Chapter 4
Bound states of the square well

One of the simplest potentials to study the properties of is the so-called square well potential,

V = 0 | x | > a V 0 | x | < a . (4.1)

square˙well


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

2 2 m d 2 d x 2 ψ ( x ) = E ψ ( x ) (4.2)

whereas in area II we have the equation

2 2 m d 2 d x 2 ψ ( x ) = ( E + V 0 ) ψ ( x ) (4.3)

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Solution to a few ODE’s

. In this class we shall quite often encounter the ordinary differential equations

d 2 d x 2 f( x)=α 2 f( x)(4.4)

which has as solution

f( x)= A 1 cos (α x)+ B 1 sin (α x)= C 1 e iα x+ D 1 e iα x,(4.5)

and

d 2 d x 2 g( x)=+α 2 g( x)(4.6)

which has as solution

g( x)= A 2 cosh (α x)+ B 2 sinh (α x)= C 2 eα x+ D 2 eα x.(4.7)
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Let us first look at E> 0. In that case the equation in regions I and I I I can be written as

d 2 d x 2 ψ ( x ) = 2 m 2 E ψ ( x ) = k 2 ψ ( x ) , (4.8)

where

k = 2 m 2 E . (4.9)

The solution to this equation is a sum of sines and cosines of k x, which cannot be normalised: Write ψ I I I( x)= A cos ( k x)+ B sin ( k x)( A, B, complex) and calculate the part of the norm originating in region I I I,

a | ψ ( x ) | 2 d x = a | A | 2 cos 2 k x + | B | 2 sin 2 k x + 2 ( A B ) sin ( k x ) cos ( k x ) d x = lim N N a 2 π k | A | 2 cos 2 ( k x ) + | B | 2 sin 2 ( k x ) = lim N N ( | A | 2 2 + | B | 2 2 ) = . (4.10)

We also find that the energy cannot be less than V 0, since we vannot construct a solution for that value of the energy. We thus restrict ourselves to V 0< E< 0. We write

E = 2 k 2 2 m , E + V 0 = 2 κ 2 2 m . (4.11)

The solutions in the areas I and III are of the form ( i= 1, 3)

ψ ( x ) = A i e k x + B i e k x . (4.12)

In region II we have the oscillatory solution

ψ ( x ) = A 2 cos ( κ x ) + B 2 sin ( κ x ) . (4.13)

Now we have to impose the conditions on the wave functions we have discussed before, continuity of ψ and its derivatives. Actually we also have to impose normalisability, which means that B 1= A 3= 0(exponentially growing functions can not be normalised). As we shall see we only have solutions at certain energies. Continuity implies that

A 1 e k a + B 1 e k a = A 2 cos ( κ a ) B 2 sin ( κ a ) A 3 e k a + B 3 e k a = A 2 cos ( κ a ) + B 2 sin ( κ a ) k A 1 e k a k B 1 e k a = κ A 2 sin ( κ a ) + κ B 2 cos ( κ a ) k A 3 e k a k B 3 e k a = κ A 2 sin ( κ a ) + κ B 2 cos ( κ a ) (4.14)

Tactical approach: We wish to find a relation between k and κ(why?), removing as manby of the constants A and B. The trick is to first find an equation that only contains A 2 and B 2. To this end we take the ratio of the first and third and second and fourth equation:

k = κ [ A 2 sin ( κ a ) + B 2 cos ( κ a ) ] A 2 cos ( κ a ) B 2 sin ( κ a ) k = κ [ A 2 sin ( κ a ) B 2 cos ( κ a ) ] A 2 cos ( κ a ) + B 2 sin ( κ a ) (4.15)

We can combine these two equations to a single one by equating the right-hand sides. After deleting the common factor κ, and multiplying with the denominators we find

[ A 2 cos ( κ a ) + B 2 sin ( κ a ) ] [ A 2 sin ( κ a ) B 2 cos ( κ a ) ] = [ A 2 sin ( κ a ) + B 2 cos ( κ a ) ] [ A 2 cos ( κ a ) B 2 sin ( κ a ) ] , (4.16)

which simplifies to

A 2 B 2 = 0 (4.17)

We thus have two families of solutions, those characterised by B 2= 0 and those that have A 2= 0.