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Chapter 8 The Simple Harmonic Oscillator
A winter rose. How can a rose bloom in December? Amazing but true, there it is, a yellow
winter rose. The rain and the cold have worn at the petals but the beauty is eternal regardless
of season. Bright, like a moon beam on a clear night in June. Inviting, like a fire in the hearth
of an otherwise dark room. Warm, like a.. .wait! Wait just a MINUTE! What is this...Emily
Dickinson? Mickey Spillane would NEVER... Misery Street.. .that’s more like it.. .a beautiful
secretary named Rose.. .back at it now.. .the mark turned yellow.. .yeah, yeah, all right.. .the
elegance of the transcendance of Euler’s number on a Parisian morning in 1873.. .what?...
The infinite square well is useful to illustrate many concepts including energy quantization
but the infinite square well is an unrealistic potential. The simple harmonic oscillator (SHO),
in contrast, is a realistic and commonly encountered potential. It is one of the most important
problems in quantum mechanics and physics in general. It is often used as a first approximation to
more complex phenomena or as a limiting case. It is dominantly popular in modeling a multitude of
cooperative phenomena. The electrical bonds between the atoms or molecules in a crystal lattice
are often modeled as “little springs,” so group phenomena is modeled by a system of coupled
SHO’s. If your studies include solid state physics you will encounter phonons, and the description of
multiple coupled phonons relies on multiple simple harmonic oscillators. The quantum mechanical
description of electromagnetic fields in free space uses multiple coupled photons modeled by simple
harmonic oscillators. The rudiments are the same as classical mechanics.. .small oscillations in a
smooth potential are modeled well by the SHO.
If a particle is confined in any potential, it demonstrates the same qualitative behavior as
a particle confined to a square well. Energy is quantized. The energy levels of the SHO will be
different than an infinite square well because the “geometry” of the potential is different. You
should look for other similarities in these two systems. For instance, compare the shapes of the
eigenfunctions between the infinite square well and the SHO.
Part 1 outlines the basic concepts and focuses on the arguments of linear algebra using raising
and lowering operators and matrix operators. This approach is more modern and elegant
than brute force solutions of differential equations in position space, and uses and reinforces Dirac
notation, which depends upon the arguments of linear algebra. The raising and lowering operators,
or ladder operators, are the predecessors of the creation and annihilation operators used in the
quantum mechanical description of interacting photons. The arguments of linear algebra provide
a variety of raising and lowering equations that yield the eigenvalues of the SHO,
En =
n +
¯hω,
and their eigenfunctions. The eigenfunctions of the SHO can be described using Hermite poly-
nomials (pronounced “her meet”), which is a complete and orthogonal set of functions.
Part 2 will explain why the Hermite polynomials are applicable and reinforce the results of
part 1. Part 2 emphasizes the method of power series solutions of a differential equation.
Chapter 5 introduced the separation of variables, which is usually the first method applied in an
attempt to solve a partial differential equation. Power series solutions apply to ordinary differential
equations. In the case the partial differential equation is separable, it may be appropriate to solve
one or more of the resulting ordinary differential equations using a power series method. We will
encounter this circumstance when we address the hydrogen atom. You should leave this chapter
understanding how an ordinary differential equation is solved using a power series solution.
We do not reach the coupled harmonic oscillator in this text. Of course, the SHO is an
important building block in reaching the coupled harmonic oscillator. There are numerous physical
systems described by a single harmonic oscillator. The SHO approximates any individual bond,
such as the bond encountered in a diatomic molecule like O 2 or N 2. The SHO applies to any
system that demonstrates small amplitude vibration.
The Simple Harmonic Oscillator, Part 1
Business suit, briefcase, she’s been in four stores and hasn’t bought a thing.. .so this mall
has got to be the meet! Now a video store. She’s as interested in videos as a cow is in eating
meat. But, right in the middle of the drama section, suddenly face to face... “Sir, do you have a
cigarette?” and walks off more briskly than Lipton ice tea. Blown. Gone. Done. Just to tell me
she knows me.. .no meet for me. I’ve got to hang up my hat, but only my hat... She doesn’t know
Charlie’s face, and maybe the meet will happen in Part 2...
- Justify the use of a simple harmonic oscillator potential, V (x) = kx
2 /2 , for a particle
confined to any smooth potential well. Write the time–independent Schrodinger equation for a
system described as a simple harmonic oscillator.
The sketches may be most illustrative. You have already written the time–independent Schrodinger
equation for a SHO in chapter 2.
The functional form of a simple harmonic oscillator from classical mechanics is V (x) =
kx
2 .
Its graph is a parabola as seen in the figure on the left. Any relative minimum in a smooth potential
energy curve can be approximated by a simple harmonic oscillator if the energy is small compared
to the height of the well meaning that oscillations have small amplitudes.
Figure 8 − 1. Simple Harmonic Oscillator. Figure 8 − 2. Relative Potential Energy Minima.
Expanding an arbitrary potential energy function in a Taylor series, where x 0 is the minimum,
V (x) = V (x 0 ) +
dV
dx
x 0
(x − x 0 ) +
d
2 V
dx^2
x 0
(x − x 0 )
2
d
3 V
dx^3
x 0
(x − x 0 )
3
defining V (x 0 ) = 0 ,
dV
dx
x 0
= 0 because the slope is zero at the bottom of a minimum, and if
E ø the height of the potential well, then x ≈ x 0 so terms where the difference (x − x 0 ) has a
[
2 m
P
2
mω
2
X
2
]
| ψ> = En | ψ> ⇐⇒ ¯hω
a
a +
| ψ> = En | ψ>.
Postscript: The Schrodinger equation is
[
P
2
2
]
| ψ> = En | ψ> , when constant factors are
excluded. The sum P^2 + X 2 = X 2 + P^2 would appear to factor as
X + iP
X − iP
, so that
[
P
2 +X
2
]
| ψ> = En | ψ> ⇒
[
X
2 +P
2
]
| ψ> = En | ψ> ⇒
X +iP
X −iP
| ψ> = En | ψ>.
This is only a qualified type of factoring because the order of the “factors” cannot be changed; X
and P are fundamentally canonical and simply do not commute. Nevertheless, the parallel with
common factoring into complex conjugate quantities is part of the motivation for the raising and
lowering operators. In fact, some authors refer to this approach as the method of factorization.
Notice that a
a =
¯hω
H −
Notice also that though X and P are Hermitian, a and a
are not.
- Show that the commutator
[
a, a
]
Problems 3 and 4 are developing tools to approach the eigenvector/eigenvalue problem of the SHO.
We want
[
a, a
]
= a a
− a
a in terms the definitions of problem 2. Letting
C =
mω
2¯h
, and D =
2 mω¯h
to simplify notation,
[
a, a
]
CX + iDP
CX − iDP
CX − iDP
CX + iDP
= C
2 X
2
− iCDX P + iDCP X + D
2 P
2
− C
2 X
2
− iCDX P + iDCP X − D
2 P
2
= 2iCD
P X − X P
= 2i
mω
2¯h
2 mω¯h
[
P, X
]
2 i
2¯h
− i¯h
= 1, since
[
P, X
]
[
X , P
]
= −i¯h.
- Show that H a
= a
H + a
¯hω.
This is a tool used to solve the eigenvector/eigenvalue problem for the SHO though it should build
some familiarity with the raising and lowering operators and commutator algebra.
H = ¯hω
a
a +
H
¯hω
= a
a +
[
a
H
¯hω
]
[
a
, a
a +
]
= a
a
a + a
− a
aa
a
= a
a
a − a
a
= −a
[
a, a
]
= −a
[
a
, H
]
= −a
¯hω ⇒ a
H − H a
, = −a
¯hω ⇒ H a
= a
H + a
¯hω.
Postscript: We will also use the fact that H a = a H − a¯hω , though its proof is posed to the
student as a problem.
- Find the effect of the raising and lowering operators using the results of problem 4.
We have written time–independent Schrodinger equation as H | ψ > = En | ψ > to this point.
Since the Hamiltonian is the energy operator, the eigenvalues are necessarily energy eigenvalues.
The state vector is assumed to be a linear combination of all energy eigenvectors. If we specifically
measure the eigenvalue En , then the state vector is necessarily the associated eigenvector which can
be written | En >. The time–independent Schrodinger equation written as H | En > = En | En >
is likely a better expression for the development that follows.
If H | En > = En | En > where En is an energy eigenvalue, then | En > = | En >
⇒ H a
| En > =
a
H + a
¯hω
| En > = a
H | En > + a
¯hω | En >
= a
En | En > + a
¯hω | En > =
En + ¯hω
a
| En > or
H
a
| En >
En + ¯hω
a
| En >
This means that a
| En > is an eigenvector of H with an eigenvalue of En + ¯hω. This is exactly
¯hω more than the eigenvalue of the eigenvector | En >. The effect of a
acting on | En > is to
“raise” the eigenvalue by ¯hω , thus a
is known as the raising operator.
Again, given that H | En > = En | En > and starting with | En > = | En >
⇒ H a | En > =
aH − a¯hω
| En > = aH | En > − a¯hω | En >
= aEn | En > − a¯hω | En > =
En − ¯hω
a | En > or
H
a | En >
En − ¯hω
a | En >
Here a | En > is an eigenvector of H with an eigenvalue of En − ¯hω. This is ¯hω less than
the eigenvalue of the eigenvector | En >. The effect of a acting on | En > is to “lower” the
eigenvalue by ¯hω , thus a is known as the lowering operator.
- What is the effect of the lowering operator on the ground state, Eg?
so a†
a† | En >
= a†a† | En > is an eigenvector of H with the eigenvalue En + 2¯hω. Suc-
cessively applying the raising operator yields successive eigenvalues. The eigenvalue of the ground
state is fixed at ¯hω/2 so all the eigenvalues can be attained in terms of the ground–state eigenvalue.
H | 0 > = E 0 | 0 > =
¯hω
| 0 > ⇒ H a
¯hω
⇒ H a
a
¯hω
⇒ H a
a
a
¯hω
| 0 > and in general
⇒ H
a
)n | 0 > =
¯hω
from which we ascertain
En =
n +
¯hω are the eigenenergies of the SHO.
Postscript: This argument does not specify the eigenvectors a
| 0 > , a
a
a
)n | 0 >.
- Find energy space representations for the eigenvectors of the SHO.
Explicit eignvalues given En =
n +
¯hω are E 0 =
¯hω , E 1 =
¯hω , E 2 =
¯hω ,
E 3 =
¯hω ,... , En =
n +
¯hω. The eigenvector/eigenvalue equations must remain
H | 0 > = E 0 | 0 > , H | 1 > = E 1 | 1 > , H | 2 > = E 2 | 2 > , H | 3 > = E 3 | 3 > ,... , H | n> = En | n>.
Combining these eigenvalue/eigenvector relations with those attained earlier using the raising op-
erator provides the ability to explicitly respresent the eigenvectors.
H | 1 > =
¯hω | 1 > = H a
| 0 > ⇒ | 1 > ∝ a
H | 2 > =
¯hω | 2 > = H a
a
| 0 > ⇒ | 2 > ∝ a
a
| 0 > ∝ a
H | 3 > =
¯hω | 3 > = H a
a
a
| 0 > ⇒ | 3 > ∝ a
a
a
| 0 > ∝ a
| 2 > and
H | n > =
n +
¯hω | n > = H
a
)n | 0 > ⇒ | n > ∝
a
)n | 0 > ∝ a
| n − 1 >
⇒ C (n) | n > = a
| n − 1 > in general, where C (n) is a proportionality constant.
Postscript: The relation of proportionality is appropriate for this argument because of the nature
of the eigenvalue/eigenvector equation. Any vector that is proportional to the eigenvector will work
in the eigenvalue/eigenvector equation. Consider
whose eigenvalues are 2 and 1 , corresponding to the eigenvectors
and
, but any vector
proportional to
also yields a true statement in the eigenvalue/eigenvector equation, e.g.,
In fact, the proportionality condition is equivalent to the normalization condition in this case.
The n in the equation C (n) | n > = a
| n − 1 > is one eigenstate higher than n − 1. The
raising operator acting on an eigenstate increases the state to the next higher eigenstate.
- Normalize C (n) | n > = a
| n − 1 >.
This problem is a good example of (a) forming adjoints and applying the normalization condi-
tion, (b) using commutator algebra, and (c) using eigenvalue/eigenvector equations. Remember
orthonormality requires that < n | n> = < n − 1 | n − 1 > = 1 , and that
[
a , a
]
(a) C (n) | n > = a
| n− 1 > ⇒ < n | C
∗ (n) = < n− 1 |
a
= < n− 1 | a is the adjoint equation
⇒ < n | C
∗ (n) C (n) | n> = < n − 1 | a a
| n − 1 > are the innner products.
We have an expression for a
a , but need to develop an expression for a a
(b) H = ¯hω
a
a +
⇒ a
a =
H
¯hω
⇒ a
a−a a
H
¯hω
−a a
subtracting a a
from both sides.
[
a , a
]
[
a
, a
]
= −1 , and a
a − a a
[
a
, a
]
[
a
, a
]
H
¯hω
−a a
H
¯hω
−a a
⇒ a a
H
¯hω
. Returning to part (a),
(c)
∣ (^) C (n)
2 < n | n> = < n − 1
H
¯hω
∣ (^) n − 1 >
∣ (^) C (n)
2 = < n − 1
H
¯hω
∣ (^) n − 1 > + < n − 1
∣ (^) n − 1 >
∣ (^) C (n)
2 = < n − 1
¯hω
n − 1 +
¯hω
∣ (^) n − 1 > +
< n − 1 | n − 1 >
Notice that all the eigenkets are of infinite dimension and that they are orthonormal. This problem
is an application of the mathematics of part 2 of chapter 1 applied to a realistic system.
The Hamiltonian is Hermitian and has unit vectors as basis vectors. The Hamiltonian must,
therefore, be diagonal with the eigenvalues on the main diagonal, i.e.,
H = ¯hω
- (a) Develop the matrix representation of the lowering operator for the SHO.
(b) Lower the third excited state of the SHO using explicit matrix multiplication, and
(c) demonstrate equivalence to a | n> =
n | n − 1 >.
An individual element of any matrix can be calculated using Dirac notation by sandwiching the
operator between the bra representing the row of interest and the ket representing the column of
interest. In general for the lowering operator,
< n | a | m> = < n |
m | m − 1 >
m < n | m − 1 >
m δn, m− 1 ,
where the lowering operator acted to the right in the first line. The Kronecker delta reflects
orthonormality. It says that the element in row n and column m − 1 is zero unless n = m − 1.
(a) Trying a few values on the main diagonal,
< 0 | a | 0 > =
0 δ 0 ,− 1 =
< 1 | a | 1 > =
1 δ 1 , 0 =
1 (0) = 0 , and
< 2 | a | 2 > =
2 δ 2 , 1 =
In fact, all elements on the main diagonal are zero. The Kronecker delta indicates that the column
must be one greater than the row to be non–zero, so
< 0 | a | 1 > =
1 δ 0 , 0 =
< 1 | a | 2 > =
2 δ 1 , 1 =
< 2 | a | 3 > =
3 δ 2 , 2 =
and the pattern continues to yield
a =
(b) a | 3 > =
(c) which is the same as a | 3 > =
3 | 2 >. (Of course, these must be the same.
The relation used for part (c) is also the relation upon which the matrix representation is built).
Postscript: The upper left element of matrix operators used to describe the SHO is row zero,
column zero. This is because the zero is an allowed quantum number for the SHO and | 0 > is
the ground state. The upper left element in most other matrices is row one, column one.
The matrix representation of the raising operator is similarly developed and is
a
- Find the matrix representation of X for the SHO.
The operators X , P , and H , correspond to position, momentum, and energy, which are dynam-
ical variables in classical mechanics, but operators in quantum mechanics. Remember
a =
mω
2¯h
X + i
2 mω¯h
P ,
a
mω
2¯h
X − i
2 mω¯h
P.
are the definition of the “ladder” operators in terms of the position and momentum operators.
Since we have matrix representations of a and a
, the matrix representation of X , P , and H ,
are a matter of chapter 1 matrix addition and multiplicative constants.
Adding the equations for a and a
a + a
mω
2¯h
X + i
2 mω¯h
P
mω
2¯h
X − i
2 mω¯h
P
mω
2¯h
X
⇒ X =
¯h
2 mω
a + a
⇒ a =
mω
2¯h
¯h
mω
y +
¯h
2 mω
mω
¯h
d
dy
⇒ a =
y +
d
dy
The eigenkets | n > in abstract Hilbert space and ψn (y) in position space are equivalent expres-
sions, and we used the fact that a | 0 > = 0 to attain eigenenergies earlier, so
| n > = ψn (y) ⇒ a | n > = a ψn (y) ⇒ a | 0 > = a ψ 0 (y) ⇒ a ψ 0 (y) = 0 , therefore
y +
d
dy
ψ 0 (y) = 0
d ψ 0 (y)
ψ 0 (y)
= −y dy
⇒ ln ψ 0 (y) = −
y
2
⇒ ψ 0 (y) = A 0 e
−y^2 / 2
where the variable of integration is absorbed into the constant A 0. Returning to the variable x ,
ψ 0 (x) = A 0 e
−mωx^2 / 2¯h
is the unnormalized ground state eigenfunction of the SHO in position space.
Postscript: Notice that the ground state eigenfunction of the SHO in position space is a Gaussian
function. The normalized ground state eigenfunction is
ψ 0 (x) =
mω
π¯h
e
−mωx
2 / 2¯h .
- (a) Find a generating function for the eigenstates of the SHO in position space in general.
(b) Find the eigenfunction for the first excited state of the SHO in position space.
Employ the result of problem 11, | n> =
n!
a
)n | 0 > , using the position space representation
of the raising operator, a
mω
2¯h
x −
¯h
2 mω
d
dx
. This is cleaner using y as defined
in problem 15. Use the result and eliminate y to express ψ 1 in terms of x for part (b).
(a) Using y as defined in problem 15, a
y −
d
dy
, so the result of problem 11 is
ψn (y) =
n!
y −
d
dy
))n
ψ 0 (y) ⇒ ψn (y) =
n!
y −
d
dy
))n( mω
π¯h
e
−y
2 / 2 .
(b) The first excited state of the SHO means n = 1 , so
ψ 1 (y ) =
mω
π¯h
y −
d
dy
e
−y^2 / 2
mω
π¯h
y e
−y^2 / 2 −
d
dy
e
−y^2 / 2
mω
π¯h
y e
−y^2 / 2 − (−y) e
−y^2 / 2
mω
π¯h
2 y e
−y^2 / 2
⇒ ψ 1 (x) =
mω
π¯h
mω
¯h
x e
−mωx^2 / 2¯h
π
mω
¯h
x e
−mωx^2 / 2¯h .
- Find the eigenfunction for the ground state and first excited state of the SHO in position space
using Hermite polynomials.
Eigenstates of the SHO can be expressed using Hermite polynomials. The n
th eigenstate is
ψn (x) =
mω
π¯h
2 n^ n!
Hn (ξ ) e
−ξ^2 / 2 where ξ =
mω
¯h
x (1)
and the Hn are Hermite polynomials. The first few Hermite polynomials are
H 0 (ξ) = 1
H 1 (ξ) = 2ξ
H 2 (ξ) = 4ξ
2 − 2
H 3 (ξ) = 8ξ
3 − 12 ξ
H 4 (ξ) = 16ξ
4 − 48 ξ
2
H 5 (ξ) = 32ξ
5 − 160 ξ
3
H 6 (ξ) = 64ξ
6 − 480 ξ
4
2 − 120
H 7 (ξ) = 128ξ
7 − 1344 ξ
5
3 − 1680 ξ
H 8 (ξ) = 256ξ
8 − 3584 ξ
6
4 − 13440 ξ
2
H 9 (ξ) = 512ξ
9 − 9216 ξ
7
5 − 80640 ξ
3
Table 6 − 1. The First Ten Hermite Polynomials.
Hermite polynomials can be generated using the recurrence relation
Hn+1 (ξ) = 2x Hn (ξ) − 2 n Hn− 1 (ξ).
The Schrodinger equation in position space for the SHO is a naturally occurring form of Hermite’s
equation. The solutions to Hermite’s equation are the Hermite polynomials. We will solve this
differential equation thereby deriving the Hermite polynomials using a power series solution in
part 2 of this chapter. Using equation (1) with the appropriate Hermite polynomial is likely the
easiest way to attain a position space eigenfunction for the quantum mechanical SHO.
ψ 0 (x) =
mω
π¯h
H 0
mω
¯h
x
e
−mωx^2 / 2¯h
mω
π¯h
e
−mωx^2 / 2¯h
mω
π¯h
e
−mωx
2 /2¯h , in agreement with our earlier calculation.
ψ 1 (x) =
mω
π¯h
H 1
(√^
mω
¯h
x
e
−mωx
2 / 2¯h =
mω
π¯h
(√^
mω
¯h
x
e
−mωx
2 / 2¯h
from a linear combination of Hermite polynomials. The Hermite polynomials, therefore, form an
orthogonal basis, and further, form an orthonormal basis when they are normalized.
The infinite set of unit vectors is orthonormal and is complete. The infinite set of sines
and cosines used for the infinite square well is orthogonal so can be made orthonormal and is
complete. The infinite set of Hermite polynomials used for the SHO is orthogonal so can be made
orthonormal and is complete. The infinite sets of Associated Laguerre polynomials, Legendre
functions, spherical harmonic functions, and numerous other sets of polynomials and functions are
orthogonal so can be made orthonormal and complete. Each of these infinite sets form a basis in
the same sense as the unit vectors of chapter 1 form a basis. These infinite sets of polynomials
and functions generally require a weighting function to demonstrate orthogonality which, again, is
a fact that is not always stated explicitly.
- Given a simple harmonic oscillator potential, graph the first six eigenenergies on an energy
versus position plot and superimpose the the first six eigenfunctions on corresponding eigenenergies
on the same plot. Plot the probability densities of the first six eigenfunctions in the same manner.
Examine the two graphs below.
Eigenenergies and Eigenfunctions Eigenenergies and Probability Density
Postscript: Like the infinite square well, it is conventional to graph energy versus position for the
eigenenergies and the eigenfunctions are conventionally located at the level of the corresponding
eigenenergies where each horizontal line represents zero amplitude for that eigenfunction.
Unlike the infinite square well, the eigenfunctions do not have an amplitude of zero at the
boundaries. The eigenfunctions approach zero asymptotically outside the potential well. Further,
eigenenergies for the SHO are evenly spaced, the ground state is at ¯hω/2 , and each successive
eigenenergy is ¯hω higher than the last. Also in contrast, the eigenenergies of the infinite square
well scale as n
2 , that is E 1 = Eg , E 2 = 4 Eg , E 3 = 9 Eg , E 4 = 16 Eg , etc.
Like the infinite square well, the probability densities are non-negative everywhere, there are
points inside the well where the probability density is zero, and there are regions of maximal and
minimal probability. Unlike the infinite square well and an additional non–classical feature of the
SHO is that there are regions of non-zero probability density outside the potential well—there is
a finite probability of finding the particle outside of the potential well. That there is a non-zero
probability density outside the wall is a feature of all but infinite, vertical potential walls.
- Sketch the linear combination Ψ (x) = 2 ψ 0 (x) + ψ 1 (x) for the SHO.
A general wavefunction of the SHO is a superposition or linear combination of its eigenfunctions,
Ψ (x) = c 0 ψ 0 (x) + c 1 ψ 1 (x) + c 2 ψ 2 (x) + c 3 ψ 3 (x) + · · · =
∑^ ∞
n = 0
cn ψn (x) , in general, or
| Ψ > = c 0 | 0 > + c 1 | 1 > + c 2 | 2 > + c 3 | 3 > + · · · =
∑^ ∞
n = 0
cn | n> for the SHO.
The linear combination given is combined graphically below.
A General State Vector that is the Superposition of two Eigenstates.
Postscript: The cn are constants that provide the relative contributions of each eigenfunction.
The cn can be any scalars so Ψ (x) can have any shape. As before, if the general wavefunc-
tion is normalized, Ψ (x) = 1 , the relative magnitudes of the cn are fixed. Also as before, the
orthogonality of the eigenfunctions of the SHO ensures that the cn are unique.
Problems 21 through 26 use the linear combination of two eigenstates,
| Ψ > = A
[
]
which is the general linear combination of eigenstates for c 0 = 2 , c 2 = 5 , and all other cn = 0.
- (a) Normalize the wavefunction | Ψ > using row and column vectors, and
(b) using Dirac notation.
- Calculate the probability of measuring E =
¯hω for the | Ψ > given.
Postulates 3 and 4. The only possible results of a measurement of energy are the eigenvalues
En =
n +
¯hω. The state vector contains only the two eigenstates | n = 0 > and | n = 2 >,
so E 0 =
¯hω and E 2 =
¯hω are the only possible results. Then P (E = En ) = | < n | Ψ > |
2 .
P (E = E 2 ) =
2
2
=
2
=
using unit vector notation. The same calculation in Dirac notation looks like
P (E = E 2 ) =
2
2
2
2
where the inner products are < i | j> = δij , meaning < 2 | 0 > = 0 and < 2 | 2 > = 1.
- Calculate the expectation value of energy for the | Ψ > given.
This problem demonstrates the calculation using the chapter 1 mathematics and is also intended
to highlight the relative ease of calculations using matrix and Dirac notation. Setup the calculation
in position space just to demonstrate the relative degree of difficulty.
= < H >ψ = < ψ | H | ψ> =
1 2
3 2
5 2
¯hω
¯hω
1 2
3 2
5 2 (5)
¯hω
¯hω.
Notice that we attain the same result if the calculation is done in a three dimensional subspace.
The same calculation in Dirac notation using the direct action of the Hamiltonian is
< E > = < ψ | H | ψ>
[
∗
∗
]
H
[
]
[
] [
¯hω
¯hω
]
[
¯hω
¯hω
]
¯hω
[
]
¯hω,
where the Hamiltonian operating to the right on the two eigenstates in equation (1) results in
the eigenvalues times the corresponding eigenstate in equation (2), and the orthonormality of
eigenstates results in equation (3) where the inner product of terms that are known to be zero.
Excluding known zeros is the norm, and in fact, eigenstates such as < 0 | 0 > = < 2 | 2 > = 1 that
are known to be one are normally not explicitly written.
Postscript: The calculation of expectation value for the given | Ψ > in position space would be
−∞
mω
π¯h
[
mω
¯h
x
2 − 1
]
e
−mωx^2 / 2¯h
[
2 m
−i¯h
d
dx
kx
2
]
×
mω
π¯h
[
mω
¯h
x
2 − 1
]
e
−mωx^2 / 2¯h dx ,
which is a statement that implies that matrix methods and Dirac notation are worthwhile, in other
words, if you want to do this integral, go ahead, we chose to avoid it in favor of modern techniques.
- Calculate the expectation value of energy for the | Ψ > given using the ladder operator
representation of the Hamiltonian.
This problem is an explicit example of ladder operators calculations. It is not the easiest way to
calculate an expectation value. The ladder operators though, are prototypes for the creation and
annihilation operators used in field theoretic descriptions of photons, electrons, phonons, etc. In
other words, the raising and lowering operators become increasingly important as you progress.
H =
a
a +
¯hω, a
| n> =
n + 1 | n + 1 >, a | n> =
n | n − 1 >, = < ψ | H | ψ>
