# Transition Between Position and Momentum Basis

I'm having some trouble following pages 55-56 of Sakurai's Modern Quantum Mechanics.

We're trying to transfer from position space into momentum space. Here's a quote:

Let us now establish the connection between the $x$-representation and the $p$-representation. We recall that in the case of the discrete spectra, the change of basis from the old set $\{ | a^{'} \rangle \}$ to the new set $\{ | b^{'} \rangle \}$ is characterized by the transformation matrix (1.5.7). Likewise, we expect that the desired information is contained in $\langle x^{'} | p^{'} \rangle$, which is a function of $x^{'}$ and $p^{'}$, usually called the transformation function from the $x$-representation to the $p$-representation. To derive the explicit form of $\langle x^{'} | p^{'} \rangle$, first recall (1.7.17), letting $|\alpha \rangle$ be the momentum eigenket $|p^{'} \rangle$, we obtain $$\langle x^{'} |P|p^{'}\rangle = -i \hbar \frac{\partial}{\partial x^{'}} \langle x^{'} | p^{'}\rangle$$ or $$p'\langle x' |p^{'}\rangle = -i \hbar \frac{\partial}{\partial x^{'}} \langle x^{'} | p^{'}\rangle$$ The solution to this differential equation for $\langle x' | p' \rangle$ is $$\langle x' | p' \rangle = N \exp \left( \frac{ip'x'}{\hbar} \right)$$

I'm not sure where our differential equation is coming from. We have our momentum operator $P$ in the position basis, acting on our eigenket $|p'\rangle$, I can't see how we can find explicitly what the inner product of the position bra $\langle x'|$ and the momentum ket $|p'\rangle$ is.

What I'm thinking now is to rearange it like so: $$\int \langle x^{'} |p^{'}\rangle dx' = \int -\frac{i\hbar}{p'} \frac{\partial}{\partial x^{'}} \langle x^{'} | p^{'}\rangle dx'$$

But it doesn't seem to lead anywhere. What am I missing?

• Hint: $\langle x|\psi \rangle = \psi(x)$ – Alfred Centauri Jun 13 '14 at 2:21
• Edit: Everything cleared up! – Astrum Jun 13 '14 at 6:31