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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 \phi and the component L_{z} of angular momentum are canonically conjugate
variables in classical mechanics. In quantum theory, the variable \phi becomes the operator of "multiplication of the wave function \psi(\phi) by \phi" and
L_{z}=-\frac{\hbar}{i}\frac{\partial}{\partial \phi}
, which implies the commutation relation
[L_{z}, \phi] =\frac{\hbar}{i} \hat{1}
These operators acting on periodic wave functions (i.e. \psi(0) = \psi(2\pi)) are Hermitian. Furthermore, L_{z} admits a complete system of orthonormal eigenfunctions \psi_{m}
L_{z}\psi_{m}(\phi) = m\hbar\psi_{m}(\phi)
, with \psi_{m}(\phi) =\frac{1}{\sqrt{2\pi}} e^{im\phi}, m\in\mathbb{Z}.
By evaluating the average value of the operator [L_{z}, \phi] in the state \psi_{m} and by taking into account the fact that Lz is Hermitian, one finds that
\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
=\langle \L_{z}^{\dagger}\psi_{m} , \phi\psi_{m}\rangle -m\hbar\langle \psi_{m},\phi \psi_{m}\rangle <br /> =(m\hbar-m\hbar)\langle \psi_{m},\phi\psi_{m} \rangle = 0 !
Hmm, paradoxical or not ? Or is it really necessary that \hbar=0 for consistency ?
Daniel.
If one introduces polar coordinates in the plane or spherical coordinates in space, then the polar angle \phi and the component L_{z} of angular momentum are canonically conjugate
variables in classical mechanics. In quantum theory, the variable \phi becomes the operator of "multiplication of the wave function \psi(\phi) by \phi" and
L_{z}=-\frac{\hbar}{i}\frac{\partial}{\partial \phi}
, which implies the commutation relation
[L_{z}, \phi] =\frac{\hbar}{i} \hat{1}
These operators acting on periodic wave functions (i.e. \psi(0) = \psi(2\pi)) are Hermitian. Furthermore, L_{z} admits a complete system of orthonormal eigenfunctions \psi_{m}
L_{z}\psi_{m}(\phi) = m\hbar\psi_{m}(\phi)
, with \psi_{m}(\phi) =\frac{1}{\sqrt{2\pi}} e^{im\phi}, m\in\mathbb{Z}.
By evaluating the average value of the operator [L_{z}, \phi] in the state \psi_{m} and by taking into account the fact that Lz is Hermitian, one finds that
\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
=\langle \L_{z}^{\dagger}\psi_{m} , \phi\psi_{m}\rangle -m\hbar\langle \psi_{m},\phi \psi_{m}\rangle <br /> =(m\hbar-m\hbar)\langle \psi_{m},\phi\psi_{m} \rangle = 0 !
Hmm, paradoxical or not ? Or is it really necessary that \hbar=0 for consistency ?
Daniel.
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DO the partial integration yourself ! Then you will see that the mistake is in the step : 
Really, then you must say that physicists who are studying non self adjoint Hamiltonians with a *real* spectrum, aren't doing physics. Neverthess, complex Hamiltonians are studied and for very good reasons. Here is a fairly recent paper by Patrick Dorey on the subject for example http://www.arxiv.org/PS_cache/hep-th/pdf/0309/0309209.pdf 
You went already too far before that, but let's not talk about it anymore. Let me tell you something : you might have noticed that I have rather ``obstinate'' ideas (I don't know if it is the right word) about many things. This is not a comfortable way of life, so one time I decided to take it easy for myself and simply defend some standard point of view in a discussion about a very delicate subject in quantum mechanics (oh yes, here she is again). The other party was kindly presenting something what many (textbooks) would label as heretic or even worse, simply impossible... Anyway, the upshot of what occurred is that I felt silly afterwards you see, because what this person told was simply correct and I did not notice it myself previously (even while studying ``the books'').
