Show that the scalar (or inner) product of basis state \(p\) and quantum state \(\phi\) with a function of the position operator \(V(x)\) can be written as an integral


 * 1) Put the product \(\langle p|\mathbf{V}(\hat{x})|\phi\rangle\) into integral form as described here: \(\langle p|\mathbf{V}(\hat{x})|\phi\rangle = \int \int dp\prime\,dy \langle p|y\rangle\langle y|\mathbf{V}(\hat{x})|p\prime\rangle\langle p\prime|\phi\rangle\)
 * 2) Use the identity \(\hat{x}|x\rangle = x|x\rangle\) to remove the operator: \(\langle p|\mathbf{V}(\hat{x})|\phi\rangle = \int \int dp\prime\,dy \langle p|y\rangle\langle y|\mathbf{V}(y)|p\prime\rangle\langle p\prime|\phi\rangle\)
 * 3) Substitute the identity \(\langle x|p \rangle=\frac{e^{i xp/\hbar}}{\sqrt{2 \pi \hbar}}\): \(\langle p|\mathbf{V}(\hat{x})|\phi\rangle = \int \int dp\prime\,dy\,\mathbf{V}(y) \frac{e^{-i py/\hbar}}{\sqrt{2 \pi \hbar}}\frac{e^{i p\prime y/\hbar}}{\sqrt{2 \pi \hbar}}\langle p\prime|\phi\rangle\)
 * 4) Simplify: \(\langle p|\mathbf{V}(\hat{x})|\phi\rangle = \int \int dp\prime\,dy\,\frac{\mathbf{V}(y)}{2\pi\hbar}e^{iy(p-p\prime)/\hbar}\langle p\prime|\phi\rangle\)
 * 5) Solve the integral with respect to \(y\), using the definition of the Fourier transform, \(\int dy \, \mathbf{V}(y)e^{iy(p-p\prime)/\hbar} = 2\pi\hbar\tilde{\mathbf{V}}(p-p\prime)\): \(\langle p|\mathbf{V}(\hat{x})|\phi\rangle = \int dp\prime\,\tilde{\mathbf{V}}(p-p\prime)\langle p\prime|\phi\rangle\)
 * 6) Put the equation in wavefunction form: \(\langle p|\mathbf{V}(\hat{x})|\phi\rangle = \int dp\prime\,\tilde{\mathbf{V}}(p-p\prime)\phi(p\prime)\)