Wave-Particle Duality and Quantum Mechanics: Exploring Matter's Wave Properties, Study notes of Physics

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PHYSICS STUDY NOTES 2025/

Major Topics Covered:

a) General Relativity

b) Quantum Theory

c) Quantum Mechanics

d) Atomic Structure

e) Mathematical Derivations.

Table of Contents

• TABLE OF CONTENTS...............................................................................................................
• I. RELATIVITY.........................................................................................................................
• A. Frames of Reference............................................................................................................................................
• B. Special Relativity..................................................................................................................................................
• C. Consequences of the Principle of Special Relativity..........................................................................................
• D. Energy and Momentum.....................................................................................................................................
• E. A Hint of General Relativity.............................................................................................................................
• II. QUANTUM THEORY....................................................................................................
• A. Black Body Radiation........................................................................................................................................
• B. Photons................................................................................................................................................................
• C. Matter Waves.....................................................................................................................................................
• D. Atoms..................................................................................................................................................................
• III. QUANTUM MECHANICS & ATOMIC STRUCTURE (ABBREVIATED).............
• A. Schrödinger Wave Equation—One Dimensional............................................................................................
• B. One-Dimensional Potentials..............................................................................................................................
• D. The Hydrogen Atom..........................................................................................................................................
• E. Multi-electron Atoms.........................................................................................................................................

In S’ we’d measure  t ’,  x ’, and u

 x  . x  t 

In S we’d measure  t ,  x , and u ^

 x . x  t

b. Galilean transformation

Implicitly, we assume that  t   t . Also, we assume that the origins coincide at t = 0. Then

x  x   vx  t 

y  y ^  vy  t 

z  z   vz  t 

 t   t 

The corresponding velocity transformations are

u 

dx 

dx ^  v  u

 v x dt dt

x x x

u 

dy 

dy ^  v  u

 v y dt dt

y y y

u 

dz 

dz ^  v  u

 v z dt dt

z z z

For acceleration

a 

dux

x (^) dt

a 

du (^) y

y dt

 ax

 ay

dvx

dt

dvy

dt

a 

duz  a

dvz

z dt

z dt

Note that for two inertial frames, the ax  ax

, ay  ay

, and az  az

Example

S-frame

. (^). du

d p

F  ma  m

dt

 , if m is constant.

dt

S.’ -frame



 dp

...^

du

dv

du

F  ma  ,^ where dt

p  mu. But u  u  v , so F ^ m^ dt

  m

dt dt

 F. That is,

a

 a

 , as they must for 2 inertial reference frames.

Notice the technique. Write the 2

nd “Law” in the S’-frame, then transform the position and

velocity vectors to the S-frame.

c

the two light rays as observed in the ether rest frame.

The sideward ray:

The time required for the light ray to travel

from the splitter to the mirror is obtained

from

( ct )

2  l

2  ( vt )

2 ⇒ t 

l 1 

2

 1 2

c

2

Now c >> v , so use the binomial theorem to simplify

l 1 v

2

t  c

(^2) c

2

2 l 1 v

2

The total time to return to the splitter is twice this: t 1  2 t 

c

(^2) c

2

For the forward light ray, the elapsed time from splitter to mirror to splitter is

l l 2 l v

2 2 l v

2 t 2  c  v

c  v

c

c

2

c

c

2

The two light rays recombine at the beam splitter with a phase difference [let 

.]:

c

 t 

c    

2 l 1 v

2 c 

l v

2

t 2 t 1

c 2 c^2   c^2

Since

 t  0 , the two light rays are out of phase even though they have traveled the same

distance. By measuring  t one could evaluate v.

However, no such phase difference was/is observed! So, there is no ether, no v

with respect to

such an ether. This null result is obtained no matter which way the apparatus is turned. The

conclusion must be that either the “Laws” of electromagnetism do not obey a Newtonian

relativity principle or that there is no universal, preferred, rest frame for the propagation of light

waves.

c. Expedients to explain the null result

length contraction—movement through the ether causes the lengths of objects to be shortened in

the direction of motion.

ether-drag theory—ether is dragged along with the Earth, so that near the Earth’s surface the

ether is at rest relative to the Earth.

 1  x 

 n

 1  nx 

n ( n 1)

x

2

v .

Ultimately, the expedients were rejected as being too ad hoc ; it’s simpler to say there is no ether.

This still implies that the “Laws” of electromagnetism behave differently under a transformation

from one reference frame to another than do the “Laws” of mechanics.

2. Postulates of Special Relativity

a. Principle of Special Relativity

It doesn’t seem sensible that one “part” of Physics should be different from another “part” of

Physics. Let’s assume that they are not different, and work out the consequences. This is what

Einstein did. He postulated that ‘All the “Laws” of Physics are the same in all inertial reference

frames.’

b. Second Postulate

The second postulate follows from the first. ‘The speed of light in a vacuum is (measured to be)

the same in all inertial reference frames.’

When the speed of light is measured in the two reference frames, it is found that c  c ^  v ,

rather c  c . Evidently, the Galilean Transformation is not correct, or anyway not exact. In any

case, we assume the postulates are true, and work out the consequences.

v

2

1  c

2

1  0.95^2

 t 

 t   2.

6 x 10

 8 sec 

2.6 x 10

 8 sec

 8.33 x 10

 8 sec.

The lifetime of a fast-moving particle is measured by noting how far it travels before decaying.

In this example l  v  t  0.95 c  8.33 x 10

 8 sec  23.7 m. In practice, we measure l and compute

 t.

c. Proper time

The proper time is the time interval measured by an observer for whom the two events occur at

the same place, so that  x ^   y ^   z ^  0.

2. Length Contraction

a. “Contraction”

Consider an object, such as a meter stick, of length L in its

own rest frame, S.

A second frame, S’, moves to the right

with a speed v relative to S.

We observe two events:

i) the point A passes the left end of

the stick

ii) the point A passes the right end

of the stick.

As measured in the S’ frame,

and  x ^  0.

L ^  v  t 

v

2

1  c

In the S frame,  x  L and  t  (^2)

 t 

. Therefore, L ^  v  t  L.

An observer in the S’ frame observes the stick to be shorter (contracted) than does the observer

in the S frame. Notice particularly that the stick is at rest in the S frame.

The contraction takes place in the direction of the relative motion. Lengths perpendicular to v

are not affected. So for instance in the situation discussed above the width and thickness of the

meter stick are still measured the same in both reference frames.

b. Proper length

The proper length of an object is that length measured in the rest frame of the object.

3. Simultaneity

a. Space-time

Each event has associated with it four numbers: x , y , z coordinates and a “value of time” which

we read off a clock located at that spatial location. There is no central universal clock, rather

there is a clock at every point in space.

b. Synchronization

We would like all clocks in a reference frame to display exactly the same reading

simultaneously, but can this be arranged? Only by the exchange of signals, which is another way

of saying only in terms of intervals. However, as we have seen, intervals are not the same for

observers in different inertial reference frames. Therefore, the concept of two events being

simultaneous has no absolute meaning.

c. Non-simultaneity

Two events viewed as simultaneous in one frame will not be seen as occurring simultaneously in

another frame.

example: a train moving with constant velocity on a straight, smooth track. One observer rides

on the train, the other observer stands beside the track.

Flashes of light are emitted at the points C 1 and C 2 when the origins (O & O’) of the two frames

coincide. To the trackside observer at O, the flashes are simultaneous. To the observer on the

train, however, the flash emitted at C’ 2 is received before the flash emitted at C’ 1. Yet both

observers measure the same speed of light, c.

v

2

1  c

2

v

2

1  c

2

 v

Notes: i) the inverse transformation is obtained by replacing v with – v.

ii) for v << c , these reduce to the Galilean transformation.

c. 4-vectors

Suppose that when O = O’, a flash of light is emitted from the origin O. In the S frame, the

distance the light wave front travels in time t is r

2  x

2  y

2  z

2  c

2 t

2

. Measured in the S’

frame, it’s r 

2  x 

2  y 

2  z 

2  c

2 t 

2

. Subtract the second expression from the first and

collect the S frame on one side of the equal sign, the S’ frame on the other side.

r

2  r 

2  c

2 t

2  c

2 t 

2

r

2  c

2 t

2  r 

2  c

2 t 

2

There is this quantity, a generalized displacement (call it s ) which is the same in the two inertial

reference frames.

s

2  s 

2

We see that the quantity ( ict ) “acts like” a component of displacement along a fourth axis. The

interval between any two events in space-time is  s

2   x

2   y

2   z

2  c

2  t

2

. The interval

is invariant under the Lorentz Transformation. That is, as measured in any two inertial frames,

 s

2   s 

2

. This is an extension of the invariance of lengths under a rotation of the coordinate

axes.

d. Transformation of velocities

Since displacements and time intervals are transformed, obviously relative velocities won’t add

simply, either.

In the S’ frame an object moves with constant velocity along the x axis; u  x

dx ^

. Transform to

dt 

the S frame; u 

dx

  dx  vdt  

dt 

ux  v (^) and similarly for the y and z x (^) v v dx v

 dt 

c

2

dx^1 ^ c

2 dt

c

2

ux

components. While dy & dz are not contracted, dt is still dilated.

example:

uA  0.5 c and uB  0.8 c , both as measured in the S frame. The S’ frame rides along with

spaceship B. Therefore, v

 u

B.

u ^ 

u (^) A  v 

 0.5 c  (0.8 c ) 

c

v u c

2 A^^1 ^

0.8 c

c

2 0.5 c^

Be careful with the directions of the velocities.

Note that when u  c and v  c , then

c 1 

v

vu  0 and u ^  u  v. On the other hand, if u  c ,

c

2

then u ^ 

c  v 

c  c.

cv

c

2

v

c

A

u

2

c

2

u

2

c

2

u

2

c

2 u

m

2 u

2

u

2

u 2

m

2  K  up

u^2 

c

u 1 2 1

c

2

u 1 u 2

 K   mc

2

u 1 u 2

mu

2  mc

2  mu

2

 K   

1

Now, if we started from rest, then u 1 = 0 and u 2 = u and  K 

mc

2

 mc

2

. Therefore, we

define the relativistic kinetic energy to be

K 

mc

2

 mc

2 .

The quantity mc

2 is called the rest energy , because it’s independent of u. The total relativistic

energy is E = K + mc

2

• V , where V is the potential energy, if any. If V = 0, then

E  K  mc

2

  mc

2 .

b. Energy-momentum relation

Take a look at the quantity ( V = 0)

E

2  m

2 c

4  m

2 c

4  m

2 c

4  m

2 c

2 u

2

 c

2 .

u

2 u

(^2) u^2 1  c

2

c

2

c

2

For photons, m = 0 and E = pc.

E

2  m

2 c

4  c

2 p

2

E

2  c

2 p

2  m

2 c

4

u

2

1  c

2

u

2

1  c

2

u

2

1  c

2

u

2

1  c

2

u

2

c

2

c

2

u

(^2) dt

1 

u

2

c

2

u

2

c

2

dt

2

^3 2

dt

u

c c

c c c

c

c. Units of mass-energy

It is convenient to express energy in units of electron-volts (eV). An electron-volt is the energy

gained by an electron upon being accelerated through a one Volt potential difference. Thus 1 eV

= 1.60x

• Joules. The rest energy of an electron is

mc

2  9.11 x 10

 31 kg  3 x 10

8 m / sec

2  8.20 x 10

 14 J  0.511 x 10

6 eV  0.511 MeV.

Often, mass is expressed in terms of MeV / c

2 so that the electron mass is 0.511 MeV / c

2 .

Sometimes, the c

2 is dropped, but it’s understood to still be there. Similarly, momentum is

expressed in terms of MeV / c , since pc = units of MeV.

3. Relativistic Mechanics

a. Force

We want the “Laws” of Mechanics to be invariant under the Lorentz Transformation. Also, we

want to recover the classical result when u << c. So, we define the relativistic force component

to be F^ ^

dpx , where x dt

px 

mux .

Let’s say the motion and force are entirely along the x-direction.

F 

d mu 

dt

m du  mu

d 1

dt

m du^1 u^2

 3 (^2 2) u du

F   mu  (^1)  2

2 dt

du u u

(^2) u

F  m^1 ^2 ^2 1 ^2

3

F  m

2

1 du

u

2 dt

Solve for the acceleration.

c

2

du F

2

dt m

2

2 ^

1 2

3 2

p

2 c

2  m

2 c

4 1

o

Knowing the momenta and masses of the decay products, we determine the mass of the incident

particle, hoping to identify it.

p  910

MeV , p  323

MeV , m c

2  m c

2  mc

2  139.6 Mev. 1 c

2 c

1 2

The energy and momenta are conserved. The total energy is

Eo  E 1  E 2  

Eo  921 MeV  352 MeV  1273 MeV

The quickest way to obtain the magnitude of the incident momentum is to use the law of cosines:

p

2 c

2  p

2 c

2  p

2 c

2  2 p p c

2

cos  1034229 MeV

2

o 1 2 1 2

poc  1017 MeV

Now that we have the total energy and the kinetic energy, the mass is obtained from

E

2  p

2 c

2  m

2 c

4

o o

m c

2 

o

 765 MeV

Evidently, the incident particle was a meson. What was its speed before it decayed? Well,

the total energy is also

m

2 c

4 2 

o o ,^ so^ solve^ that^ for^ u. u

2

c

2

u  1 

c

d. Mass-energy equivalence

When we speak of the total energy being conserved that includes the total rest energy. For

instance, consider the decay of a neutron that is initially at rest.

n  p  e 

The neutron decays into a proton, an electron and an anti-neutrino. The three product particles

are observed to have total kinetic energy of K = 0.781 MeV. The initial energy is just the rest

energy of the neutron, Ei = 939.57 MeV. The total final energy is

E (^) f  mpc^2  mec^2  K  938.28 MeV  0.511 MeV  0.781 MeV  939.57 MeV

Notes: i) The rest energy of the anti-neutrino is too small to bother with.

ii) Keep in mind the rounding of numbers and significant digits when substituting

numerical values into the formulae.

iii) Notice that mn  mp  me. A portion of the neutron’s rest energy has been

converted into kinetic energy.

p

2 c

2  m

2 c

4 2

E  p c

2 2 2 o o

m

2 c

4

o

E

2 o

E

E. A Hint of General Relativity

1. Equivalence

In Special Relativity it is asserted that all inertial reference frames are equivalent—the “laws” of

physics are the same in all inertial reference frames. No experiment done in one frame can

detect its uniform motion relative to another frame. Can the same be said for reference frames

that have a relative acceleration?

a. Elevator

Recall the past discussion of a person standing in an

elevator. If the elevator moves perfectly smoothly and

there are no floor indicator lights, then the person inside

will have no perception of the elevator’s motion, except

for feeling perhaps the elevator floor pressing upward

on his or her feet. [Keep in mind: the person gets no

information from any source outside the reference frame

of the elevator.] Contrast this situation with that of

another person standing in a similar elevator, but this

elevator is simply resting level on the Earth’s surface.

The person in this elevator also feels the floor pressing

upward on his or her feet, also has no perception of the

elevator’s motion. We, as omniscient external

observers, know that this second elevator is resting on the surface of a planet, and that what the

person inside is experiencing is the gravitational force exerted by that planet. The point is that

there is no experiment that either of the persons inside the elevators could perform that would

distinguish between the two situations. Pendula would swing back and forth just the same;

projectiles would follow the same kinds of arcs, etc.

b. Light and gravity

Imagine ourselves as observers far from any source of gravitational force. Nearby, we observe a

closed “elevator” which is accelerating, relative to us, at a constant rate, a

o.^ A^ person^ standing

inside the “elevator” sends a series of light pulses

toward one wall—he or she and we see the light pulses

dropping toward the floor as they approach the wall.

The light follows a curved path inside the elevator.

The Postulate of General Relativity asserts that the

“laws” of physics have the same form for observers in

any frame of reference, regardless of its acceleration

relative to another frame. We have seen that an

accelerated frame is equivalent to one in a gravitational

field. It follows that the force of gravity must affect a

beam of light just as it affects the motion of a massive

projectile. Indeed, experiment has shown that it does.

But, light has no mass.