Representation of conjugate momentum

[AlonsoMcLaren]

We know that in Cartesian position basis the representation of momentum is -ihbar (d/dx)

Consider a cylindrical/spherical/whatever curvilinear coordinates. To make life simple, consider a particle constrained to move on a circle so that its position can described by θ only. Suppose we express the wavefunction as a function of θ, not x. The system has an Lagrangian from which we can find the conjugate momentum pθ

Can we thus declare that pθ can be represented by -ihbar (d/dθ) in the θ basis?

 

[Matterwave]

Canonical quantization sort of only works well in rectilinear coordinates iirc. Quantizing generalized coordinates is a messy subject. However, in the case of cylindrical coordinates in one dimension, All you have to do is turn x into a periodic coordinate and it should work fine.

 

[vanhees71]

The momentum operator in position representation is always
$$\hat{\vec{p}}=-\mathrm{i} \hbar \vec{\nabla}.$$
Since $\vec{\nabla}$ is a vector operator, it can be expressed in any coordinates you like. It always has the same meaning, but perhaps I don't understand the question right.

 

[moss]

Quantization will work in any coordinate system as long as there exists coordinate transformation between different system, that is the whole pt. and the physics does not depend on a specific coordinate system. Generalized coordinates are used all the time in QFT, QM and Stat.Physics, you name it.

On a circle, theta can be replaced by x, just consider the circle is very large and is almost a straight line btw 2 pts. A & B. Why not? There is only one basis vector in theta-basis, since 1-dim. If the radius of the circle is changing then use r and theta as basis.

You can also transform the coordinates from x-quantized system to theta system with simple coordinate transformation and that's it, you have a theta-quantized system.

Bottom line is that you quantize in one system of coordinates and just transform into any coordinates, the system remain quantized.