Atawe get
C 3 e−κa=Acos(ka) +Bsin(ka)
−κC 3 e−κa=−kAsin(ka) +kBcos(ka).Divide these two pairs of equations to get two expressions forκ.
κ =kAsin(ka) +kBcos(ka)
Acos(ka)−Bsin(ka)−κ =−kAsin(ka) +kBcos(ka)
Acos(ka) +Bsin(ka)Factoring out thek, we have two expressions for the same quantity.
κ
k=
Asin(ka) +Bcos(ka)
Acos(ka)−Bsin(ka)
κ
k=
Asin(ka)−Bcos(ka)
Acos(ka) +Bsin(ka)If we equate the two expressions,
Asin(ka) +Bcos(ka)
Acos(ka)−Bsin(ka)=
Asin(ka)−Bcos(ka)
Acos(ka) +Bsin(ka)and cross multiply, we have
(Asin(ka) +Bcos(ka))(Acos(ka) +Bsin(ka))
= (Acos(ka)−Bsin(ka))(Asin(ka)−Bcos(ka)).TheA^2 andB^2 terms show up on both sides of the equation and cancel. What’s left is
AB(sin^2 (ka) + cos^2 (ka)) = AB(−cos^2 (ka)−sin^2 (ka))
AB = −ABEitherAorB, but not both, must be zero. We have parity eigenstates, again, derived from the
solutions and boundary conditions.
This means that the states separate into even parity and odd parity states. We could have guessed
this from the potential.
Now lets use one equation.
κ=Asin(ka) +Bcos(ka)
Acos(ka)−Bsin(ka)
kk If we setB= 0, the even states have the constraint on the energy that
κ= tan(ka)kand, if we setA= 0, the odd states have the constraint
κ=−cot(ka)k.