physics 4617/5617: quantum physics course lecture...
TRANSCRIPT
Physics 4617/5617: Quantum PhysicsCourse Lecture Notes
Dr. Donald G. LuttermoserEast Tennessee State University
Edition 5.1
Abstract
These class notes are designed for use of the instructor and students of the course Physics 4617/5617:Quantum Physics. This edition was last modified for the Fall 2006 semester.
III. The Time-Independent Schrodinger Equation
A. Stationary States.
1. How does one determine the wave function Ψ(x, t) from first prin-
ciples?
a) It is solved through a knowledge of V (i.e., the potential
energy function) in the Schrodinger equation.
b) In this class, we will assume that the potential V is in-
dependent of t =⇒ then the Schrodinger equation can be
solved by the method of separation of variables:
Ψ(x, t) = ψ(x) f(t) , (III-1)
where ψ (lowercase) is a function of x alone and f is a
function of t alone.
c) For separable solutions we have
∂Ψ
∂t= ψ
df
dt,
∂2Ψ
∂x2 =d2ψ
dx2 f (III-2)
(ordinary derivatives now) and the Schrodinger equation
(Eq. II-1) becomes
ihψdf
dt= − h2
2m
d2ψ
dx2 f + V ψf . (III-3)
d) Dividing through by ψf gives
ih1
f
df
dt= −
h2
2m
1
ψ
d2ψ
dx2 + V . (III-4)
e) The left side is a function of t alone, and the right side, a
function of x alone =⇒ the only way this can be true is
III–1
if both sides of the equation are constant (which we will
set equal to E). Then
ih1
f
df
dt= E , (III-5)
ordf
dt= −
iE
hf , (III-6)
and
−h2
2m
1
ψ
d2ψ
dx2 + V = E , (III-7)
or
−h2
2m
d2ψ
dx2 + V ψ = Eψ . (III-8)
2. Separation of variables has turned a partial differential equation
into two ordinary differential equations (Eqs. III-6 & III-8).
a) Eq. (III-6) is easy to solve by just rearranging terms and
integrating:
f(t) = Ce−iEt/h . (III-9)
Typically the integration constant C is absorbed into the
solution for ψ, hence we state the solution to Eq. (III-6)
as
f(t) = e−iEt/h . (III-10)
b) Meanwhile, Eq. (III-8) is referred to as the time-independent
Schrodinger equation (in one dimension). To solve it,
we must specify the potential V (x).
3. When is a separable solution to the Schrodinger equation valid?
a) Stationary states:
i) These are states which even though the wave func-
tion depends upon time:
Ψ(x, t) = ψ(x) e−iEt/h , (III-11)
III–2
the probability density does not:
|Ψ(x, t)|2 = Ψ∗Ψ = ψ∗e+iEt/hψe−iEt/h = |ψ(x)|2 .(III-12)
ii) In a stationary state, 〈x〉 is constant and 〈p〉 =
0 (from Eq. II-45) =⇒ nothing ever happens in a
stationary state.
b) States of Definite Total Energy:
i) In classical physics, the total energy (kinetic plus
potential) is called the Hamiltonian:
H(x, p) =p2
2m+ V (x) . (III-13)
ii) The corresponding Hamiltonian operator is de-
fined by
H = − h2
2m
∂2
∂x2 + V (x) , (III-14)
where the “hat” (i.e., ) means this is an operator.
iii) Thus the time-independent Schrodinger equation
(Eq. III-8) can be written as
Hψ = Eψ . (III-15)
iv) The expectation value of the total energy is
〈H〉 =∫ +∞
−∞ψ∗Hψ dx = E
∫ +∞
−∞|ψ|2 dx = E .
(III-16)
v) Moreover,
H2ψ = H(Hψ) = H(Eψ) = E(Hψ) = E2ψ ,
(III-17)
III–3
and hence
〈H2〉 =∫ +∞
−∞ψ∗H2ψ dx = E2
∫ +∞
−∞|ψ|2 dx = E2 .
(III-18)
vi) So the standard deviation in H is given by
σ2H = 〈H2〉 − 〈H〉2 = E2 − E2 = 0 . (III-19)
vii) If σ = 0, then every member of the sample must
share the same value =⇒ the distribution has zero
spread.
viii) Hence, a separable solution has the property
that every measurement of the total energy is cer-
tain to return the value E.
c) The general solution of the time-independent Schrodinger
equation is a linear combination of separable solutions.
i) The time-independent Schrodinger equation yields
an infinite collection of solutions (ψ1(x), ψ2(x), ψ3(x),
. . .), each with its associated value of the separa-
tion constant (E1, E2, E3, . . .) =⇒ thus there is a
different wave function for each allowed energy:
Ψ1(x, t) = ψ1(x)e−iE1t/h, Ψ2(x, t) = ψ2(x)e
−iE2t/h, . . .
(III-20)
ii) Hence the general solution of the wave function
for these cases is
Ψ(x, t) =∞∑
n=1cn ψn(x) e
−iEnt/h . (III-21)
iii) Every solution to the time-independent Schrodinger
III–4
equation can be written in this form — it is simply
a matter of finding the right constants (c1, c2, . . .).
B. Infinite Square Well.
1. Assume we have a potential of the following form:
V (x) =
0, if 0 ≤ x ≤ a,
∞, otherwise(III-22)
as shown in Figure III-1.
V(x)
xa
Figure III–1: The infinite square well potential (Eq. III-22).
a) A particle in this potential is completely free, except at
the two ends (x = 0 and x = a), where an infinite force
prevents it from escaping.
b) Outside the well, ψ(x) = 0 =⇒ the probability of finding
the particle there is zero.
c) Inside the well, V = 0, and the time-independent Schrodinger
equation becomes
−h2
2m
d2ψ
dx2 = Eψ , (III-23)
ord2ψ
dx2 = −k2ψ, where k ≡√
2mE
h. (III-24)
III–5
d) Eq. (III-24) is just the (classical) simple harmonic os-
cillator equation with the general solution of
ψ(x) = A sin kx +B cos kx , (III-25)
where A andB are constants that are fixed by the bound-
ary conditions of the problem:
i) Both ψ and dψ/dx are continuous.
ii) Continuity of ψ(x) requires that
ψ(0) = ψ(a) = 0 , (III-26)
so as to join the solution outside the well, hence,
ψ(0) = A sin 0 +B cos 0 = B = 0 , (III-27)
so
ψ(x) = A sin kx . (III-28)
iii) At the other boundary, ψ(a) = A sin ka = 0.
Since ψ must be normalizable, A 6= 0, so
ka = 0,±π,±2π,±3π, . . . (III-29)
iv) But k = 0 is no good since that would give
ψ(x) = 0, and the negative solutions give nothing
new, since sin(−θ) = − sin(θ) and we can absorb
the minus sign into A. So the distinct solutions are
kn =nπ
a, with n = 1, 2, 3, . . . (III-30)
v) Note that the boundary condition at x = a does
not determine A, but rather the constant k, and
hence the possible values of E:
En =h2k2
n
2m=n2π2h2
2ma2 . (III-31)
III–6
Exercise: Prove Eq. (III-25) is the solution to Eq. (III-24).
Exercise: Prove Eq. (III-31).
e) Unlike the classical case, a quantum particle in the infinite
square well cannot have just any old energy — it can
only have the allowed values of Eq. (III-31) =⇒ allowed
states.
f) As usual, we find the amplitude A from the normalization
criterion:
1 =∫ a
0ψ∗ψ dx
=∫ a
0|A|2 sin2(kx) dx
= |A|2∫ a0
1
2[1 − cos(2kx)] dx
=|A|2
2
[∫ a0dx −
∫ a0
cos(2kx) dx]
=|A|2
2
[x|a0 −
1
2ksin(2kx)|a0
]
=|A|2
2
{a−
1
2k
[sin 2
(nπ
a
)a− sin 0
]}
=|A|2
2
(a−
1
2ksin 2nπ
)
= |A|2a
2
A =
√√√√2
a(III-32)
(we are only worried about the magnitude of A here), so
our final wave function becomes
ψn(x) =
√√√√2
asin
(nπ
ax
). (III-33)
2. As can be seen, for an infinite square well, there are an infinite
amount of solutions to the time-independent Schrodinger equa-
III–7
tion. Figure III-2 shows the first 3 of these stationary states —
as can be seen, these are just standing waves.
a) The lowest energy state, ψ1 is called the ground state.
b) The higher energy states scale as n2 and are called ex-
cited states.
c) These states alternate from even to odd, with respect to
the center of the well. (ψ1 is even, ψ2 is odd, ψ3 is even,
and so on.)
d) As you go up in energy, each successive state has one more
node (zero crossing). ψ1 has none (the end points don’t
count), ψ2 has 1, ψ3 has 2, etc.
e) They are mutually orthogonal (i.e., the wave functions
are said to be orthonormal), in the sense that∫ψ∗m(x)ψn(x) dx = δmn , (III-34)
where δmn is the Kronecker delta:
δmn =
0, if m 6= n;
1, if m = n.(III-35)
f) They are complete, in the sense that any other function,
f(x), can be expressed as a linear combination of them:
f(x) =∞∑
n=1cnψn(x) =
√√√√2
a
∞∑
n=1cn sin
(nπ
ax
). (III-36)
Proof of this completion criterion requires knowledge of
Fourier series (see below).
Exercise: Prove Eq. (III-34) with the wave function in Eq. (III-33).
3. The time-dependent form of the solution to this Schrodinger
III–8
a x
Ψ1(x)
a
x
Ψ2(x)
a x
Ψ3(x)
Figure III–2: The first 3 stationary states of the infinite square well.
equation is thus
Ψ(x, t) =∞∑
n=1cn
√√√√2
asin
(nπ
ax
)e−i(n
2π2h/2ma2)t . (III-37)
a) Then our initial condition equation becomes
Ψ(x, 0) =∞∑
n=1cnψn(x) . (III-38)
b) By making use of the orthonormality of the solutions, the
initial condition equations then give us the mechanism for
finding the coefficients of the series:
cn =
√√√√2
a
∫ a0
sin
(nπ
ax
)Ψ(x, 0) dx . (III-39)
Example III–1. A particle in an infinite square well has the initial wave
function
Ψ(x, 0) = Ax(a− x) .
(a) Normalize Ψ(x, 0). Graph it. Which stationary state does it most closelyresemble? On that basis, estimate the expectation value of the energy.
(b) Compute 〈x〉, 〈p〉 and 〈H〉, at t = 0. (Note : This time you cannot get〈p〉 by differentiating 〈x〉, because you only know 〈x〉 at one instant intime.) How does 〈H〉 compare with your estimate in (a)?
III–9
Solution (a):
Ψ(x, 0) = Ax(a− x)
Ψ∗(x, 0) = Ax(a− x)
1 =∫ a0
Ψ∗Ψ dx = A2∫ a
0x(a− x)x(a− x) dx
= A2∫ a0x2(a2 − 2ax+ x2
)dx
= A2∫ a0
(a2x2 − 2ax3 + x4
)dx
= A2[1
3a2x3 − 1
2ax4 +
1
5x5]a
0
= A2[1
3a5 − 1
2a5 +
1
5a5]
=A2
30
[10a5 − 15a5 + 6a5] =
A2
30a5
A =
√√√√30
a5 .
To graph this function, note that at the endpoints (i.e., x = 0 and x = a),
Ψ(x, 0) = 0. Also, expanding out the integrand gives ax−x2, which clearly
indicates that this wave function has a parabolic shape. We can find the
extrema of this parabola by taking the derivative and setting it equal to
0: a− 2x = 0, or x = a/2. The second derivative tells us the orientation
of the parabola: −2, so the curve is concave downward as shown on
the graph on the next page. As one can see from an inspection of this
curve (see figure on next page), this wave function resembles the harmonic
oscillator solution in the ground state (n = 1) since it resembles a sine
function (note that we will be investigating simple harmonic oscillators in
section III.E). As such, we can guess the expectation value of the energy
will follow Eq. (III-31):
E =π2h2
2ma2 = 4.93h2
ma2 .
III–10
ax
Ψ(x,0)
The wave function Ψ(x, 0) = x(a− x).
Solution (b):
〈x〉 =∫ a0
Ψ∗xΨ dx = A2∫ a0
[x(a− x)x2(a− x)
]dx
= A2∫ a0
[x3(a− x)2
]dx
= A2∫ a0
[x3(a2 − 2ax + x2)
]dx
= A2∫ a0
(a2x3 − 2ax4 + x5)
)dx
= A2[1
4a2x4 − 2
5ax5 +
1
6x6]a
0
= A2[1
4a6 − 2
5a6 +
1
6a6]
=30
60a5
(15a6 − 24a6 + 10a6
)
=1
2a5a6 =
a
2.
Which is just what you would expect from the appearance of the graph.
〈p〉 =∫ a0
Ψ∗h
i
∂
∂xΨ dx = A2h
i
∫ a
0
{x(a− x)
∂
∂x[x(a− x)]
}dx
III–11
= A2h
i
∫ a
0[x(a− x)(a− 2x)] dx
= A2h
i
∫ a
0
[x(a2 − 3ax + 2x2)
]dx
= A2h
i
∫ a
0
[a2x− 3ax2 + 2x3
]dx
= A2h
i
[1
2a2x2 − ax3 +
2
4x4]a
0
= A2 h
2i
(a4 − 2a4 + a4) = 0 .
The particle initially has an expectation of being at rest.
〈H〉 =∫ a
0Ψ∗
− h2
2m
∂2
∂x2 + V
Ψ dx .
Now,∂2
∂x2Ψ =∂2
∂x2 [x(a− x)] =∂
∂x(a− 2x) = −2 ,
so
〈H〉 = − h2
2m
∫ a
0Ψ∗∂
2Ψ
∂x2 dx + V∫ a0
Ψ∗Ψ dx
=h2
mA2
∫ a
0x(a− x) dx + V
=h2
mA2
[1
2ax2 −
1
3x3]a
0+ V
=h2
mA2
(1
2a3 − 1
3a3)
+ V
=h2
6mA2
(3a3 − 2a3
)+ V
=h2
6m
30
a5a3 + V
=5h2
ma2 + V .
From this, it is clear that
E = 5h2
ma2 ,
III–12
which is close to the solution found for the harmonic oscillator.
C. Fourier Analysis.
1. As we continue on with our work with the Schrodinger equation,
we will often encounter wave functions that take the form
Ψ(x, t) =1√2π
∫ ∞
−∞A(k) ei(kx−ωt) dk . (III-40)
a) In the wave function, x and t represent their usual mean-
ings of the position and time independent variables, re-
spectively. The new variables introduced here in the in-
tegral are:
i) The wave number, k, which is inversely related
to the wavelength, λ, of a wave through
k ≡ 2π
λ.
Using Eq. (I-77) we can show that the wave number
is related to the particle/wave’s linear momentum
via
p = hk .
ii) The angular frequency, ω, which is directly re-
lated to the frequency, ν, of a wave through
ω ≡ 2πν .
Using Eq. (I-77) we can show that the angular fre-
quency is related to the particle/wave’s energy via
E = hω .
iii) Finally note that one can write a dispersion re-
lation for a wave with these new variables. Using
III–13
Eq. (I-75) for a non-relativistic free particle (i.e.,
V = 0), the total energy of the wave is
E =p2
2m.
Making use of our relations above, this energy-
momentum equation becomes the following disper-
sion relation:
ω(k) =hk2
2m.
b) At t = 0, this equation takes on a form that may be
familiar:
Ψ(x, 0) =1√2π
∫ ∞
−∞A(k) eikx dk . (III-41)
c) Eq. (III-41) reveals that the amplitude function A(k) is
the Fourier transform of the wave function Ψ(x, t) at t
= 0 =⇒ the amplitude function is related to the wave
function at t = 0 by a Fourier integral.
2. Fourier analysis — the generation and deconstruction of Fourier
series and integrals are the mathematical methods that underlies
the construction of wave packets by superposition.
a) Mathematicians commonly use Fourier analysis to rip func-
tions apart, representing them as sums or integrals of sim-
ple component functions, each which is characterized by
a single frequency.
b) This method can be applied to any function f(x) that is
piecewise continuous — i.e., that has at most a finite num-
ber of finite discontinuities. As we have already noted,
wave functions must be continuous, so as such, satisfies
this condition and so are prime candidates for Fourier
analysis.
III–14
c) Whether we represent f(x) via a Fourier series or Fourier
integral depends on whether or not this function is pe-
riodic =⇒ any function that repeats itself is said to be
periodic.
d) More precisely, if there exists a finite number L such that
f(x + L) = f(x), then f(x) is periodic with period L.
e) We can write any function that is periodic (or that is
defined on a finite interval) as a Fourier series.
f) However if f(x) is non-periodic or is defined on the infinite
interval from −∞ to +∞, we must use a Fourier integral.
3. Fourier Series. Fourier series are not mere mathematical de-
vices; they can be generated in the laboratory (or telescope) =⇒a spectrometer decomposes an electromagnetic wave into spectral
lines, each with a different frequency and amplitude (intensity).
Thus, a spectrometer decomposes a periodic function in a fashion
analogous to the Fourier series.
a) Suppose we want to write a periodic, piecewise continuous
function f(x) as a series of simple functions. Let L denote
the period of f(x), and choose as the origin of coordinates
the midpoint of the interval defined by this period−L/2 ≤x ≤ L/2.
b) If we let an and bn denote (real) expansion coefficients, we
can write the Fourier series of this function as
f(x) = a0 +∞∑
n=1
[an cos
(2πn
x
L
)+ bn sin
(2πn
x
L
)].
(III-42)
III–15
c) We calculate the coefficients in Eq. (III-42) from the func-
tion f(x) as
a0 =1
L
∫ L/2−L/2
f(x) dx , (III-43)
an =2
L
∫ L/2
−L/2f(x) cos
(2πn
x
L
)dx (n = 1, 2, . . .)
(III-44)
bn =2
L
∫ L/2
−L/2f(x) sin
(2πn
x
L
)dx (n = 1, 2, . . .) .
(III-45)
d) Notice that the summation in Eq. (III-42) contains an
infinite number of terms. In practice we retain only a
finite number of terms =⇒ this approximation is called
truncation.
i) Truncation is viable only if the sum converges to
whatever accuracy we want before we chop it off.
ii) Truncation is not as extreme an act as it may
seem. If f(x) is normalizable, then the expansion
coefficients in Eq. (III-42) decrease in magnitude
with increasing n, i.e.,
|an| → 0 and |bn| → 0 as n→ ∞ .
iii) Under these conditions, which are satisfied by
physically admissible wave functions, the sum in
Eq. (III-42) can be truncated at some finite max-
imum value nmax of the index n. (Trial and error
is typically needed to determine the value of nmax
that is required for the desired accuracy.)
iv) If f(x) is particularly simple, all but a small, fi-
nite number of coefficients may be zero. One should
III–16
always check for zero coefficients first before evalu-
ating the integrals in Eqs. (III-43, 44, 45).
4. The Power of Parity. One should pay attention as to whether
one is integrating an odd or an even function. Trigonometric
functions have the well-known parity properties:
sin(−x) = − sin x (odd) (III-46)
cos(−x) = + cos x (even) . (III-47)
As such, if f(x) is even or odd, then half of the expansion coef-
ficients in its Fourier series are zero.
a) If f(x) is odd [f(−x) = −f(x)], then
an = 0 (n = 0, 1, 2, 3, . . .)
f(x) =∞∑
n=1bn sin
(2πn
x
L
).
(III-48)
b) If f(x) is even [f(−x) = +f(x)], then
bn = 0 (n = 1, 2, 3, . . .)
f(x) =∞∑
n=0an cos
(2πn
x
L
).
(III-49)
c) If f(x) is either an even or and odd function, it is then
said to have definite parity.
5. The Complex Fourier Series: If f(x) does not have a definite
parity, we can expand it in a complex Fourier series.
a) To derive this variant on the Fourier series in Eq. (III-
42), we just combine the coefficients an and bn so as to
introduce the complex exponential function ei2πnx/L:
f(x) =∞∑
n=−∞cne
i2πnx/L . (III-50)
III–17
b) Note carefully that in the complex Fourier series in Eq.
(III-50) the summation runs from −∞ to ∞. The expan-
sion coefficients cn for the complex Fourier series are
cn =1
L
∫ L/2
−L/2f(x) e−i2πnx/L dx . (III-51)
Exercise: Derive Eqs. (III-50) and (III-51) and thereby determine
the relationship of the coefficients cn of the complex Fourier series
of a function to the coefficients an and bn of the corresponding real
series.
6. Fourier Integrals: Any normalizable function can be expanded
in an infinite number of sine and cosine functions that have in-
finitesimally differing arguments. Such an expansion is called a
Fourier integral.
a) A function f(x) can be represented by a Fourier integral
provided the integral∫∞−∞ |f(x)|2 dx exists =⇒ all wave
functions satisfy this condition for they are normalizable.
b) The Fourier integral has the form
f(x) =1√2π
∫ ∞
−∞g(k)eikx dk , (III-52)
which is the inverse Fourier transform.
c) The function g(k) plays the role analogous to that of the
expansion coefficients cn in the complex series (Eq. III-50).
The relationship of g(k) to f(x) is more clearly exposed
by the inverse of Eq. (III-52),
g(k) =1√2π
∫ ∞
−∞f(x)e−ikx dx , (III-53)
which is the famed Fourier transform equation. In math-
ematical parlance, f(x) and g(k) are said to be Fourier
transforms of one another.
III–18
d) More precisely, g(k) is the Fourier transform of f(x),
and f(x) is the inverse Fourier transform of g(k).
e) When convenient, we will use the shorthand notation
A(k) = F [Ψ(x, 0)] and Ψ(x, 0) = F−1 [A(k)]
(III-54)
to represent Eqs. (III-53) and (III-52), respectively.
f) Many useful relationships follow from the intimate rela-
tionship between f(x) and g(k). For our purposes, the
most important is the Bessel-Parseval relationship:∫ ∞
−∞|f(x)|2 dx =
∫ ∞
−∞|g(k)|2 dk . (III-55)
D. The Gaussian Wave Packet.
1. In Eqs. (III-40) and (III-41), we stated that there will often be
time when the wave function will be expressed in terms of an in-
tegral of an amplitude function =⇒ calculating the wave function
is the inverse Fourier transform of the amplitude function.
2. Hence, to calculate the amplitude function, one merely takes
the Fourier transform of the wave function:
A(k) = F [Ψ(x, 0)] =1√2π
∫ ∞
−∞Ψ(x, 0)e−ikx dx . (III-56)
3. The Bessel-Parseval relationship (Eq. III-55) guarantees that the
Fourier transform of a normalized function is normalized:∫ ∞
−∞|A(k)|2 dk =
∫ ∞
−∞|Ψ(x, 0)|2 dx = 1 . (III-57)
Example III–2. The Amplitude Function for a Gaussian. The
Gaussian function is a wave packet with a well-defined center and a single
peak. Its amplitude function has the same properties. Initially (i.e., t = 0),
III–19
the Gaussian function contains one parameter, a real number L that governs
its width. Show how the amplitude function of a Gaussian scales with L and
prove that both the Gaussian wave function and its amplitude function obey
normalization rules.
Solution:
The most general form of such a function has a center at x◦ 6= 0 and
corresponds to an amplitude function that is centered at k◦ 6= 0:
Ψ(x, 0) =
(1
2πL2
)1/4eik◦xe−[(x−x◦)/(2L)]2 .
For simplicity though, assume the Gaussian is centered at x◦ = 0 with an
amplitude function centered at k◦ = 0, i.e.,
Ψ(x, 0) =
(1
2πL2
)1/4e−x
2/(4L2) .
-3 -2 -1 0 1 2 3 x
Ψ(x,0)
L Ψ(x,0) for L = 0.5
Ψ(x,0) for L = 1.0
|Ψ(x,0)|2 for L = 1.0
Two Gaussian wave packets of the form in Example III-2; parameter L for
these functions takes on the values L = 0.5 and L = 1.0 (solid curves). The
corresponding probability density for L = 1.0 is shown as a dashed curve.
In the figure above, you’ll find two such wave packets (with different
values of L). Each exhibits the characteristic shape of a Gaussian function
III–20
(i.e., a Bell-shaped curve): Each has a single peak and decreases rather
sharply as |x| increases from zero. But because the Gaussian function
decays exponentially, it never actually equals zero for finite |x|. (It is,
nonetheless, normalizable.) You’ll also find in this figure above that the
probability density |Ψ(x, 0)|2 for a Gaussian. This figure illustrates one
of the special properties of a Gaussian wave function =⇒ its probability
is also a Gaussian function, one that has the same center but is narrower
than the state function from which it is calculated.
To find the amplitude function for this Gaussian, make sure that both the
state function and the amplitude function are normalized as requested in
the problem. We begin by substituting the state function listed above
into Eq. (III-56) for the amplitude function:
A(k) =1√2π
(1
2πL2
)1/4 ∫ ∞
−∞exp
[− 1
4L2x2 − ikx
]dx .
We note that a Table of Integrals give
∫ ∞
−∞e−αx
2−βx dx =
√π
αeβ
2/(4α) (α > 0) .
Let α = 1/(4L2) and β = ik, then we see that
A(k) =
(2
πL2)1/4
e−k2L2
.
Comparing the mathematical form of the amplitude function and the
initial wave function, we see that the Fourier transform of a Gaussian
function of variable x is a Gaussian of variable k.
Now, check for normalization
∫ ∞
−∞Ψ∗Ψ dx =
(1
2πL2
)1/2 ∫ ∞
−∞e−x
2/(2L2) dx
=
(1
2πL2
)1/22∫ ∞
0e−x
2/(2L2) dx
=
(1
2πL2
)1/22
1
2/√
2L
√π
III–21
=
(1
2πL2
)1/2 √2πL
=
2πL2
2πL2
1/2
=√
1 = 1 . QED
∫ ∞
−∞A(k)∗A(k) dk =
(2
πL2)1/2 ∫ ∞
−∞e−2k2L2
dk
=
(2
πL2)1/2
2∫ ∞
0e−2k2L2
dk
=
(2
πL2)1/2
21
2√
2L
√π
=
(2
πL2)1/2 √ π
2L2
=
2πL2
2πL2
1/2
=√
1 = 1 . QED
E. The Quantum Simple Harmonic Oscillator.
1. From classical mechanics, the equation of motion of a mass con-
nected to a spring is govern by Hooke’s Law:
F = −kx = md2x
dt2,
where k is the spring constant.
2. The solution to Hooke’s Law gives a simple harmonic oscillation
in position:
x(t) = A sin(ωt) +B cos(ωt) ,
III–22
where A and B are constants determined from either initial con-
ditions or boundary conditions, and
ω =
√√√√ k
m(III-58)
is the (angular) frequency of oscillation. The potential energy is
V (x) =1
2kx2 . (III-59)
3. As can be seen by the above equation, the potential for Hooke’s
law is parabolic. Practically any potential can be approximated
by a parabola, if one takes the limits small enough in the neigh-
borhood of a local minimum.
a) One can expand any function about a local minimum in
a Taylor series. In the case of the potential:
V (x) = V (x◦)+1
1!V ′(x◦)(x−x◦)+
1
2!V ′′(x◦)(x−x◦)2 + · · ·
(III-60)
where x◦ is the location of the local minimum.
b) We can set V (x◦) = 0 since the value of this constant
is irrelevant to the force. Also V ′(x◦) = 0 since x◦ is
a minimum. If we drop terms higher than the second
derivative, we get
V (x) ∼=1
2V ′′(x◦)(x− x◦)
2 , (III-61)
which describes simple harmonic oscillation about the point
x◦, with an effective spring constant k = V ′′(x◦).
c) It should be noted from this Taylor series approximation
that any oscillatory motion is approximately simple har-
monic, as long as the amplitude is small.
III–23
4. The quantum problem is to solve the Schrodinger equation for
the potential
V (x) =1
2mω2x2 . (III-62)
a) With this potential, the time-independent Schrodinger’s
equation becomes
− h2
2m
d2ψ
dx2 +1
2mω2x2ψ = Eψ . (III-63)
b) There are 2 ways to solve this equation: The power se-
ries expansion method and the ladder operator method.
This second method is primarily just clever algebra and
we will go through that solution first.
5. The Ladder Operator (Algebraic) Method.
a) Let’s rewrite Eq. (III-63) as
1
2m
(h
i
d
dx
)2
+ (mωx)2ψ = Eψ . (III-64)
b) We must factor the term in the square brackets. We can
do this by realizing that if these terms were numbers, we
could factor them as
u2 + v2 = (u − iv)(u+ iv) .
However in our case, u and v are operators and operators
do not, in general, commute (i.e., uv 6= vu).
c) Let us introduce the a-operators as
a± ≡ 1√2m
(h
i
d
dx± imωx
). (III-65)
Let’s check the various products of these operators:
i) First,
(a−a+)f(x) =1
2m
(h
i
d
dx− imωx
) (h
i
d
dx+ imωx
)f(x)
III–24
=1
2m
(h
i
d
dx− imωx
) (h
i
df
dx+ imωxf
)
=1
2m
−h2d
2f
dx2 + hmωd
dx(xf)
−hmωxdfdx
+ (mωx)2f
]
=1
2m
−h2d
2f
dx2 + hmωxdf
dx+ hmωf
−hmωxdfdx
+ (mωx)2f
]
=1
2m
(h
i
d
dx
)2
+ (mωx)2 + hmω
f(x)
a−a+ =1
2m
(h
i
d
dx
)2
+ (mωx)2+
1
2hω . (III-66)
ii) Pulling the (1/2)hω term to the other side of
the equation, we get the following for Schrodinger’s
equation:(a−a+ − 1
2hω
)ψ = Eψ . (III-67)
iii) From a similar treatment, one can easily show
that
a+a− =1
2m
(h
i
d
dx
)2
+ (mωx)2−
1
2hω , (III-68)
and the Schrodinger equation becomes(a+a− +
1
2hω
)ψ = Eψ . (III-69)
iv) Also note that
a−a+ − a+a− = hω . (III-70)
d) If ψ satisfies the Schrodinger equation with energy E, then
a+ψ satisfies the Schrodinger equation with energy (E +
III–25
hω):(a+a− +
1
2hω
)(a+ψ) =
(a+a−a+ +
1
2hωa+
)ψ
= a+
(a−a+ +
1
2hω
)ψ = a+
[(a−a+ − 1
2hω
)ψ + hωψ
]
= a+(Eψ + hωψ) = (E + hω)(a+ψ) . QED(III-71)
e) By the same token, a−ψ is a solution with energy (E−hω):(a−a+ − 1
2hω
)(a−ψ) =
(a−a+a− − 1
2hωa−
)ψ
= a−
(a+a− − 1
2hω
)ψ = a−
[(a+a− +
1
2hω
)ψ − hωψ
]
= a−(Eψ − hωψ) = (E − hω)(a−ψ) . QED(III-72)
f) We call the a-operators the ladder operators: a+ is
called the raising operator, and a−, the lowering op-
erator =⇒ if we can find one solution with energy E, we
can find all other solutions by using these operators.
g) There exist a lowest energy state for the quantum har-
monic oscillator which satisfies
a−ψ◦ = 0 , (III-73)
where ψ◦ is referred to as the ground state.
i) As a result, we can write the lowering operator for
this ground state as
1√2m
(h
i
dψ◦
dx− imωxψ◦
)= 0 , (III-74)
ordψ◦
dx= −
mω
hxψ◦ . (III-75)
ii) The solution to this differential equation is trivial:
∫ dψ◦
ψ◦= −
mω
h
∫x dx
III–26
lnψ◦ = −mω
2hx2 + constant
ψ◦ = A◦e−(mω/2h)x2
. (III-76)
iii) Plugging this into Schrodinger equation, (a+a−+
(1/2)hω)ψ◦ = E◦ψ◦ and noting Eq. (III-73), we see
that
E◦ =1
2hω . (III-77)
Note that
ω = 2πν
h =h
2π,
so
hω =h
2π· 2πν = hν .
h) From this ground state, we can easily calculate the excited
states with the raising operator:
ψn(x) = An(a+)ne−(mω/2h)x2
, with En =
(n +
1
2
)hω .
(III-78)
i) For instance, the first excited state is
ψ1 = A1a+e−(mω/2h)x2
= A11√2m
(h
i
d
dx+ imωx
)e−(mω/2h)x2
=A1√2m
[h
i
(−mω
hx
)e−(mω/2h)x2
+ imωxe−(mω/2h)x2].
(III-79)
ii) This simplifies to
ψ1 = (iA1ω√
2m)xe−(mω/2h)x2
. (III-80)
Note that since ψ must be real, the amplitude A1
must be imaginary (see Example III-3 below).
III–27
Example III–3. The raising and lowing operators:
(a) The raising and lowering operators generate new solutions tothe Schrodinger equation, but these new solutions are not cor-rectly normalized. Thus a+ψn is proportional to ψn+1, anda−ψn is proportional to ψn−1, but we would like to know theprecise proportionality constants. Use integration by parts andthe Schrodinger equation (Eqs. III-67 and III-69) to show that
∫ ∞
−∞|a+ψn|2 dx = (n+ 1) hω (III-81)
∫ ∞
−∞|a−ψn|2 dx = nhω , (III-82)
and hence (with i’s to keep the wave function real)
a+ψn = i√
(n+ 1)hω ψn+1 (III-83)
a−ψn = −i√nhω ψn−1 . (III-84)
(b) Use Eq. (III-83) to show that the normalization constant An
in Eq. (III-78) is
An =
(mω
πh
)1/4 (−i)n√n!(hω)n
. (III-85)
Solution (a):
∫ ∞
−∞|a+ψn|2 dx =
∫ ∞
−∞(a+ψn)
∗(a+ψn) dx
=1
2m
∫ ∞
−∞
[(h
i
d
dx+ imωx
)ψn
]∗
×[(h
i
d
dx+ imωx
)ψn
]dx
=1
2m
∫ ∞
−∞
(−hi
dψ∗n
dx− imωxψ∗
n
)
×(h
i
dψndx
+ imωxψn
)dx
III–28
=1
2m
∫ ∞
−∞
[h2dψ
∗n
dx
dψndx
− hmωxdψ∗
n
dxψn
−hmωxψ∗n
dψndx
+ (mωx)2ψ∗nψn
]dx .
At this point, note that
d
dx(ψ∗
nψn) = ψ∗n
dψndx
+dψ∗
n
dxψn .
As such, we can simplify the integral above as
∫ ∞
−∞|a+ψn|2 dx =
1
2m
∫ ∞
−∞
[h2dψ
∗n
dx
dψndx
+ (mωx)2ψ∗nψn
−hmωx ddx
(ψ∗nψn)
]dx
=1
2m
[∫ ∞
−∞h2dψ
∗n
dx
dψndx
dx +∫ ∞
−∞(mωx)2ψ∗
nψn dx
−∫ ∞
−∞hmωx
d
dx(ψ∗
nψn) dx
]. (A)
The first integral in Eq. (A) is found by using integration by
parts, letu = dψn
dxdv = dψ∗
n
du = d2ψn
dx2 dx v = ψ∗n,
then∫ ∞
−∞h2dψ
∗n
dx
dψndx
dx =∫ ∞
−∞h2dψn
dx
dψ∗n
dxdx
= h2∫ ∞
−∞
dψndx
dψ∗n
= h2ψ∗
n
dψndx
∣∣∣∣∣
∞
−∞−∫ ∞
−∞ψ∗n
d2ψndx2 dx
= −h2∫ ∞
−∞ψ∗n
d2ψndx2 dx , (A1)
where ψ∗n(dψn/dx)|∞−∞ → 0 since the wave function must be nor-
malizable. The second integral in Eq. (A) can be written as∫ ∞
−∞(mωx)2ψ∗
nψn dx =∫ ∞
−∞ψ∗n(mωx)
2ψn dx . (A2)
III–29
The third and final integral in Eq. (A) in once again solved with
integral by parts, let
u = x dv = d(ψ∗nψn)
du = dx v = ψ∗nψn ,
then
hmω∫ ∞
−∞xd
dx(ψ∗
nψn) dx = hmω∫ ∞
−∞xd(ψ∗
nψn)
= hmω[xψ∗
nψn|∞−∞ −
∫ ∞
−∞ψ∗nψn dx
]
= −hmω , (A3)
which results once again since xψ∗nψn|∞−∞ → 0 in order for the
wave function to be normalizable.
Now plugging Eqs. (A1), (A2), and (A3) back into Eq. (A) gives∫ ∞
−∞|a+ψn|2 dx =
1
2m
∫ ∞
−∞
−h2ψ∗
n
d2ψndx2 + ψ∗
n(mωx)2ψn
dx
− 1
2m(−hmω) ,
or with use of the Schrodinger equation (Eq. III-64), we get∫ ∞
−∞|a+ψn|2 dx =
1
2m
∫ ∞
−∞
−h2ψ∗
n
d2ψndx2 + ψ∗
n(mωx)2ψn
dx
+1
2hω
=∫ ∞
−∞ψ∗n
− h2
2m
d2
dx2 +1
2mω2x2
ψn dx +
1
2hω
=∫ ∞
−∞ψ∗nEnψn dx +
1
2hω
= En∫ ∞
−∞ψ∗nψn dx +
1
2hω
= En +1
2hω = (n+
1
2) hω +
1
2hω
= (n+ 1) hω . QED
Note that we could have solved this much more simply by using
operator algebra and noting that a∗+ = a− (see next section):∫ ∞
−∞(a+ψn)
∗(a+ψn) dx =∫ ∞
−∞(ψ∗
na−)(a+ψn) dx
III–30
=∫ ∞
−∞ψ∗na−a+ψn dx =
∫ ∞
−∞ψ∗n(En +
1
2hω)ψn dx
= (En +1
2hω)
∫ ∞
−∞ψ∗nψn dx = En +
1
2hω = (n+
1
2) hω +
1
2hω
= (n+ 1) hω . QED
Now for the proof of the second integral relation
∫ ∞
−∞(a−ψn)
∗(a−ψn) dx =1
2m
∫ ∞
−∞
[(h
i
d
dx− imωx
)ψn
]∗
×[(h
i
d
dx− imωx
)ψn
]dx
=1
2m
∫ ∞
−∞
(−hi
dψ∗n
dx+ imωxψ∗
n
)
×(h
i
dψndx
− imωxψn
)dx
=1
2m
∫ ∞
−∞
[h2dψ
∗n
dx
dψndx
+ hmωxdψ∗
n
dxψn
+hmωxψ∗n
dψndx
+ (mωx)2ψ∗nψn
]dx
=1
2m
∫ ∞
−∞
[h2dψ
∗n
dx
dψndx
+ (mωx)2ψ∗nψn
+hmωxd
dx(ψ∗
nψn)
]dx . (B)
The first two integrals in Eq. (B) are identical to those in Eq.
(A) and the third is just the negative of the third integral in Eq.
(A). As such, Eq. (B) reduces to
∫ ∞
−∞|a−ψn|2 dx =
1
2m
∫ ∞
−∞
−h2ψ∗
n
d2ψndx2 + ψ∗
n(mωx)2ψn
dx
−1
2hω
=∫ ∞
−∞ψ∗n
− h2
2m
d2
dx2 +1
2mω2x2
ψn dx − 1
2hω
=∫ ∞
−∞ψ∗nEnψn dx −
1
2hω
III–31
= En∫ ∞
−∞ψ∗nψn dx −
1
2hω
= En +1
2hω = (n+
1
2)hω − 1
2hω
= nhω . QED
We next have to prove the solutions of the raising and lowering
operators. Assume that a+ψn = cψn+1, for some constant c.
With ψn and ψn+1 normalized,∫ ∞
−∞|a+ψn|2 dx = |c|2
∫ ∞
−∞|ψn+1|2 dx = |c|2 = (n+ 1) hω ,
so c =√
(n+ 1)hω. Note however, to achieve consistency with
the integrals above, c = i√
(n+ 1)hω, so
a+ψn = i√
(n+ 1)hω ψn+1 .
Similarly, a−ψn = bψn−1, for some constant b, so∫ ∞
−∞|a−ψn|2 dx = |b|2
∫ ∞
−∞|ψn−1|2 dx = |b|2 = nhω ,
so b =√nhω. Once again, to achieve consistency, b = −i
√nhω,
so
a−ψn = −i√nhω ψn−1 .
Solution (b): From Eqs. (III-76) and (III-78),
ψ◦ = A◦e−(mω/2h)x2
, e−(mω/2h)x2
=ψ◦
A◦
ψn = An(a+)ne−(mω/2h)x2
=An
A◦(a+)nψ◦
=An
A◦(a+)n−1 (a+ψ◦)︸ ︷︷ ︸
i√hωψ1
=An
A◦(i√hω)(a+)n−2 (a+ψ1)︸ ︷︷ ︸
i√
2hωψ2
= · · ·
=An
A◦(i√hω)(i
√2hω)(i
√3hω) · · · (i
√nhω)ψn
=An
A◦in√n!(hω)n ψn,
An =(−i)n
√n!(hω)n
A◦ .
III–32
Normalizing ψ◦ gives
1 = |A◦|2∫ ∞
−∞e−mωx
2/h dx = |A◦|2√√√√ πh
mω
or
A◦ =
(mω
πh
)1/4,
so
An =
(mω
πh
)1/4 (−i)n√n!(hω)n
.
6. The Power Series Expansion (Analytic) Method.
a) Let
ξ ≡√mω
hx and
dξ
dx=
√mω
h(III-86)
in the Schrodinger equation
−h2
2m
d2ψ
dx2 +1
2mω2x2ψ = Eψ . (III-87)
Then using the chain rule,
d2ψ
dx2 =d
dx
(dψ
dx
)=
d
dξ
(dψ
dx
)dξ
dx=
d
dξ
(dψ
dξ
dξ
dx
)dξ
dx
=d
dξ
(dψ
dξ
√mω
h
)√mω
h=mω
h
d2ψ
dξ2 , (III-88)
plugging this back into Eq. (III-87) gives
− h2
2m
mω
h
d2ψ
dξ2 +1
2mω2
(h
mω
)ξ2ψ = Eψ
−hω2
d2ψ
dξ2 +hω
2ξ2ψ = Eψ
d2ψ
dξ2 − ξ2ψ = −2E
hωψ
III–33
d2ψ
dξ2 =
(ξ2 − 2E
hω
)ψ
d2ψ
dξ2 =(ξ2 −K
)ψ , (III-89)
where
K ≡2E
hω. (III-90)
Finally, we can rewrite Eq. (III-89) as
d2
dξ2 − ξ2 +K
ψ(ξ) = 0 , (III-91)
which is called Weber’s equation.
b) To solve this DE, let’s first look at the asymptotic limit
of very large ξ (hence x) such that ξ2 � K, then
d2ψ
dξ2 ≈ ξ2ψ , (III-92)
which has the approximate solution
ψ(ξ) ≈ Ae−ξ2/2 +Be+ξ2/2 . (III-93)
i) The B term is clearly not normalizable since ψ
blows up as ξ → ∞.
ii) The realistic solution thus has the asymptotic form
ψ(ξ) → Ae−ξ2/2, at large ξ . (III-94)
c) Since we know the asymptotic form of the solution, we can
re-examine Eq. (III-89 or III-91) and assume the solution
to this equation has the functional form
ψ(ξ) = Ah(ξ) e−ξ2/2 , (III-95)
where h is some function of ξ that contains the constant
energy term K and the coefficient A is determined from
III–34
the normalization condition. Since we insist that the wave
function for a real wave be normalizable, then
limξ→±∞
(h(ξ) e−ξ
2/2)→ 0 . (III-96)
i) Note that h(ξ) must depend upon the energy term
K since the asymptotic form Ae−ξ2/2 does not.
ii) Also note that since the asymptotic form is an
even function, the h(ξ) portion of the wave func-
tion will dictate the parity of the complete wave
function =⇒ if h(ξ) is an even function, then the
wave function will have even parity, and if h(ξ) is
an odd function, then the wave function will have
odd parity.
d) We now need to come up with the functional form of h(ξ).
i) The first derivative of Eq. (III-95) is
dψ
dξ= A
(dh
dξ− ξh
)e−ξ
2/2 . (III-97)
ii) The second derivative is thus
d2ψ
dξ2 = A
d
2h
dξ2 − 2ξdh
dξ+ (ξ2 − 1)h
e−ξ
2/2 .
(III-98)
iii) Using these derivatives in the Schrodinger equa-
tion (e.g., Eq. III-91) gives
d2h
dξ2 − 2ξdh
dξ+ (K − 1)h = 0 . (III-99)
e) Now, let’s assume that h(ξ) can be expressed as a power
series
h(ξ) = a0 + a1ξ + a2ξ2 + · · · =
∞∑
j=0ajξ
j . (III-100)
III–35
(Note that the ‘a’ parameters here are series coefficients
and not the raising and lowering operators we were dis-
cussing earlier.)
i) The first derivative of this series is
dh
dξ= a1 + 2a2ξ + 3a3ξ
2 + · · · =∞∑
j=0jajξ
j−1.
(III-101)
ii) The second derivative is
d2h
dξ2 = 2a2 + 2 · 3a3ξ + 3 · 4a4ξ2 + · · ·
=∞∑
j=0(j + 1)(j + 2)aj+2ξ
j . (III-102)
iii) Putting these into Eq. (III-99), we find
∞∑
j=0[(j + 1)(j + 2)aj+2 − 2jaj + (K − 1)aj ] ξ
j = 0 .
(III-103)
iv) It follows from the uniqueness of the power series
expansion that the coefficient of each power of ξ
must vanish,
(j + 1)(j + 2)aj+2 − 2jaj + (K − 1)aj = 0 ,
and hence that
aj+2 =(2j + 1 −K)
(j + 1)(j + 2)aj . (III-104)
v) This recursion formula is entirely equivalent to
the Schrodinger equation itself. Given a0 we can in
principle derive a2, a4, a6, ..., and given a1 it gener-
ates a3, a5, a7, ....
III–36
f) Let us therefore write
h(ξ) = he(ξ) + ho(ξ) , (III-105)
where he(ξ) are the even terms of the series expansion
(i.e., an even function):
he(ξ) ≡ a0 + a2ξ2 + a4ξ
4 + · · · (III-106)
and ho(ξ) are the odd terms of the series expansion (i.e.,
an odd function):
ho(ξ) ≡ a1ξ + a3ξ3 + a5ξ
5 + · · · (III-107)
g) Our next job is to figure out the functional form of the
‘a’ coefficients. Essentially, we will guess at a form for
aj and test it to see if it is consistent with Eqs. (III-91)
and (III-99). But first, let’s note the series expansions of
exponential functions.
i) For the exponential function, we have
e±ξ = 1±ξ+1
2!ξ2± 1
3!ξ3+ · · · =
∞∑
j=0bjξ
j, (III-108)
where
bj = (±1)j1
j!. (III-109)
ii) If we now substitute ξ2 for ξ in the above series
expansion, we get
eξ2
= 1 + ξ2 +1
2!(ξ2)2 +
1
3!(ξ2)3 +
1
4!(ξ2)4 + · · ·
= 1 + ξ2 +1
2ξ4 +
1
6ξ6 +
1
24ξ8 + · · ·
=∞∑
j = 0(even)
bjξj, (III-110)
where
bj =1
(j/2)!. (III-111)
III–37
iii) We next ask if the series expansion above con-
verges. To test this, we use the d’Alembert ratio
test which takes the limit of the ratio of two suc-
cessive terms as the counter goes to infinity:
bj+2ξj+2
bjξj=
1[(j+2)/2]!ξ
j+2
1(j/2)!ξ
j=
1
(j/2) + 1ξ2 =
2
j + 2ξ2 .
Now as taking the limit of this ratio as j → ∞, we
get
limj→∞
2
j + 2ξ2 ≈ lim
j→∞
2
jξ2 → 0 , (III-112)
thus this series converges.
h) We are now faced with a wave function solution to the
Schrodinger equation of the form
ψ(ξ) = Ah(ξ) e−ξ2/2 = A
∞∑
j=0ajξ
j
e−ξ
2/2 (III-113)
= (infinite series in ξ) (decaying Gaussian).
i) For this wave function to be physically admissible,
it must go to zero as ξ → ∞. But does the infinite
sum produce a finite number? Let’s assume that
our wave function has even parity. Then we will
only look at he(ξ) in this case. (We could make a
similar argument for the ho(ξ) odd parity solution.)
ii) For very large j, we see from Eq. (III-104) that the
recursion formula takes on the approximate form of
aj+2 ≈2
jaj . (III-114)
iii) But what is aj when j is large? Let us take the
ratio of two successive terms and run the d’Alembert
III–38
ratio test. Using Eq. (III-104), this ratio is
aj+2ξj+2
ajξj=
2j + 1 −K
(j + 2)(j + 1)ξ2 .
Now as taking the limit of this ratio as j → ∞, we
get
limj→∞
2j + 1 −K
(j + 2)(j + 1)ξ2 ≈ lim
j→∞
2j
j2 ξ2 = lim
j→∞
2
jξ2 ,
(III-115)
which is precisely the same limit for large j as eξ2
!
From this coincidence, we can make use of Eq. (III-
111) to write that
aj ≈C
(j/2)!(for even j) , (III-116)
for some constant C at large j.
iv) This yields (at large ξ, where the higher powers
dominate)
h(ξ) ≈ C∑ 1
(j/2)!ξj ≈ C
∑ 1
k!ξ2k ≈ Ceξ
2
.
(III-117)
v) If h goes like eξ2
, then ψ goes like eξ2/2 (see Eq.
III-113), which is precisely the asymptotic behavior
we don’t want!
vi) There is only one way out of this dilemma =⇒the power series must terminate ! There must occur
some highest j (call it n = jmax) such that an+2 = 0.
i) We see from Eq. (III-104) that this will occur when
K = 2n+ 1 , (III-118)
III–39
or using Eq. (II-90), we see that the requirement that for
h(ξ) to be a finite polynomial, the energies of a simple
harmonic oscillator are restricted to the form2E
hω= 2n+ 1
En =hω
2(2n+ 1)
=
(n +
1
2
)hω n = 0, 1, 2, . . . (III-119)
Hence we recover, by a completely different method, the
fundamental quantization condition found in Eq. (III-78).
j) As a result of this, we see that for any energy not in
the form of Eq. (III-119), the wave function of the sim-
ple harmonic oscillator blows up at large ξ and the wave
function is no longer un-normalizable, hence unphysical.
From this we see that the boundary conditions demand
energy quantization.
k) The recursion formula now reads
aj+2 =−2(n − j)
(j + 1)(j + 2)aj (j ≤ n) . (III-120)
i) If n = 0 (i.e., the ground state), there is only one
term in the series (we must pick a1 = 0 to kill ho,
and j = 0 in Eq. (III-120) yields a2 = 0):
h0(ξ) = a0 , (III-121)
and hence
ψ0(ξ) = A0 a0 e−ξ2/2 , (III-122)
which reproduces Eq. (III-76).
ii) For n = 1 we pick a0 = 0, and Eq. (III-120) with
j = 1 yields a3 = 0, so
h1(ξ) = a1ξ , (III-123)
III–40
Table III–1: The first few Hermite polynomials, Hn(x).H0 = 1H1 = 2xH2 = 4x2 − 2H3 = 8x3 − 12xH4 = 16x4 − 48x2 + 12H5 = 32x5 − 160x3 + 120x
and hence
ψ1(ξ) = A1 a1 ξ e−ξ2/2 (III-124)
confirming Eq. (III-80).
iii) For n = 2, j = 0 yields a2 = −2a0, and j = 2
gives a4 = 0, so
h2(ξ) = a0(1 − 2ξ2) (III-125)
and
ψ2(ξ) = A2 a0 (1 − 2ξ2) e−ξ2/2 , (III-126)
and so on.
l) hn(ξ) is thus a polynomial of degree n in ξ, involving even
powers only, if n is an even integer, and odd powers only,
if n is an odd integer =⇒ Hermite polynomials (des-
ignated by Hn(ξ)) share these same characteristics (see
Table III-1). Note that in Hermite polynomials, the
coefficient of the term with the maximum power
is arbitrarily set to 2n.
m) Normalization.
i) Finally, we need to figure out the functional form
of the normalization constant An. We need to con-
vert back to the x label (from ξ, see Eq. III-86). To
III–41
simplify matters, let
β =
√mω0
hso ξ = β x (III-127)
for the ground state.
ii) For the n = 0 ground state, normalization gives
1 =∫ ∞
−∞ψ∗ψ dx
=∫ ∞
−∞A2
0e−β2x2
dx
= A20
√√√√ π
β2
A0 =
β
2
π
1/4
. (III-128)
iii) We can continue this for some of the higher-order
wave functions (I leave it to the students to do this
on their own). We can compare these results to
derive the following recursion relationship for the
normalization constant:
An =1√2nn!
A0 =1√2nn!
β
2
π
1/4
. (III-129)
iv) With this, the normalized stationary states for
the simple harmonic oscillator are
ψn(ξ) =
β
2
π
1/41√2nn!
Hn(ξ)e−ξ2/2 , (III-130)
which is identical to Eq. (III-78).
v) Finally, we can write the complete wave equation
by including the time dependence (i.e., Eqs. III-10,
III–42
III-119, III-129, and III-130) as
Ψn(x, t) =
β
2
π
1/41√
2nn!
Hn(βx) e
−β2x2/2e−i(2n+1)ω0t/2 ,
(III-131)
where β =√mω0/h (see Eq. III-127).
Example III–4. In the ground state of the harmonic oscillator, what is
the probability (correct to 3 significant digits) of finding the particle outside
the classically allowed region? Hint: Look in a math table under “Normal
Distribution” or “Error Function.”
Solution:
For the ground state, the wave function is
ψ0 =
(mω
πh
)1/4e−ξ
2/2 ,
so the probability is
P = 2
√mω
πh
∫ ∞
x0
e−ξ2
dx = 2
√mω
πh
√√√√ h
mω
∫ ∞
ξ0e−ξ
2
dξ .
The classically allowed region extends out to: 12mω
2x20 = E0 = 1
2hω, or
x0 =√h/mω, so ξ0 = 1. Therefore,
P =2√π
∫ ∞
1e−ξ
2
dξ = 2[1 − F (√
2)] = 0.157 ,
where F (z) is the notation used in the CRC Tables.
Example III–5. In this problem we explore some of the more useful the-
orems (stated without proof) involving Hermite polynomials.
(a) The Rodrigues formula states that
Hn(ξ) = (−1)neξ2(d
dξ
)ne−ξ
2
.
Use it to derive H3 and H4.
III–43
(b) The following recursion relation gives you Hn+1 in terms of the 2 pre-ceding Hermite polynomials:
Hn+1(ξ) = 2ξHn(ξ)− 2nHn−1(ξ) .
Use it, together with your answer to (a), to obtain H5 and H6.
(c) If you differentiate an n-th order polynomial, you get a polynomial oforder (n− 1). For the Hermite polynomials, in fact:
dHn
dξ= 2nHn−1(ξ) .
Check this by differentiating H5 and H6.
(d) Hn(ξ) is the n-th z-derivative, at z = 0, of the generating functionexp(−z2 + 2zξ); or, to put it another way, it is the coefficient of zn/n!in the Taylor series expansion for the function:
e−z2+2zξ =
∞∑
n=0
zn
n!Hn(ξ) .
Use this to rederive H0, H1, and H2.
Solution (a):
d
dξ(e−ξ
2
) = −2ξe−ξ2
(d
dξ
)2
e−ξ2
=d
dξ(−2ξe−ξ
2
) = (−2 + 4ξ2)e−ξ2
(d
dξ
)3
e−ξ2
=d
dξ[(−2 + 4ξ2)e−ξ
2
] = [8ξ + (−2 + 4ξ2)(−2ξ)]e−ξ2
= (12ξ − 8ξ3)e−ξ2
(d
dξ
)4
e−ξ2
=d
dξ[(12ξ − 8ξ3)e−ξ
2
]
= [12 − 24ξ2 + (12ξ − 8ξ3)(−2ξ)]e−ξ2
= (12 − 48 ξ2 + 16 ξ4)e−ξ2
H3(ξ) = −eξ2(d
dξ
)3
e−ξ2
= −12ξ + 8ξ3
III–44
H4(ξ) = −eξ2(d
dξ
)4
e−ξ2
= 12 − 48 ξ2 + 16 ξ4 .
Solution (b):
H5(ξ) = 2ξH4 − 8H3 = 2ξ(12 − 48 ξ2 + 16 ξ4) − 8(−12ξ + 8ξ3)
= 120ξ − 160ξ3 + 32ξ5
H6(ξ) = 2ξH5 − 10H4
= 2ξ(120ξ − 160ξ3 + 32ξ5) − 10(12 − 48 ξ2 + 16 ξ4)
= −120 + 720ξ2 − 480ξ4 + 64ξ6 .
Solution (c):
dH5
dξ= 120 − 480ξ2 + 160ξ4 = 10(12 − 48ξ2 + 16ξ4)
= (2)(5)H4√
dH6
dξ= 1440ξ − 1920ξ3 + 384ξ5 = 12(120ξ − 160ξ3 + 32ξ5)
= (2)(6)H5 .√
Solution (d):d
dz(e−z
2+2zξ) = (−2z + 2ξ)e−z2+2zξ;
putting in z = 0 gives
H0(ξ) = 2ξ .
(d
dz
)2
(e−z2+2zξ) =
d
dz[(−2z + 2ξ)e−z
2+2zξ] = [−2 + (−2z + 2ξ)2]e−z2+2zξ;
putting in z = 0 gives
H1(ξ) = −2 + 4ξ2 .
(d
dz
)3
(e−z2+2zξ) =
d
dz{[−2 + (−2z + 2ξ)2]e−z
2+2zξ}
= {2(−2z + 2ξ)(−2) +
[−2 + (−2z + 2ξ)2](−2z + 2ξ)}e−z2+2zξ ;
III–45
putting in z = 0 gives
H2(ξ) = −8ξ + (−2 + 4ξ2)(2ξ) = −12ξ + 8ξ3 .
F. The Free Particle
1. A free particle is simply one where V (x) = 0 for all x. The
time-independent Schrodinger equation (TISE) becomes
− h2
2m
d2ψ
dx2 = Eψ , (III-132)
ord2ψ
dx2 = −k2ψ, where k ≡√
2mE
h. (III-133)
a) Instead of using sines and cosines, we will express the
solution in exponential notation:
ψ(x) = Aeikx +Be−ikx . (III-134)
b) Since there are no boundary conditions, k (hence E) can
take on any positive value =⇒ a continuum of possible
energies.
c) Taking on the standard time dependence gives
Ψ(x, t) = Aeik(x−hk2mt) +Be−ik(x+
hk2mt). (III-135)
2. Any function of x and t that depends upon these variables in the
special combination (x ± vt) (for some constant v) represents a
wave of fixed profile, traveling in the ∓x-direction, at speed v.
a) As such, the first term in Eq. (III-135) represents a wave
traveling to the right.
b) The second term, a wave traveling to the left.
III–46
c) Since the only difference between the 2 terms in Eq. (III-
135) is the sign of k, we can write
Ψk(x, t) = Aeik(x−hk2m t) , (III-136)
and let k run negative (as well as positive) to cover the
case of waves traveling to the left:
k ≡ ±√
2mE
h, with
k > 0 =⇒ traveling to the right,
k < 0 =⇒ traveling to the left.(III-137)
3. The speed of the wave is
vquantum =h|k|2m
=
√√√√ E
2m. (III-138)
a) On the other hand, the classical speed of a free particle
with energy E is given by E = (1/2)mv2 (pure kinetic
since V = 0), so
vclassical =
√√√√2E
m= 2vquantum . (III-139)
b) The quantum mechanical wave function travels a half the
speed of the particle it is suppose to represent!
c) This wave function is not normalizable! =⇒ in other
words, there is no such thing as a free particle with a
definite energy.
4. The general solution to the TISE is still a linear combination
of separable solutions (only this time, it’s an integral over the
continuous variable k, instead of a sum over the discrete index
n):
Ψ(x, t) =1√2π
∫ +∞
−∞φ(k)ei(kx−
hk2
2m t) dk . (III-140)
III–47
a) Now this wave function can be normalized [for appropri-
ate φ(k)], but it necessarily carries a range of k’s, and
hence a range of energies and speeds.
b) As such, this is called a wave packet.
c) The initial wave function is thus
Ψ(x, 0) =1√2π
∫ +∞
−∞φ(k)eikx dk . (III-141)
5. This is a classic problem of Fourier analysis (see §III.C); the
answer is provided by Plancherel’s theorem:
f(x) =1√2π
∫ +∞
−∞F (k)eikx dk ⇐⇒ F (k) =
1√2π
∫ +∞
−∞f(x)e−ikx dx .
(III-142)
a) F (k) is called the Fourier transform of f(x).
b) f(x) is called the inverse Fourier transform of F (k).
c) The necessary and sufficient condition on f(x) is that∫+∞−∞ |f(x)|2 dx be finite.
6. The solution to the generic, free-particle quantum problem is
thus
φ(k) =1√2π
∫ +∞
−∞Ψ(x, 0)e−ikx dx . (III-143)
a) A wave packet is a sinusoidal function whose amplitude
is modulated by φ (Figure III-3); it consists of ripples
contained within an envelope.
b) What corresponds to the particle velocity is not the speed
of the individual ripples (the so-called phase velocity),
but rather the speed of the envelope (the group velocity)
III–48
x
Ψvg
vp
Figure III–3: A wave packet. The envelope travels at the group velocity; the ripples travel at thephase velocity.
— which, depending upon the nature of the waves, can
be greater than, less than, or equal to the velocity of the
ripples that go to make it up.
c) For a free particle in quantum mechanics, the group veloc-
ity is twice the phase velocity as shown in Eq. (III-139).
7. But what are the respective group and phase velocities of the
wave packet with the general form
Ψ(x, t) =1√2π
∫ +∞
−∞φ(k) ei(kx−ωt) dk ? (III-144)
a) Here the dispersion relation, that is, the formula for ω
as a function of k, is ω = (hk2/2m).
b) The amplitude function φ(k) insures that the integrand in
Eq. (III-144) is negligible except in the vicinity of k◦ (the
position in k-space of the peak value of φ(k)). As such,
III–49
we can Taylor expand:
ω(k) ∼= ω◦ + ω′◦(k − k◦) , (III-145)
where ω′◦ is the derivative of ω with respect to k at point
k◦.
c) Let s ≡ k − k◦, then
Ψ(x, t) ∼=1√2π
∫ +∞
−∞φ(k◦ + s) ei[(k◦+s)x−(ω◦+ω′
◦s)t] ds .
(III-146)
d) At t = 0,
Ψ(x, 0) =1√2π
∫ +∞
−∞φ(k◦ + s) ei(k◦+s)x ds (III-147)
and at later times
Ψ(x, t) ∼=1√2πei(−ω◦t+k◦ω′
◦t)∫ +∞
−∞φ(k◦+s) e
i(k◦+s)(x−ω′◦t) ds .
(III-148)
e) Except for the shift from x to (x− ω′◦t), this last integral
is the same as the one for Ψ(x, 0), thus
Ψ(x, t) ∼= ei(−ω◦t+k◦ω′◦t) Ψ(x− ω′
◦t, 0) . (III-149)
f) Apart from the phase factor in front (which won’t affect
|Ψ|2 in any event), the wave packet moves along at speed
vgroup =dω
dk
∣∣∣∣∣k◦
=hk
m, (III-150)
which is contrasted with the ordinary phase velocity
vphase =ω
k=hk
2m. (III-151)
g) Hence, the group velocity is twice as great as the phase
velocity:
vclassical = vgroup = 2vphase. (III-152)
III–50
Example III–6. A free particle has the initial wave function
Ψ(x, 0) = Ae−ax2
,
where A and a are constants (a is real and positive).
(a) Normalize Ψ(x, 0).
(b) Show that Ψ(x, t) can be written in the form
Ψ(x, t) =
(2a
π
)1/4 e−ax2/[1+(2ihat/m)]
√1 + (2ihat/m)
.
Hint : Integrals of the form∫ +∞
−∞e−(ax2+bx) dx
can be handled by “completing the square.” Let y ≡√a [x + (b/2a)],
and note that (ax2 + bx) ≡ y2 − (b2/4a).
(c) Find |Ψ(x, t)|2. Express your answer in terms of the quantity w ≡√a/[1 + (2hat/m)2. Sketch |Ψ|2 (as a function of x) at t = 0, and
again for some very large t. Qualitatively, what happens to |Ψ|2 as timegoes on?
(d) Find 〈x〉, 〈p〉, 〈x2〉, 〈p2〉, σx, and σp. Hint : You will need to show that〈p2〉 = ah2, but it may take some algebra to reduce it to this simpleform.
(e) Does the uncertainty principle hold? At what time t does the systemcome closest to the uncertainty limit?
Solution (a):
1 = |A|2∫ +∞
∞e−2ax2
dx = |A|2√π
2aor
A =
(2a
π
)1/4.
III–51
Solution (b):
Using Eq. (III-143), we can write
φ(k) =1√2πA∫ +∞
−∞e−ax
2
e−ikx dx =1√2πA∫ +∞
−∞e−(ax2+ikx) dx .
Using the substitution variables in the question (i.e., y from the last page
and let b = ik), the integral above as the functional form
∫ +∞
−∞e−(ax2+bx) dx =
∫ +∞
−∞e−y
2+(b2/4a) 1√ady
=1√aeb
2/4a∫ +∞
−∞e−y
2
dy =
√π
aeb
2/4a . (A)
So,
φ(k) =1√2π
(2a
π
)1/4√π
ae−k
2/4a
=1
(2πa)1/4 e−k2/4a .
Now using φ(k) in Eq. (III-140), we can solve for Ψ,
Ψ(x, t) =1√2π
1
(2πa)1/4
∫ +∞
−∞e−k
2/4a ei(kx−hk2t/2m) dk
=1√2π
1
(2πa)1/4
∫ +∞
−∞e−
k2
4a+ixk− ihk2t
2m dk
=1√2π
1
(2πa)1/4
∫ +∞
−∞e−[k2( 1
4a+ iht2m )−ixk] dk .
At this point, let
β =
(1
4a+iht
2m
)and γ = −ix .
Then making use of Eq. (A) above, the integral in the equation above
becomes∫ +∞
−∞e−[k2( 1
4a+ iht2m )−ixk] dk =
∫ +∞
−∞e−(βk2+γk) dk
III–52
=
√√√√π
βeγ
2/4β
=
√π
√14a + iht
2m
e−x2/4( 1
4a+ iht
2m) .
Substituting this integral solution into the equation for Ψ above, we get
Ψ(x, t) =1√
2π(2πa)1/4
√π
√14a + iht
2m
e−x2/4( 1
4a+ iht
2m)
=
(2a
π
)1/4 e−ax2/(1+2ihat/m)
√1 + 2ihat/m
.
Solution (c): Let θ = 2hat/m, then
|Ψ|2 =
√√√√2a
π
e−ax2/(1+iθ) e−ax
2/(1−iθ)√
(1 + iθ)(1 − iθ).
The exponent is
−ax2
(1 + iθ)−
ax2
(1 − iθ)= −ax2 (1 − iθ + 1 + iθ)
(1 + iθ)(1 − iθ)=
−2ax2
1 + θ2 ,
and
|Ψ|2 =
√√√√2a
π
e−2ax2/(1+θ2)√
(1 + θ2).
Or, with w ≡√
a1+θ2 , we get
|Ψ|2 =
√√√√2
πw e−2w2x2
.
x x
|Ψ |2 |Ψ |2
t = 0 t >> 0
As t increases, the graph of |Ψ|2 flattens out and broadens, as can be
III–53
seen by examining w (i.e., proportional to the peak of the probability) as
t � 0,
w =
√a
1 + θ2 =
√√√√ a
1 + (2hat/m)2
≈√√√√ a
(2hat/m)2 =
√√√√ m2
4h2at2=
m
2ht√a.
As t gets very large, w goes to zero.
Solution (d):
〈x〉 =∫ +∞
−∞x |Ψ|2 dx =
∫ +∞
−∞x
√√√√2
πw e−2w2x2
dx = 0 (odd integrand)
〈p〉 = md〈x〉dt
= 0
〈x2〉 =
√√√√2
πw∫ +∞
−∞x2e−2w2x2
dx =
√√√√2
πw
1
4w2
√π
2w2 =1
4w2
〈p2〉 = −h2∫ +∞
−∞Ψ∗d
2Ψ
dx2 dx .
Write Ψ = Be−bx2, where B ≡
(2aπ
)1/4 1√1+iθ and b ≡ a
1+iθ . Then
d2Ψ
dx2 = Bd
dx(−2bxe−bx
2
) = −2bB(1 − 2bx2) e−bx2
Ψ∗ d2Ψ
dx2 = −2b|B|2(1 − 2bx2) e−(b+b∗)x2
;
b+ b∗ =a
1 + iθ+
a
1 − iθ=
2a
1 + θ2 = 2w2 .
|B|2 =
√√√√2a
π
1√1 + θ2
=
√√√√2
πw .
So
Ψ∗ d2Ψ
dx2 = −2b
√√√√2
πw (1 − 2bx2) e−2w2x2
〈p2〉 = 2bh2
√√√√2
πw∫ +∞
−∞(1 − 2bx2) e−2w2x2
dx
III–54
= 2bh2
√√√√2
πw
{√π
2w2 − 2b1
4w2
√π
2w2
}
= 2bh2(1 − b
2w2
).
But
1 − b
2w2 = 1 −(
a
1 + iθ
)1 + θ2
2a
= 1 − 1 − iθ
2=
1 + iθ
2=
a
2b,
so
〈p2〉 = 2bh2 a
2b= h2a .
The uncertainty for x and p is thus
σx =√〈x2〉 − 〈x〉2 =
1
2w;
and
σp =√〈p2〉 − 〈p〉2 = h
√a .
Solution (e):
σx σp =1
2wh√a =
h
2
√1 + θ2
=h
2
√√√√√1 +
(2hat
m
)2
≥ h
2.
√
The closest that this system gets to the Heisenberg uncertainty limit oc-
curs at t = 0 , at which time it is right at the uncertainty limit.
G. The Delta-Function Potential.
1. In classical mechanics, there are two distinct states associated
with a particle in a potential field:
III–55
a) Bound state: The particle is stuck in the potential well
and oscillates back and forth between the turning points
(a, b) where E = V (a) = V (b), E > V (x), a < x < b, and
E < V (x), for x < a, and x > b (e.g., harmonic oscillator).
b) Scattering state: E > V (x) on one side of a given point
and E < V (x) on the other side. Then, a particle ap-
proaches the turning point where E = V from infinity,
then bounces and returns to infinity.
c) Some potential fields permit both of kinds states.
2. We have just derived the quantum mechanics analogies to these
two states:
a) A linear combination (i.e., summation) of normalizable,
stationary states of specific energy En (i.e., infinite po-
tential well, harmonic oscillators) =⇒ bound states.
b) An integral over non-normalizable, continuous values of
k-states (i.e., free particles) =⇒ scattering states.
c) Because of the quantum effect of tunneling (see below),
allows a bound particle to “leak” through a finite potential
barrier, the type of state a particle is actually in is only a
function of the potential at infinity:E < V (−∞) and V (+∞) =⇒ bound state
E > V (−∞) or V (+∞) =⇒ scattering state(III-153)
d) Since most potentials in physics go to zero at infinity, the
criterion is simplified even further:E < 0 =⇒ bound state
E > 0 =⇒ scattering state(III-154)
III–56
e) Because the infinite potential well and harmonic oscillator
potentials go to infinity as x → ±∞, they admit only
bound states.
f) Because the free particle potential is zero everywhere, it
allows only scattering states.
3. The Dirac delta function, δ(x), gives rise to both kinds of
states and is informally defined as
δ(x) =
0, if x 6= 0
∞ if x = 0
, with
∫ +∞
−∞δ(x) dx = 1. (III-155)
It is an infinitely high, infinitesimally narrow spike at the origin,
whose area is 1.
a) Technically, it is not a function at all, since it is not finite
at x = 0 (mathematicians call it a generalized function,
or distribution).
b) It can be thought of as the limit of a sequence of functions,
such as rectangles (or triangles) of ever-increasing height
and ever-decreasing width.
c) It is an extremely useful construct in theoretical physics
=⇒ in electrodynamics, the charge density of a point
charge is a delta function.
d) Notice that δ(x − a) would be a spike of area 1 at the
point a.
e) The delta function interacts with other functions as fol-
lows:
f(x) δ(x − a) = f(a) δ(x− a), (III-156)
because the product is zero anyway except at point a. In
III–57
particular,∫ +∞
−∞f(x) δ(x − a) dx = f(a)
∫ +∞
−∞δ(x − a) dx = f(a).
(III-157)
Under the integral sign, the delta function serves to pick
out the value of f(x) at the point a.
4. Let’s consider the potential of the form
V (x) = −αδ(x), (III-158)
where α is some constant. The TISE then is
−h2
2m
d2ψ
dx2 − αδ(x)ψ = Eψ. (III-159)
a) This potential yields both bound states (E < 0) and scat-
tering states (E > 0).
b) In the region x < 0, V (x) = 0, so
d2ψ
dx2 = −2mE
h2 ψ = κ2ψ, (III-160)
where
κ ≡√−2mE
h(E < 0). (III-161)
c) The solution to Eq. (III-160) is
ψ(x) = Ae−κx +Beκx, (III-162)
but the first term blows up as x → −∞, so A must be
zero:
ψ(x) = Beκx (x < 0). (III-163)
d) For x > 0, V (x) = 0 once again and from the same logic
above (except now ψ(x) blows up as x → +∞), we get
ψ(x) = Fe−κx (x > 0). (III-164)
III–58
x
ψ(x)
B
Figure III–4: The wave function for a δ function potential.
e) We now use the standard boundary conditions for ψ:
1. ψ is always continuous, and
2. dψ/dx is continuous except at pointswhere the potential is infinite.
(III-165)
f) In this case, the first boundary condition tells us that
B = F , so
ψ(x) =
Beκx (x ≤ 0),
Be−κx (x ≥ 0).(III-166)
The 2nd boundary condition tells us nothing in this case.
g) This solution for the wave function has a discontinuity in
the derivative at x = 0 as shown in Figure III-4. The δ
function must determine the discontinuity in the deriva-
tive of ψ at x = 0.
i) First, integrate the TISE from −ε to ε, then take
the limit as ε→ 0:
−h2
2m
∫ ε
−ε
d2ψ
dx2 dx+∫ ε
−εV (x)ψ(x) dx = E
∫ ε
−εψ(x) dx .
(III-167)
III–59
The first term in this equation can be rewritten as∫ ε
−ε
d2ψ
dx2 dx =∫ ε
−ε
d
dx
(dψ
dx
)dx =
∫ ε−εd
(dψ
dx
)
=dψ
dx
∣∣∣∣∣
ε
−ε= ∆
(dψ
dx
). (III-168)
The integral on the RHS of the TISE equation
above is zero,
limε→0
∫ ε−εψ(x) dx = 0 , (III-169)
since this integral is the area of a sliver with van-
ishing width and finite height. As such under these
circumstances, Eq. (III-167) becomes
∆
(dψ
dx
)=
2m
h2 limε→0
∫ ε−εV (x)ψ(x) dx . (III-170)
ii) The potential energy integral on the RHS of Eq.
(III-170) can be determined with the help of Eqs.
(III-157) and (III-158):∫ ε
−εV (x)ψ(x) dx =
∫ ε
−ε[−αδ(x)] ψ(x) dx
= −α∫ ε−εδ(x)ψ(x) dx
= −α ψ(0) , (III-171)
which is independent of ε, hence taking the limit
will not change the form of this solution.
iii) We next need to take the derivative of Eq. (III-
166):
dψ
dx=
Bκeκx (x < 0),
−Bκe−κx (x > 0).(III-172)
So, if we let these derivatives get infinitesimally
close to x = 0, we see that
(dψ/dx)|+ = −Bκ(dψ/dx)|− = +Bκ ,
(III-173)
III–60
hence
∆
(dψ
dx
)=
(dψ
dx
)∣∣∣∣∣+−(dψ
dx
)∣∣∣∣∣−= −2Bκ . (III-174)
iv) Meanwhile ψ(0) = B, so using Eqs. (III-171) and
(III-174) in Eq. (III-170) gives
−2Bκ = −2m
h2 αB
or
κ =mα
h2 , (III-175)
which gives an energy (via Eq. III-161) of
E = −h2κ2
2m= −mα
2
2h2 . (III-176)
h) Finally, normalizing ψ gives:
∫ +∞
−∞|ψ(x)|2 dx = 2|B|2
∫ ∞
0e−2κx dx =
|B|2
κ= 1 ,
(III-177)
or
B =√κ =
√mα
h. (III-178)
i) Hence, the delta-function well, regardless of its strength
α, has exactly one bound state:
ψ(x) =
√mα
he−mα|x|/h
2
; E = −mα2
2h2 . (III-179)
5. For scattering states with E > 0 and x < 0, the TISE reads
d2ψ
dx2 = −2mE
h2 ψ = −k2ψ , (III-180)
where
k ≡√
2mE
h(E > 0). (III-181)
III–61
a) The solution to Eq. (III-180) is
ψ(x) = Aeikx +Be−ikx (x < 0), (III-182)
which is a trigonometric function =⇒ neither term blows
up.
b) Similarly for x > 0,
ψ(x) = Feikx +Ge−ikx (x > 0). (III-183)
c) Continuity of ψ(x) at x = 0 requires that
F +G = A+B . (III-184)
d) The derivatives are
dψ
dx= ik
(Feikx − Ge−ikx
), for (x > 0)
dψ
dx= ik
(Aeikx − Be−ikx
), for (x < 0). (III-185)
Let x → 0 on either side of zero, then
dψ
dx|+ = ik(F − G)
dψ
dx|− = ik(A− B) , (III-186)
and hence ∆(dψ/dx) = ik(F − G− A+B).
e) Meanwhile, ψ(0) = (A+B), so the second boundary con-
dition says
ik(F − G− A+B) = −2mα
h2 (A+B) , (III-187)
or more compactly
F −G = A(1 + 2iβ) − B(1 − 2iβ), where β ≡ mα
h2k.
(III-188)
III–62
f) Having imposed the boundary conditions, we are left with
2 equations (i.e., Eqs. III-184 & III-188) with 4 unknowns
— 5 if you count k.
i) Normalization won’t help since this isn’t a normal-
izable state.
ii) Remember that the term eikx corresponds to par-
ticles propagating to the right in Eqs. (III-182 &
III-183) and e−ikx to particles traveling to the left.
iii) It follows that A corresponds to the amplitude of
a wave coming from the left, andB is the amplitude
of the wave returning to the left in Eq. (III-182).
iv) Likewise, F is the amplitude of the wave traveling
off to the right, and G is the amplitude of the wave
coming in from the right in Eq. (III-183).
g) Let’s insist that in our scattering experiment, we only fire
particles in from the left, then
G = 0 (for scattering from the left). (III-189)
i) A is then the amplitude of the incident wave.
ii) B is the amplitude of the reflected wave.
iii) F is the amplitude of the transmitted wave.
h) Solving Eqs. (III-184) & (III-188) for B and F we get
B =iβ
1 − iβA , F =
1
1 − iβA . (III-190)
i) The probability of finding the particle at a specified loca-
tion is given by |Ψ|2, so the relative probability that an
III–63
incident particle will be reflected back is
R ≡ |B|2
|A|2=
β2
1 + β2 , (III-191)
where R is called the reflection coefficient.
j) The probability of transmission is given by the transmis-
sion coefficient:
T ≡ |F |2
|A|2=
1
1 + β2 , (III-192)
and, as you can see, the total probability of finding a
particle somewhere is then T +R = 1.
k) In terms of energy
R =1
1 + (2h2E/mα2), T =
1
1 + (mα2/2h2E).
(III-193)
The higher the energy, the greater the probability of trans-
mission.
l) This is not the end of the story, however, since this po-
tential gives a wave function that is not normalizable, so
they don’t represent possible particle states.
i) The resolution to this problem is to form normal-
izable linear combinations of the stationary states
(as in the case of the free particle).
ii) However, the math is extremely complicated and
it will not be reproduced here. It is best solved
numerically with a computer.
iii) Since it is impossible to create a normalizable free
particle wave function without involving a range of
III–64
energies, R and T should be interpreted as the ap-
proximate reflection and transmission probabilities
for particles in a narrow energy range about E.
m) Note that unlike classical mechanics, having E < Vmax
will result in T > 0 =⇒ in quantum mechanics some of
the particles get past the barrier. This phenomenon is
called tunneling.
H. The Finite Square Well.
1. As was introduced for the “free particle” (§III.F), the wave func-
tion for traveling waves can always be described by
ψ(x) = Aeiαx +B e−iαx .
a) Here, the A-coefficient term corresponds to a wave trav-
eling to the right.
b) The B-coefficient term corresponds to a wave traveling to
the left.
2. Meanwhile, a standing wave is characterized by the following
wave function:
ψ(x) = C sin(αx) +D cos(αx) .
Here, the sine and the cosine terms do not have any specific
directional information associated with them, which is easy to
understand since we are talking about a standing wave.
Exercise: Mathematically prove the statement just made about
the directionality about the sine and cosine terms in the equation
above. Hint: Note that
e±iα = cosα ± i sinα ,
sinα =ei α − e−i α
2i, and
III–65
xa-a
E
-Vo
1 2 3
4 5 6
Scattering StatesE > 0
Bound StatesE < 0
Figure III–5: The finite square well potential (Eq. III-196).
cosα =ei α + e−i α
2.
3. With this information in mind, consider the finite square well:
V (x) =
−V◦, for − a < x < a,
0, for |x| > a,(III-194)
where V◦ is a (positive) constant as shown in Figure III-5.
a) Bounds states will occur when E < 0 (as indicated by
Regions 1, 2, and 3 in Figure III-5).
b) Scattering states will occur when E > 0 (as indicated
by Regions 4, 5, and 6 in Figure III-5).
4. Bound States (E < 0) of the finite square well.
a) Region 1: For bound states in the region x < −a,V (x) = 0 and the TISE becomes
−h2
2m
d2ψ
dx2 = Eψ, ord2ψ
dx2 = κ2ψ , (III-195)
III–66
where
κ ≡√−2mE
h(III-196)
is real and positive. The general solution is identical to
Eq. (III-149) but, as before, the first term blows up (as
x → −∞), so A = 0 and
ψ(x) = Beκx, for (x < −a). (III-197)
b) Region 2: In the region −a < x < a, V (x) = −V◦ and
the TISE becomes
− h2
2m
d2ψ
dx2 − V◦ψ = Eψ , ord2ψ
dx2 = −`2ψ , (III-198)
where
` ≡√
2m(E + V◦)
h, (III-199)
hence, E + V◦ is the energy that the wave is above the
minimum potential.
i) Although E is negative, for a bound state, it must
be greater than −V◦, otherwise, the bound-state
wave function would not be normalizable =⇒ as
such, ` is also real and positive.
ii) The general solution is
ψ(x) = C sin(`x) +D cos(`x) , for (−a < x < a),
(III-200)
where C and D are arbitrary constants.
Exercise: Prove Eq. (III-200) by taking its double derivative
and plugging ψ and d2ψ/dx2 back into Eq. (III-198).
c) Region 3: For bound states in the region x > a, the
potential is again zero and the general solution has the
form of ψ(x) = Fe−κx +Geκx, but the second term blows
III–67
up (as x→ ∞), so
ψ(x) = Fe−κx, for (x > a). (III-201)
d) The final solution for the bound states (E < 0) of
the finite square well: Imposing the standard boundary
conditions: ψ and dψ/dx continuous at −a and a. Also,
realizing that the potential is an even function, the wave
function will either be odd or even, and we only need to
look at the boundary conditions only on one side (say, at
+a) — the other side is automatic since ψ(−x) = ±ψ(x).
We will do the even solutions here and do the odd solu-
tions in the next exercise (note that the cosine is an even
function, and the sine, an odd function. The even solution
for E < 0 is
ψ(x) =
Fe−κx, for (x > a),
D cos(`x), for (0 < x < a),
ψ(−x), for (x < 0).
(III-202)
i) The continuity of ψ(x), at x = a, says
Fe−κa = D cos(`a) . (III-203)
ii) The continuity of dψ/dx says
−κFe−κa = −`D sin(`a) . (III-204)
iii) Dividing Eq. (III-204) by (III-203) gives
κ = ` tan(`a) . (III-205)
e) Since ` and κ are both functions of E, we can determine
a formula for E by letting
z ≡ `a, and z◦ ≡a
h
√2mV◦ . (III-206)
III–68
zπ/2 π 3π/2 2π 5π/2
tan (z)
√(zo/z)2 - 1_______
Figure III–6: Graphical solution to Eq. (III-209) for z◦ = 8 (even states). Black dots indicateintersection points.
Then according to Eqs. (III-196) and (III-199), (κ2+`2) =
2mV◦/h2, so κa =
√z2◦ − z2 and Eq. (III-205) becomes
tan z =√
(z◦/z)2 − 1 . (III-207)
i) This is a transcendental equation for z (and hence
for E) as a function of z◦ (which is a measure of
the size of the well).
ii) It can be solved numerically by plotting tan z
and√
(z◦/z)2 − 1 on the same grid and looking for
points of intersection (see Figure III-6).
f) Two limiting cases are of special interest:
i) Wide, deep well: If z◦ is very large, the inter-
sections occur just slightly below z◦ = nπ/2, with
n odd; it follows that
En + V◦ ∼=n2π2h2
2m(2a)2 . (III-208)
III–69
• Here, (En + V◦) is the energy above the bottom
of the well.
• On the right, we have the solution for an infinite
square well energies of width 2a, or rather, halfof them, since n is odd.
• So a finite square well goes to an infinite squarewell as V◦ → ∞. However, for any finite V◦ there
are only a finite number of bound states.
ii) Shallow narrow well: As z◦ decreases, there
are fewer and fewer bound states, until finally (for
z◦ < π/2, where the lowest odd state disappears)
only one remains.
• Note that there is always, at least, one bound
state despite how weak the well becomes.
5. Scattering States (E > 0) of the finite square well.
a) Region 4: For the scattering (E > 0) states, to the left,
where V (x) = 0, we have
ψ(x) = Aeikx +Be−ikx , for (x < −a), (III-209)
where
k ≡√
2mE
h. (III-210)
Exercise: Prove Eq. (III-209) by taking its double deriva-
tive and plugging ψ and d2ψ/dx2 back into Eq. (III-198).
(Remember, V = 0 here.)
b) Region 5: Inside the well, where V (x) = −V◦,
ψ(x) = C sin(`x) +D cos(`x) , for (−a < x < a),
(III-211)
III–70
where
` ≡√
2m(E + V◦)
h. (III-212)
Exercise: Prove Eq. (III-211) by taking its double derivative
and plugging ψ and d2ψ/dx2 back into Eq. (III-198). Explain
why the C andD terms do not correspond to waves traveling
in any specific direction.
c) Region 6: To the right, assuming no incoming wave from
that region, we have
ψ(x) = Feikx . (III-213)
d) As described at the beginning of this subsection, in Equa-
tions (III-209) and (III-213), A is the incident amplitude,
B is the reflected amplitude, and F is the transmitted
amplitude of a wave traveling in the vicinity of this finite
square well.
e) There are 4 boundary conditions for these three equations
(Eqs. III-209, III-211, and III-213):
i) Continuity of ψ(x) at −a gives
Ae−ika +Beika = −C sin(`a) +D cos(`a) .
(III-214)
ii) Continuity of dψ/dx at −a gives
ik[Ae−ika − Beika] = `[C cos(`a) +D sin(`a)] .
(III-215)
iii) Continuity of ψ(x) at +a gives
C sin(`a) +D cos(`a) = Feika . (III-216)
III–71
iv) Continuity of dψ/dx at +a gives
`[C cos(`a)−D sin(`a)] = ikFeika . (III-217)
f) We can use two of these to eliminate C and D, and solve
the remaining two for B and F :
B = isin(2`a)
2k`(`2 − k2)F, (III-218)
F =e−2ikaA
cos(2`a) − i[sin(2`a)/2k`] (k2 − `2).(III-219)
g) The transmission coefficient (T = |F |2/|A|2) is then
T−1 = 1 +V 2◦
4E(E + V◦)sin2
(2a
h
√2m(E + V◦)
).
(III-220)
Exercise: Show how Eq. (III-220) results from Eq. (III-219).
h) Notice that T = 1 (i.e., the well becomes transparent)
whenever the argument of the sine is zero, or
2a
h
√2m(E + V◦) = nπ , (III-221)
where n is any integer.
i) The energies for perfect transmission are given by
En + V◦ =n2π2h2
2m(2a)2 , (III-222)
which happens to be precisely the allowed energies for the
infinite square well.
III–72
Example III–7. Analyze the odd bound-state wave functions for the fi-
nite square well. Derive the transcendental equation for the allowed energies,
and solve it graphically. Examine the two limiting cases. Is there always at
least one odd bound state?
Solution:
In place of Eq. (III-202), we have
ψ(x) =
Fe−κx, for (x > a),D sin(`x), for (0 < x < a),
−ψ(−x), for (x < 0).
Continuity of ψ(x) gives:
Fe−κa = D sin(`a) ;
Continuity of dψ/dx gives:
−κFe−κa = D` cos(`a) .
Dividing these two gives:
−κ = ` cot(`a), or
−κa = `a cot(`a) =⇒√z2◦ − z2 = −z cot z ,
or
cot z =√
(z◦/z)2 − 1 .
III–73
Wide, deep well: Intersections are
at π, 2π, 3π, etc. Energies are the
same as Eq. (III-208), but now for
n even. This fills the rest of thestates for the infinite square well.
Shallow, narrow well: If z◦ < π/2,
there is no odd bound state. The
corresponding condition on V◦ is
V◦ <π2h2
8ma2 .zπ/2 π 3π/2 2π 5π/2
zo
cot (z)
√(zo/z)2 - 1_______
III–74