J.J Sakurai shows in the section of ' Momentum operator in the position basis' as

$P$$\lvert\alpha\rangle$=$\int dx^{'}\lvert\ x{'}\rangle\Bigl(-i{h\over 2\pi}$ $\partial\over\partial x{'}$$ \langle\ x{'}\rvert \alpha\rangle \Bigr)$

this gives

$\langle\ x{'}\rvert P\lvert\alpha\rangle$=$-i{h\over 2\pi}$ $\partial\over\partial x{'}$$ \langle\ x{'}\rvert \alpha\rangle $

I don't get this step can anyone please help me?

A little help would really be appreciated.


\begin{eqnarray} P | \alpha \rangle &=& -i\hbar \int{\rm d}y |y\rangle \frac{\partial}{\partial y}\langle y| \alpha\rangle \\ \langle x' | P | \alpha \rangle &=& -i\hbar \langle x'| \int{\rm d}y |y\rangle \frac{\partial}{\partial y}\langle y| \alpha\rangle \\ &=& -i\hbar \int{\rm d}y \color{blue}{\langle x'|y\rangle} \frac{\partial}{\partial y}\langle y| \alpha\rangle \\ &=& -i\hbar \int{\rm d}y \color{blue}{\delta(x' - y)}\frac{\partial}{\partial y}\langle x| \alpha\rangle \\ &=& -i\hbar \frac{\partial}{\partial x'}\langle x'| \alpha\rangle \end{eqnarray}

  • $\begingroup$ caverac I have an error in question , I used x instead of x^', now i corrected question , can u please make this corrections to your answer @ caverac $\endgroup$ – Robin Raj May 19 at 18:52
  • $\begingroup$ @RobinRaj Sure, done! $\endgroup$ – caverac May 19 at 21:08
  • $\begingroup$ thanks for your precious help caverac @ caverac $\endgroup$ – Robin Raj May 21 at 7:24

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