m = -j\:,\:-j+1\:,\:\cdots \:,\:j-1\:,\:j} \nonumber\\ \end{equation} c) A complete set of common eigenfunctions of $L^2$ and $L_3$ is enumerated by the pair $(j,m)$.

(Note : it has been proved that the orbital angular momentum has only integer values of $j$ while the half-integer values are due exclusively to spin. Moreover, spin can take integer values too).

The non-commutativity of the position and momentum operators along any axis yields the non-commutativity of the components of angular momentum which in turn gives its quantization. This same position-momentum non-commutativity is responsible for the quantization of the energy of the linear harmonic oscillator.

I don't think that these results are a matter of intuition.

  • $\begingroup$ But it doesn't answer the question. For example, we can understand quantisation in energies using the argument of continuity at the boundary values of arbitrary wavefunction $\endgroup$
    – Prabhat
    Jan 30 at 6:51