Talk:Canonical commutation relation
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Question
Anyone read Dirac's book on QM? I'm missing something on this derivation. -ub3rm4th
Quoted (section 22, page 93):
- <math> (\frac{\partial}{\partial q_r})^* \left.\psi\right\rangle^*
= e^{-i\gamma}\frac{\partial}{\partial q_r} \left.\psi\right\rangle
= e^{-i\gamma}\frac{\partial}{\partial q_r} e^{i\gamma}\left.\psi\right\rangle^*<math>
showing that
- <math> (\frac{\partial}{\partial q_r})^* = e^{-i\gamma}\frac{\partial}{\partial q_r} e^{i\gamma}<math>
or, with the help of <math>\frac{\partial}{\partial q_r}f - f\frac{\partial}{\partial q_r} = \frac{\partial f}{\partial q_r}<math>,
- <math>(\frac{\partial}{\partial q_r})^* = \frac{\partial}{\partial q_r} + i\frac{\partial\gamma}{\partial q_r}<math>
Question, more urgent!
- <math>\frac{\partial}{\partial q_r} \left. \psi\right\rangle = 0<math>
Why?