# Overlap of Canonically Conjugate Variables

If we make a change of basis the scalar product $\langle x | p \rangle$ arises. We found that the momentum operator in the position basis is \begin{align} \langle x|\hat{p}|\psi (t) \rangle = -i \hbar \frac{\partial}{\partial x} \psi(x,t) \end{align} for any $|\psi (t) \rangle$. So we choose \begin{align} \langle x|\hat{p}|p \rangle = -i \hbar \frac{\partial}{\partial x} \langle x|p \rangle \end{align} but $|p \rangle$ satisfies the eigenvalue equation $\hat{p}|p \rangle= p|p \rangle$, so \begin{align} \langle x|\hat{p}|p \rangle = p \langle x|p \rangle \end{align} The last two equations together give the ODE \begin{align} -i \hbar \frac{\partial}{\partial x} \langle x|p \rangle = p \langle x|p \rangle \end{align} which has the solution

\begin{align} \langle x|p \rangle \propto e^{i \, p \cdot x / \hbar } \end{align}

Similarly

\begin{align} \langle p|x \rangle \propto e^{-i \, p \cdot x / \hbar } \end{align}

In general the position is desribed by a vector $\mathbf{r}$ and the momentum by ${\mathbf{p}}$. In three dimensions

\begin{align} \langle \mathbf{r}|\mathbf{p} \rangle = \frac{1}{(2 \pi \hbar)^{3/2}} e^{i \, {\mathbf{p}} \cdot \mathbf{x} / \hbar } \end{align} and \begin{align} \langle \mathbf{p}|\mathbf{r} \rangle = \frac{1}{(2 \pi \hbar)^{3/2}} e^{-i \, {\mathbf{p}} \cdot \mathbf{x} / \hbar } \end{align} with the normalisation factor $\frac{1}{(2 \pi \hbar)^{3/2}}$