Mathematical Methods for Physics and Engineering : A Comprehensive Guide

25.7 WKB METHODS

for some constantA. This form is only valid for negativezclose toz= 0 and is not
appropriate within the well as a whole, where the approximation (25.55) leading to Stokes’
equation is not valid. However, it does allow us to determine the correct combination of
the WKB solutions found earlier for the proper continuation inside the well of the solution
found forz>0. This is

ψ 1 (x)=

A

√

k(x)

sin

(∫a

x

k(u)du+

π

4

)

.

A similar argument gives the continuation inside the well of the evanescent solution
required in the regionx<−aas

ψ 2 (x)=

B

√

k(x)

sin

(∫x

−a

k(u)du+

π

4

)

.

However, for a consistent solution to the problem, these two functions must match, both
in magnitude and slope, at any arbitrary pointxinside the well.We therefore require both
of the equalities

A

√

k(x)

sin

(∫a

x

k(u)du+

π

4

)

=

B

√

k(x)

sin

(∫x

−a

k(u)du+

π

4

)

(i)

and

−

1

2

Ak′

√

k^3 (x)

sin

(∫a

x

k(u)du+

π

4

)

+

A

√

k(x)

[−k(x)]cos

(∫a

x

k(u)du+

π

4

)

=−

1

2

Bk′

√

k^3 (x)

sin

(∫x

−a

k(u)du+

π

4

)

+

B

√

k(x)

[k(x)]cos

(∫x

−a

k(u)du+

π

4

)

. (ii)

The general condition for the validity of the WKB solutions is that the derivatives of the
function appearing in the phase integral are small in some sense (see subsection 25.7.3 for
a more general discussion); here, ifk′/

√

k^3 k/

√

k,i.e.k′k^2 , then we can ignore the
k′terms in equation (ii) above. In fact, for this particular situation, this approximation
is not needed since the first of the equalities, equation (i), ensures that thek′-dependent
terms in the second equality (ii) cancel. Either way, we are left with a pair of homogeneous
equations forAandB. For them to give consistent values for the ratioA/B,itmustbe
that
A
√
k(x)

sin

(∫a

x

k(u)du+

π

4

)

×

B

√

k(x)

[k(x)]cos

(∫x

−a

k(u)du+

π

4

)

=

A

√

k(x)

[−k(x)]cos

(∫a

x

k(u)du+

π

4

)

×

B

√

k(x)

sin

(∫x

−a

k(u)du+

π

4

)

.

This condition reduces to

sin

[(∫a

x

k(u)du+

π

4

)

+

(∫x

−a

k(u)du+

π

4

)]

=0,

sin

[(∫a

−a

k(u)du+

π

2

)]

=0,

⇒

∫a

−a

k(u)du=(n+^12 )π.

Sincek(x)>0 in the range−a<x<a,nmay take the values 0, 1, 2,....
IfV(x)hastheformV(x)=V 0 x^2 sthen, for thenth allowed energy level,En=V 0 a^2 ns
and

k^2 (x)=

2 m


^2

(En−V 0 x^2 s).