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PHYSICS 430
Lecture Notes on Quantum Mechanics
J. Greensite
Physics and Astronomy Department
San Francisco State University
Fall 2003
Copyright (C) 2003 J. GreensiteChapter 1
The Classical State
Inthefirstquarterofthiscentury,itwasdiscoveredthatthelawsofmotionformulated
byGalileo,Newton, Lagrange,Hamilton, Maxwell, andmanyothers, wereinadequate
to explain a wide range of phenomena involving electrons, atoms, and light. After a
greatdealofeffort,anewtheory(togetherwithanewlawofmotion)emergedin1924.
Thattheoryisknownasquantummechanics, anditisnowthebasicframeworkfor
understanding atomic, nuclear, and subnuclear physics, as well as condensed-matter
(or ”solid-state”) physics. The laws of motion (due to Galileo, Newton,...) which
preceded quantum theory are referred to as classical mechanics.
Although classical mechanics is now regarded as only an approximation to quan-
tum mechanics, it is still true that much of the structure of the quantum theory is
inherited from the classical theory that it replaced. So we begin with a lightning
review of classical mechanics, whose formulation begins (but does not end) with
Newton’s law F =ma.
1.1 Baseball, F = ma, and the Principle of Least
Action
Take a baseball and throw it straight up in the air. After a fraction of a second, or
perhaps a few seconds, the baseball will return to your hand. Denote the height of
the baseball, as a function of time, as x(t); this is the trajectory of the baseball. If
we make a plot ofx as a function oft, then any trajectory has the form of a parabola
(in a uniform gravitational field, neglecting air resistance), and there are an infinite
number of possible trajectories. Which one of these trajectories the baseball actually
follows is determined by the momentum of the baseball at the moment it leaves your
hand.
However, if we require that the baseball returns to your hand exactlyΔt seconds
after leaving your hand, then there isonly one trajectorythat the ball can follow. For
a baseball moving in a uniform gravitational field it is a simple exercise to determine
78 CHAPTER 1. THE CLASSICAL STATE
this trajectory exactly, but we would like to develop a method which can be applied
to a particle moving in any potential field V(x). So let us begin with Newton’s law
F =ma, which is actually a second-order differential equation
2
d x dV
m =− (1.1)
2
dt dx
It is useful to reexpress this second-order equation as a pair of first-order equations
dx p
=
dt m
dp dV
= − (1.2)
dt dx
where m is the mass and p is the momentum of the baseball. We want to find the
solution of these equations such that x(t ) = X and x(t +Δt) = X , where X
0 in 0 f in
and X are, respectively, the (initial) height of your hand when the baseball leaves
f
1
it, and the (final) height of your hand when you catch the ball.
With the advent of the computer, it is often easier to solve equations of motion
numerically, rather than struggle to find an analytic solution which may or may not
exist (particularly when the equations are non-linear). Although the object of this
section is not really to develop numerical methods for solving problems in baseball,
we will, for the moment, proceed as though it were. To make the problem suitable
for a computer, divide the time intervalΔt into N smaller time intervals of duration
=Δt/N, and denote, for n = 0,1,...,N,
t ≡t +n ,
n 0
x =x(t ) , p =p(t ) ,
n n n n
(1.3)
x =X , x =X
0 in N f
An approximation to a continuous trajectory x(t) is given by the set of points x
n
connected by straight lines, as shown in Fig. 1.1. We can likewise approximate
derivatives by finite differences, i.e.
"
dx x(t )−x(t ) x −x
n+1 n n+1 n
→ =
dt
t=tn
"
dp p(t )−p(t ) p −p
n+1 n n+1 n
→ =
dt
t=t
n
" " "
2
d x 1 dx dx
→ −
2
dt dt dt
t=t t=t t=t
n n n−1
)
1 (x −x ) (x −x )
n+1 n n n−1
→ − (1.4)
1
We will allow these positions to be different, in general, since you might move your hand to
another position while the ball is in flight.1.1. BASEBALL, F =MA, AND THE PRINCIPLE OF LEAST ACTION 9
and integrals by sums
+
N−1
t +Δt
0 ,
dtf(t)→ f(t ) (1.5)
n
t
0
n=0
where f(t) is any function of time. As we know from elementary calculus, the right
hand side of (1.4) and (1.5) equals the left hand side in the limit that → 0, which
is also known as the continuum limit.
Wecannowapproximate thelawsofmotion,byreplacingtime-derivatives in(1.2)
by the corresponding finite differences, and find
- .
p
n
x = x +
n+1 n
m
"
dV(x )
n
p = p − (1.6)
n+1 n
dx
n
These are iterative equations. Given position x and momentum p at time t = t ,
n
we can use (1.6) to find the position and momentum at time t = t . The finite
n+1
difference approximation of course introduces a slight error;x andp , computed
n+1 n+1
2
from x and p by (1.6) will differ from their exact values by an error of order .
n n
This error can be made negligible by taking sufficiently small.
It is then possible use the computer to find an approximation to the trajectory in
one of two ways: (i) the ”hit-or-miss” method; and (ii) the method of least action.
• The Hit-or-Miss Method
The equations of motion (1.2) require as input both an initial position, in this
case x =X , and an initial momentum p which is so far unspecified. The method
0 in 0
is to make a guess for the initial momentum p =P , and then use (1.2) to solve for
0 0
x ,p , x ,p , and so on, until x ,p . If x ≈ X , then stop; the set x is the
1 1 2 2 N N N f n
"
(approximate) trajectory. If not, make a different guess p =P , and solve again for
0
0
x ,p . By trial and error, one can eventually converge on an initial choice for p
n n 0
such that x ≈ X . For that choice of initial momentum, the corresponding set of
N f
points x , connected by straight-line segments, gives the approximate trajectory of
n
the baseball. This process is illustrated in Fig. 1.2.
• The Method of Least Action
Lets return to the 2nd-order form of Newton’s Laws, written in eq. (1.1). Again
using (1.4) to replace derivatives by finite differences, the equation F = ma at each
time t becomes
n
/ 0
m x −x x −x dV(x )
n+1 n n n−1 n
− =− (1.7)
dx
n
The equations have to be solved for n = 1,2,...,N−1, with x =X and x = X
0 in N f
kept fixed. Now notice that eq. (1.7) can be written as a total derivative
)
2 2
d 1 (x −x ) 1 (x −x )
n+1 n n n−1
m + m −V(x ) = 0 (1.8)
n
dx 2 2
n10 CHAPTER 1. THE CLASSICAL STATE
so thatF =ma can be interpreted as a condition that a certain function ofx should
n
be stationary. Let us therefore introduce a very important expression, crucial in both
classical and quantum physics, which is known as the ”action” of the trajectory.
The action is a function which depends on all the points x , n = 0,1,...,N of the
n
trajectory, and in this case it is
1 2
N−1 2
,
1 (x −x )
n+1 n
Sx≡ m −V(x ) (1.9)
i n
2
n=0
Then Newton’s Law F = ma can be restated as the condition that the action func-
tional Sx is stationary with respect to variation of any of the x (except for the
i i
endpoints x and x , which are held fixed). In other words
0 N
1 2
N−1
2
,
d d 1 (x −x )
n+1 n
Sx = m −V(x )
i n
dx dx 2
k k
n=0
)
2 2
d 1 (x −x ) 1 (x −x )
k+1 k k k−1
= m + m −V(x )
k
dx 2 2
k
= −ma(t )+F(t )
k k
= 0 for k = 1,2,...,N−1 (1.10)
This set ofconditionsis known asthePrincipleofLeast Action. It istheprinciple
that the action S is stationary at any trajectory x satisfying the equations of
n
motion F =ma, eq. (1.7), at every time t .
n
The procedure for solving for the trajectory of a baseball by computer is to pro-
gram the computer to find the set of points x which minimizes the quantity
n
"
2
,
∂S
Q = (1.11)
∂x
k
k
The minimum is obtained at Q = 0, where S is stationary. This set of points, joined
by straight-line segments, gives us the approximate trajectory of the baseball.
Problem: Do it on a computer by both methods.1.1. BASEBALL, F =MA, AND THE PRINCIPLE OF LEAST ACTION 11
2
Problem: Dyre’s Dilemma
In discussing the motion of the baseball, we have been ignoring a lot of details
about baseballs, such as the composition of the interior, the pattern of the stitching,
and the brand-name printed on the surface. Instead, the baseball has been treated
as though it were essentially a structureless point of mass m. It is necessary to
make idealizations like this in physics; the real world is otherwise too complicated to
describe. But sometimes an idealization misses something crucial. See if you can find
what goes wrong in the following argument, which tries to prove that a rolling wheel
(or, for that matter, a rolling baseball) can never come to rest through friction with
the ground.
”Proof”: As shown in Fig. 1.3, the forward momentum of a wheel in the positive
x-direction can only be eliminated by a force applied in the opposite direction. But
the only place this force could be applied by friction is the point where the wheel
touches the ground. And a force in the negative x-direction, applied at this point,
will have the effect of making the wheel spin faster Therefore, the wheel will never
come to rest due to friction. QED.
Is this reasoning correct? Can you solve Dyre’s Dilemma?
2
I owe this exercise to Dr. Jeppe Dyre, Roskilde University, Denmark.12 CHAPTER 1. THE CLASSICAL STATE
1.2 Euler-Lagrange and Hamilton’s Equations
In brief, the Euler-Lagrange equations are the second-order form of the equations of
motion(1.1),whileHamilton’sequationsarethefirst-orderform(1.2). Ineitherform,
the equations of motion can be regarded as a consequence of the Principle of Least
Action. We will now re-write those equations in a very general way, which can be
applied to any mechanical system, including those which are much more complicated
than a baseball.
We begin by writing
N−1
,
Sx = Lx ,x˙ (1.12)
i n n
n=0
where
1
2
Lx ,x˙ = mx˙ −V(x ) (1.13)
n n n
n
2
and where
x −x
n+1 n
x˙ ≡ (1.14)
n
Lx ,x˙ is known as the Lagrangian function. Then the principle of least action
n n
requires that, for each k, 1≤k≤N−1,
N−1
,
d d
0 = Sx = Lx ,x˙
i n n
dx dx
k k
n=0
N−1
,
∂ ∂Lx ,x˙ dx˙
n n n
= Lx ,x˙ + (1.15)
k k
∂x ∂x˙ dx
k n k
n=0
and, since
1
n =k−1
dx˙
n
1
= − n =k (1.16)
dx
k
0 otherwise
this becomes
)
∂ 1 ∂ ∂
Lx ,x˙ − Lx ,x˙ − Lx ,x˙ = 0 (1.17)
k k k k k−1 k−1
∂x ∂x˙ ∂x˙
k k k−1
Recalling that x =x(t ), this last equation can be written
n n
" ) " "
∂Lx,x˙ 1 ∂Lx,x˙ ∂Lx,x˙
− − = 0 (1.18)
∂x ∂x˙ ∂x˙
t=t t=t t=t −
n n n
This is the Euler-Lagrange equation for the baseball. It becomes simpler when we
take the → 0 limit (the ”continuum” limit). In that limit, we have
x −x dx
n+1 n
x˙ = → x˙(t) =
n
dt
+
N−1
t +Δt
, 0
S = Lx ,x˙ → S = dtLx(t),x˙(t) (1.19)
n n
t
0
n=11.2. EULER-LAGRANGE AND HAMILTON’S EQUATIONS 13
where the Lagrangian function for the baseball is
1
2
Lx(t),x˙(t) = mx˙ (t)−Vx(t) (1.20)
2
and the Euler-Lagrange equation, in the continuum limit, becomes
∂L d ∂L
− = 0 (1.21)
∂x(t) dt∂x˙(t)
For the Lagrangian of the baseball, eq. (1.20), the relevant partial derivatives are
∂L dVx(t)
= −
∂x(t) dx(t)
∂L
= mx˙(t) (1.22)
∂x˙(t)
which, when substituted into eq. (1.21) give
2
∂ x dV
m + = 0 (1.23)
2
∂t dx
This is simply Newton’s law F =ma, in the second-order form of eq. (1.1).
We now want to rewrite the Euler-Lagrange equation in first-order form. Of
course, we already know the answer, which is eq. (1.2), but let us ”forget” this
answer for a moment, in order to introduce a very general method. The reason the
Euler-Lagrange equation issecond-order in the time derivatives isthat ∂L/∂x˙ is first-
order in the time derivative. So let us define the momentum corresponding to the
coordinate x to be
∂L
p≡ (1.24)
∂x˙
This gives p as a function of x and x˙, but, alternatively, we can solve for x˙ as a
function of x and p, i.e.
x˙ =x˙(x,p) (1.25)
Next, we introduce the Hamiltonian function
Hp,x =px˙(x,p)−Lx,x˙(x,p) (1.26)
Since x˙ is a function of x and p, H is also a function of x and p.
The reason for introducing the Hamiltonian is that its first derivatives with re-
spect to x and p have a remarkable property; namely, on a trajectory satisfying the
Euler-Lagrange equations, the x and p derivatives of H are proportional to the time-
derivatives of p and x. To see this, first differentiate the Hamiltonian with respect to
p,
∂H ∂x˙(x,p) ∂L∂x˙(p,x)
= x˙ +p −
∂p ∂p ∂x˙ ∂p
= x˙ (1.27)14 CHAPTER 1. THE CLASSICAL STATE
where we have applied (1.24). Next, differentiating H with respect to x,
∂H ∂x˙(x,p) ∂L ∂L∂x˙(p,x)
= p − −
∂x ∂x ∂x ∂x˙ ∂x
∂L
= − (1.28)
∂x
Using the Euler-Lagrange equation (1.21) (and this is where the equations of motion
enter), we find
∂H d ∂L
= −
∂x dt∂x˙
dp
= − (1.29)
dt
Thus, with the help of the Hamiltonian function, we have rewritten the single 2nd
order Euler-Lagrange equation (1.21) as a pair of 1st order differential equations
dx ∂H
=
dt ∂p
dp ∂H
= − (1.30)
dt ∂x
which are known as Hamilton’s Equations.
Forabaseball, theLagrangianisgivenbyeq. (1.20),andthereforethemomentum
is
∂L
p = =mx˙ (1.31)
∂x˙
This is inverted to give
p
x˙ =x˙(p,x) = (1.32)
m
and the Hamiltonian is
H = px˙(x,p)−Lx,x˙(x,p)
4 5
p 1 p
2
= p − m( ) −V(x)
m 2 m
2
p
= +V(x) (1.33)
2m
Note that the Hamiltonian for the baseball is simply the kinetic energy plus the
potential energy; i.e. the Hamiltonian is an expression for the total energy of the
baseball. Substituting H into Hamilton’s equations, one finds
1 2
2
dx ∂ p p
= +V(x) =
dt ∂p 2m m
1 2
2
dp ∂ p dV
= − +V(x) =− (1.34)
dt ∂x 2m dx
which is simply the first-order form of Newton’s Law (1.2).1.3. CLASSICAL MECHANICS IN A NUTSHELL 15
1.3 Classical Mechanics in a Nutshell
AllthemachineryoftheLeastActionPrinciple, theLagrangianFunction,andHamil-
ton’sequations, isoverkillinthecaseofabaseball. Inthatcase,weknewtheequation
of motion from the beginning. But for more involved dynamical systems, involving,
say, wheels, springs, levers, and pendulums, all coupled together in some complicated
way, the equations of motion are often far from obvious, and what is needed is some
systematic way to derive them.
i
For any mechanical system, the generalized coordinates q are a set of vari-
ables needed to describe the configuration of the system at a given time. These could
be a set of cartesian coordinates of a number of different particles, or the angular
displacement of a pendulum, or the displacement of a spring from equilibrium, or all
of the above. The dynamics of the system, in terms of these coordinates, isgiven by a
i
Lagrangian function L, which depends on the generalized coordinates q and their
i
first time-derivatives q˙ . Normally, in non-relativistic mechanics, we first specify
1. The Lagrangian
i i
Lq q˙ = Kinetic Energy−Potential Energy (1.35)
One then defines
2. The Action
+
i i
S = dtLq ,q˙ (1.36)
From the Least Action Principle, following a method similar to the one we used for
the baseball (see Problem 4), we derive
3. The Euler-Lagrange Equations
∂L d ∂L
− = 0 (1.37)
i i
∂q dt∂q˙
These are the 2nd-order equations of motion. To go to 1st-order form, first define
4. The Generalized Momenta
∂L
p ≡ (1.38)
i
i
∂q˙
i
which can be inverted to give the time-derivatives q˙ of the generalized coordinates
in terms of the generalized coordinates and momenta
i i n
q˙ =q˙ q ,p (1.39)
n
Viewing q˙ as a function of p and q, one then defines16 CHAPTER 1. THE CLASSICAL STATE
5. The Hamiltonian
,
i n i i
Hq ,p≡ p q˙ −Lq ,q˙ (1.40)
i n
n
Usually the Hamiltonian has the form
Hp,q = Kinetic Energy+Potential Energy (1.41)
Finally, the equations of motion in 1st-order form are given by
6. Hamilton’s Equations
∂H
i
q˙ =
∂p
i
∂H
p˙ = − (1.42)
i
i
∂q1.3. CLASSICAL MECHANICS IN A NUTSHELL 17
Example: The Plane Pendulum
Our pendulum is a mass m at the end of a weightless rigid rod of length l, which
pivots in a plane around the point P. The ”generalized coordinate”, which specifies
the position of the pendulum at any given time, is the angle θ (see Fig. 1.4).
1. Lagrangian
1
2 2
˙
L = ml θ −(V −mglcos(θ)) (1.43)
0
2
where V is the gravitational potential at the height of point P, which the pendulum
0
reaches at θ = π/2. Since V is arbitrary, we will just set it to V = 0.
0 0
2. The Action
4 5
+
t
1
1
2 2
˙
S = dt ml θ +mglcos(θ) (1.44)
t 2
0
3. Euler-Lagrange Equations
We have
∂L
= −mglsin(θ)
∂θ
∂L
2
˙
= ml θ (1.45)
˙
∂θ
and therefore
2
¨
ml θ+mglsin(θ) = 0 (1.46)
is the Euler-Lagrange form of the equations of motion.
4. The Generalized Momentum
∂L
2
˙
p = =ml θ (1.47)
˙
∂θ
5. The Hamiltonian
Insert
p
˙
θ = (1.48)
2
ml
into
4 5
1
2 2
˙ ˙
H =pθ− ml θ +mglcos(θ) (1.49)
2
to get
2
1 p
H = −mglcos(θ) (1.50)
2
2ml18 CHAPTER 1. THE CLASSICAL STATE
6. Hamilton’s equations
∂H p
˙
θ = =
2
∂p ml
∂H
p˙ = − =−mglsin(θ) (1.51)
∂θ
which are easily seen to be equivalent to the Euler-Lagrange equations.
Problem - Two pointlike particles moving in three dimensions have masses m and
1
m respectively, and interact via a potential V(x% −x% ). Find Hamilton’s equations
2 1 2
of motion for the particles.
Problem - Suppose, instead of a rigid rod, the mass of the plane pendulum is
connected to point P by a weightless spring. The potential energy of the spring is
1
2
k(l−l ) , where l is the length of the spring, and l is its length when not displaced
0 0
2
byanexternal force. Choosingl andθ asthe generalized coordinates, find Hamilton’s
equations.
1.4 The Classical State
Prediction is rather important in physics, since the only reliable test of a scientific
theory is the ability, given the state of affairs at present, to predict the future.
Stated rather abstractly, the process of prediction works as follows: By a slight
disturbance known as a measurement, an object is assigned a mathematical repre-
sentation which we will callitsphysical state. The laws ofmotion aremathematical
rules by which, given a physical state at a particular time, one can deduce the phys-
ical state of the object at some later time. The later physical state is the prediction,
which can be checked by a subsequent measurement of the object (see Fig. 1.5).
From the discussion so far, its easy to see that what is meant in classical physics
by the ”physical state” of a system is simply its set of generalized coordinates and
a
the generalized momenta q ,p . These are supposed to be obtained, at some time
a
t , by the measurement process. Given the physical state at some time t, the state
0
at t+ is obtained by the rule:
"
∂H
a a
q (t+) =q (t)+
∂p
a
t
"
∂H
p (t+) =p (t)− (1.52)
a a
a
∂q
t
In this way, the physical state at any later time can be obtained (in principle) to an
arbitrary degree of accuracy, by making the time-step sufficiently small (or else, if1.4. THE CLASSICAL STATE 19
a
possible, by solving the equations of motion exactly). Note that the coordinates q
alone are not enough to specify the physical state, because they are not sufficient to
predict the future. Information about the momenta p is also required.
a
a
The space of all possible q ,p is known as phase space. For a single particle
a
moving in three dimensions, there are three components of position and three compo-
nentsofmomentum,sothe”physicalstate”isspecifiedby6numbers(x,y,z,p ,p ,p ),
x y z
which can be viewed as a point in 6-dimensional phase space. Likewise, the physical
state of a system of N particles consists of 3 coordinates for each particle (3N co-
ordinates in all), and 3 components of momentum for each particle (3N momentum
components in all), so the state is given by a set of 6N numbers, which can be viewed
as a single point in 6N-dimensional space.
As we will see in the next lectures, classical mechanics fails to predict correctly
the behavior ofboth light and matter atthe atomic level, and isreplaced by quantum
mechanics. But classical and quantum mechanics have a lot in common: they both
assign physical states to objects, and these physical states evolve according to 1st-
order differential equations. The difference lies mainly in the contrast between a
physical state as understood by classical mechanics, the ”classical state”, and its
quantum counterpart, the ”quantum state”. This difference will be explored in the
next few lectures.20 CHAPTER 1. THE CLASSICAL STATEChapter 2
Origins of Quantum Mechanics
Where do correct ideas come from? Do they drop from the sky? No Are they innate
in the mind? No They come from social practice, and from it alone.
- Mao Tse-Tung
The suggestion that all matter is composed of atoms, and the name ”atom” itself,
are due to the Greek thinker Democritus, who lived five centuries before Christ. Not
until the 19thcentury, however, did evidence for this hypothesis begin to accumulate,
particularly from thermodynamics. The evidence was indirect, but compelling: as-
suming that gases are composed of atoms, one could derive analytically the equation
of state for ideal gases PV = nRT, which had been discovered empirically by Boyle
and others. In addition, assuming that solids as well as gases are composed of atoms,
one could deduce their specific heats, which agreed fairly well with the experimental
values at high temperatures . By the early 20th century, improvements in technology
and the discovery of radioactivity had enabled physicists to study in some detail the
internal structure of atoms, the mass and charge of the electron, and the interaction
of atoms with light.
Certain aspects of atomic physics which emerged from these early investigations
were puzzling and even paradoxical, in the sense that the observed behavior of elec-
trons, atoms, and light seemed in contradiction to the known laws of mechanics and
electromagnetism. These aspects fell roughly into three categories:
1. The Particle-like Behavior of Light Waves
Black-body radiation, the photoelectric effect, the Compton effect.
2. The Puzzling Stability of the Atom
Why doesn’t the electron fall into the nucleus? What is the origin of atomic
spectra?
3. The Wave-like Behavior of Particles
Electron diffraction.
2122 CHAPTER 2. ORIGINS OF QUANTUM MECHANICS
Quantum mechanics emerged as an attempt to explain these phenomena and, as
in the bible, the story begins with light.
2.1 Black-Body Radiation
Isaac Newton believed that light is composed of particles, and he had good reason
to think so. All wave motion exibits interference and diffraction effects, which are
the signature of any phenomenon involving waves. Newton looked for these effects
by passing light through small holes, but no diffraction effects were observed. He
concluded that light is a stream of particles.
One ofNewton’s contemporaries, Christian Huygens, wasan advocate ofthe wave
theory of light. Huygens pointed out that the refraction of light could be explained
if light moved at different velocities in different media, and that Newton’s inability
to find diffractive effects could be due simply to the insensitivity of his experiments.
Interference effects are most apparent when wavelengths are comparable to, or larger
than, the size of the holes. If the wavelength of light were very small compared to the
size of the holes used by Newton, interference effects would be very hard to observe.
Huygens turned out to be right. More sensitive optical experiments by Young
(1801) and Fresnel demonstrated the interference and diffraction of light, and mea-
surements by Foucault (1850) showed that the speed of light in water was different
from the speed of light in air, as required to explain refraction. Then Maxwell, in
1860, by unifying and extending the laws of electricity and magnetism, demonstrated
that electric and magnetic fields would be able to propagate through space as waves,
√
traveling with a velocity v = 1/ µ , which turned out to equal, within experimen-
0 0
tal error, the known velocity of light. Experimental confirmation of the existence of
electromagnetic waves followed shortly after, and by the 1880s the view that light is
a wave motion of the electromagnetic field was universally accepted.
It is a little ironic that following this great triumph of the wave theory of light,
evidence began toaccumulate thatlightis, afterall, astreamofparticles(or, atleast,
light has particle properties which somehow coexist with its wave properties). The
first hint of this behavior came from a study of black-body radiation undertaken by
Max Planck, which marks the historical beginning of quantum theory.
Any object, at any finite temperature, emits electromagnetic radiation at all pos-
sible wavelengths. The emission mechanism is simple: atoms are composed of nega-
tively charged electrons and positively charged nuclei, and upon collision with other
atoms these charges oscillate in some way. According to Maxwell’s theory, oscillating
charges emit (and can also absorb) electromagnetic radiation. So it is no mystery
that if we have a metallic box whose sides are kept at some constant temperature
T, the interior of the box will be filled with electromagnetic radiation, which is con-
stantly being emitted and reabsorbed by the atoms which compose the sides of the
box. There was, however, some mystery in the energy distribution of this radiation
as a function of frequency.2.1. BLACK-BODY RADIATION 23
The energy density of radiation in the box, as a function of frequency, is easily
workedoutusingtheequipartitionprincipleofstatisticalmechanics. Thetotalenergy
is
1
E = no. of degrees of freedom× kT
rad
2
1
= 2×(no. of standing waves)× kT (2.1)
2
where k is Boltzman’s constant and T is the temperature of the box. An electro-
magnetic field in a box can be thought of as a superposition of an infinite number of
standing waves; the ”degrees offreedom” arethe amplitudes of each distinct standing
wave. The factor of 2 comes from the fact that each standing wave can be in one of
two possible polarizations.
As we will see in a later lecture, the number of standing waves that can exist in
a cubical box of volume V, for frequencies in the interval f,f +Δf, is
4π
2
N(f)Δf =V f Δf (2.2)
3
c
Then the energy of radiation in this range of frequencies will be
2
1 4πkTf
ΔE = 2N(f)Δf× kT = VΔf (2.3)
rad
3
2 c
The energy density per unit frequency E(f,T)is therefore
2
ΔE 4πkTf
rad
E(f,T)≡ = (2.4)
3
VΔf c
which is known as the Rayleigh-Jeans law.
The Rayleigh-Jeans law can be tested by making a hole in the box, and measur-
ing the intensity of radiation emitted from the box as a function of frequency; this
intensity is directly proportional to E(f,T). Radiation from a small hole in a cavity
is known as ”black-body radiation”, because any radiation falling into the hole is not
reflected out the hole, but is ultimately absorbed by the walls. The experimental
result, compared to the prediction, is shown in Fig. 2.1. Theory disagrees with
experiment at high frequencies. In fact, it is clear that there had to be something
wrong with theory, because the total energy is predicted to be
1
E = 2×(no. of standing waves)× kT
rad
2
= ∞ (2.5)
simply because the range of frequencies is infinite, so there is an infinite number of
different standing waves that can be set up in the box. The energy of a box is finite24 CHAPTER 2. ORIGINS OF QUANTUM MECHANICS
(otherwise its mass, according to special relativity, would be infinite), so this result
cannot possibly be correct.
Planck’s contribution to this problem was a masterpiece of what is known in
physics asphenomenology. The first step of phenomenology is to stare at the data,
in this case the experimental curve shown in Fig. 2.1, and try to find some simple
analytical expression that fits it. Planck found that
3
8πhf 1
E(f,T)= (2.6)
3 hf/kT
c e −1
would do nicely, if one chose the constant h to be
−34
h = 6.626×10 J-s (2.7)
The second step is to try to derive the analytical expression for the data, starting
from some simple physical assumptions about the system. Planck took aim at the
equipartition principle. This principle is only valid if the energy associated with each
degree of freedom can take on any value between 0 and∞, depending on the physical
state. In electromagnetism, the energy of a standing wave of a given wavelength is
proportional to the square of itsamplitude, which can certainly be any number in the
range 0,∞. Planck’s suggestion was that, for some unknown reason, the oscillating
charges in the walls could only emit or absorb energy in multiples of hf, where f is
the frequency of the oscillator. This means that the energy of radiation of frequency
f in the box could only have the possible values
E =nhf (2.8)
n
where n is an integer. This assumption, combined with the rules of statistical me-
chanics, is enough to deduce the Planck distribution (2.6).
Note the appearance in Planck’s formula of the constant h, known as Planck’s
constant. It is one of the three most fundamental constants in physics, sharing the
honorwithc,thespeedoflight, andG,Newton’sconstant. Alltheoreticalpredictions
of quantum physics, to the extent that they disagree with classical physics, have
Planck’s constant h appearing somewhere in the expression.
2.2 The Photoelectric Effect
Thesuccess ofPlanck’s ideaimmediatelyraisesthequestion: whyisitthatoscillators
in the walls can only emit and aborb energies in multiples of hf? The reason for this
was supplied by Albert Einstein in 1905, in connection with his explanation of the
photoelectric effect.
It was found by Lenard, in 1900, that when light shines on certain metals, the
metals emit electrons. This phenomenon is known as the photoelectric effect, and2.2. THE PHOTOELECTRIC EFFECT 25
what is surprising about it is that the energy of the emitted electrons is independent
of the intensity of the incident light.
The experimental setup, used to measure the energy of the most energetic elec-
trons, is shown in Fig. 2.2. The electrons are emitted at the photocathode, fall
on a collecting plate, and complete a circuit. The resulting current is measured by
an ammeter. A battery can be used to put a potential difference V across the gap
between the cathode and the plate, which tends to repel the electrons back towards
the cathode. For an electron to reach the collecting plate, it must have an energy of
at least eV, where e is the magnitude of the charge of an electron. As the repelling
potential is increased, fewer electrons have sufficient energy to reach the plate, and
the current measured by the ammeter decreases. Let V denote the voltage where
s
the current just vanishes; i.e. V is the lowest voltage such that, at any V V , the
s s
current is zero. This means that the maximum energy of the emitted electrons is just
E =eV (2.9)
max s
It is found experimentally that E , contrary to expectation, is independent of
max
the intensity of the light falling on the photocathode. As the intensity increases, for
fixed light frequency, the number of emitted electrons increases, butE is constant.
max
On the other hand, when E is plotted as a function of frequency, the result is as
max
shown in Fig. 2.3. For frequencies f greater than some minimum f = E /h, the
0 0
data forE vs. f fits a straight line, whose slope equals Planck’s constant In other
max
words
E =hf−E (2.10)
max 0
Einstein’s explanation of this formula is quite simple: Light is composed of parti-
cles called photons. Each photon has an energy
E =hf (2.11)
photon
Suppose thattheminimum energyrequired tofree anelectron fromthephotocathode
is E , and an electron absorbs energy from the light by absorbing a photon. Then
0
the maximum possible energy for an emitted electron is the energy of the absorbed
photon, less the minimum energy required to free the electron from the metal, i.e.
E =hf−E (2.12)
max 0
Raising the intensity of the light simply increases the number of photons hitting the
metal per second, which increases the number of emitted electrons per second. But
the energy of each electron is dependent only on the energy of the absorbed photon,
and this energy depends only on frequency.
Einstein’stheoryofthephotoncompositionoflightimmediatelyexplains Planck’s
condition that the energy of electromagnetic radiation of frequency f, in a box, is
restricted to the values
E =nhf (2.13)
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