This document derives the energy of the first excited state of the harmonic oscillator using the Schrödinger equation. It first shows that the wave function for the first excited state is a1xe−ax2, where a is a constant. It then substitutes this wave function and its derivatives into the Schrödinger equation for the harmonic oscillator to obtain an expression for the energy E1 in terms of the angular frequency ω. Solving this expression yields the result that the energy of the first excited state is E1 = (3/2)ħω.

1. Chapter 35
Applications of the Schrödinger Equation
The Harmonic Oscillator
5 •• Use the procedure of Example 35-1 to verify that the energy of the
first excited state of the harmonic oscillator is 02
3
1 ωh=E . (Note: Rather than
solve for a again, use the step-8 result h02
1
ωma = obtained in Example 35-1.)
Picture the Problem We can differentiate ψ(x) twice and substitute in the
Schrödinger equation for the harmonic oscillator. Substitution of the given value
for a will lead us to an expression for E1.
The wave function for the first
excited state of the harmonic
oscillator is:
( )
2
11
ax
xeAx −
=ψ
Compute dψ1(x)/dx: ( ) [ ] 22
11
1 axax
eAxeA
dx
d
dx
xd −−
==
ψ
Compute d2
ψ1(x)/dx2
:
( ) [ ]
( ) 2
2222
1
32
1
32
1112
1
2
64
442
ax
axaxaxax
eAaxxa
eAxaeaxAeaxAeA
dx
d
dx
xd
−
−−−−
−=
+−−==
ψ
Substitute in the Schrödinger equation to obtain:
( )[ ] 222
111
22
02
1
1
32
2
64
2
axaxax
xeAExeAxmeAaxxa
m
−−−
=+−− ω
h
Dividing out (one can do this because the exponential function is never
zero) yields:
2
1
ax
eA −
( )[ ] xExmaxxa
m
1
32
02
132
2
64
2
=+−− ω
h
or
( ) ( ) xExmax
m
xa
m
1
32
02
1
2
32
2
6
2
4
2
=++− ω
hh
8

2. Applications of the Schrödinger Equation 9
Substitute for a to obtain:
xExmx
m
m
x
m
m
1
32
02
10
2
3
2
0
2
2
6
22
4
2
=+⎟
⎠
⎞
⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
− ω
ωω
h
h
h
h
Solve for E1 to obtain:
002
3
1 3EE == ωh
Reflection and Transmission of Electron Waves: Barrier
Penetration
13 •• A particle that has mass m is traveling in the direction of increasing x.
The potential energy of the particle is equal to zero everywhere in the region
0<x and is equal to U0 everywhere in the region x > 0, where U0 > 0. (a) Show
that if the total energy is E = αU0, where α ≥ 1, then the wave number k2 in the
region is given by0>x k2 = k1 α −1( ) α , where k1 is the wave number in the
region . (b) Using a spreadsheet program or graphing calculator, graph the0<x
reflection coefficient R and the transmission coefficient T as functions of α, for
.51 ≤≤ α
Picture the Problem We can use the total energy of the particle in the region
0>x to express k2 in terms of α and k1. Knowing k2 in terms of k1, we can use
( )
( )2
21
2
21
kk
kk
R
+
−
= to find R and T = 1 − R to determine the transmission coefficient T.
(a) Using conservation of energy,
express the energy of the particle in
the region x > 0:
00
2
2
2
2
UU
m
k
α=+
h
Solving for k2 gives: ( )
h
12 0
2
−
=
αmU
k
From the equation for the total
energy of the particle: h
0
1
2 Um
k
α
=

3. Chapter 3510
Express the ratio of k2 to k1: ( )
α
α
α
α
1
2
12
0
0
1
2 −
=
−
=
h
h
Um
mU
k
k
and 12
1
kk
α
α −
=
(b) The reflection coefficient R is
given by:
( )
( )2
21
2
21
kk
kk
R
+
−
=
Factor k1 from the numerator and
denominator to obtain:
2
1
2
2
1
2
1
1
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
=
k
k
k
k
R
Substitute the result from Part (a)
for k2/k1:
2
2
2
1
1
1
1
1
1
1
1
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎝
⎛
−
+
−
−
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
−
=
α
α
α
α
α
α
α
α
R
The transmission coefficient is given
by:
2
1
1
1
1
11
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎝
⎛
−
+
−
−
−=−=
α
α
α
α
RT
A spreadsheet program to plot R and T as functions of α is shown below. The
formulas used to calculate the quantities in the columns are as follows:
Cell Content/Formula Algebraic Form
A2 1.0 α
B2 (1−SQRT((A2−1)/A2))/
(1+SQRT((A2−1)/A2))^2
2
1
1
1
1
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎝
⎛
−
+
−
−
α
α
α
α

4. Applications of the Schrödinger Equation 11
C2 1−B2 2
1
1
1
1
1
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎝
⎛
−
+
−
−
−
α
α
α
α
A B C
1 α R T
2 1.0 1.000 0.000
3 1.2 0.298 0.702
4 1.4 0.198 0.802
5 1.6 0.149 0.851
18 4.2 0.036 0.964
19 4.4 0.034 0.966
20 4.6 0.032 0.968
21 4.8 0.031 0.969
22 5.0 0 29.0 0.971
The following graph was plotted using the data in the above table:
0.0
0.2
0.4
0.6
0.8
1.0
1 2 3 4 5
α
R
T
15 •• A 10-eV electron (an electron with a kinetic energy of 10 eV) is
incident on a potential-energy barrier that has a height equal to 25 eV and a width
equal to 1.0 nm. (a) Use Equation 35-29 to calculate the order of magnitude of the
probability that the electron will tunnel through the barrier. (b) Repeat your
calculation for a width of 0.10 nm.

5. Chapter 3512
Picture the Problem The probability that the electron with a given energy will
tunnel through the given barrier is given by Equation 35-29.
(a) Equation 35-29 is: a
eT α2−
=
where
( ) ( )
hh
EUmEUm −
=
−
= 0
2
0
22
α
Multiply the numerator and
denominator of α by c to obtain:
( )
c
EUmc
h
−
= 0
2
2
α
where
mMeV10974.1 13
⋅×= −
ch
Using evaluate T:keV,5112
e =cm
( ) ( )( ) 1718
13
9
10109.5
mMeV10974.1
eV10eV25keV5112
m100.12exp −−
−
−
≈×=
⎪⎭
⎪
⎬
⎫
⎪⎩
⎪
⎨
⎧
⋅×
−
×−=T
(b) Repeat with a = 0.10 nm:
( ) ( )( ) 22
13
9
10109.1
mMeV10974.1
eV10eV25keV5112
m1010.02exp −−
−
−
≈×=
⎪⎭
⎪
⎬
⎫
⎪⎩
⎪
⎨
⎧
⋅×
−
×−=T
General Problems
29 •• Eight identical non-interacting fermions are confined to an infinite
two-dimensional square box of side length L. Determine the energies of the three
lowest-energy states. (See Problem 22.)
Picture the Problem We can determine the energies of the state by identifying
the four lowest quantum states that are occupied in the ground state and
computing their combined energies. We can then find the energy difference
between the ground state and the first excited state and use this information to
find the energy of the excited state.
Each n, m state can accommodate
only 2 particles. Therefore, in the
ground state of the system of 8
fermions, the four lowest quantum
states are occupied. These are:
(1,1), (1,2), (2,1) and (2,2)
Note that the states (1,2) and (2,1) are
distinctly different states because the x
and y directions are distinguishable.

6. Applications of the Schrödinger Equation 13
The energies are quantized to the
values given by:
( )2
2
2
12
2
,
8
221
nn
mL
h
E nn +⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
The energy of the ground state is the sum of the energies of the four lowest
quantum states:
( ) ( ) ( )
( )
( ) 2
2
2
2
22
2
2
22
2
2
22
2
2
22
2
2
2,21,22,11,10
5
8552
8
2
22
8
2
12
8
221
8
211
8
2
mL
h
mL
h
mL
h
mL
h
mL
h
mL
h
EEEEE
=+++⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+
+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
++⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
++⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
+++=
The next higher state is achieved by
taking one fermion from the (2, 2)
state and raising it to the next higher
unoccupied state. That state is the
(1, 3) state. The energy difference
between the ground state and this
state is:
( ) ( )
2
2
22
2
2
22
2
2
2,23,1
4
22
8
31
8
mL
h
mL
h
mL
h
EEE
=
+−+=
−=Δ
Hence, the energies of the degenerate
states (1,3) and (3,1) are:
2
2
2
2
2
2
01,33,1
4
21
4
5
mL
h
mL
h
mL
h
EEEE
=+=
Δ+==
The three lowest energy levels are
therefore: 2
2
0
5
mL
h
E =
and two states of energy
2
2
21
4
21
mL
h
EE ==
37 ••• In this problem, you will derive the ground-state energy of the
harmonic oscillator using the precise form of the uncertainty principle,
2hpx x ≥ΔΔ , where Δx and Δpx are defined to be the standard deviations
Δx( )2
= x − x( )
2
and Δpx( )2
= px − px( )
2
. Proceed as follows:
1. Write the total classical energy in terms of the position x and momentum px
using U x( )= 1
2
mω0
2
x2
and K = 1
2
px
2
m .

7. Chapter 3514
2. Show that Δx( )2
= x − x( )
2
= x2
− x
2
and
Δpx( )2
= px
− px( )
2
= px
2
− px
2
. (Hint: See Equations 17-34a and 17-
34b.)
3. Use the symmetry of the potential energy function to argue that x and px
must be zero, so that Δx( )2
= x2
and Δpx( )2
= px
2
.
4. Assume that 2hpx x =ΔΔ to eliminate px
2
from the average energy
E = 1
2
px
2
m + 1
2
mω0
2
x2
= 1
2
px
2
m + 1
2
mω0
2
x2
and write E as
( ) ZmmZhE 2
02
12
8 ω+= , where Z = x2
.
5. Set d E dZ = 0 to find the value of Z for which E is a minimum.
6. Show that the minimum energy is given by 02
1
min
ωhE += .
Picture the Problem We can follow the step-by-step procedure outlined in the
problem statement to show that 02
1
min
ωhE += .
1. The total classical energy is:
( ) 22
02
1
2
2
xm
m
p
xUKE x
ω+=+=
Hence the average classical energy
is given by:
22
02
1
2
2
xm
m
p
E x
ω+= (1)
2. Express the standard deviation of
Δpx:
( ) ( )
22
2
xx
2
2
x
2
22
2
2
xx
xx
xxx
xxx
pp
pppp
pppp
ppp
−=
+−=
+−=
−=Δ
Proceeding similarly for the
standard deviation of Δx gives:
( ) 222
xxx −=Δ
3. From the symmetry of the
potential energy function we can
conclude that x and xp must
be zero. Hence:
( ) 22
xx =Δ
and
( ) 22
xx pp =Δ

8. Applications of the Schrödinger Equation 15
4. Rewrite equation (1) in terms of
to obtain:( )2
xpΔ
( ) 22
02
1
2
22
02
1
2
2
2
xm
m
p
xm
m
p
E
x
x
ω
ω
+=
+=
Δ
Using the uncertainty principle
(Δx Δpx = h/2) to eliminate 2
xp
gives:
( )
22
02
1
2
2
22
02
1
2
2
22
02
1
2
8
8
2
2
xm
xm
h
xm
xm
h
xm
m
x
h
E
ω
ω
ω
+=
+=
+
⎟
⎠
⎞
⎜
⎝
⎛
=
Δ
Δ
Letting 2
xZ = yields:
Zm
mZ
h
E 2
02
1
2
8
ω+=
5. Differentiate E with respect to
Z and set this derivative equal to
zero:
extremafor0
8
8
2
02
1
2
2
2
02
1
2
=
+−=
⎥
⎦
⎤
⎢
⎣
⎡
+=
ω
ω
m
mZ
h
Zm
mZ
h
dZ
d
dZ
Ed
Solve for Z to find the value of Z that
minimizes E (see the remark
below):
02 ωm
h
Z =
6. Evaluating E when
02 ωm
h
Z =
gives:
02
1
0
2
02
10
2
min
2
2
8
ω
ω
ω
ω
h
m
h
m
h
m
m
h
E
+=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+⎟
⎠
⎞
⎜
⎝
⎛
=
Remarks: All we’ve shown is that Z = h/(2mω0) is an extreme value, i.e.,
either a maximum or a minimum. To show that Z = h/(2mω0) minimizes <E>
we must either 1) show that the second derivative of <E> with respect to Z
evaluated at Z = h/(2mω0) is positive, or 2) confirm that the graph of <E> as a
function of Z opens upward at Z = h /(2mω0).