Approximation Methods I: Time-Independent Perturbation Theory ! 205
(0)
eigenstate φn , while its unperturbed eigenstate is φn , is given by
!
!2
!
!
!
!
2
(1)
! PHYS
!
!Hrn
!
"
HW
502
2018
Solutions
1
!
!
|bn |2 = !1 −
#
$2 ! .
!
!
2
(0)
(0)
r̸=n Er − En
!
!
Problem 1
13.3 A harmonic oscillator potential is perturbed by the term λbx2 . Calculate
the first-order and second-order corrections to the energy eigenvalues.
(1)
The first-order correction to energy eigenvalues is En = ⟨n|H (1) |n⟩
where H (1) = bx2 . We have the following relations for linear harmonic
oscillator:
%
√
√
a|n⟩ = n|n − 1⟩, a† |n⟩ = n + 1|n + 1⟩, x = !/(2mω) (a + a† ).
Now
En(1)
=
=
=
⟨n|H (1) |n⟩
b!
⟨n|(a + a† )2 |n⟩
2mω
b!
⟨n|a2 + a†2 + aa† + a† a|n⟩ .
2mω
We note that
⟨n|am |n⟩
⟨n|am a†l |n⟩
⟨n|a†m al |n⟩
= 0 , ⟨n|a†m |n⟩ = 0 ,
= 0 if m ̸
=l,
= 0 if m ̸
=l.
(1)
Hence, in the expression for En , ⟨n|a2 |n⟩ = 0 and ⟨n|a†2 |n⟩ = 0. Then
En(1)
=
=
=
=
=
b!
⟨n|aa† + a† a|n⟩
2mω
'
√
b! & √
⟨n|a n + 1|n⟩ + ⟨n|a† n|n − 1⟩
2mω
b!
[⟨n|n + 1|n⟩ + ⟨n|n|n⟩]
2mω
b!
(2n + 1)⟨n|n⟩
2mω (
)
b!
1
!ω .
n+
mω 2
2
Next, consider
En(2)
=
"
m̸=n
*
! (1) !2
!Hnm !
(0)
(0)
En − Em
+
.
1
2
206 ! Solutions to the Exercises in Quantum Mechanics I: The Fundamentals
We obtain
(1)
Hnm
=
=
b!
2mω
b!
2mω
=
b!
2mω
=
b!
2mω
Then
En(2)
!
"
⟨n|a2 + a†2 + aa† + a† a|m⟩
√
!
√
⟨n|a m|m − 1⟩ + ⟨n|a† m + 1|m + 1⟩
√
"
√
+⟨n|a m + 1|m + 1⟩ + ⟨n|a† m|m − 1⟩
√
√
! √ √
⟨n| m m − 1|m − 2⟩ + ⟨n| m + 1 m + 2|m + 2⟩
√
√
"
√ √
+⟨n| m + 1 m + 1|m⟩ + ⟨n| m m|m⟩
#$
$
m(m − 1) δn,m−2 + (m + 1)(m + 2) δn,m+2
%
+(m + 1)δn,m + mδn,m .
=
&
=
=
(0)
(0)
(En − Em )
' (1) '2
'
& 'Hnm
(n − m)!ω
m̸
=n
(
)
b 2 !2
(n + 2)(n + 1) (n − 1)n
+
22 m2 ω 2 !ω
−2
2
+
*
1
b2
!ω .
n+
−
2
4
2m ω
2
m̸
=n
=
' (1) '2
'Hnm '
13.4 Calculate the first and second-orders corrections to the energy eigenvalues of a linear harmonic oscillator with the cubic term −λµx3 added to
the potential. Discuss the condition for the validity of the approximation.
The Hamiltonian of the perturbed system is H = H (0) + λH (1) where
1 2 1 2 (1)
p + kx , H = −µx3 . The first-order correction to energy
H (0) =
2m x 2
eigenvalues is given by
En(1) = ⟨n| − µx3 |n⟩ = −µ
*
!
2mω
+ 3 /2
⟨n|(a + a† )3 |n⟩ .
The expansion of (a + a† )3 is
a3 + a2 a† + aa† a + a† a2 + a†2 a + a† aa† + aa†2 + a†3 .
†
The eigenvalues and the eigenfunctions of the unperturbed system are
!
"nπx #
!2 π 2 n2
2
(0)
(0)
sin
.
, n = 1, 2, · · · , φn =
En =
4mL2
L
L
(1)
The first-order correction En is
Problem
2122: ! Solutions to the Exercises
in Quantum Mechanics I: The Fundamentals
$
2eE L
= −
x sin2 (nπx/L) dx
L
13.10 If the Hamiltonian of a particle
in a box of length L is subjected to a
0
$ by
L H (1) = −eEx calculate the first-order
uniform electric field given
eE
[x − x cos(2nπx/L)] dx
− eigenvalues.
correction to the =
energy
L 0
% 2
&L '
L2 of the unperturbed
L
&
The eigenvalues and theeE
eigenfunctions
system are
− 2 2 cos(2nπx/L)&
= −
!
L
2
4n π
0
"
!2 π 2 n2
2
nπx #
(0)
(0)
sin
.
, eEL
n = 1, 2, · · · , φn =
En =
.
4mL2=
L
L
2
(1)
The first-order correction En is
13.11 A particle in a box potential of
width L is perturbed by the term V0 for
$ L
0 < x < L/2 and
the first-order correction to
2eE Compute
2
(1)zero otherwise.
E
=
−
x
sin
(nπx/L)
dx
n
the energy eigenvalues.
L 0
$
eE L of the unperturbed system are
The eigenvalues and =
eigenfunctions
[x − x cos(2nπx/L)] dx
−
L 0
!
"nπx
2 2 2
%
!
2
π
n
&L '#
2
2
(0)
(0)
L
eE
L
sin &
.
En =
, n = 1, 2, · · · , φn =
− 2 2 cos(2nπx/L)
&
4mL2 = −
L
L
L
2
4n π
0
Then
eEL
.$
=
2 0 L/2
2V
(1)
En
=
sin2 (nπx/L) dx
L of0 width L is perturbed by the term V0 for
13.11 A particle in a box potential
$
0 < x < L/2 and zero otherwise.
V0 L/2 Compute the first-order correction to
=
[1 − cos(2nπx/L)] dx
the energy eigenvalues.
L 0
Problem 3
V0
The eigenvalues and=eigenfunctions
of the unperturbed system are
.
2
!
"nπx #
2 2 2
!
2
π
n
(
(0)
(0)
sin
.
En =
, n = 1, 2, · · · , φn =
−a/x, xL> 0
4mL2
L
13.12 An electron in the potential V (x) =
is in its ground
∞,
x≤0
Then
$
state characterized by the eigenfunction
φ0 = N xe−βx . The particle is
2V0 L/2
(1)
2
subjected to an applied
electric field Esin
the x-direction.
Calculate the
En
=
dx
e in(nπx/L)
L state
0
first-order correction to ground
energy
eigenvalue.
$
V0 L/2
=
[1 − cos(2nπx/L)]
dx
First, calculate β. Substituting
solution
in the Schrödinger
equation
L the
0
!2 V0
a
−= ψxx. − ψ = Eψ
2m 2
x
(
−a/x, x > 0
13.12 An electron in the potential V (x) =
is in its ground
∞,
x≤0
state characterized by the eigenfunction φ0 = N xe−βx . The particle is
subjected to an applied electric field Ee in the x-direction. Calculate the
first-order correction to ground state energy eigenvalue.
En(1)
First, calculate β. Substituting the solution in the Schrödinger equation
2
3
4
Approximation Methods I: Time-Independent Perturbation Theory ! 213
we get
"
!2 !
−
−2β + β 2 x − a = Ex .
2m
Equating the coefficients of x0 and x equal to zero separately we get
E=−
ma2
,
2!2
β=
am
.
!2
Next, from the normalization condition 1 = N 2
N = 2β 3 /2 .
#∞
0
x2 e−2 βx dx we find
Next, the first-order correction to the ground state energy can be cal(1)
culated from E0 = ⟨H (1) ⟩. The perturbation H (1) due to the applied
electric field is eEe x. Then
$ ∞
(1)
2
= N eEe
x3 e−2 βx dx
E0
0
$ ∞
2
N eEe
y 3 e−y dy
=
4
16 β
%0
'
$ ∞
&
N 2 eEe
3 −y &∞
2 −y
=
−y e 0 + 3
y e dy
16 β 4
0
3 N 2 eEe
=
8 β4
3 eEe
=
2β
3 eEe !2
.
=
2ma
13.13 Assume that the perturbation added to the infinite height potential is
(
0,
0 < x < L/2
Vp (x) =
V0 (x − L/2), L/2 < x < L.
Calculate the first-order correction to the energy eigenvalues.
The eigenvalues and eigenfunctions of the unperturbed system are
)
*nπx +
2 2 2
!
π
n
2
(0)
sin
.
,
n
=
1,
2,
·
·
·
,
φ
=
En(0) =
n
4 mL2
L
L
Then
En(1)
(1)
= Hnn
=
2V0
$
L
(x − L/2) sin2 (nπx/L) dx.