Question 3.4: A particle of mass m is bound in the delta function well V(x...

The (position space) wave function is (Equation 2.132)

\Psi (x)=\frac{\sqrt{m\alpha } }{\hbar } e^{-m\alpha |x|/\hbar^2 }; E=-\frac{m\alpha ^2}{2\hbar ^2}             (2.132)

\psi (x,t)=\frac{\sqrt{m\alpha } }{\hbar } e^{-m\alpha |x|/\hbar ^{2}}e^{-iEt/\hbar }

(where E=-m\alpha^{2} /2\hbar ^{2} ). The momentum space wave function is therefore

\phi (p,t)=\frac{1}{\sqrt{2\pi \hbar } } \frac {\sqrt {m \alpha } }{\hbar } e^{-iEt/\hbar }\int_{-\infty } ^{\infty }{e^{-ipx/\hbar }e^{-m\alpha |x|/\hbar ^{2}}}dx= \sqrt {\frac{2}{\pi } }\frac{p^{3/2}_{0} e^{-iEt/\hbar }}{p ^{2}+p^{2}_{0}}

(I looked up the integral). The probability, then, is

\frac{2}{\pi } p^{3}_{0}\int_{p_{0}}^{\infty }{\frac {1}{(p^{2}+p^{2}_{0})^2} }dp=\frac{1}{\pi}\left [\frac {pp _{0}}{p^{2}+p^{2}_{0}}+\tan ^{-1} \left (\frac{p}{p_ {0} } \right) \right]\mid ^{\infty }_ {p_ {0} }= \frac {1}{4}-\frac{1}{2\pi} =0.0908

(again, I looked up the integral).