PHYS 610A
Spring 2004
HW 11 – SOLUTIONS
Shankar 16.1.2 Let V = 0 for -a  x  a, =  for | x | > a. Let Ψ(x) be (x+a)(x-a)
= x2 – a2 (unnormalized). First, normalize: < Ψ | Ψ > = -aa dx | Ψ | 2 =
-aa dx ( x2 – a2 )2 = -aa dx ( x4 – 2 a2 x2 + a4 ) = (16/15) a5, so
Ψ(x) = (15/16a5)1/2 (x2 – a2). Now that Ψ, compute E = < Ψ | H | Ψ >/ < Ψ | Ψ >
2
= < Ψ | H | Ψ >. But Hˆ   2m d 2 dx 2 , and (d/dx)2 (x2 –a2) = 2, so
E = < Ψ | H | Ψ > = (15/16a5)(- 2/2m) -aa dx ( x2 – a2 ) 2 = (15/16a5)(- 2/2m)(-8a3/3)
= (5/4) ( 2 / ma2); this is the variational estimate for the ground state energy. The exact
g.s. energy we already know: 2 π2 / (2m(2a)2) = (π2 /8) ( 2 / ma2).
Now (π2 /8) 1.23 < 5/4, so variational theorem is satisfied. Notice that even those we
didn’t actually vary any parameters, this crude trial wfn gave a very good estimate for the
g.s. energy.
ma 2V02
2 d 2
Shankar 16.1.3 Hˆ  
.
E
(exact)
=
. Let

aV

(
x
)

g.s.
0
2m dx 2
2 2
 
trial ( x)  
1/ 4
exp( x 2 / 2) which is normalized. In class we already showed
2 d 2
 2

 2

.
Now
,
so




aV

(
x
)



aV
E
(

)

 aV0
0
0
2
2m dx
4m

4m

2
aV0


To find minimum, take derivative:
E ( ) 

 0 , or, solving,

4m 2 
 
2
ma 2V0
4(maV0 ) 2
 min 
, Then, substituting, E ( min )  
, which is slightly higher in
 2
 4
energy than the exactly solution given above.
2 d 2
exp( ar )
 V0
Extra credit Johnson 11.2 The radial Hamiltonian is Hˆ r  
.
2
2m dr
r

Let u(r) = C r exp ( -a r/2). We will need the following integral:  dr  r n exp( ar ) 
0
n!
a n1

Find C by normalization:
 dr  C
2
r 2 exp( ar )  2 / a 3 so C = (a3/2)1/2.
0
You want to compute < u | H | u > (because we assume the angular momentum l=0, we
have already integrated out that part of the wfn. You should understand this...)

 2 d 2 
u (r ) and u”(r) = -a(1-ar/4)e-ar/2
First, the kinetic energy < KE > =  dr  u (r ) 
2 
 2m dr 
0

a3  2 
2a4  1 a 2  2a2
 
(a)  dr  (r  14 ar 2 ) exp( ar ) 
and < KE > =
 

2  2m 
4m  a 2 4 a 3  8m
0
PHYS 610A
HW 10 SOLUTIONS
p. 2
(extra credit, continued)
The potential energy < PE > =

  dr  u (r ) V0 exp( r / R) / r  V0
2
0

 

a3
a3
1
1
dr

r
exp

r
a

R


V
0

2 0
2 (a  R 1 ) 2
2a2
a3
1
The total energy E(a) 
and to find the minimum,
 V0
8m
2 (a  R 1 ) 2
1

2a
2
a  3R 1
2 1 a  3R
or
; this is probably as far as
0
E (a)
 V0 a
a
a
4m
2 (a  R 1 ) 3
2mV0
(a  R 1 ) 3
most of you got. It does require solving a cubic equation for the most general case. As
you can see, there are very few analytically solvable problems in quantum mechanics,
even with approximate methods such as variational.
We can, however, look at two limits:
m 2V02
2
1
; for the hydrogen atom
 and E(amin) = 
2mV0 a
2h 2
V0 = e2 and we regain our usual result.
We can let R  , so that
We can let R  0, which is like a delta- function; then we have to have V0   and
simultaneously have V0 R2 be constant ; then rewrite the variational condition as

2a
1 aR  3
2
2
3
2
or
or

V
R
a
0
E (a)
 V0 R 2 a 2
 a and
0
a
4m
2 (aR  1) 3
4m
2
6 m V0 R 2

E(amin) = 
6

.
144m(V0 R 2 ) 2


