0
0
-V 0
Energy
V(x)
x
E
−κ
C e Ae + Beκx ikx -ikx x
1 C e 3The calculation (See section 9.7.3) shows that eitherAorBmust be zero for a solution. This
means that thesolutions separate into even parity and odd parity states. We could have
guessed this from the potential.
Theeven stateshave the (quantization) constraint on the energy that
κ= tan(ka)k
√
− 2 mE
̄h^2= tan
√
2 m(E+V 0 )
̄h^2a
√
2 m(E+V 0 )
̄h^2
√
−E
E+V 0= tan
√
2 m(E+V 0 )
̄h^2a
and theodd stateshave the constraint
κ=−cot(ka)k
√
−E
E+V 0
=−cot
√
2 m(E+V 0 )
̄h^2a
These are transcendental equations, so we will solve them graphically. The plot below compares
the square root on the left hand side of the transcendental equations to the tangent on the right
for the event states and to “-cotangent” on the right for odd states. Where the curves intersect
(not including the asymptote), is an allowed energy. There is always one even solution for the 1D
potential well. In the graph shown, there are 2 even and one odd solution. The wider and deeper
the well, the more solutions.