Inside [ βπ,π] the TISE reads
ββ22ππ2ππ₯2π=πΈπor
π2ππ₯2π=βπ2π,π=β2ππΈβnoting that πΈ <0 states are forbidden (which would come up in the solutions anyway).
The general solution is then
π(π₯)=Λπ΄sinβ‘ππ₯+Λπ΅cosβ‘ππ₯we take boundary conditions π( βπ) =π(π) =0.
Now
π(βπ)=π΄sinβ‘(βππ)+π΅cosβ‘(βππ)=βπ΄sinβ‘ππ+π΅cosβ‘ππgiving solutions giving solutions for ππ =ππ2π with π΄ =0 for odd π and π΅ =0 for even π.
The π =0 solution is not normalizable and hence is rejected as unphysical,
and negative π gives a rescaling of a positive π solution.
Thus the energies are
πΈπ=β2π2π2π=π2π2β28ππNormalisation for odd π gives
1=β¨ππ|ππβ©=|π΅|2β«πβπcos2β‘πππ₯2πππ₯=|π΅|2β«π0(1+cosβ‘πππ₯π)ππ₯=|π΅|2[π₯+πππsinβ‘πππ₯π]π₯=ππ₯=0=|π΅|2πLikewise for even π we have
1=β¨ππ|ππβ©=|π΄|2β«πβπsin2β‘πππ₯2πππ₯=|π΄|2β«π0(1βcosβ‘πππ₯π)ππ₯=|π΄|2[π₯βπππsinβ‘πππ₯π]π₯=ππ₯=0=|π΄|2πSo π΄ =π΅ =β1/π.