# Momentum operator in position basis

On Sakurai page 52 the momentum operator in the position basis is defined but I'm having trouble understanding one line of the derivation.

\begin{align} \left(1-\frac{ip\Delta x'}{\hbar}\right)|\alpha\rangle & = \int dx' T(\Delta x')|x'\rangle\langle x'|\alpha\rangle \\ & = \int dx'|x'+\Delta x'\rangle\langle x'|\alpha\rangle \\ & = \int dx'|x'\rangle\langle x'-\Delta x'|\alpha\rangle \\ & = \int dx'|x'\rangle \left(\langle x'|\alpha\rangle-\Delta x'\frac{\partial}{\partial x'}\langle x'|\alpha\rangle \right) \end{align} How do we get from the second to last line to the last line? I understand that the partial derivative would come from expressing momentum in the position basis, but how is that coming from the second to last line? Any help would be greatly appreciated

• From the second to the third just change variable: $x'\rightsquigarrow x'-\Delta x'$, any constant shift of the variable doesn't contribute to the Jacobian of the change of variables, obviously. Commented Jul 25, 2018 at 20:21

1. A change of integration variable: i.e. setting $x''=x'+\Delta x'$, so that you get $$\int dx'|x'+\Delta x'\rangle\langle x'|\alpha\rangle = \int dx''|x''\rangle\langle x''-\Delta x''|\alpha\rangle,$$ and then dropping the primes for convenience.
2. A Taylor expansion of the wavefunction: when you do $$\int dx'|x'\rangle\langle x'-\Delta x'|\alpha\rangle = \int dx'|x'\rangle \left(\langle x'|\alpha\rangle-\Delta x'\frac{\partial}{\partial x'}\langle x'|\alpha\rangle \right),$$ you're basically doing $$\psi_\alpha(x'-\Delta x') = \psi_\alpha(x') - \Delta x' \psi'_\alpha(x'),$$ and re-casting that in Dirac-notation shape.