PH 216, Problem set 1, Due 4/6
1. In class, we estimated the groundstate energy of the 3D SHO (see Shankar 12.6.42)
using a ‘guess’ of
ψ
0
(
a
;
r
) =
R
(
a
;
r
)
Y
0
0
(
θ
,
φ
)
,
by varying
a
and minimizing
ψ
(
a
)

H

ψ
(
a
)
.
Suppose we try to get the first energy
E
1
using the trial function
ψ
1
=
R
(
a
;
r
)
Y
0
1
,
Where the
Y
enforces orthogonality with our
ψ
0
. A trial function with zero nodes that
has the right asymptotics (
∝
r
+1
as
r
→
0 and
→
0 as
r
→ ∞
) is
ψ
2
,
1
,
0
=
N
(
r/a
0
) exp(
−
r/
2
a
0
) cos
θ
.
what then is
E
1
, and how does this compare with the first level of the actual 3D SHO?
Solution:
We have
E
(
a
) =
ψ
(
a
)

H

ψ
(
a
)
=
ψ
(
a
)

T

ψ
(
a
)
+
ψ
(
a
)

V

ψ
(
a
)
.
As in class, we can use the quantum virial theorem,
T
=
−
1
2
V
→
T
+
V
=
E
n
=
−
2
/
2
ma
2
n
2
=
−
T
for an energy state of energy
E
n
. So we just have to evaluate
V
=
1
32
π
a
3
∞
0
r
2
dr
d
Ω
(
ω
r
2
/
2)(
r/a
)
2
exp(
−
r/a
) cos
2
θ
,
where I have taken the normalization of the wavefunction from Shankar 13.1.27. But
since
V
does nothing to angles, the angular integral just falls out as 4
π
, so we get
V
=
m
ω
2
48
π
a
5
∞
0
dr r
6
exp(
−
r/a
) =
m
ω
2
48
π
a
5
×
720
a
6
=
15
a
2
m
ω
2
π
.
Combining our results leads to
E
(
a
) written as:
E
(
a
) =
2
8
ma
2
+ 15
a
2
m
ω
2
With this in hand, we minimize the energy with respect to
a
:
dE
da
=
−
2
4
ma
3
+ 30
am
ω
2
= 0
,
which gives
a
min
=
2
120
m
2
ω
2
1
/
4
.
1
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Thus, we get
E
1
=
E
(
a
min
) =
15
2
ω
≈
2
.
74
ω
.
The exact result for the nth energy level of the 3D SHO is given by
E
n
= (
n
+ 3
/
2)
ω
,
and with
n
= 1,
E
1
= 2
.
5
ω
. This gives
∆
E
1
≈
0
.
24
ω
which is greater than zero as
required by the variational principle.
2. (Shankar 16.2.8) Consider the
= 0 radial equation for the threedimensional Coulomb
problem. Since
V
(
r
) is singular at the turning point
r
= 0, we cannot use
n
+
3
4
as in
eq. (16.2.42) of Shankar.
(a) Will the additive constant be more or less than
3
4
? (Only heuristic reasoning is
required, but you are welcome to derive the result formally if you like.)
(b) By analyzing the exact equation near
r
= 0, it can be shown that the constant
equals 1. Using this constant, show that the WKB energy levels agree with the
exact results
Solution:
(a) Heuristically, as discussed in class if the wavefunction must vanish at both ends,
the accumulated phase must add up to
n
halfcycles, so (
n
+ 1)
π
.
But if the
wavefunction need not vanish, then you fit somewhat less, and the WKB formal
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 Spring '09
 Energy, dz dz dz

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