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In this case A1 = - B3, and the relation between k and
is
modified to
ka = - acot a. |
(4.18) |
From the graphical solution, in Fig. 4.3
we see that this type of solution only occurs
for
a greater than
/2.
Figure 4.3:
The graphical solution for the odd states of the square well.
|
|
In the middle region all these solutions behave like sines, and you
will be asked to show that the solutions turn into minus themselves
when x goes
to - x. (We say that these functions are odd.)