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Here's something interesting i found:

If one introduces polar coordinates in the plane or spherical coordinates in space, then the polar angle [itex] \phi [/itex] and the component [itex] L_{z} [/itex] of angular momentum are canonically conjugate

variables in classical mechanics. In quantum theory, the variable [itex] \phi [/itex] becomes the operator of "multiplication of the wave function [itex] \psi(\phi) [/itex] by [itex] \phi [/itex]" and

[tex] L_{z}=-\frac{\hbar}{i}\frac{\partial}{\partial \phi} [/tex]

, which implies the commutation relation

[tex] [L_{z}, \phi] =\frac{\hbar}{i} \hat{1} [/tex]

These operators acting on periodic wave functions (i.e. [itex] \psi(0) = \psi(2\pi)[/itex]) are Hermitian. Furthermore, [itex] L_{z} [/itex] admits a complete system of orthonormal eigenfunctions [itex] \psi_{m} [/itex]

[tex] L_{z}\psi_{m}(\phi) = m\hbar\psi_{m}(\phi) [/tex]

, with [itex] \psi_{m}(\phi) =\frac{1}{\sqrt{2\pi}} e^{im\phi}, m\in\mathbb{Z} [/itex].

By evaluating the average value of the operator [itex] [L_{z}, \phi] [/itex] in the state [itex] \psi_{m} [/itex] and by taking into account the fact that Lz is Hermitian, one finds that

[tex] \frac{\hbar}{i} =\langle \psi_{m}, \frac{\hbar}{i}\hat{1}\psi_{m}\rangle = \langle \psi_{m} , L_{z}\phi\psi_{m}\rangle -\langle \psi_{m},\phi L_{z}\psi_{m}\rangle [/tex]

[tex] =\langle \L_{z}^{\dagger}\psi_{m} , \phi\psi_{m}\rangle -m\hbar\langle \psi_{m},\phi \psi_{m}\rangle

=(m\hbar-m\hbar)\langle \psi_{m},\phi\psi_{m} \rangle = 0 !! [/tex]

Hmm, paradoxical or not ? Or is it really necessary that [itex] \hbar=0 [/itex] for consistency ?

Daniel.

If one introduces polar coordinates in the plane or spherical coordinates in space, then the polar angle [itex] \phi [/itex] and the component [itex] L_{z} [/itex] of angular momentum are canonically conjugate

variables in classical mechanics. In quantum theory, the variable [itex] \phi [/itex] becomes the operator of "multiplication of the wave function [itex] \psi(\phi) [/itex] by [itex] \phi [/itex]" and

[tex] L_{z}=-\frac{\hbar}{i}\frac{\partial}{\partial \phi} [/tex]

, which implies the commutation relation

[tex] [L_{z}, \phi] =\frac{\hbar}{i} \hat{1} [/tex]

These operators acting on periodic wave functions (i.e. [itex] \psi(0) = \psi(2\pi)[/itex]) are Hermitian. Furthermore, [itex] L_{z} [/itex] admits a complete system of orthonormal eigenfunctions [itex] \psi_{m} [/itex]

[tex] L_{z}\psi_{m}(\phi) = m\hbar\psi_{m}(\phi) [/tex]

, with [itex] \psi_{m}(\phi) =\frac{1}{\sqrt{2\pi}} e^{im\phi}, m\in\mathbb{Z} [/itex].

By evaluating the average value of the operator [itex] [L_{z}, \phi] [/itex] in the state [itex] \psi_{m} [/itex] and by taking into account the fact that Lz is Hermitian, one finds that

[tex] \frac{\hbar}{i} =\langle \psi_{m}, \frac{\hbar}{i}\hat{1}\psi_{m}\rangle = \langle \psi_{m} , L_{z}\phi\psi_{m}\rangle -\langle \psi_{m},\phi L_{z}\psi_{m}\rangle [/tex]

[tex] =\langle \L_{z}^{\dagger}\psi_{m} , \phi\psi_{m}\rangle -m\hbar\langle \psi_{m},\phi \psi_{m}\rangle

=(m\hbar-m\hbar)\langle \psi_{m},\phi\psi_{m} \rangle = 0 !! [/tex]

Hmm, paradoxical or not ? Or is it really necessary that [itex] \hbar=0 [/itex] for consistency ?

Daniel.

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