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How would we treat an operator of the form $ \hat{\mathbf{A}} \propto \hat{\mathbf{r}} \cdot \hat{{\mathbf{p}}} $ ?

Would it have eigenstates that are also eigenfunctions of position and/or momentum?

EDIT:

I am interested in knowing if there exists a common set of eigenstates of both r and p, which normally would not commute

SuperCiocia
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    Have you looked at its commutator with position/momentum for your latter question? And what about its "treatment" is unclear to you - it's just another operator! – ACuriousMind Dec 29 '14 at 18:43
  • I am interested in knowing if there exists a common set of eigenstates of both $r$ and $p$, which normally would not commute – SuperCiocia Dec 29 '14 at 19:32
  • Comment to the last question (v3): If two operators $\hat{A}$ and $\hat{B}$ have a common basis of eigenvectors, it means that they can be diagonalized simultaneously, and commute $[\hat{A},\hat{B}]=0$. See also e.g. this and this SE posts. – Qmechanic Dec 30 '14 at 17:35

1 Answers1

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If you keep the order when expanding the inner product you get terms of the form $x^ip_i$ which are not self-adjoint, due to the fact that $x^i$ doesn't commute with $p_i$. Hence the spectrum of such operator won't lie in $\mathbb R$, nor there is any hope in finding simultaneous eigenstates for both $x^i$ and $p_i$.

Expression like this are usually quantised by symmetrization: some people would simply write $$\widehat{\mathbf r\cdot\mathbf p} = \frac12(\hat{\mathbf r}\cdot\hat{\mathbf p} + \hat{\mathbf p}\cdot\hat{\mathbf r})$$ which is now self-adjoint.

Phoenix87
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  • But now you've obtained an operator, which, although Hermitian, still has high chances of being rubbish for actual problem involving that classical observable... – Ruslan Dec 30 '14 at 07:40
  • I don't think this is really a problem, as expressions like this don't usually occur in many relevant situations – Phoenix87 Dec 30 '14 at 09:07