Roger Griffith
Physics 137B
hw. # 5
Proffessor Clarke
Problem # 1
Consider a 1-D simple harmonic. Use pertubation theory to calculate the shift in the ground state
energy introduce if one uses the relativistic expression (correct to first order) for the kinetic energy. [Use
the same trick as for the hydrogen atom!]
We know that the ground state wavefunction for the 1-D simple harmonic oscillator is given by
ψ0 =
! mω "1/4
mω 2
e− 2! x
π!
we also know that the first order relativistic correction in terms of the kinetic energy is
p4
and p2 = −!2 ∇2
8m3 c2
thus the first order correction to the ground state wavefunction is given by
"
Hr = −
E01 = #ψ0 |H " |ψ0 $ = −
!4
#∇2 ψ0 |∇2 ψ0 $
3
2
8m c
where
∇2 ψ0 =
so we find
E01
=
=
=
=
E01
=
! mω "1/4 ∂2
"
! mω "1/4 ! mω " mω 2 ! mω
2
− 2! x
2
− mω
2! x ) =
e
x
−
1
(e
π!
∂x2
π!
!
!
"2
!4 ! mω "2 ! mω 2
− 3 2
x − 1 ψ0 |ψ0 $
#
8m c
!
!
# Z ∞ #!
$
$
Z ∞
mω x2
mω "2 4
mω 2 − mω x2
!4 ! mω "2 ! mω "1/2
−
2
x e ! dx +
x −2
e ! dx
− 3 2
8m c
!
π!
!
!
0
−∞
%
&
'
&
'
&
' (
!2 m2 ω2 ! mω "1/2 √ 3 m2 ω2
! 5/2 mω ! 3/2
! 1/2
−
−
+
π
8m3 c2 π!
4 !2
mω
! mω
mω
% &
(
'
&
'
&
'
! 1/2
! 1/2
!2 ω2 ! mω "1/2 3 ! 1/2
−
+
−
8mc2 !
4 mω
mω
mω
3 !2 ω2
−
32 mc2
1
Problem # 2
Calculate the wavelength, in centimeters, of the photon emitted under a hyperfine transition in the
ground state (n = 1) of deuterium . Deuterium is “heavy” hydrogen, with an extra neutron in the nucleus.
gd e
Sd ;
The proton and neutron bindtogether to form a deuteron, with spin 1 and magnetic moment µd = 2m
d
the deuteron g-factor gd = 1.71. [Hint: This is intended to be a very short problem-just scale down from
hydrogen with appropriate g-factors and masses.]
we know
gd e
Sd and gd = 1.71
2md
and the wavelength of the emmited photon under a hyperfine splitting is given as
µd =
λd =
hc
ΔEd
and λd =
ΔEh
λh
ΔEd
ΔEh =
µ0 g p e2
3πm p me a3
λh = 21 cm
to find the hyperfine splitting for the deuteron we use
Eh1 f
µ0 gd e2
=
#Sd · Se $
3πmd me a3
but we know
1
Sd · Se = (S2 − Se2 − Sd2 )
2
we have two special cases
S
2
=
)
3 2
4!
15 2
4!
when s = − 12
when s = 21
Sn2 = !2 sn (sn + 1)
3
S2e = − !2
4
S2d = 2!2
and so we find
Eh1 f
µ0 gd e2 !2
=
3πmd me a3
)
−1
+ 12
(singlet)
(triplet)
therefore
µ0 gd e2 !2
2πmd me a3
therefore the wavelength of the emmited photon is
ΔEd =
λd =
2g p md
2(2)(5.91)
λh =
λh = 4.36(21 cm) = 91.53 cm
3gd m p
3(1.71)
\
2
Problem # 3
Write down the Slater determinent for three non-interacting fermions. The individual wavefunction
for each particle is assumed to be normalized
The Slater determinent is given as
*
*


* ψa (1) ψa (2) ψa (3) *
ψ
(1)ψ
(2)ψ
(3)
−ψ
(1)ψ
(3)ψ
(2)
a
c
a
c
b
b
*
1
1 *
ΨA = √ ** ψb (1) ψb (2) ψb (3) ** = √  −ψa (2)ψb (1)ψc (3) +ψa (2)ψb (3)ψc (1) 
6 * ψ (1) ψ (2) ψ (3) *
6 +ψ (3)ψ (1)ψ (2) −ψ (3)ψ (2)ψ (1)
a
c
a
c
c
c
c
b
b
(i). Verify that the normalization of the overall antisymmetric wavefunction Ψa is correct.
#ΨA |ΨA $ = 1
for normalization
2
ψa (1)ψb (2)ψc (3) −ψa (1)ψb (3)ψc(2)
1
−ψa (2)ψb(1)ψc (3) +ψa (2)ψb (3)ψc(1)  = 1
6
+ψa (3)ψb(1)ψc (2) −ψa (3)ψb (2)ψc(1)

all the cross-terms will be zero becauses
Z
)
1
ψa (i)ψb (i)dx = δab =
0
a=b
a &= b
so we find
1
[(ψa (1)ψb (2)ψc (3))2 + (ψa (1)ψb (3)ψc(2))2 + (ψa (2)ψb (1)ψc (3))2
6
+(ψa (2)ψb (3)ψc (1))2 + (ψa (3)ψb (1)ψc (2))2 + (ψa (3)ψb (2)ψc(1))2 ] = 1
therefore
6
1
[1 + 1 + 1 + 1 + 1 + 1] = = 1
6
6
therefore the normalization constant is correct.
(ii). Demonstrate that ΨA is indeed antisymmetric with respect to an exchange in labels for two particles.
if we exchange two particle labels we get




ψa (2)ψb (1)ψc(3) −ψa (2)ψb (3)ψc (1)
ψa (2)ψb (1)ψc (3) −ψa (1)ψb (2)ψc (3)
1
1
√  −ψa (1)ψb (2)ψc (3) +ψa (1)ψb (3)ψc (2)  = √  +ψa (1)ψb (3)ψc (2) −ψa (2)ψb (3)ψc (1) 
6 +ψ (3)ψ (2)ψ (1) −ψ (3)ψ (1)ψ (2)
6 +ψ (3)ψ (2)ψ (1) −ψ (3)ψ (1)ψ (2)
a
c
a
c
a
c
a
c
b
b
b
b
and it is therefore antisymmetric under particle exchange
(iii). Show that ΨA vanishes if any two fermions are in the same quantum state.
if we let ψa = ψb then we find
3


ψa (2)ψa (1)ψc (3) −ψa (2)ψa (3)ψc (1)
1
√  −ψa (1)ψa (2)ψc (3) +ψa (1)ψa (3)ψc (2)  = 0
6 +ψ (3)ψ (2)ψ (1) −ψ (3)ψ (1)ψ (2)
a
a
c
a
a
c
so if two fermions are in the same quatum state then ΨA = 0.
Problem # 4
(a). A cubical box (V = 0 for|x|, |y|, |z| < a,V = ∞ otherwise) contains 5 neutrons and 3 protons.
Asumme each particle has a mass m, and neglect all interactions between particles. Calculate the ground
state energy.
since all the particles are fermions, the Pauli exclusion principle states that only two identical fermions
are allowed to occupy the same state. Since we have 5 neutrons and 3 protons (which have half-integer
spin and are fermions) than the Pauli exclusion principle applies to these particles. We have
π 2 !2
8ma2
We know that for the ground state (which is 2-fold degenerate) we find 2 protons and 2 neutrons and
Exyz = (n2x + n2y + n2z )E1
E1 ≡
so
n = 1 nx = 1 ny = 1 nz = 1
and if we have 2 neutrons and 2 protons in the ground state we find
E111 = (2n + 2p)3E 1 = 12E1
in the first excited state (which is 6-fold degenerate) we have 3 neutrons and 1 proton and we find
n=2
nx = 1 ny = 1 nz = 2
nx = 1 ny = 2 nz = 1
nx = 2 ny = 1 nz = 1
therefore
E112 = E121 = E211 = (3n + p)6E1 = 24E1
thus
Eg = (2n + 2p)3E1 + (3n + p)6E1 = 36E1 =
36π2 !2
9π2 !2
=
8ma2
2ma2
(b). The same box now contains 8 interacting bosons of mass m . Calculate the ground state energy.
Since we know that bosons are not subject to the Pauli exclusion principle, then we can pack all 8
bosons into the ground state and we find that the ground state energy is given by
3π2 !2
24π2 !2
=
Eg = (8bosons )3E1 = 24E1 =
8ma2
ma2
4