Suppose we have a particle of mass $m$ confined to the surface of a sphere of radius $R$. The classical Lagrangian of the system is

$$L = \frac{1}{2}mR^2 \dot{\theta}^2 + \frac{1}{2}m R^2 \sin^2 \theta \dot{\phi}^2 $$

The canonical momenta are $$P_\theta = \frac{\partial L }{\partial \dot{\theta }} = m R^2 \dot{\theta }$$ and $$P_\phi = \frac{\partial L }{\partial \dot{\phi }} = m R^2 \sin^2 \theta \dot{\phi }$$

The Hamiltonian is

$$H = \frac{P_\theta^2}{2 m R^2} + \frac{P_\phi^2}{2 m R^2 \sin^2\theta }$$

Now start to quantize the system. We replace $P_\theta $ and $P_\phi $ as $-i\hbar \frac{\partial}{\partial \theta}$ and $-i\hbar \frac{\partial}{\partial \phi} $, respectiely, to obtain

$$H = -\frac{ \hbar^2 \partial^2}{2 m R^2 \partial \theta^2} - \frac{\hbar^2 \partial^2 }{2 m R^2 \sin^2\theta \partial \phi^2 } $$

This is apparently wrong, it should be the total angular momentum!

So what is the right procedure to quantize a system, especially a system in curvilinear coordinates?


2 Answers 2


In a nutshell, the problem with OP's choice of operators $\hat{p}_j$ and $\hat{H}$ is that they are not selfadjoint wrt. to the pertinent measure $\mu$. In other words, the usual integration by parts method to prove selfadjointness does not work.

Here are some more details. Let us put the constants $m=1=R$ for simplicity. Then OP's Lagrangian becomes of the form

$$\tag{1} L~=~\frac{1}{2}g_{ij}~\dot{x}^i\dot{x}^j,$$

with coordinates $x^1\equiv\theta$, $x^2\equiv\phi$, and metric tensor

$$\tag{2} g_{ij}~=~ \begin{pmatrix} 1 & 0 \\ 0 &\sin^2\theta \end{pmatrix}. $$

Classically, the Lagrangian momenta are

$$\tag{3} p_i ~=~g_{ij}~\dot{x}^j,$$

and the Hamiltonian is

$$\tag{4} H~=~\frac{1}{2} g^{ij} ~p_i p_j. $$

The volume form in configuration space is

$$\tag{5} \mu ~=~ \sqrt{g}~ \mathrm{d}x^1 \wedge \mathrm{d}x^2 ~=~ \sin{\theta} ~\mathrm{d}\theta \wedge \mathrm{d}\phi. $$

The Hilbert space is $L^2(S^2,\mu)$. What is the Schrödinger representation of the momentum operators? Well, now we run into operator ordering ambiguities. The momentum operators should as a minimum satisfy (i) the CCR, and (ii) be selfadjoint wrt. to the measure (5). One idea to ensure this is to use

$$\tag{6} \hat{p}_j~=~ \frac{\hbar}{i\sqrt[4]{g}} \frac{\partial}{\partial x^j} \sqrt[4]{g}. $$

Similarly, we can choose a selfadjoint Hamiltonian operator to be the Laplace-Beltrami operator:

$$\tag{7} \hat{H}~=~-\frac{\hbar^2}{2}\Delta ~=~ -\frac{\hbar^2}{2\sqrt{g}}\frac{\partial}{\partial x^i}\sqrt{g}~ g^{ij} \frac{\partial}{\partial x^j}~=~ \frac{1}{2\sqrt[4]{g}} \hat{p}_i\sqrt{g}~ g^{ij} ~\hat{p}_j\frac{1}{\sqrt[4]{g}} . $$

In case of the two-sphere $S^2$, this Hamiltonian operator (7) leads to the square of the angular momentum $\hat{\bf L}^2$. Classically, the operators (6) and (7) reduce to the functions (3) and (4), respectively.


  1. Bryce DeWitt, Supermanifolds, Cambridge Univ. Press, 1992; Section 6.7.
  • 1
    $\begingroup$ I do not understand the notion of "being selfadjoint wrt. to a measure". What does it mean exactly ? $\endgroup$
    – Trimok
    Commented Jul 6, 2014 at 12:38
  • 1
    $\begingroup$ Do you understand what it means intuitively? $\endgroup$
    – Qmechanic
    Commented Jul 6, 2014 at 12:42
  • 1
    $\begingroup$ Well, not sure..., does this mean that the measure has to be part of the practical definition of the inner product ? $\endgroup$
    – Trimok
    Commented Jul 6, 2014 at 12:51
  • 4
    $\begingroup$ $\uparrow$ Yes. $\endgroup$
    – Qmechanic
    Commented Jul 6, 2014 at 12:52
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    $\begingroup$ @Qmechanic: Don't mention it. Indeed, as you are saying, 9 out of 10 quantizers will always take the hamiltonian to be the quadratic Casimir of the symmetry algebra, so it automatically commutes with all of its generators and realizes the symmetry. The very same structure presents with this principle in phase-space quantization, where its is easiest for me to monitor the quantization, T L Curtright and C K Zachos, New Journal of Physics 4.1 (2002) 83 (dx.doi.org/10.1088/1367-2630/4/1/383) but that's probably just me... $\endgroup$ Commented Sep 26, 2015 at 23:52

Conventional wisdom (as stated in the textbooks of Shankar, or Griffiths, for example) says to avoid quantizing operators in curvilinear coordinates whenever possible. Better to quantize the cartesian operators $p_x, p_y, p_z$ and then switch to curvilinear coordinates in the quantum theory. Refer to this discussion by Professor Robert Jaffe at MIT:[1]

The canonical quantization method becomes complicated and subtle when one tries to apply it to coordinate systems that include singular points. A familiar example is spherical polar coordinates $(r,\theta,\phi)$. The origin, $r=0$, is a singular point for spherical polar coordinates---for example, $\theta$ and $\phi$ are not defined at $r=0$. If you follow the canonical formalism through from Lagrangian to canonical momenta $(p_r,p_{\theta},p_{\phi})$ to Hamiltonian, to canonical commutators, a host of difficulties arise. Although it is possible to sort them out by insisting that all the canonical momenta be Hermitian operators, it is considerably easier to quantize the system in Cartesian coordinates and make the change to spherical polar coordinates at the quantum level. This is the path taken in most elementary treatments of quantum mechanics in three dimensions: the operator $p^2 = p_1^2 + p_2^2 +p_3^2$ is recognized as the Laplacian in the coordinate representation $(p_j \to -i\hbar > \partial/\partial x_j \implies p^2 \to \nabla^2)$ and the transformation to polar coordinates is made by writing the Laplacian and the wavefunction in terms of $r$, $\theta$, and $\phi$. As a rule of thumb, the canonical approach becomes cumbersome when the classical coordinates and/or momenta do not range over the full interval from $-\infty$ to $+\infty$.

[1] R.L. Jaffe, "Canonical Quantization and Application to the Quantum Mechanics of a Charged Particle in a Magnetic Field", Supplementary Notes for MIT’s Quantum Theory Sequence, February 2007


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