HarmonicOscillator (1).pdf

Harmonic oscillator
Notes on Quantum Mechanics
http://quantum.bu.edu/notes/QuantumMechanics/HarmonicOscillator.pdf
Last updated Thursday, November 30, 2006 13:50:03-05:00
Copyright © 2006 Dan Dill (dan@bu.edu)
Department of Chemistry, Boston University, Boston MA 02215
à Classical harmonic motion
The harmonic oscillator is one of the most important model systems in quantum mechanics. An
harmonic oscillator is a particle subject to a restoring force that is proportional to the displacement of
the particle. In classical physics this means
F = m a = m
„2 x
ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
„t2
= -k x
The constant k is known as the force constant; the larger the force constant, the larger the restoring
force for a given displacement from the equilibrium position (here taken to be x = 0). A simple
solution to this equation is that the displacement x is given by
x = sinI
è!!!!!!!!!!
k ê m tM,
since
m
„2 x
ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
„t2
= m
„2
ÅÅÅÅÅÅÅÅ
ÅÅÅÅ
„t2
sinI
è!!!!!!!!!!
k êm tM
= -mI
è!!!!!!!!!!
k ê m M
2
sinI
è!!!!!!!!!!
k ê m tM
= -k sinI
è!!!!!!!!!!
k ê m tM
= -k x.
The quantity
è!!!!!!!!!!
k êm plays the role of an angular frequency,
w = 2 p n =
è!!!!!!!!!!
k ê m .
The larger the force constant, the higher the oscillation frequency; the larger the mass, the smaller the
oscillation frequency.
à Schrödinger equation
The study of quantum mechanical harmonic motion begins with the specification of the Schrödinger
equation. The linear restoring forces means the classical potential energy is
V = -‡ F „ x = -‡ H-k xL „ x =
1
ÅÅÅÅÅ
2
k x2
,

and so we can write down the Schrödinger equation as
i
k
j
j
j-
Ñ2
ÅÅÅÅÅÅÅÅ
ÅÅÅÅ
2 m
„2
ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
„x2
+
1
ÅÅÅÅÅ
2
k x2y
{
z
z
z yHxL = E yHxL.
Next, it will be helpful to transform this equation to dimensionless units. We could use the same
length and energy units that we have used for the particle in a box and for the one-electron atom, but
there is a different set of units that is more natural to harmonic motion.
Since harmonic motion has a characteristic angular frequency, w =
è!!!!!!!!!!
k ê m , it makes sense to
measure energy in terms of w. It turns out that the choice Ñ wê 2 works well. (The choice Ñw would
seem more obvious, but the factor of 1ê2 simplifies things somewhat.). Next, we can use the energy
unit to determine the length unit. Specifically, let's use for the unit of length the amount by which the
oscillator must be displace from equilibrium (x = 0) in order for the potential energy to be equal to
the energy unit. That is, the unit of length, x0, satisfies
Ñ w
ÅÅÅÅÅÅÅÅ
ÅÅÅÅ
2
=
1
ÅÅÅÅÅ
2
k x0
2
=
1
ÅÅÅÅÅ
2
m w2
x0
2
and so
x0 = $%%%%%%%%%%%
Ñ
ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
m w
This means we can express energy as
E =
Ñ w
ÅÅÅÅÅÅÅÅ
ÅÅÅÅ
2
e
in terms of dimensionless multiples e of Ñ wê 2, and length as
x = $%%%%%%%%%%%
Ñ
ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
m w
r
in terms of dimensionless multiples r of
è!!!!!!!!!!!!!!
Ñêm w . In these dimensionless units, the Schrödinger
equation becomes
„2
ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅ
„ r2
yHrL = -tHrL yHrL,
in terms of the dimensionless kinetic energy
tHrL = e - r2
.
Verify that the Schrödinger equation has this form in the dimensionless units of energy
and length that we have chosen.
Show that the length unit, x0 =
è!!!!!!!!!!!!!!
Ñê m w , can be written alternatively as
è!!!!!!!!!!!!!
Ñ wê k and
"###################
Ñë
è!!!!!!!!
k m .
Here is a plot of the dimensionless potential energy.
2 Harmonic oscillator
Copyright © 2006 Dan Dill (dan@bu.edu). All rights reserved

-4 -2 2 4
r
5
10
15
20
25
vHrL
Harmonic potential energy, in units Ñ wê2. Length r is in units
è!!!!!!!!
!!!!!!!
Ñêm w .
à Energies and wavefunctions
It turns out that the quantal energies in the harmonic potential are
ej = 2 j - 1,
where j is the number of loops in the wavefunction. Here is the lowest energy wavefunction—the
wavefunction with one loop. (This and the following example wavefunctions in this part are
determined by Numerov integration of the Schrödinger equation.)
-6 -4 -2 2 4 6
r
0.1
0.2
0.3
0.4
0.5
0.6
0.7
y e = 1
Lowest energy harmonic oscillator wavefunction. The energy is 2μ1 - 1 = 1, in units Ñ wê2. Displacement r from equilibrium is in
units
è!!!!!!!!
!!!!!!!
Ñêm w . The vertical lines mark the classical turning points.
The vertical lines mark the classical turning points, that is, the displacements for which the harmonic
potential equals the energy.
turn@j_D := ρ ê. Solve@ρ2
== 2 j − 1, ρD êê Evaluate;
turn@jD
9−
è!!!!!!!!
!!!!!!!!!!
−1 + 2 j ,
è!!!!!!!!
!!!!!!!!!!
−1 + 2 j =
Here is the sixth lowest energy wavefunction,
Harmonic oscillator 3

-6 -4 -2 2 4 6
r
-0.4
-0.2
0.2
0.4
y e = 11
Sixth lowest energy harmonic oscillator wavefunction. The energy is 2μ6 - 1 = 11, in units Ñ wê2. Displacement r from equilibrium
is in units
è!!!!!!!!
!!!!!!!
and here is the 20th lowest energy wavefunction,
-7.5 -5 -2.5 2.5 5 7.5
r
-0.4
-0.2
0.2
0.4
y e = 39
20th lowest energy harmonic oscillator wavefunction. The energy is 2μ6 - 1 = 11, in units Ñ wê2. Displacement r from equilibrium
is in units
è!!!!!!!!
!!!!!!!
This wavefunction shows clearly the general feature of harmonic oscillator wavefunctions, that the
oscillations in wavefunction have the smallest amplitude and loop length near r = 0, where the
kinetic energy is largest, and the largest amplitude and loop length near the classical turning points,
where the kinetic energy is near zero.
Finally, here are the seven lowest energy wavefunctions.

-6 -4 -2 2 4 6
r
-0.6
-0.4
-0.2
0.2
0.4
0.6
y
Seven lowest energy harmonic oscillator wavefunctions. The energies are 2μ j - 1 = 1, 3, …, 13, in units Ñ wê2. Displacement r
from equilibrium is in units
è!!!!!!!!
!!!!!!!
à Absolute units
We have expressed energy as
E =
Ñ w
ÅÅÅÅÅÅÅÅ
ÅÅÅÅ
2
e,
in terms of dimensionless multiples e of Ñ wê 2, and length as
x = $%%%%%%%%%%%
Ñ
ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
m w
r,
in terms of dimensionless multiples r of
è!!!!!!!!!!!!!!
Ñêm w . To get a feeling for these units, let's see how they
translate into actual energies and length for particular molecules.
The atoms in hydrogen halide molecules, HF, HCl, etc., vibrate approximately harmonically about
their equilibrium separation. The mass undergoing the harmonic motion is the reduced mass of the
molecule,
m =
ma mb
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅÅ
ma + mb
(I remember that the product of the masses goes in the numerator since the ratio must have units of
mass.) To calculate the reduced mass we need to determine the mass of each atom, and to do this, we
need to know which isotope of each atom is present in the molecule.
Recall that isotope masses are given in units of atomic mass, u. The atomic mass unit is defined such
that the mass of exactly one gram of carbon 12 is Avogadro's number times u. This means that the
atomic mass unit is
u =
1 Gram ê Mole
AvogadroConstant
ê. Gram → 10−3
Kilogram
1.66054× 10−27
Kilogram
Let's calculate the reduced mass for HCl. If we use the most stable isotope of each atom, 1 H and
35
Cl, the result is

μ1H35Cl = i
k
j
j
mH mCl
mH + mCl
y
{
z
z í AvogadroConstant êê. 8
mH → 1.0078 Gram ê Mole,
mCl → 34.9688 Gram ê Mole,
Gram → 10−3
Kilogram
<
1.62661× 10−27
Kilogram
Here are the reduced masses for other combinations of isotopes, together with that for 1 H 35Cl.
1H35Cl 1.62661× 10−27
Kilogram
1H37Cl 1.62908× 10−27
Kilogram
2H35Cl 3.1622× 10−27
Kilogram
2H37Cl 3.17153× 10−27
Kilogram
Confirm that these results are correct.
The effect a change in the lighter isotope is larger than the effect of a change in the heavier isotope.
Show why this is so.
The next step is to determine the harmonic angular frequency, w = 2 p n. This is done by measuring
the frequency of light that causes the molecule to change its vibrational wavefunction by one loop,
since
D Ematter = Ñ w = h n.
For 1 H 35Cl the measured value is n
è = 2990 cm-1. The unit n
è is the reciprocal wavelength,
corresponding to the frequency,
n
è =
1
ÅÅÅÅÅ
l
=
1
ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
cê n
=
n
ÅÅÅÅÅ
c
.
This means that angular frequency is related to wavenumber as
w = 2 p n = 2 p
c
ÅÅÅÅÅ
l
= 2 p c n
è.
Hence, the angular frequency of harmonic motion in 1 H 35Cl is
5.63212× 1014
Second
Verify this result.
Note that this value properly corresponds to the IR spectral region.
Having determined the oscillator mass and angular frequency, we can evaluate its length unit,
x0 =
è!!!!!!!!!!!!!!
Ñê m w .
0.10729 ﬁ
To interpret this result, recall that we have defined the unit of length so that the when the oscillator is
displaced this distance from its equilibrium point, the potential energy equals the zero-point energy.
That is, x0 is the classical turning point of the oscillation when the oscillator wavefunction has 1
loop. This means that when 1 H 35Cl is in its ground state its classically allowed region is
2 x0 = 0.21458 Þ wide. The equilibrium internuclear distance of HCl is 1.27 Þ, and so ground state
harmonic motion expands and compresses the bond by a bit less than 10%.

Evaluate x0 for 1
H 81
Br (n
è = 2650 cm-1
) and 1
H 127
I (n
è = 2310 cm-1
), and analyze your
results in comparison to the value for 1 H 35Cl.
Here are the answers I get.
μ ω x0
1H35Cl 1.62661× 10−27
5.63212× 1014
0.10729 fi
1H81Br 1.6529× 10−27
4.99168× 1014
0.113055 fi
1H127I 1.66031× 10−27
4.35124× 1014
0.12082 fi
Comparison of HCl, HBr and HI
Plot, on the same set of axes, the harmonic potential for HCl, HBr, and HI, Measure length
in Þ. Indicate the first four energy levels of each potential curve. Do this using horizontal
lines spanning the allowed region at each energy on each curve. Measure energy in units
of the zero-point energy of HCl. In a separate table give energies (in units of the zero-point
energy of HCl) and the right side (x > 0) classical turning point (in Þ) for the first four
energy levels of each molecule.
Here are the expressions I get for the potential curve, with distance, x, in Þ and energy in units of the
zero-point energy of HCl.
1H35Cl 86.872 x2
1H81Br 69.3414 x2
1H127I 52.9255 x2
Verify that, for HCl, when the displacement is its distance unit, x0 = 0.10729 Þ, the
potential energy is 1, since we are using as energy unit the zero-point energy of HCl. Hint:
Evaluate k ê 2 in J m-2, divide it by the zero point energy, Ñ wê 2 in J, and then convert the
result from m-2 to Þ-2
.
These potential energy expressions show that the force constant, k, decreases going form
HCl to HI. Evaluate the force constant for each molecules, in J m-2 = kg s-2. Answer:
516.0, 411.9, 314.4.
Here is the plot of the results I get.

-0.3 -0.2 -0.1 0.1 0.2 0.3
Hx-xeLêÞ
2
4
6
8
EêHÑwHClê2L
Harmonic potential energy curves and lowest four harmonic energy levels (horizontal lines) for 1
H 35
Cl (n
è = 2990 cm-1
), 1
H 81
Br
(n
è = 2650 cm-1
) and 1
H 127
I (n
è = 2310 cm-1
). Energy is in units of the zero-point energy of 1
H 35
Cl, 2.969μ10-20
J.
Here is the tabulation of energies and right side turning points for the lowest four levels of each
molecule.
loops HjL EjêH—ωHClê2L xtpêﬁ
1H35Cl
1
2
3
4
1.
3.
5.
7.
0.10729
0.185832
0.239908
0.283863
1H81Br
1
2
3
4
0.886288
2.65886
4.43144
6.20401
0.113055
0.195818
0.252799
0.299116
1H127I
1
2
3
4
0.772575
2.31773
3.86288
5.40803
0.12082
0.209266
0.270161
0.319659
Lowest four harmonic energy levels and right side classical turning points for 1
H 35
Cl (n
è = 2990 cm-1
), 1
H 81
Br (n
è = 2650 cm-1
) and
1
H 127
I (n
è = 2310 cm-1
). Energy is in units of the zero-point energy of 1
H 35
Cl, 2.969μ10-20
J.
The results reflect the effects of the decreasing harmonic frequency going from HCl to HI: The force
constant decreases, and so the harmonic potential energy curve rises less steeply on either side of its
minimum, with the result that turning points are farther apart and so a wider allowed region at a
given total energy. The effect is an increase in loop length and so a lowering of energy for a given
number of loops, analogous to the energy lowering in an infinite well when the well width is
increased.
Now, here are two final questions to consider.
Show that the decrease in harmonic frequency, w, and so in the force constant, k, going
from HCl to HI cannot be due to the increasing reduced mass alone. Hint: Compare the
change in harmonic frequency expected due to mass alone to the actual change in
harmonic frequency. Answer: Relative frequency expected due to reduced mass: 1, 0.9920,
0.9898; actual relative frequency: 1, 0.8863, 0.7726.
What do you suppose the decrease in force constant is due to?
à Analytic wavefunctions
It turns out that the harmonic oscillator Schrödinger equation can be solved analytically. The wave
functions have the general form

yjHrL =
1
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅ
ÅÅÅ
"################
#################
#
2j-1 H j - 1L!
è!!!!
p
Hj-1HrL ‰-r2ê2
in terms of Hermite polynomials, Hj-1HrL. Here are the first few Hermite polynomials.
loops Hermite polynomial
1 1
2 2 ρ
3 −2 + 4 ρ2
4 −12 ρ + 8 ρ3
5 12 − 48 ρ2 + 16 ρ4
6 120 ρ − 160 ρ3 + 32 ρ5
7 −120 + 720 ρ2 − 480 ρ4 + 64 ρ6
First several Hermite polynomials Hj-1HrL. When the polynomials are multiplied by the factor ‰-r2ë2
the resulting function has the
number of loops given in the first column. Polynomial corresponding to even numbers of loops are even about r = 0; polynomials
corresponding to odd numbers of loops are odd about r = 0.
Mathematica knows about Hermite polynomials, and so it is easy to construct a function for
harmonic oscillator wavefunctions. Here is the Mathematica function for the wavefunction with j
loops.
ψ@j_, ρ_D :=
1
"################
################
######
2j−1 Hj − 1L!
è!!!!
π
HermiteH@j − 1, ρD −ρ2ê2
These wavefunctions are normalized to 1; for example,
‡
−∞
∞
ψ@6, ρD2
ρ
1
They are also orthogonal, as must be so since they are eigenfunctions of the harmonic oscillator
Hamiltonian operaotr which is hermitian; for example
‡
−∞
∞
ψ@6, ρD ψ@3, ρD ρ
0
Here is a plot of y6HrL.
-6 -4 -2 2 4 6
r
-0.4
-0.2
0.2
0.4
y6
Analytic harmonic oscillator wavefunction y6HrL. The function is normalized to 1. Displacement r from equilibrium is in units
è!!!!!!!!
!!!!!!!
Ñêm w

This is the same as the function that we obtained earlier using Numerov integration, to within the
accuracy of the numerical implementation.
à Quantal harmonic motion
To treat harmonic motion quantum mechanically, we need to construct wavepackets. A general
expression for a wavepacket is
YHrL = N ‚
j
gj yjHrL,
in terms of relative weights gj and the normalization constant
N = 1ì$%%%%%%%%%%%%
⁄
j
gj
2 .
For example, a packet composed of waves with 1, 2 and 3 loops, with relative weights 25%, 50% and
25% is
YHrL =
1
ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
"################################
########
0.252
+ 0.52
+ 0.252
80.25 y1HrL + 0.5 y2HrL + 0.25 y3HrL
Use of the orthonormality of the component waves, that is, that Ÿ yjHrL ykHrL „ r = dj k, to
confirm that this wavepacket is normalized, that is, that Ÿ » YHrL »2 „ r = 1.
Let's use Mathematica to construct and plot harmonic oscillator wavepacket probability densities.
First, we can define the list of weights gj. For the example above, this is
g = 80.25, 0.50, 0.25<;
Next, we can construct a list of wavefunctions. For the example above, this is
f = ψ@#, ρD & ê@ 81, 2, 3<
9
− ρ2
2
π1ê4 ,
è!!!!
2 − ρ2
2 ρ
π1ê4 ,
− ρ2
2 H−2 + 4 ρ2L
2
è!!!!
2 π1ê4
=
The normalization factor,
1
ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
"################################
########
0.252
+ 0.52
+ 0.252
,
evaluates to
norm =
1
è!!!!!!!!!
g.g
1.63299
The sum of the products of the functions times their weights,
g1 y1HrL + g2 y2HrL + g3 y3HrL,
is

g.f
0.187781 − ρ2
2 + 0.531126 − ρ2
2 ρ + 0.0663907 − ρ2
2 H−2 + 4 ρ2
L
Putting everything together, the normalized wavepacket is
packet = norm g.f
1.63299 J0.187781 − ρ2
2 + 0.531126 − ρ2
2 ρ + 0.0663907 − ρ2
2 H−2 + 4 ρ2
LN
We can check that this packet is normalized, as
‡
−∞
∞
packet2
ρ
1.
Here is what the probability density (the square of the wavepacket) looks like.
-4 -2 2 4
r
0.1
0.2
0.3
0.4
0.5
0.6
0.7
»Y»2
Probability density » YHrL »2
of a three component harmonic oscillator wavepacket YHrL. The probability density is normalized to 1.
Displacement r from equilibrium is in units
è!!!!!!!!
!!!!!!!
Ñêm w
The packet is localized near the right classical turning point. If we were to add more waves to the
packet, the width of the localized region would become smaller, in accordance with the Heisenberg
indeterminacy relation.
The next thing we need to do is to make the wavepacket move. To do this we need only to insert the
time dependent phase factors,
‰-Â Ej têÑ
= ‰-Â Ñ w
ÅÅÅÅÅÅÅÅÅ
2 H2 j-1L têÑ
= ‰-Â wH j- 1
ÅÅÅÅ
2 L t
,
to the components of the packet. For our example packet, the list of these phase factors is
phase = − ω I#−
1
2
M t
& ê@ 81, 2, 3<
9 − 1
2 t ω
, − 3
2 t ω
, − 5
2 t ω
=
The list of functions times their phase factors is
f phase
9
− ρ2
2 − t ω
2
π1ê4 ,
è!!!!
2 − ρ2
2 − 3 t ω
2 ρ
π1ê4 ,
− ρ2
2 − 5 t ω
2 H−2 + 4 ρ2L
2
è!!!!
2 π1ê4
=

The sum of the products of the functions times their phase factors and their weights, and multiplied
by the normalization factor,
1
ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
"##############
⁄j gj
2
Ig1 y1HrL ‰-Â wH1- 1
ÅÅÅÅ
2 L t
+ g2 y2HrL ‰-Â wH2- 1
ÅÅÅÅ
2 L t
+ g3 y3HrL ‰-Â wH3- 1
ÅÅÅÅ
2 L t
M,
is
packetWithTime = norm g.Hf phaseL
1.63299 J0.187781 − ρ2
2 − t ω
2 + 0.531126 − ρ2
2 − 3 t ω
2 ρ + 0.0663907 − ρ2
2 − 5 t ω
2 H−2 + 4 ρ2
LN
Show that this wavepacket is normalized to 1, for every value of time, t.
At t = 0 the time-dependent wavepacket, YHr, 0L, is identical to the time independent wavepacket,
YHrL, and so gives the same probability density.
Verify that this statement is correct.
Here is what the probability density (the square of the wavepacket) looks like at time t = 0.25ê n, one
quarter of the way through one oscillation period.
-4 -2 2 4
r
0.05
0.1
0.15
0.2
0.25
»Y»2
Probability density » YHr, tL »2
of a three component harmonic oscillator wavepacket YHr, tL at time t = 0.25ên. The probability
density is normalized to 1. Displacement r from equilibrium is in units
è!!!!!!!!
!!!!!!!
Ñêm w
Here is what the probability density looks like at time t = 0.5ên, half way through one oscillation
period.

-4 -2 2 4
r
0.1
0.2
0.3
0.4
0.5
0.6
0.7
»Y»2
of a three component harmonic oscillator wavepacket YHr, tL at time t = 0.5ê n. The probability density
is normalized to 1. Displacement r from equilibrium is in units
è!!!!!!!!
!!!!!!!
Ñêm w
Finally, here is what the probability density looks like at time t = 2 pê w = 1ê n, after one oscillation
period.
-4 -2 2 4
r
0.1
0.2
0.3
0.4
0.5
0.6
0.7
»Y»2
of a three component harmonic oscillator wavepacket YHr.tL at time t = 1ê n. The probability density is
normalized to 1. Displacement r from equilibrium is in units
è!!!!!!!!
!!!!!!!
Ñêm w
The packet has returned to its original form and location at t = 1ê n.
à Energy of wavepackets: expectation values
The energy of a wave packet is defined to be
XH = ‡
-¶
¶
Y*
Hr, tL H YHr, tL „ r.
The notation X… denotes "average" or "expectation" and such an expression is known as the average
value of the expectation value of the physical quantity corresponding to the operator that appears
between the brackets. For example, the expectation value of position would be
Xr = ‡
-¶
¶
Y*
Hr, tL r YHr, tL „ r.
Since the operator for position is just "multiply by position," we can rearrange this expression as

Xr = ‡
-¶
¶
r À YHr, tL À2
„ r.
In this form the average value of the position is seen to be just the average of all possible positions,
weighted by the probability that the particle is at each position. It is this interpretation that led of the
name average value or expectation value. If the operator is more complicated than "multiply by",
then we cannot rearrange things in this way, but we still interpret the expression in the same way.
Show that the expression for the expectation value of the squared momentum, in
dimensionless units, is Xp2 = -Ÿ Y*Hr, tL ∑2 ê ∑ r2 YHr, tL „ r.
Taking account of the orthonormality of the harmonic oscillator wavefunctions,
‡
-¶
¶
yjHrL ykHrL „ r = dj k,
it is not too difficult to show that
XH =
1
ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
⁄j gj
2
‚
k
gk
2
Ek,
an average of the component energies, each weighted by its relative contribution to the probability
density.
Show that this expression is correct for any wavepacket composed of orthonormal
components ykHrL.
Why do you suppose this expression is independent of time?
Here is a Mathematica function that takes as input a list of component wavefunctions (in terms of the
number of loops of each component) and a list of the corresponding weights, and computes the
dimensionless energy expectation value of the packet.
εAvg@j_List, g_ListD :=
1
g.g
g2
.HH2 # − 1L & ê@ jL
For example, the average energy of a wavepacket consisting of an equal mixture of the one-loop and
two-loop wavefunctions is
εAvg@81, 2<, 81, 1<D
2
This is what we expect, since y1 has energy e1 = 1 and y2 has energy e2 = 3. A packet composed of
10% y1 and 90% y2 has instead the average energy
2.97561
This is very close to e2, as we would expect, since the packet is most y2.
à Wavepacket machine
Collecting everything together, a general time-dependent harmonic oscillator wavepacket is

YHr, tL =
1
ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅ
"##############
⁄k gk
2
‚
j
gj yjHrL ‰-Â wH j- 1
ÅÅÅÅ
2 L t
.
Show that this expression is normalized to 1 for all values of t.
Show that » YHr, tL »2 oscillates with period 1ê n.
Here is a Mathematica function that makes a wavepacket, for a specified choice of component
wavefunctions, yj, and weights, gj.
HOPacket@j_List, g_ListD :=
1
è!!!!!!!!!
g.g
g.Jψ@#, ρD − I#−
1
2
M 2 π t
& ê@ jN
In this function, time is measured in dimensionless units of the oscillation period, 2 pê w = 1ê n; this
means that t = 1 corresponds to elapsed time 1 ê n = 2 p
è!!!!!!!!!!
m ê k . As example of this function, here is
the two-component wavepacket composed of equal contributions of one-loop and two-loop
components.
−3 π t− ρ2
2 H 2 π t +
è!!!!
2 ρL
è!!!!
2 π1ê4
Verify that this expression is correct, by constructing the wavepacket by hand.
Now that we have a tool to construct harmonic oscillator wavepackets, let's explore the properties of
different packets.
à Gaussian wavepacket
We can construct wavepackets of essentially arbitrary shape by appropriate choice of weights gj.
Experimentally, this amounts to appropriate excitation of the oscillator into a coherent superposition
of wavefunctions yjHrL. (Coherent means there is a definite phase relation between the components
of the packet.) One common superposition results in a Gaussian distribution of weights. The
Gaussian distribution centered at m and with mean squared deviation )variance) s is
gauss@σ_, x_, μ_D :=
1
è!!!!!!!!
2 π σ
−Hx−μL2êH2 σ2L
The Gaussian distribution is normalized to 1; for example
‡
−∞
∞
gauss@1, x, 0D x
1
Here is the Gaussian distribution centered about 0 with mean squared deviation 1 (the so-called
normal density function).

-4 -2 2 4
x
0.1
0.2
0.3
0.4
probability
Gaussian probability distribution with mean 0 and variance 1. The filled circles marked values of the distribution at 7 equally
spaced values of x centered on the mean.
We can select weights gj that approximate this Gaussian distribution by evaluating the distribution at
a range of points centered about the mean. For example, here is a set of thirteen such weights chosen
to span the distribution.
g = gauss@1, #, 0D & ê@ Range@−3, 3, 1D êê N êê sf2
80.0044, 0.054, 0.24, 0.4, 0.24, 0.054, 0.0044<
These weights are indicated as the filled circled on the plot of the Gaussian distribution above.
Here is a plot of a Gaussian wavepacket consisting of the first 13 harmonic oscillator wavefunctions
yjHrL, throughout one cycle of oscillation.
-10
-5
0
5
10
r
0
0.2
0.4
0.6
0.8
1
t
0
0.2
0.4
0.6
0.8
»YHr,tL»2
-10
-5
0
5
10
r
Harmonic oscillator Gaussian wavepacket probability density throughout one cycle of oscillation. The packet is composed of the
first 13 wavefunctions yj HrL.

Show that the dimensionless energy expectation value of this wavepacket is XH=13.
Could you have predicted this result without doing a detailed calculation?
Here is a plot of a Gaussian wavepacket consisting of 13 harmonic oscillator wavefunctions yjHrL
centered about j = 27, throughout one cycle of oscillation.
-10
-5
0
5
10
r
0
0.2
0.4
0.6
0.8
1
t
0
0.25
0.5
0.75
1
»YHr,tL»2
-10
-5
0
5
10
r
Harmonic oscillator Gaussian wavepacket probability density throughout one cycle of oscillation. The packet is composed of 13
wavefunctions yj HrL centered at j = 27.
Here is a plot of a Gaussian wavepacket consisting of 13 harmonic oscillator wavefunctions yjHrL
centered about j = 47, throughout one cycle of oscillation.
-10
-5
0
5
10
r
0
0.2
0.4
0.6
0.8
1
t
0
0.2
0.4
0.6
0.8
»YHr,tL»2
-10
-5
0
5
10
r
Harmonic oscillator Gaussian wavepacket probability density throughout one cycle of oscillation. The packet is composed of 13
wavefunctions yj HrL centered at j = 47.

These three Gaussian harmonic oscillator wavepackets all have the same number of
adjacent components, but the center component is successively higher. Use the surface
plots of the probability densities to see what difference this corresponds to physically.
à Localization of harmonic motion
We have seen that adding more components to a wavepacket localizes the probability density to a
smaller region of space. We can illustrate this by constructing Gaussian packets centered at the same
yj, but with differing numbers of adjacent components. Here are the packets consisting of 5, 13 and
21 components, for times corresponding to the first quarter of the period, 1ê n.
Harmonic oscillator Gaussian wavepacket probability densities throughout one quarter cycle of oscillation. The packets are each
centered about j = 30; the left column is the 5-component packet, the middle column is the 13-component packet, and the right
column is the 21-component packet.
The 5-component packet is poorly localized, the 13-component packet is fairly localized, and the
21-component packet is highly localized.

HarmonicOscillator (1).pdf

More Related Content

Similar to HarmonicOscillator (1).pdf (20)

Recently uploaded (20)

HarmonicOscillator (1).pdf