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One of the simplest potentials to study the properties of is the so-called
square well potential,
Figure 4.1:
The square well potential
|
|
We define three areas, from left to right I, II and III. In areas I and
III we have the Schrödinger equation
whereas in area II we have the equation
-   (x) = (E + V0) (x) |
(4.3) |
Let us first look at E > 0. In that case the equation in regions
I and III can be written as
 (x) = - E (x) = - k2 (x), |
(4.8) |
where
k = . |
(4.9) |
The solution to this equation is a sum of sines and cosines of kx,
which cannot be normalised:
Write
(x) = Acos(kx) + Bsin(kx) (A, B,
complex) and calculate the part of the norm originating in region
III,
| (x)|2dx |
= |
| A|2cos2kx + | B|2sin2kx + 2 (AB*)sin(kx)cos(kx)dx |
|
| |
= |
N | A|2cos2(kx) + | B|2sin2(kx) |
|
| |
= |
N(| A|2/2 + | B|2/2) = . |
(4.10) |
We also find that the energy cannot be less than - V0, since we
vannot construct a solution for that value of the energy. We thus
restrict ourselves to - V0 < E < 0. We write
E = - , E + V0 = . |
(4.11) |
The solutions in the areas I and III are of the form (i = 1, 3)
(x) = Aiekx + Bie-kx. |
(4.12) |
In region II we have the oscillatory solution
(x) = A2cos( x) + B2sin( x). |
(4.13) |
Now we have to impose the conditions on the wave functions we have
discussed before, continuity of
and its derivatives. Actually we
also have to impose normalisability, which means that B1 = A3 = 0
(exponentially growing functions can not be normalised). As we shall see
we only have solutions at certain energies.
Continuity implies that
| A1e-ka + B1eka |
= |
A2cos( a) - B2sin( a) |
|
| A3eka + B3e-ka |
= |
A2cos( a) + B2sin( a) |
|
| kA1eka - kB1eka |
= |
A2sin( a) + B2cos( a) |
|
| kA3eka - kB3e-ka |
= |
- A2sin( a) + B2cos( a) |
(4.14) |
After imposing the normalisability condition we realise we can take the
ratio of the first and third and second and fourth equation:
| k |
= |
![$\displaystyle {\frac{\kappa [A_2\sin(\kappa a) + B_2\cos(\kappa a)]}{A_2 \cos(\kappa a) - B_2 \sin(\kappa a)}}$](img91.gif) |
|
| k |
= |
![$\displaystyle {\frac{\kappa [A_2\sin(\kappa a) - B_2\cos(\kappa a)]}{A_2 \cos(\kappa a) + B_2 \sin(\kappa a)}}$](img92.gif) |
(4.15) |
We can combine these two equations to a single one by equating the
right-hand sides. After deleting the common
factor
, and multiplying with the denominators we find
[A2cos( a)+B2sin( a)][A2sin( a)-B2cos( a)] & = [A2sin( a) + B2cos( a)][A2cos( a) - B2sin( a)], |
which simplifies to
We thus have two families of solutions, those characterised by B2 = 0
and those that have A2 = 0.
Next: 4.1 B2 = 0
Up: Quantum Mechanics I
Previous: 3.3 Analysis of the