Question 3.9: Derive the transformation from the position-space wave funct...

We want to find \Phi (p,t)=〈p|S(t)〉  given \psi(x,t)=〈x|S(t)〉 . We can relate the two by inserting a resolution of the identity:

\Phi (p,t)=〈p|S(t)〉                (3.108)

=\left\langle \begin{matrix} p \begin {vmatrix} \left(\int{dx |x〉〈x| }\right) \\ \end{vmatrix} S(t)\\ \end{matrix} \right\rangle

=\int{〈p|x〉 } 〈x|S(t)〉 dx.

=\int{〈p|x〉 } \Psi (x,t) dx.

Now,  〈x|p〉 is the momentum eigenstate (with eigenvalue p) in the position basis—what we called , in Equation 3.32 [f_{p}(x)= \frac{1} {\sqrt{2\pi \hbar }  } e^{ipx/\hbar}].So

〈p|x〉 =〈x|p〉 ^{*}=[f_{p}(x)]^{*}=\frac{1}{\sqrt{2\pi h} } e^{-ipx/\hbar }

Plugging this into Equation 3.108 [\Phi (p,t)=〈p|S(t)〉  ] gives

\Phi (p,t)=\int{\frac{1}{\sqrt{2\pi \hbar } } } e^{-ipx/\hbar }\Psi (x,t)dx

which is precisely Equation 3.54 [\Phi (p,t)=\frac{1}{\sqrt{2\pi \hbar }}\int_{-\infty }^{\infty }{ } e^{-ipx/\hbar }\Psi (x,t) dx].