Download Lectures in physics Part 2: Electricity, magnetism and quantum mechanics Przemysław Borys 20.05.2014

Lectures in physics
Part 2: Electricity, magnetism and quantum mechanics
Przemysław Borys
20.05.2014
13. Vector field operators.
13.1. The nabla operator
Much of the vector operations on fields in 3D is done using the nabla operator. It is a special type of
vector:
∇= i
∂  ∂  ∂
j
k
∂x
∂y
∂z
13.2. The gradient
Probably the simplest operation that makes use of the nabla operator is the gradient. In this
operation the nabla acts on a scalar field and returns a vector that gives the rate of change of the
field in the direction of the greatest change. It is written as:
grad f  x , y , z =∇ f  x , y , z =i
∂f ∂f ∂f
j
k
∂x
∂y
∂z
A two dimensional example of the gradient operation is shown below:
13.2. The divergence
The divergence stands for a dot product between the nabla operator and a vector field:
div f  x , y , z =∇⋅f =
∂f ∂f ∂f


∂x ∂ y ∂z
To understand what does the divergence represent by itself, consider the following figure:
We can see the vector f entering the cube on each side. It is possible to calculate whether the
same amount of the vector exits the cube as enters or whether there is an accumulation. We measure
this by a flux of the vector over all sides (the flux is the dot product of the vector times the surface
vector through which the vector enters/escapes the volume—because the larger the entering surface,
the more vector enters/escapes the cube—you could imagine this for example considering the flow
of water):
J V = [ f x  x  x− f x  x  ]  y  z[ f y  y y − f y  y ]  x  z [ f z  z  z − f z  z  ]  y  x
Because of the dot product between the surface element and the vector, the components at
x x ,
y y ,
z  z enter to the equation with a positive sign (the same alignment of
the vectors), and the components at x, y, z enter the equation with negative sign, because the surface
element vector points opposite to the f vector.. If we now divide the total flux by the volume of the
cube, V = x  y  z , we obtain:
J V f x  x x− f x  x  f y  y y − f y  y  f z  z  z − f z  z 
=


V
x
y
z
which in the limit of  x , y , z  0 gives:
JV ∂ f x ∂ f y ∂ f z
=


=div f
V
∂x
∂y
∂z
So, divergence measures how much of the vector escapes out of the volume through the side
walls. So it says whether the volume is a source of vectors or it is a sink. It is possible to sum the
above equation over a larger volume that is partitioned into adjacent boxes. In such case it will
correspond to a sum (an integral) of divergences on one side and a sum of flux contributions on the
second side. The flux contributions are such, that they cancel to zero on internal box walls, and only
the summation of external walls of the volume gives nonzero contrubution. Finally, a more general
theorem that resembles the above reasoning is known in mathematics as the Gauss-Ostrogradski
Theorem:
∯S f⋅d s =∰V div f dxdydz
13.3. The curl
Since the divergence is a dot product of the nabla and the vector field, a natural extension of the
vector field operations would be to consider a cross product of nabla and the vector field. Such a
cross product is called the curl, or rotation in some languages.
∣
i
∂
curl f =∇ × f =
∂x
fx
j
∂
∂y
fy
k
∂
∂z
fz
∣
Why is this supposed to have something to do with a rotation? Let us expand the determinant:
curl f = i

 
 
∂fz ∂fy
∂fy ∂fz  ∂fy ∂fx
−
 j
−
k
−
∂y
∂z
∂z
∂x
∂x
∂y

To simplify, let us consider a vector field that is perpendicular to the z axis, i.e. the cross product
returns then only the k component (perpendicular to the field). If we expand the partial
derivatives to deltas, we obtain:

f  x  x − f y  x f x  y y− f x  y
curl f = k y
−
x
y

To make it more clear, let us multiply both sides by the surface of  x  y which contains the
considered field element:
curl f  S =k [ f y  x x − f y  x  y f x  y− f x  y y  x ]
And now take a look on the following vector, that circulates along the frame built between points
 x , y  and  x x , y y :
And now just think: would the curl be nonzero if the field would not rotate? For example if the
parallel vectors would point in the same direction? So here is the answer on how does it work. In
case of rotating vectors, each of the terms on the right hand side is positive and contributes to the
nonzero value of the curl.
If the rotation takes place around a different axis than z, we will obtain the terms related to these
axes, for example the i and k terms of a resulting vector. The resulting vector then gives the
rotation axis of the field, and its value informs about the magnitude of the rotation.
As we did in the case of divergence, the equation that we have written above for curl f  S can
be generalized. We can sum the curls over some specified plane, like below:
In such case, the edge contributions of the vector field from internal walls come in pairs: one from
the first elementary square, another from the adjacent square. These contributions have opposite
signs and reduce to zero, so sum of the curls over a fixed plane relates to the sum of the vector field
along the external edges of the plane. It forms then the, so called, Stokes theorem:
 ∮ f⋅dl

∯S curl f⋅ds=
L
i.e. it allows to change the surface integral of a curl into a line integral around the contour of the
chosen surface.
Questions:
1. Provide the interpretation to the divergence and describe the Gauss-Ostrogradski theorem.
2. Provide the interpretation to the curl and describe the Stokes theorem.
3. Provide the interpretation of a gradient.
14. The Electric Field
14.1 The Coulomb's law
The Coulomb's law describes the force that occurs between two electrical charges. What are
charges? They constitute one of the properties (like mass) of matter. The matter is built of atoms,
while the atoms contain elementary particles like protons, electrons, neutrons. Protons are the
particles that exhibit a positive charge, electrons posess a negative charge, while the neutrons are
electrically neutral, as the name suggests. Charges are measured in Coulombs [C], and the smallest
quantity of a charge is the elementary charge e=1.6⋅10−19 C characterizes a single electron or a
single proton.
The charges can be easily generated by rubbing. For example, when you comb your hair, the hair
charges up, and your hair can become lifted up into the direction of the comb. This happens because
the comb removes the negative electrons from the hair and leaves it positively charged (the atoms
still contain protons, but certain amount of electrons is removed). Then, because the comb contains
extra negative electron charges, while the hair contains positive charges, an interaction occurs in
form of a Coulomb force:
=
F
k q1 q 2
r2
r
The force acts in the direction of the vector that connects the positions of the two interacting
charges and can be attracting if the two charges have opposite signs, or repulsing, if the charges
have identical signs. The proportionality constant equals to:
k=
[ ]
1
Nm 2
≈9⋅109
4 0
C2
−12
because the vacuum electrical permittivity 0=8.854⋅10
[ ]
C2 N
m2
.
14.2. The superposition principle
In case when we wish to establish the trajectory of a charge Q, and there is n charges qi that
surround the considered charge, the resulting force acting on Q is simply the vectorial sum of the
forces from individual charges, i.e.:
n
 =∑
F
i=1
kQqi
r i2
ri
This is called the superposition principle and allows to consider the additive effect of each of the
surrounding charges.
14.3. The electric field
It would be tempting to divide
 from the previous equation by Q to obtain something that is
F
independent of the considered charge Q, and can characterize the properties of the space. Such an
object is called the electric field and represents the force that an unit charge would experience in
the space:

n
kqi
 = F =∑
E
ri
i=1
Q
r 2i
The unit of the electric field is
[
N V
=
C m
]
where the alternative definition
V
N
makes use of the
unit [V], which stands for „volts” and will be defined in the next paragraph.
14.4. The electric field potential.
It is instructive to calculate the energy which needs to be supplied to a charge Q to separate it from
its pair charge q. This is equivalent to moving one of the charges to the infinity:
∞
∞
W=
∫r
 ⋅dr

W =∫r F
0
kQq
0
r2
∞
∣
kQq
kQq
dr=−
=
r r
r0
0
This is the amount of work that needs to be done to „free” the charge. So the energy of the electric
field is the negative of this value, i.e.
W =−
kQq
r
This energy can be expressed as „an energy per a test charge Q”. Such quantity is called the
potential around charge q:
V=
The potential is measured in volts
[
V=
J
C
]
W
kq
=−
Q
r
With the use of potential it is easy to calculate the
energy required to move the charge to some distance from its pair:
W =Q [V r 2−V r 1]=QU
where the potential difference U =V 2−V 1 is called the voltage.
An important relation holds between the potential and the electric field. Since the potential
∞
 
W −∫r F⋅dr
V= =
Q
Q
∞

−∫r F

V=
dr
Q
∞

V =−∫ 
E dr
0
0
r0
Differentiating this (in 3D this corresponds to taking of a gradient), we obtain:
 =−grad V
E
14.5. The Gauss law
The most general formulation of the electric field was done in the form of the Gauss law, which
constitutes the third Maxwell law:
 =Q
0∯S 
E⋅dS
Which says that the integral of the dot product between the electric field vector and the surface
normal vector over a closed surface multiplied by the vacuum electrical permittivity equals to the
charge enclosed by the considered surface.
This law is sometimes written in the differential form as:
0 div 
E =Q
If we integrate both sides over a closed volume, we obtain:
0∰V div 
E dxdydz=∰V V dxdydz
Using the Gauss-Ostrogradski theorem on the left hand side, and integrating the charge density on
the right hand side, we obtain:
 =Q
0∯S 
E⋅dS
14.6. The Coulomb law from the Gauss law
Consider a point charge surrounded by a spherically-symmetrical integration surface. We say:
 =Q
 0∯ 
E⋅dS
Now, the field lines emanate from the charge center at straight angle to the spherical integration
surface (recall what Coulomb law states about the direction of the electrical force). Thus, the angle
between the normal surface vector and the field lines is 0o or 180o, i.e. these vectors are parallel or
anti-parallel. Consider a parallel orientation for a positive test charge. Then the cosine of the angle
between vectors reduces to „1” and we can write:
0∯ EdS=Q
I.e. without the vector arrows. Now due to the symmetry of the sphere, the charge in its center looks
exactly the same from any point on the sphere. Therefore, we can expect an identical electric field E
on each of the points equidistant from the sphere center. Thus, taking the integration surface to be
the sphere, we obtain:
0 E ∯ dS =Q
0 E⋅4  r 2=Q
1 Q
E=
4 0 r 2
Which resembles the Coulomb law because
 =q 
F
E .
Questions
1. State the Coulomb's law and the superposition principle.
2. Define the electric potential and relate it to the work
3. Describe the Gauss law and derive the Coulomb law from it.
15. Current flow
The current is an ordered movement of the electrical charges within an electrically conductive
medium. Electrical conductors are typically metals, where the electrons of the most distant orbitals
become freed from the individual atoms and form an electron gas which fills the volume of the
conductor. Such electrons can freely move in the applied electric field and thus the flow of the
current is possible.
15.1. The Ohm's law
One of the basic properties that characterizes the flow of a current is the Ohm's law. Different
materials require different amounts of work to be done upon the chrages in order to move through
the material. The matter produces a „resistance” to the movement of the electrons and it is therefore
called a resistor from the point of view of electrotechnics. We speak about the resistance R which
is defined in the Ohm's law as the ratio of the voltage difference between two ends of the resistor to
the current that flows through:
R=
The resistance is measured in Ohms
[
=
V
A
]
U
I
. The resistance of different types of materials can
be of different magnitudes. For example the resistance of skin is measured as about a mega Ohm,
while the resistance of a metal wire is a very small fraction of an Ohm. Therefore, given a fixed
voltage applied to the skin and to the metal wire, the current will be much greater in the wire which
exhibits less resistance.
15.2. The Kirchhoff laws
The Kirchhoff laws state certain conservation properties of the circuits. First of such properties is
expressed by the First Kirchhoff Law:
n
∑i=1 I i=0
which means that the sum of currents entering a node of the circuit must be equal to zero, i.e. the
current cannot be generated in a node and the same amount of current enters the node as it leaves
the node.
The Second Kirchhoff Law states that the potential drop along a closed loop of circuit equals zero.
So if we have some voltage sources and resistors in a loop of circuit, then going along a specific
direction, the sum of potential drops must be equal to zero:
n
∑i=0 U i =0
In the above figure, the second Kirchhoff's law would read:
E−U 1−U 2 −U 3=0 . This law is a
direct consequence of the fact that electric field is a potential field where the energy of a charge
(and consequently its potential) does not depend upon the path that it travels.
15.3. Serial and parallel connection of resistors.
If you consider two resistors connected in series:
By definition, the net resistance R can be defined as
R=
U
I
The voltage U consists of two voltages occuring on both resistors. From the Ohm's law we can
have:
U =U 1U 2= IR1 IR2
Substituting to the previous equation this results in:
R= R1R2
And this is the way we handle serial connection of the resistors. The resistances simply add
together.
Now let us consider a parallel connection of the resistors:
This time the voltage on both resistors is identical, but the currents differ. The currents can be
expressed by the Ohm's law as:
U
R1
U
I 2=
R2
I 1=
By the First Kirchhoff's law we know that
I =I 1 I 2 and so we can calculate the equivalent
resistance of the circuit as:
R=
U
U
1
=
=
I U U
1
1


R1 R 2 R1 R2
We can alternatively say that:
1 1
1
= 
R R1 R2
Which means that the inverses of the resistance (which are called conductances G=
summation in this type of connection.
1
) undergo
R
15.4. Capacitors
Capacitors are electric circuit elements that store charges in response to the applied voltage. The
definition of the capacitance, a quantity which characterizes the capacitors represents the ration of
the stored charge to the applied voltage:
C=
Q
U
It is convenient to write it in the „see you” form:
Q=CU
The latter can be differentiated to obtain a relation between the current and the voltage (the Ohm's
law) for the capacitor:
dQ
dU
=I =C
dt
dt
dU 1
= I
dt C
Evidently, the ratio
1
plays here a role thet is normally played by the resistance, i.e. the
C
proportionality coefficient between the voltage and the current. However, this time the Ohm's law is
a differential equation, and not an algebraic equation. If there are no voltage changes along the
capacitor, there is no current (i.e. the capacitor just stores the charge and nothing happens).
However, when the voltage changes, the capacitor must expell or accept some charges to satisfy
Q=CU and there occurs a flow of current.
15.5. Derivation of the flat capacitor equation
The flat capacitor consists of two plates of a conductor, filled with surface charge Q which are
separated by a distance d and produce an electric field E between the plates.
Choosing the Gaus surface, we can write the Gauss Law for the dashed surfaces:
 =Q
 0∯ 
E⋅dS
The integration takes part along all sides of the dashed cuboid. If the surface A is much larger than
the side surfaces, we can neglect the contribution of sides to the integral. Idealizing the surface A to
the infinite value, we can also postulate that due to the translational symmetry (no matter which
point we choose on a plate, everything looks the same from any perspective, one always can see an
infinite surrounding) the elecric field between the plates is constant. The same applies to the electric
field above the upper plate, and in addition, we can speculate that this field is close to zero, as the
contributions of the field lines from the top positive charge and bottom negative charge cancel (see
the directions of arrows in the figure close to the letter „A”). In addition, the electric field vectors
are aligned with the normal surface vector of the plates which allows to evaluate the Gauss Law as:
0 [ EA0A ] =Q
Q
E=
0 A
Because the electric field is related to the potential by
field gives
E=
 = grad V , which in 1D constant electric
E
V U
=
, we can rewrite the flat capacitor equation as:
d
d
d
Q
Q=
0 A
C
0 A
C=
d
U=
15.6. Serial and parallel connections of capacitors
The capacitors can be connected in series (on the left) or in parallel (on the right of the figure
below).
If the capacitors are connected in parallel, the voltage U on each of the capacitors is identical. The
resulting charges are Q1=C 1 U and Q2 =C 2 U . Thus, the equivalent capacitance of the parallel
connection reads:
C=
Q 1Q 2
=C 1C 2
U
In a series connection on the other hand, each single capacitor must store the same charge on both
plates, but each of these charges has an opposite sign. This, due to the electroneutrality of the
conductors which connect the capacitors, implies an equal charge on all capacitors in series. Thus,
the voltages on these capacitors can be calculated by U 1=
capacitance reads:
Q
Q
, U 2=
. The equivalent
C1
C2
C=
Q
Q
=
U 1U 2 Q Q

C1 C2
1
C=
1
1

C1 C2
1 1
1
= 
C C1 C 2
Questions
1. Define the Ohm's and Kirchhoff's laws.
2. Discuss the serial and parallel connections of resistors.
3. Discuss the serial and parallel connections of capacitors.
4. Derive the flat capacitor equation from the Gauss law
16. The magnetic field
16.1. The Lorentz Force
Consider an infinitely long wire which is negatively charged like this:
The electrons move from left to right, and there is more of them in the wire, so the wire becomes
negatively charged. If we would put a test positive charge close to the wire, it would become
attracted to the wire.
BUT! What happens if we observe the wire from a moving reference frame???
The stationary distances between protons undergo Lorentz contraction, in turn the distances
between electrons become expanded as they are observed at smaller velocity. In a reference frame
that is stationary to the electrons, the wire in the above figure becomes... positively charged! (at a
smaller velocity of the reference frame, the wire could also become neutral). But this would mean
that the positive test charge in this reference frame is repulsed from the wire. This is impossible, as
the description of physical phenomena in all coordinate frames should predict the same behaviour.
The Lorentz force is found to manage this difficulty. It turns out that when the charges are in
motion, another field appears, which can be called the magnetic field B. This field acts on a
moving charge with a force equal to:
 =q v × 
F
B
or, taking into account that a charge moving to a distance L with velocity v gives a current
q qv
I= =
t L
so qv =IL , and
 =I 
F
L×
B
This force can compensate the relativistic effects to the electrostatics and is called the Lorentz
force.
16.2. Ampere's law
The Ampere's law is another basic law of the electromagnetic field theory after the Gauss Law. It is
also known as the second law of Maxwell:

∮L B⋅dl=
0I
This law relates the magnetic field B along the closed loop of integration L with the electric
current I enclosed by the integration contour. The proportionality constant 0=1.256⋅10
−6
N
A2
is the vacuum magnetic permeability, or just „the magnetic constant”.
The Ampere's law can be expressed in a differential form. This is the typical occurence in the list
of Maxwell laws:
curl 
B =0 
J
where

J is the current density, i.e. the current per unit area of the cross-section. If we integrate
the above with respect to
∯S dS
and we apply the Stokes theorem to the curl, we recover the
Ampere's law in the integrated form.
16.3 The magnetic field around an infinitely long wire
This situation is the simplest one to consider using the Ampere's law. Take a look on the following
figure:
In this figure, the situation in radius R around the wire is symmetrical and one can expect the same
value of B along the perimeter of the circle. The dot product between the contour element and the
field lines is done for zero angle so the cosine equals „1”. This results in:

∮ B⋅dl=
0I
∮ Bdl =0 I
B ∮ dl=0 I
B 2  R=0 I
 I
B= 0
2R
16.4 The magnetic field in a solenoid
Consider a very long solenoid, so long that it can be considered almost infinite. Then, assume the
following integration contour:
In this figure I have depicted the integration contour with a dashed line, the field lines around turns
with dotted lines. What can be seen is that the contrubitions of the lower turn sections cancel the
contributions of the upper turn sections outside of the solenoid. This gives the chance to treat the
magnetic field outside the solenoid being equal to zero. In turn inside of the solenoid, the
contributions from the bottom and from the top add together. The field is nonzero. If the solenoid is
long enough, we can assume a symmetry with respect to translations (no matter which point of the
solenoid we consider, we see „infinitely” many turns to the left and to the right). This gives rise to
the assumption that the magnetic field along the length of the solenoid inside of the turns should be
constant. Taking this all together:

∮ B⋅dl=
0n I
BL B L d 0LB R d =0 n I
BL≈0 n I
nI
B=0
L
16.5. Faradays induction
The Faraday's induction requires the introduction of a magnetic field flux  B which can be
defined by:

 B=∬S 
B⋅dS
i.e. this is the flow of the magnetic field lines through a given surface. The Faraday's law of
induction can then be stated as:
E=−
∂
∂t
where E is the electromotive force induced along the perimeter of S by the changes in  . This
force grows with the magnitude of the magnetic flux (which is the larger, the larger is the
considered area and the magnitude of B) and with the rate of change of the magnetic flux.
It is possible to formulate the Faraday's law in a differential form, whcih forms the first Maxwell
law:
 =−
curl E
∂
B
∂t
16.6. Gauss Law for the magnetic field. The 4th Maxwell law.
According to the Maxwell laws, the divergence of the electric field is the charge density. In case of
the magnetic field, there are not „magnetic monopoles” that would produce the field. The field is
always produced by tiny circulating charge circuits. Therefore, the divergence of the magnetic field
is said to be equal zero:
 =0
div B
16.7. Maxwell equations for steady fields and for the time varying fields
The Maxwell equations that we have learned so far can be summarized in a following way for fields
that do not change in time, i.e. where
∂
B ∂
E
=
=0 :
∂t ∂ t
1. curl 
E =0
2. curl 
B =0 
J
 Q
3. div E=
0
4. div 
B =0
These equations can be further generalized to the fields that vary with time. Actually, the first
Maxwell law (the Faraday's Law of induction) already included the possibility of time variability in
B (which is zero for a steady field). But this is not all. It was noticed by Maxwell that the Ampere's
law is inconsistent. Taking the divergence of this equation one obtains that:
 =0 div J
div curl B
0=div J
because the divergence of a curl is zero for any vector. The second equation is a continuity equation
for the current, which says that the current cannot vanish. However, the charge can be accumulated
in a given point of space (as happens in a capacitor), giving rise to a more general continuity
equation:
div J 
∂ Q
=0
∂t
The second term can be substituted using the Gauss Law differentiated with respect to time:
[
]

∂E
div 
J 0
=0
∂t
Modifying the RHS of the Ampere's law to produce such a continuity equation results in:
curl 
B =0 J 0 0

∂E
∂t
This correction can be understood by the notion of a capacitor. We know, the field E between the
plates of a flat capacitor is (recall the derivation):
E=
Q
0 A
Differentiating this equation with respect to time we obtain:
∂E I 1
=
∂ t A 0
∂E J
=
∂t  0
∂E
J = 0
∂t
Maxwell called this current the displacement current.This is not a real current, associated with the
motion of charges but it produces a magnetic field and causes the Maxwell equation to be
consistent. The resulting set of Maxwell equations for time varying fields reads:
1. curl 
E =−
∂
B
∂t

∂E
2. curl 
B =0 
J 0  0
∂t

= Q
3. div E
0
4. div 
B=0
16.8. Electromagnetic wave equation
To derive the electromagnetic wave equation, we start with the Maxwell Equations for the vacuum.
In the vacuum, there are no charges, so both divergences are set to zero, and current is also
nonexistent. The laws then become very symmetrical:

 =− ∂ B
1. curl E
∂t
∂
E
2. curl 
B =0 0
∂t
 =0
3. div E
4. div 
B =0
Differentiate the second equation in time and substitute the first equation for the derivative of B:
∂ B
∂2 E
curl =0 0 2
∂t
∂t
Now it is necessary to use an identity of the vector calculus, which can be easily checked by
expanding the operators (which is quite tedious):
2
curl curl 
E= grad div 
E−∇ E
A short derivation:
∂ Ez ∂ E y
∂ Ex ∂ E z  ∂ E y ∂ E x
∇× 
E =i
−
 j
−
k
−
∂y
∂z
∂z
∂x
∂x
∂y

 

 

 
∂2 E y ∂ 2 E x ∂ 2 E x ∂ 2 E z
∂2 E z ∂2 E y ∂2 E y ∂2 E x



∇× ∇× E = i
−
−

j
−
−

∂ x ∂ y ∂ y2
∂ y ∂ z ∂ z2
∂ z2 ∂ x ∂ z
∂ x2 ∂ y ∂ x
k

∂ 2 E x ∂ 2 E z ∂ 2 E z ∂2 E y
−
−

∂ z ∂ x ∂ x2
∂ y2 ∂ z ∂ y
Now in the brackets we add and subtract terms
∇× ∇× 
E = i



∂2 E i
∂i 2
to obtain:
∂2 E x ∂2 E y
∂ 2 E z ∂2 E x ∂2 E x ∂2 E x


−
−
−
∂ x2 ∂ x ∂ y ∂ x ∂ z ∂ x2
∂ y2
∂ z2





∂ 2 E z ∂2 E y ∂2 E x ∂2 E y ∂ 2 E y ∂ 2 E y

j


−
−
−
∂ y ∂ z ∂ y2 ∂ y ∂ x ∂ x2
∂ y2
∂ z2
∂2 E x
∂2 E y ∂2 E z ∂2 E z ∂2 E z ∂ 2 E z


−
−
−
∂ z ∂ x ∂ z ∂ y ∂ z2
∂ x2
∂ y2
∂ z2
which resembles the desired relation (the first three terms in each bracket form the gradient of a
divergence; the second three terms form the laplacian)..
k
Because the divergence of E is zero (3rd Maxwell equation) we can eventually say that:

∂2 E

∇ E =0 0 2
∂t
2
Which is the electromagnetic wave equation, where, as you shall notice after recalling the
structure of the mechanical wave equation, the term 0 0=
determined by c=
1
, so the speed of light can be
v2
1
. And it turns out to be the case! Moreover, this speed of light is
  0 0
measured with respect to no particular reference frame (the wave spreads in vacuum) which
suggests that it should have the same value in any inertial reference frame.
16.9. The Biot-Savart Law
The Biot-Savart law states that:
 Id l ×r
d
B= 0
3
4 r
where dB is the magnetic induction contribution that originates from a current I element of length
dl, which is located in distance r from the considered point of space.
The derivation of thes law requires the use of special relativity, specifically the relativistic
transformation of forces, which we have not done yet.
To start with, we need a transformation law for a general force F, which can be expressed as the
derivative of the momentum p=mv, which contains two terms that undergo the transformation:
relativistic mass and velocity. Let us recall the idea of the relativistic mass that can be assigned to a
particle moving at velocity u or u' in the primed reference frame:
mu =
m

1−
u2 or
c2
mu ' =
m

1−
u'2
2
c
We can try to describe this relativistic mass from point of view of the moving reference frame,
approaching the body at velocity v. The body has a resultant velocity in this new frame (recall
relativistic summation of velocities between u and -v), which in the x direction is equal to:
u x '=
u x −v
u v
1− x 2
c
In the y and z direction, the transformed velocity components are easily derived from the Lorentz
transforms (see previous chapters, or next two pages, formula (*)) as:
u ' y=
In this case, we can write
uy
uz
u ' z=
u v and
u v
1− x2
1− x2
c
c
2
2
2
u ' 2 u x −v  u y u z
=
2
c2
ux v
2
c 1− 2
c


and:
2
u'
1− 2 =
c
1−
2u x v
c
2

u2x v 2
c
4
−

u2x
c
2
1−

2u x v
c
2

ux v
c
2
2
2
2
v uy uz
− 2− 2 − 2
c
c c
2
which reduces to:
  
u 2x
v2
v 2 u2
1−
1−
1 4 − 2 − 2
c2
c2
u'2
c
c c
1− 2 =
=
2
2
c
ux v
ux v
1− 2
1− 2
c
c
u 2x v 2




Substituting this to the relativistic mass formula at velocity u', we obtain:
mu ' =

u'2
c2
ux v
=
1−
1−
mu ' =mu
mu
m

c2
v2
1− 2
c


u2
1− 2
c
1−

u '2
c2
= mu 1−
uxv
c2

Which gives the transformation law for a motion on straight line. Fortunatelly, this transformation
looks the same also in 3D if we replace u by ux in the final formula. But since this time we will need
to consider a 3D picture of the situation, a full momentum vector is needed, so we need to know
how to express the velocity components in the inertial reference frame that moves at velocity v (in
the x direction).
Recall that from the Lorentz transformations we had:
x ' = x −vt
y '= y
z '= z
v
t ' = t− 2 x
c

with
=

1

1−
v 2 . This formulation results in infinitesimals:
2
c
dx ' =dx −vdt
dy '=dy
dz '=dz
v
dt ' = dt− 2 dx
c


The transformed velocity vector reads:
dx '  dy '  dz '
j
k
dt '
dt '
dt '
i u x −v  j u y k u z (*)


u' =

v
1−u x 2
c
u
 ' = i
To calculate the momentum, we must multiply the obtained velocity by the relativistic mass

mu ' = mu 1−
ux v
c2

:
p u ' =mu [  u x −v  iu y ju z k ]
that is:
p u ' = p x −mu v  i p y j  p z k
Now we are close to the end. The force is a time derivative of the momentum:
 ' = d p ' dt =
F
dt dt '
1
u v
1− x2
c
[
F x−

dm u
F
F
v i  y j z z


dt
]
In this equation we must substitute the derivative of the relativistic mass. We can do so by
considering the kinetic energy derivative which gives the power required for acceleration. The
kinetic energy is calculated from the Einstein's formula E=mc2. This energy is velocity dependent
(through the relativistic mass m), so the difference of E at different velocities gives the kinetic
energy. We will return to this problem when we start the quantum mechanics and derive this
equation. Anyway, the kinetic energy reads:
2
E K =m u c −m 0 c
2
the power, i.e. the time derivative of the kinetic energy is:
 ⋅
p= F
u=
dE k dmu 2
=
c
dt
dt
Calculating from the above the dmu/dt and substituting in the force equation, we end up in:
 '=
F
1−
1
uxv
c
2
[
F x−
F
F
v
F x u x F y u y F z u z ] i  y j z k
2 [


c

]
`
Now there is a huge amount of work in expressing the ui by the ui'. From the equation (*) on the
previous page we have:


u y = 1−
u z = 1−
ux v
c2
ux v
c2


u' y
u' z
Which allows to write the F'x as:
F ' x =F x −
v
 F y u ' y F z u ' z 
c2
1
To transform Fy and Fz we need to calculate ux and substitute in the

 1−
uxv
c
2

:

u x −v= 1−
ux v

u'x
c2
vu ' x
u x=
u' v
1 x2
c
2
c vu ' x 
u x= 2
c u ' x v
or, even better if we have:
ux v
c2
=
v 2u ' x v
c 2u ' x v
1
Now, let us substitute this to the

 1−

uxv
c
2

:

v2
v2
1−
c2
c2
= 2
u v c u ' x v−v 2u ' x v
1− x2
c
c 2u ' x v

1−
 
 
v2
v2
1−
u'x v
u' xv
c2
c2
=
1
=
1
2
u v
c
c2
v2
1− x2
1− 2
c
c
1−
 

After this is done, a formula is obtained:
u' v
 ' = F x i  F y j F z k − v2 [ F y u ' y  F z u ' z ] i  x2 [  F y j F z z ]
F
c
c
which can be, after some tricky observations reduces to:
 '=F
 ' 1 F
 '2
F
 ' 1=F x i  F y j F z k
F
v 

F ' 2=u ' × 2 × F
'1
c
[
]
You can check this out easily by performing the cross products using

A×
B= i  A y B z −A z B y  j  A z B x −A x B z  z  A x B y − A y B x  , taking v =[−v ,0 ,0] directed
along the x axis (containing only the vx component):
v × F 1=−v  F z jv  F y z
u ' ×v × F1 =i u ' y v  F y −u ' z v  F z  j u ' x v  F y  z u ' x v  F z

Which reproduces the desired terms (you still need to divide them by c2). Applying this relation to
the Coulomb law:
=
F
q0 q
4  0 r
3
r
yields for low velocities (   1 ) simply the unchanged Coulomb force as
 ' 1=
F
q0 q
4 0 r 3
 '1 :
F
r '
But there occurs a second, u' velocity dependent component in the moving coordinate frame:
 ' 2 =q 
F
u '×
[
q0 v ×r '
2
4 0 c r '
Recall now that the Lorentz force (magnetic force) is
3
]
 =q u ' × 
F
B . We can rearrange the terms in
 ' 2 in such way to put q u ' befor the cross product, and rest of the terms after the cross
F
product, identifying the magnetic field as:

B=
q 0 v ×r '
4 0 c 2 r ' 3
Because as the charge moves an infinitely small distance dl, it can be considered to be a current
 , and since c 2= 1 (recall the electromagnetic wave equation), we can finally state
q 0 v =I dl
0 0
that:

 0 I dl×r '
dB=
4 r ' 3
which is the Biot-Savart law! Difficult, isn't it? But you don't have to remember it all. Just
remember that the magnetic field occurs as a relativistic transformation of the electric field.
Reference: Artice. M. Davis: From Coulomb to Biot-Savart via relativity. San Jose University,
http://www.engr.sjsu.edu/adavis/Web02/EE140.htm
Questions
1. What is the relativistic reason to introduce the Lorentz force? Define the Lorentz force.
2. Discuss the Ampere's law in both forms and derive the equation for a magnetic field around
an infinitely long wire.
3. Derive the fromula for a magnetic field in an infinitely long solenoid.
4. Define the law of Faraday's induction and define the magnetic flux.
5. *Derive the electromagnetic wave equation (do not prove the vector identity).
6. Write down the Biot-Savart law and discuss qualitatively how does it relate to the
Coulomb's law (without calculations).
17. Quantum optics.
Major reference: Z. Kleszczewski, Fizyka kwantowa, atomowa i ciała stałego, Wyd. Pol. Śl, 1997.
17.1. The Rayleigh-Jeans distribution of radiation modes
Consider a volume element in form of a cube with side length L.
How many waves fit to the cube betwen any given wavelength  and infinity? All of the waves
must have nodes on the edges of the cube; otherwise a destructive interference will attenuate them
after reflections. The condition for such a wave is:
L=
n 
2
The number of waves between the given wavelength and infinity follows from the above equation
(because it is equal to the number of half-wavelengths of the considered wave which fit in the
cube):
n=

2L

Since the wave can be a superposition of any modes of radiation in 3D, the total number of possible
waves with wavelength between  and infinity equals:
N = nx ny nz =
8L 3
3

The number of modes per unit volume follows after dividing by
n=
L3 and equals:
8
3

But this estimate is not entirely valid for the number of wave between  and infinity because
superposition of two (or three) waves with wavelength equal to  results in a net wavelength that
is less then  ! To understand this, consider a wave equation:
A r = Amax sin


2
2
2
x
y
z = Amax sin  k x x k y y k z z = Amax sin  k⋅r 
x
y
z
So instead of representing the wave in terms of the wavelengths, it is more convenient to consider it
in terms of their inverses, which form such called wave vector k. One immediately recognizes that
the resulting wave oscillates in the direction parallel to the wave vector k, where the dot product
reduces to zero, and the resultant wave vector is:
∣k∣2=k 2x k 2y k 2z
which implies by k =
2
that

1 1
1 1
= 2  2 2
2
 x  y z
Thus, a wave that has components  x = y = z = would have a net wavelength of =

,
3
and this is the origin of our difficulties. Instead of taking a volume of a cube in the space of inverses
of the waveleght, we should rather take the volume of a positive quarter of the sphere, which
assures that the net wavelength does not exceed  . Correcting our expression for n we obtain:
3
n=

14 2
4

=
83 
3 3
This should be multiplied by two because the wavelength can be polarized in two independent
directions, so:
n=
8
3 3
Because we are not actually interested in the number of radiation modes between  and infinity,
but rather in radiation modes at given wavelength, we shall differentiate this equation to obtain an
expression for the number of radiation modes at wavelength  :
dn=
Observing that =
8
d
4
c
c
and d =− 2 d  we can write:


dn=
8 2
d
3
c
We ignore the „minus” sign because it only denotes the „direction” of counting: either from top to
bottom, or from bottom to top. The sign has no importance for us.
Now, if we would assume classically that each mode of radiation contains an amount of energy
equal to kT, we would end up in an energy density formula derived by Raileygh and Jeans:
8 2 kT
 , T =
d
c3
The energy contained in a body in the radiation modes is proportional to  2 and is not limited by
anything... For large frequencies it becomes infinite! This is an unphysical effect that is called an
ultraviolet catastrophy.
17.2. The Planck's radiation formula
The Rayleigh-Jeans formula was derived under the assumption that each mode of the radiation can
be associated with an energy portion of kT. However, the quantum mechanics predicts that light is
quantized, and each single photon conveys an energy portion equal to E=hv (or E=ħω; h=6.64
10-34Js; ħ=h/2π). If so, then not all of the quants of the light are equally probable, because
probability to find a highly energetical particle is less than that, to find a low energetical particle.
This was known many years earlier and was formalized in the Boltzmann formula:
−E
p  E ~e kT
The energy emitted in portions (quants) of hv can contain at once either 0 quants or single quant hv,
or two quants 2hv, or three quants 3hv, ... n quants nhv. The average of these quantities equals to:
∞
〈 E 〉=
∑n=0 nh  e
∞
∑n=0 e
Denoting
x=e
−h 
kT
−nh 
kT
−nh 
kT
∑n=0 nh   e
∞
=
∑n=0  e
∞
−h  n
kT

−h  n
kT

, we have (notice that for n=0 the term in the summation of the numerator is
zero):
〈 E 〉=
h  x 12x3x 2...
1x x 2 x3 ...
Where in the numerator we have taken „x” in fron of the summation. The denominator is a
geometric series with sum equal to
such series
1
. The numerator in turn can be seen as a product of two
1−x
1x x 2 x 3...2 =12x3x 2... and thus has a sum equal to
1
.
1− x2
Taking this all together, the average energy reads:
〈 E 〉=
h x
h
= h
1− x
e kT −1
Multiplying this average energy per radiation mode by the density of radiation modes gives the
Planck formulua for the energy density of modes in a black body:
 , T =
8 h 3
c3
1
h
kT
e −1
For low frequencies, the denominator of the second term behaves like
reduces to that, predicted by Rayleigh-Jeans.
h
and the formula
kT
17.3. Emissive power
The Rayleigh-Jeans and Planck's formulas typically do not consider the energy density of the black
body but rather the emissive power of a body which measures the amount of energy [J] emitted in
an unit of time [s] by a surface S [m2]. Consider the following figure:
An element dV contains radiation in the amount of  dV , which spreads as quants of photons at
the speed of light c in all directions. Part of this radiation ( =
S cos 
 dV ) enters a pinhole and
4 r 2
exits from the black body. (a container with a pinhole is a great approximation of a black body: the
pinhole accepts all incoming wavelengths, just as an ideal black body should).
In a time interval equal to t, the radius r=ct (while dr=cdt). Therefore, substituting for r, and
expressing dV in terms of the spherical coordinate system (see figure), we obtain:
S cos 
 r d r sin  d dr
4 r2
S cos 
dE=
 d sin  d  dr
4
dE =
integrating the outcoming energy over all points within the black body that contribute to the energy
flux (i.e. which are inside of the sphere of radius r=ct), we obtain the total emitted energy equal to:

2
r
S
E=
∫0 d  ∫02 cos sin  d  ∫0 dr
4

S
cos2 
E=
2 −
4
2
Sr Sct
E= =

4
4
∣

2
r
0
Now it is the right time to introduce the emissive power:
=
E c
= 
St 4
With this relation, the Planck's Law reads:
3
 ,T =
2h
c2
1
h
kT
e −1
Remember this.
Reference: „emmisive power” term: Kirchhoff’s Law of Thermal Emission: 150 Years ,
Pierre-Marie Robitaille, Progr. Phys. 4, 2009.
17.4. The Wien's law
It is possible to rewrite the emmisive power in terms of the wavelength instead of in terms of the
frequency. We can utilize the relations:
c=
c
=

−c
d = 2 d 

Where we need the differential relation as well because the emmisive power deals with power
emitted at frequency ±d  (recall the derivation of the energy density—it was a derivative with
respect to lambda and we did there a similar change in the variables). Substituting, we obtain:
 , t=
2 h c 2
5
1
e
hc
kT 
−1
The Wien's law determines the wavelength of maximum emmission in terms of this equation. We
can now substitute
x=
hc
and then calculate a derivative
kT 
d  x
=0 . It should be compared
dx
to zero because we are searching for the maximum emission wavelength. The equation reads:
5
5
'
 
5
2k T
x
=0
4 3
x
h c
e −1
5x4 e x −1−x 5 e x
=0
x
2
e −1
x
x
5e −5− xe =0
Where in the second equation we remove the „temperature dependent” constant (we are interested
in the maximum wavelength at fixed temperature), and we have removed the x=0 term that
corresponds to an infinite wavelength. To solve this equation it is possible to use the Newton's
method that is based upon the Taylor's series expansion:
f  x 1 = f  x 0  f '  x 0  x
If we want to find a solution of f(x)=0, we can put f(x1)=0 in the above, so that:
0= f  x 0  f '  x 0 x1− x 0 
f  x0 
x 1= x0 −
f '  x0 
And in general, because the Taylor series is just an approximation, we can expect that f(x1) will not
exactly be equal to zero, so we repeat the procedure until a sufficient precision is obtained:
x n1= x n−
f  x n
f '  xn 
In our case,
f  x=e x 5− x−5
f '  x=e x 5− x−e x =e x 5− x−e x
Substituting to the Newton's formula gives (simplifying by the exponential):
5− x−5 e−x
x n1= x n−
4− x
so, using the calculator you first enter some initial value of x, for example 10:
[1][0][=]
Then, you start iterations of the expression:
[Ans]-[(][5][-][Ans]-[5][x][shift][ln][Ans][)][:][(][4][-][Ans][)]
Now, after you press a few times the [=] button (10 times in my calculations), you will finally
obtain x=4,965. This corresponds to
 max T =b
b=2.898⋅10−3 mK
Which is the Wiens's law. It is of great use in astronomy to determine the temperature of stars, but
also in the temperature measurements in the modern infra-red thermometers.
17.8 The Stefan-Boltzmann law
Let us start again from the Planck's law. This time we can use the frequency form. We will
substitute:
x=
h
kT
kT
dx=d 
h
Substituting this to the Planck's distrubution we obtain:
1 dP 2  x 3 k 4 T 4 1
=
2 3
x
S dc
c h
e −1
1 dP 2 k 3 T 3 x 3
= 2 2
x
S dx
c h
e −1
 x , T =
Integrating this expression with respect to dx, it is possible to calculate the total power emitted by
the black body in all frequencies:
P
2  k 4 ∞ x3
=T 4 2 3 ∫0 x
dx
S
c h
e −1
The integral is just a constant (which can be evaluated numerically, for example in the C
programming language:
#include<stdio.h>
#include<math.h>
void main()
{
double integral=0,x,dx=0.01;
for(x=dx;x<1000;x+=dx) integral+=x*x*x/(exp(x)-1)*dx;
printf(„s=%e\n”,integral*2*3.1415*pow(1.38E-23,4)/pow(3E8,2)/pow(6.64E-34,3));
}
Which should give approximately the Stefan-Boltzmann constant, equal to:
−8
=5.67⋅10
W
.
m2 K 4
If you don't like programming, you can estimate the integral using your calculator by evaluating the
value of the integrand for integer values between 0 and, say, 12, and then sum up the values. The
result is already quite accurate.
Finally, the Stefan Boltzmann law states that the power emitted from a black body per unit
emission surface equals to:
P
= T 4
S
Questions
1. How would you calculate the possible number of waves with wavelength between  and
infinity?
2. Describe the UV catastrophy in the Rayleigh-Jeans law. What was the idea of Planck to
avoid the catastrophy (do not derive the Planck's formula here!)?
3. *Derive the Planck's radiation formula.
4. Explain the meaning of emissive power.
5. *Derive the Wien's displacement law from the Planck's formula (without the numerical
details like Newton-Raphson method).
6. Derive the Stefan-Boltzmann law from the Planck's formula.
18. Quantum mechanics
18.1. The de'Broglie hypothesis
One of the most important discoveries that started the development of quantum mechanics was the
de Broglie hypothesis that associates a wave to each mass particle:
=
h
p
Where  is the wavelength of this „wave of matter”, h is the Planck's constant and p is the
momentum of the considered mass particle. This relation finds great confirmation in diffraction
experiments, where for example the electrons scattered on two slits produce interference patterns,
just like if they would be waves.
18.2. The Schroedinger equation
If we assume that the matter behaves as a wave, we should be able to associate a wave function to
the particle:
= x cos t
We can now introduce this equation to the wave equation,
∂2  1 ∂2 
=
∂ x 2 v 2 ∂t 2
which results in (assuming v = f and =2  f ):
∂2 
2
=− 2 
2
∂x
v
2
∂ 
4 2 f 2
=−

∂ x2
2 f 2
∂ 2  −4 2
=

∂ x2
2
Now we will employ the de'Broglie formula, which gives:
∂ 2  −4 2 p2
=

∂ x2
h2
In classical regime we can relate the momentum to the kinetic energy (K=E-U, where U is the
potential energy). Because
mv 2 p 2
,
K=
=
2
2m
2
∂ 2  −2m 4   E−U 
=

∂ x2
h2
or:
2
ℏ2 d  x
 E −U   x =0
2m ∂ x 2
which is the stationary Schroedinger equation. Please note that we have replaced
= x cos t by single  x  after dividing the equation by cos  t which results in
elimination of the time dependency. Therefore, we call this equation a stationary version.
Furthermore we have replaced the Planck's constant with the „h-bar” constant, ℏ=
h
.
2
18.3. The intepretation of the wavefunction
The wavefunction by itself is not an observable. It cannot be directly observed. It can be considered
a sort of a method of description of the quantum phenomena, but not a directly measurable quantity.
However, there is a quantity directly related to the wavefunction, which has a measurable
interpretation and was proposed by Born. It states that the square of the wavefunction, i.e.:
p= x ∗  x 
measures the probability density to find a quantum particle at position x. Because this is a
probability density, its integral over the available space must normalize to 1 (the particle must exist
somewhere in the space):
∞
∫−∞   x ∗  x  dx=1
18.4. The Bohr model of the atom
The Schroedinger equation is very useful to obtain exact solutions to the quantum problems and
valuable theoretical formulas. There is however also an older quantum model that deserves
attention. It is the Bohr's model of the hydrogen atom. It fails for other atoms than hydrogen, but it
gives a great insight into the nature of behaviour of the electrons on the atomic orbitals. It gives a
sort of intuition that gives you the understanding of several phenomena that you encounter in the
future and which would be very difficult to understand in terms of a differential equation like the
Schroedinger equation (even though it is far more precise than the Bohr's model).
Bohr considered that electrons of an atom follow circular orbits around the nucleus. Today we know
this is not true—the electron does not circulate in classical meaning, but it behaves like a standing
wave of probability and it can be found anywhere along the orbital at given time moment. However
it has certain properties of the Bohr model, like angular momentum, etc.
Bohr considered that electrons are attracted to the nucleus by the electrical force
F=
kZe 2
r2
and
mv 2
. Additionally, he was forced to
F=
r
this force is balanced with the centrifugal force
introduce quantized angular momentum: mvr =n ℏ .Balance of forces leads to the allowed orbit
radii:
∣
mv 2 kZe 2
= 2 ⋅m r 3
r
r
2
2
2
m v r =mkZe2 r
n 2 ℏ2
r=
mk Z e 2
It is possible to calculate the energy of electrons following such orbits:
mv 2 kZe 2
−
2
r
2
mv r kZe 2
E=
−
r 2
r
2
kZe kZe 2
E=
−
2r
r
−kZe 2
E=
2r
E=
In the third line we made use of the force equality, considered above. Substituting the formula for r:
E=
−mk 2 Z 2 e 4
2 n2ℏ2
This is sometimes written in a more useful form using the Rydberg Energy as:
Z2
E=−R E 2
n .
R E =13.6eV
18.4. The particle in the box
Consider the following energy landscape:
This energy landscape of a quantum particle corresponds to a „particle in the box”. The particle can
reside in any position of the interval (0,L) (a box), but it cannot fall outside due to the large (in
theoretical approximation—infinite) potential barriers.
Now let us take a look on the Schroedinger equation:
2
ℏ2 d  x
 E −U   x =0
2m ∂ x 2
We can rewrite it in a more convenient form of a haromic oscillator:
d 2  x  −2m
= 2  E−U  x
d x2
ℏ
where the term −
2m
 E−U  corresponds to the angular velocity  20 , or rather, because this
ℏ2
is an x-dependent problem (not time dependent), it corresponds to the wave vector k 2 . The
periodic solution can be found in terms of sines, cosines or in terms of complex exponentials
(because e i =cos −i sin  ). This time, we prefer to search for the solution in terms of sines and
cosines:
 x = Acos kxB sin kx
2m
2m E
k=
 E−U = 
2
ℏ
ℏ

where we could omit U because it equals 0 in the considered interval where the particle exists. It
goes without saying that the wavefunction should vanish on the interval boundaries. This sets A=0.
Then, to assure the vanishing of  L , a condition must be met that:
kL=n 
 2mE L=n 
ℏ
n 2 ℏ 2 2
E=
2m L2
So, the particle in the box displays quantized energy levels, just like the hydrogen atom of Bohr!
It is also possible to determine the constant B using the normalization condition:
B
2
L
∫0 sin 2 kx dx=1


L
x sin 2kx
B
−
=1
2
4k 0
L
B 2 =1
2
2
B=
L
2

Where in the integral of the sine we made use of the fact that it vanishes for x=0 and for x=L (due to
quantization condition).
18.5. The tunnel effect
The tunnel effect predict a possibility of transmission of a quantum particle through a potential
barrier of finite width that is greater than the energy of the particle. In classical terms this would
mean—for example, that a body that has kinetic energy
mv 2
mgh on a ground level, can pass
2
with nonzero probability a mountain of height h. Of course, for classical problems this probability is
extremely small (far beyond the scale of measurement). To start, let us sketch the considered energy
landscape which divides into three parts:
One can see that the wavefunction should oscillate in regions I and III, while it should be damped in
the region II. This follows immediately from consideration of the concavity/convexity of the
wavefunction (measured by the second derivative to remind the math) for negative or positive
values of  taking into account the sign of the (E-U) term. To solve this problem, we consider
continuity of the wavefunction for x=0 and x=L. This means the value of  and its
derivativeshould be equal on both sides of neighbouring regions.
This time for each region we will consider an exponential solution of form:
 x =A e−ikx B e ikx
 2m E−U 
k=
ℏ
The negative exponent corresponds to a wave propagating to the left, and the positive exponent
corresponds to a wave that propagates to the right (to understand this, consider the full, time
dependent picture e−i kx  t  ). This leads to:
 I =A1 e−ikx B 1 e ikx , k = 
2mE
ℏ
 2m U −E 
 II = A2 e− x B2 e  x , =
ℏ
2mE
 III = A3 e ikx B3 e ikx , k = 
ℏ
In region III there is no reflection, so A3=0. The continuity equations read:
1 A1 B1= A2B 2
2 A2 e − L B 2 e L= B3 eikL
3 −ikA 1ikB 1=− A2 B2
4 − A2 e− L  B2 e  L =ik B3 eikL
We have 5 unknowns but only 4 equations. Fortunatelly, we don't need all of the constants to have
∗
 
B 3 B3
the transmission coefficient T =
B 1 B1
divided coefficients will be denoted:
. So we can divide each of the equations by
Ai
= i and
B1
A1 .The
Bi
=i :
B1
1 11= 22
2  2 e− L 2 e  L =3 e ikL
3 −ik 1ik =− 2 2
− L
L
ikL
4 − 2 e  2 e =ik 3 e
We can eliminate 1 from (3) after substituting from (1) the expression 1=2 2−1 :
2  2 e− L  2 e  L =3 e ikL
3 −ik  2−ik 22ik=− 2 2
− L
L
ikL
4 − 2 e  2 e =ik 3 e
Now we can substitute 2 4 2 and 2− 4 4 :
 2 2  2 e  L=ik  3 eikL
3 −ik  2−ik 22ik=− 2 2
− L
ikL
 4 2  2 e =−ik  3 e
The first and the last equation allow to calculate 2 and 2 to substitute to (3), which we
rearrange before doing that:
ik

2 3
3 2ik=2 ik −2 ik
ikL L −ik
4  2=e

2 3
 2 2=eikL− L
After the substitution, we will have only one unknown, 3 , which being squared gives us the
transmission coefficient:
[
2ik= −e ikL L
2
2
]
−ik 
ik 
e ikL− L
3
2
2
This results in:
3=
4ik  e−ikL
e− L ik 2−e  L −ik 2
Squaring, and taking a limit where the negative exponential in the denomiator can be neglected,
being small compared to the other term, we obtain (remember about the sign change close to i in the
complex conjugated denominator):
2
2
 
2L
16 k 
E
E −ℏ
T =3  = 2 2 2 e−2  L =16
1−
e
U
U
 k  
∗
3
 2m U − E 
18.6 The hydrogen atom s orbital
We are now interested in a solution of the 3D Schroedinger equation for a symmetrical
wavefunction of the hydrogen atom (the s-orbital). The Schroedinger equation in 3D is just a
straightforward generalization of the 1D equation. Instead of:
2
ℏ2 d  x
 E −U   x =0
2m ∂ x 2
we write:
(*)
ℏ2 2
∇  x , y , z  E−U  x , y , z =0
2m
To consider the s orbital of the hydrogen atom (a spherically symetrical wavefunction), it is
convenient to introduce spherical coordinates to describe the problem. To proceed we need to
2
convert the Laplacian ( ∇ =
∂
∂
∂
 2  2 ) to the spherical coordinates, which is a bit
2
∂ x ∂ y ∂z
complicated. The spherical coordinates look like the following:
The coordinates obey the following relations:
x=r sin cos 
y=r sin sin 
z=r cos 
r = x 2  y 2z 2
(**)
z
cos =
r
y
tan =
x
Now, let us consider a single x derivative of some function,
∂ f ∂ f ∂ r ∂ f ∂  ∂ f ∂
=


∂ x ∂r ∂ x ∂ ∂ x ∂ ∂ x
We need to have the derivatives of the polar coordinates. By the last three formulas we have:
∂r
2x
x
=
= =sin cos 
2
2
2
∂ x 2  x  y z r
then to obtain a derivative of
∂
, we can calculate the derivative of the 5-the equation of (**):
∂x
z
 x  y 2z 2
∂  −2xz
−sin 
=
∂ x 2 r3
∂  r 2 sin  cos cos 
sin 
=
∂x
r3
∂  cos cos 
=
∂x
r
cos =
To obtain
2
∂
we make use of the last equation of (**):
∂x
y 1 ∂ − y
=
x cos 2  ∂ x x 2
1 ∂  −r sin sin 
=
cos2  ∂ x r 2 sin 2  cos2 
∂  −sin 
=
∂ x r sin 
tan =
The next step is to consider the:
∂ f ∂ f ∂ r ∂ f ∂ ∂ f ∂ 
=


∂ y ∂r ∂ y ∂  ∂ y ∂  ∂ y
the derivatives can be found again by the equations of (*). From the 4-th equation of (**) we have
(per analogy to the x derivative of r):
∂r y
= =sin sin 
∂y r
From the 5-th equation of (**) we have
z
 x  y 2z 2
∂ −2yz
−sin 
=
∂x
2r 3
∂  cos  sin 
=
∂x
r
cos =
2
From the 6-th equation of (**) we have:
tan =
y
x
1 ∂ 1
1
= =
2
cos  ∂ y x r sin cos 
∂ cos 
=
∂ y r sin 
Finally, consider the:
∂ f ∂ f ∂r ∂ f ∂ ∂ f ∂
=


∂ z ∂r ∂ z ∂  ∂ z ∂  ∂ z
From the 4-th equation of (**) we have:
∂r z
= =cos 
∂z r
From the 5-th equation of (**) we have:
z
 x  y 2 z 2
∂ 1 2z 2
−sin  = − 3
∂z r 2r
∂
−1
cos 2 
=

∂ z r sin  r sin 
∂ −sin2  −sin 
=
=
∂ z r sin 
r
cos =
2
where the last transition was done using the trigonometrical identity sin 2 cos 2 =1 .
Now having these formulas for the transformation of derivatives to spherical coordinates, we can
transform the laplacian of our wavefunction  x , y , z . Assuming it is symmetrical, the
derivatives with respect to angles vanish, and we have:
∂ ∂
=
sin cos 
∂ x ∂r
∂ ∂
=
sin sin 
∂ y ∂r
∂ ∂
=
cos 
∂z ∂r
To calculate the second order derivatives, we must substitute the above equations to the relations for
a first derivative. This time the angular derivatives do not vanish, because we have sines and
cosines!
∂2 
∂2  ∂ cos 2  cos2  ∂ sin 2 
2
2
=sin
cos



r
∂r r
∂ x2
∂ r2 ∂ r
2
2
2
2
∂ 
∂  ∂  cos sin  ∂  cos2 
2
2
=sin
sin



r
r
r
∂ y2
∂ r2 ∂ r
2
2
2
∂ 
∂  ∂  sin 
=cos 2 

2
2
∂r r
∂z
∂r
summing up, and making use of the trigonometrical unity relation, we end up with:
2
∇ =
∂ 2  2 ∂

2
r ∂r
∂r
or:
2
∇ =
1 ∂ 2 ∂
r
r2 ∂ r ∂ r
or:
∇ 2 =
1 ∂2
r
r ∂ r2
(you can easily check the alternative forms by applying the differentiations). If we would not
assume that  is independent of the angles, the Laplacian would be more complicated:
∇ 2 =
{
1 ∂2
1 1 ∂
∂
1 ∂2 
r

sin


r ∂r 2
∂  sin 2  ∂ 2
r 2 sin  ∂ 


}
The Schroedinger equation (*) with the simplified version of a spherical Laplacian reads now:
1 d2
2m
r  2  E−U =0
2
r dr
ℏ
Which after substitution of the Coulombic energy U =
−kZe 2
gives:
r


1 d2
−2m
kZe 2
r = 2 E

r dr 2
r
ℏ
We will now rescale the variables to simplify the equation. We will express r by the multiplicity of a
Bohr radius r B and E by the multiplicity of Bohr's energy of the first orbit ( E R , the Rydberg
constant):
2
r =r B = ℏ 2 
mkZe
k 2 Z 2 me 4
E= E R =

2 ℏ2
After the substitution we have:
[
]
2


mkZe 2 1 d 2
−2m k 2 Z 2 me4
mk 2 Z 2 e 4

=


2
2
2
2
 d 2
ℏ
ℏ
2ℏ
ℏ 
Which reduces easily to:
2
 
1 d
2
=−  
 d 2

or:
 
d
2
=−   (****)
2

d
clearly eq. (****) is an equation for
f =  , so we rewrite it as:
d2 f
2
=−  f
2

d
 
The above formula is clearly solved for some product of an exponential term with another function,
since the exponential term after differentiation will cancel on both sides. So we take
f =e−  g  , Substituting to the equation we obtain:


d2 g
dg
2
−2 
  2 g=0
2

d
d
 is a free parameter, so we can set it equal to 2=− which will reduce the equation to the
form:
d2 g
dg 2
−2 
 g =0 (*****)
2
d 
d
This equation is suitable for a solution by means of the power series. Assume that
∞
g  =∑ k=1 a k k , so
∞
dg
=∑k =1 a k k k −1
d
and
∞
∞
d2 g
=∑k =2 a k k k −1 k−2=∑ k=1 a k 1 k 1 k  k−1 . Substituting to (*****) we have:
2
d
∞
∑k =1  k 1 k a k1 k −1−2  k a k k −12 a k k−1=0
The terms with the same power in  must cancel to zero, which gives a condition:
k 1 k a k1−2  k −1a k =0
2  k −1
a k1=
a
k k 1 k
2
(******)
ak≈
a
k k
2 k
ak≈
k!
Where the last two relations are written for k sufficiently large. The last formula shows that the
power series is just an expansion of a function
g  =e 2   . This gives a wavefunction
e−  e 2   e  
=
=


Such wavefunction grows with the radius  , predicting the probability to find an electron being
the largest at infinitely large distance from the nucleus! An error? No. Everything depends upon
 . If we set it equal to =
1
, where n is a positive integer, then the iterative formula number
n
2 of (******) will produce zero at k=n. The series will become truncated, and the exponential term
will not be formed! Now, remembering that =− 2 (where  is the multiplicity of the
hydrogen atom energy with respect to the energy of the base state of Bohr model), we can see that
the energy must be quantized to obtain a physically meaningful solution!
HOW TO MEMORIZE THE KEY POINTS OF THE DERIVATION?
●
We write down the Schroedinger equation in spherical coordinates with potential energy
resembling the Coulomb interaction between the electron and nucleus,
●
We transform the equation to new variables by assuming r =r b  and
●
We observe that the equation does not depend on  alone, but rather on   , so we
make a substitution
E= E R  ,
f =  and we obtain an equation for f.
●
We assume the solution in form
●
The resulting equation for
f =g e− 
g   is solved by assuming a series solution for
∞
g =∑ k=0 a k k
●
Substituting this expansion to the differential equation, we must set =
1
(with n integer)
n
to truncate the expansin which otherwise converges to an exponential, yielding non physical
solution (with most probable position infinitely away from the nucleus).
●
The condition for  yields the quantization rule for energy states,
E=
ER
2
n
18. 7. The angular dependency of the hydrogen atom
[the start of this reasoning after the lecture notes for course SS 06-20-102 Freie Universitat Berlin,
I could not establish the author.]
To handle the angular dependency of the hydrogen atom, we need to consider the full form of
Laplacian in spherical coordinates:
2
∇ =
{
1 ∂2
1 1 ∂
∂
1 ∂2 
r

sin


r ∂ r2
∂
r 2 sin  ∂
sin 2  ∂ 2


}
The Schroedinger equation reads now:
{

}

1 ∂2
1
1 ∂
∂
1 ∂2  −2m
kZe 2
r  2
sin 
 2
= 2 E

r ∂ r2
∂
r
r sin  ∂
sin  ∂ 2
ℏ


Assuming that the wavefunction is composed of three independent component functions,each
depending only on a single variable, r ,  , =R r  P  F  , we can separate the variables
in Schroedinger equation. First multiply both sides by r 2 :
{
2
r

}
2

∂
1 ∂
∂
1 ∂  −2m  2
r 
sin 
 2
= 2 Er kZe2 r 
2
2
sin

∂
∂
∂r
sin  ∂
ℏ
We can see that the large brace contains terms that do not depend on radius! Let us rearrange the
terms:
r
{
∂2
2m  2
1 ∂
∂
1 ∂2 
2

r

Er
kZe
r
=−
sin


sin  ∂
∂
∂ r2
ℏ2
sin 2  ∂ 2


}
Now, introducing the desired form of the wavefunction, we obtain:
PF r
{
∂2
2m  2
F ∂
∂P
P ∂2 F
2

r
R
Er
kZe
r
RPF
=−R
sin


sin  ∂ 
∂ sin 2  ∂ 2
∂ r2
ℏ2


}
Dividing by PFR, we separate the variables as:
{
2
PF r


2
∂
2m
F ∂
∂P
P ∂ F
r R 2  Er 2 kZe 2 r =−R
sin 
 2
2
2
sin

∂

∂
∂r
ℏ
sin  ∂
}
The left hand side depends on different variables than the right hand side. When one is changing
variables on the left hand side, the variables on the right hand side may remain unchanged. The only
possibility to satisfy this equation is thus to have the left hand side (and the right hand side) equal to
a constant:
2
2
r ∂
2m
L
r R 2  Er 2kZe 2 r  = 2
R ∂r2
ℏ
ℏ
(*)
2
1
∂
∂P
1
∂ F −L2
sin 

= 2
P sin  ∂
∂
F sin 2  ∂ 2
ℏ


If we divide the upper equation by
2mr 2
, the „L” term will gain the units of energy, and we will
ℏ2
be able to provide an interpretation for this term:
2
2
2
1 ∂
kZe
L
r R E
= 2 (**)
2
rR ∂ r
r
mr
If one would assume
L=mvr (the classical analogy to angular momentum), then on the right
hand side the term would correspond to the kinetic energy of rotational motion. So L represents the
angular momentum. If L=0, one obtains the equation for an s-orbital that we have considered in the
previous paragraph.
The second equation of (*) can be again divided into two equations. The only dependency upon
 is found in the second term on the left hand side. It is obvious that in case of fixed  the
value of the  -dependent term cannot change if it is supposed to satisfy the equation. So we
must set it equal to a constant, and we will call the constant ml . In such way, the second equation
of (*) divides into:
2
ml
1
∂
∂P
−L 2
sin 
−
=
P sin  ∂
∂  sin2  ℏ 2
(***)
1 ∂2 F
2
=−ml
F ∂ 2


The second of these equations can be written in form of a Simple Harmonic Oscillator:
2
∂ F
=−m2l F
2
∂
Which provides a periodic solution such as
F =F 0 ei m  . Please note that in order to have
l
F(Φ)=F(Φ+2π) we must have integer m, i.e. the angle of F must be an integer multiplicity of 2π.
In this way we introduce the quantization of m!
The first equation is more difficult to handle. It results in a differential equation that is called a
Legendre Differential Equation. To see this we substitute
(***):
x=cos  in the second equation of
x=cos 
dx=−sin d 
−dx
d =
sin 
d
d
=−sin 
d
dx
m2
1 d
dP
−L 2
1− x 2 
− l 2= 2
P dx
dx 1−x
ℏ
Rearrainging the terms, we obtain the associated Legendre differential equation:


m2l
d
L2
2 dP
1−x  P 2 −
=0
dx
dx
ℏ 1−x 2
To start the analysis, we will consider a case with ml =0 , which results in the primary (not
associated) Legendre differential equation:
d
dP
L2
1−x 2 
=P 2 =0
dx
dx
ℏ
To make this suitable for a solution using power series method, we evaluate the derivative:
1−x 2 
d2P
dP L 2
−2x
 P=0
dx ℏ 2
dx 2
Here we can substitute the power series representation:
∞
P=∑i =0 a i x i
2x
∞
∞
dP
=2x ∑ i=1 i a i x i−1=∑i=1 2 i a i x i
dx
∞
∞
d2 P
=∑ i=2 ii −1a i x i−2=∑i =0 i2i1 a i2 x i
dx
2
∞
2 d P
−x
=∑i =2 −i i−1 ai xi
dx
Which results in:
∞
∞
∞
∞
∑i=2 i2i1 ai 2 xi −∑i=0 i i−1a i x i−∑i=1 2i a i x i∑i=0
2
L
a x i=0
2 i
ℏ
For i>1 all the sums provide a term, and we can compare the coefficients under the same power of
x i which gives a recursion formula:
L2
i2i1a i2 −ii−1 a i−2ia i  2 a i =0
ℏ
2
L
ii−12i− 2
ℏ
a i2=a i
i2i1
L2
i i1− 2
ℏ
a i2=a i
i2i1
Obviously, the series of coefficients a will be finite ONLY if the numerator of this expression
takes the value of 0 at some point. So we have a condition for the possible values of the
constant:
L2
=l l1
2
ℏ
We cannot determine the coeffcients a0 and a1 with the above formula. These are the integration
constants. Assuming the first equal to zero, we obtain an odd power series, otherwise the series will
be even. Two linearly independent solutions, exactly as we expect for the second order differential
equation.
Assuming noninteger constants for
L2
ℏ2
results in a series that diverges at
x=±1 (at angles of
0 or 180 degrees). You could check this by considering the behaviour of the following ratio for large
i:
2
L
i i− 2
a i2
ℏ
i 2i
2i−2
2
= 2
≈ 2
=1− 2
≈1−
ai
i
i 3i2 i 3i2
i 3i2
2
This behaviour is identical to the harmonic series! In the harmonic series we have:
a i2
i
2
=
=1−
a i i2
i
Thus, the considered series diverges like the half of harmonic series if the constant does not cut the
series. [this proof after the notes of Colin Wilkin course Mathematical Methods in Physics and
Astronomy, Chapter 4 Legendre Functions; Univeristy College, London]
HOORAY! We have quantized the angular momentum! But this is not the whole solution of the
angular problem. In general we must handle the associated Legendre differential equation, where
2
ml 0 . To handle this we first write a handful formula for the Legendre polynomial (Rodriguez
formula):
P 0=1
n
2
n
1 d  x −1
P n  x = n
2 n!
dx n
Which shows that the expression to be differentiated is a polynomial of order 2n. After n-fold
differentiation, we obtain a polynomial of degree n. Remember this. Now back to the associated
Legendre equation:


m2l
d
L2
2 dP
1−x  P 2 −
=0
dx
dx
ℏ 1−x 2
For integer values of m this equation is solved by an associated Legendre function, that reads:
m
P l ,m  x=1− x 2 2
dm
Pl  x 
dx m
i.e. it is constructed out of the ordinary Legendre polynomials.The structure of these functions is
given in tables, but that is not important now. Important is that we differentiate m-times a
polynomial of degree n! m cannot exceed n in order to have a nonzero solution! Another known
result of the quantum mechanics of hydrogen atom!
The first few associated Legendre polynomials read:
P 0,0  x=1
P 1,0  x=x =cos 
P1,1  x=− 1− x 2=−sin 
P 2,0  x=0.5 3x 2−1=0.53cos 2 −1
P 2,1  x =−3x  1−x 2 =−3 cos  sin 
P 2,2  x=31−x 2 =3 sin2 
Recall that the full angular wavefunction is a product of Y l , m = F m  P l , m  , where
l
F =e i m  . Thus:
l
l
l
Y 0,0  x =1
Y 1,0  x=cos 
Y 1,1  x =−sin  ei 
P 2,0  x=0.5 3cos 2 −1
i
P 2,1  x=−3 cos sin  e
P 2,2  x=3 sin 2  ei 2 
To be able to draw the plots of the wavefunction we need the R component that describes the radial
dependency. The equation was:
r ∂2
2m  2
L2
2

r
R
Er
kZe
r
=
2
2
R ∂ r2
ℏ
ℏ
1 ∂2
2m
kZe 2 L2
r
R
E
− 2 R=0
r ∂ r2
r
ℏ2
r


Using the same substitution that we made previously ( E= E R , r = r B ), we obtain:


1 d2
2 L2
=−

−

 d 2
 2
The only difference is the introduction of the term
2
∞
∞
−l l1
−L
=
 −l l 1 ∑k=1 a k  k−2=−l l1 ∑ k=0 a k1 k−1
2
2


This term needs to be added to (******) in paragraph 18.6, resulting in:
[−l l1k 1 k ]a k1−2 k −1 a k =0
2 k−1
a k1=
a
k  k1−ll1 k
recalling that =1 /n , we have:
a k 1=
2 k /n−1
a (****)
k k 1−l l1 k
In addition, there is only one term that is multiplied by −1 ( a 1 −1 ), which indicates that
a 1=0 in the expansion. Therefore by eq. (****) all following a k =0 . However, for k=l the
denominator becomes infinite and the a l 1 cannot be determined by this formula [0/0 symbol]. It
is the only point where the coefficients can start be nonzero. Thus, the only nonzero coefficients of
the expansion in the radial part of the wavefunction are between l+1 and n. This gives a condition
for the angular momentum quantization: l must be less than n, since otherwise the
wavefunction is zero. Let's calculate the few radial wavefunctions (remember, the polynomial is
multiplied by e
−
n
and divided by  ):
R 1,0~e−
R2,0 ~1−/2 e
R2,1 ~ e
−
2
−
2
Combining R and Y into a complete wavefunction allows to plot „clouds” of the orbitals (below—
the vertical axis is the x axis, the horizontal axis is the z axis; the cloud is independent of 
because e−i m  e i m =1 ). The presented orbitals were generated using a computer program:
#include<pgm.h>
#include<math.h>
#include<stdlib.h>
double PSI100(double r, double teta){return pow(exp(-r),2);}
double PSI200(double r, double teta){return pow(exp(-r/2)*(1.0-r/2),2);}
double PSI210(double r, double teta){return pow(r*exp(-r/2)*cos(teta),2);}
double PSI211(double r, double teta){return pow(r*exp(-r/2)*sin(teta),2);}
void main(int argc, char *argv[])
{ gray **img;
double x,y,z,r,teta,fi,p;
int rows=600,cols=800,i,j; long t=0;
pgm_init(&argc,argv); img=pgm_allocarray(cols,rows);
for(i=0;i<rows;i++) for(j=0;j<cols;j++) img[i][j]=0;
for(t=0;t<1E6;t++) {
teta=6.282*rand()/RAND_MAX; r=9.9*rand()/RAND_MAX;
p=PSI200(r,teta)*1000;
y=r*cos(teta)*30; x=r*sin(teta)*30;
if(p+img[300-(int)y][400+(int)x]>255) img[300-(int)y][400+(int)x]=255;
else img[300-(int)y][400+(int)x]+=(gray)p;
}
pgm_writepgm(stdout,img,cols,rows,255,0);
}
A SHORT OVERVIEW OF THE KEY STEPS IN THE DERIVATION
●
We write down the full Schroedinger equation in spherical coordinates with Coulomb
potential energy term for the interaction between electron and the nucleus
●
We assume the solution in form =Rr  F  P  and substitute this to the
Schroedinger equation. This allows to separate the angular dependent terms from the radius
dependent terms (because two sides of the equation, each dependent on different variable
must equal to a constant,
●
L2
).
ℏ2
Analysing the meaning of the above constant in the radial equation, we conclude that
L2
is similar to the square of angular momentum.
●
The angular part can be further divided into separate equations for  and  after
introduction of a constant −m 2 for the  -dependent term.
●
The equation for  is a simple harmonic oscillator equation with frequency m. To assure
the constancy of the wavefunction under 360-degree rotation, m must be integer
(quantization of the magnetic quantum number).
●
The equation for  can be reduced by substitution of
x=cos  into associated Legendre
equation.
●
Solution of the associated Legendre equation in case of m=0 (which gives normal Legendre
equation), using power series method, gives a condition that to truncate the series of solution
coefficients
●
L2
=l l 1 , where l is integer.
2
ℏ
For m nonzero, the solution is given by an associated Legendre function which is nonzero
only if m<n. This gives the condition for the relation between m and n.
18.8. Magnetic moment of the orbiting electron in hydrogen atom
Recall the F component of the wavefunction for the hydrogen atom:
F =F 0 ei m 
l
Clearly, the higher value of m, the more „waves” are produced while travelling the  direction.
This means that momentum is circulating in this direction (de Broglie hypothesis!), producing
angular momentum aligned to the z axis.
p =
h

(*)
hr
hr
L z = p  r= =
=ml ℏ
 2 r /ml
The wavelength above was calculated by knowing the number of periods in the F wavefunction,
that is proportional to ml .
It is interesting to note that for ml =l the value of
angular momentum
L z =l ℏ is not equal to the length of the full
L= L2 = l l1 ! This is because of the uncertainty principle which does
not allow us to fully localize the momentum. The rest of the angular momentum stems from the Lx
and Ly components.
By the uncertainty principle of quantum mechanics (see further chapters) we cannot obtain any
other component of the angular momentum. Thus we can only have L (the length of the total
angular momentum) nad its projection to a single axis (most commonly Lz). The other components
attain various configuration (which however result in the length of angular momentum equal to L).
In general therefore the angular momentum is often visualised by a cone of possible orientations:
A circular loop of wire produces magnetic moment, that is defined by the formula:
pm= I 
A (**)
where I is the current in the loop and A is the area covered by the loop. Such dipole is a source of
magnetic field. For example consider the field produced by a loop that we have calculated on
exercises using the Biot-Savart law:
B=
 0 I R2
2 R2 z 2 
=
3/ 2
 0 pm
2 R 2z 23/ 2
where R is the circuit radius and z is the distance from the dipole in the direction parallel to the
normal to its surface. As R goes to zero, we obtain:
B=
0 p m
2 z 3
if you would put a dipole in external magnetic field, then it would receive a force that tends to turn
it around. Consider the following picture:
The loop is under the action of the Lorentz force:
 =I 
F
L×
B
Any possible instability in the position of the considered loop of current (solid line) causes a net
torque
=
2⋅L
ILB sin 90o sin 
2
=IL 2 Bsin 
that tends to turn the it upside down, until the force vectors start pointing outwards. The work done
under rotation equals to:


W =∫0 IL 2 B sin  d =−IL 2 B cos ∣0 =2I L 2 B=2p e B
So the energy change from the high energy state to the low energy state equals 2p e B . Therefore
B .
we say that the energy in magnetic field can be calculated as W = pe⋅
To calculate the magnetic moment of an orbiting electron, we will make use of definition (**).
p e =IA=
qv
q v  r 2 qvr q⋅mvr
q
r 2=
=
=
=
L (***)
s
2 r
2
2m
2m
Because we already know from the solution of the Schroedinger equation for hydrogen atom that
L2
=l l1 , and thus
ℏ2
L=ℏ  l l1 , so:
p e=
where B =
qℏ
−e ℏ
l l 1=

 l l 1=−B  l l1
2m
2m
eℏ
is the Bohr magneton, the basic unit of magneticity in atomic scale.
2 me
To calculate the energy of a magnetic dipole in external field, say in the z direction, one needs to
calculate the dot product:
W =− pe B̇z
W =−B l l1cos  B z
where  is the angle of the angular momentum L to the Z axis. But the projection of angular
momentum to the z axis is given by the magnetic number m, i.e.
L z =L cos  or
m=  l l1 cos  . Therefore, the energy of the orbiting electron in the external magnetic field
(we skip the index z for generality) reads:
W =−B ml B
Which justifies the name of this quantum number as the „magnetic quantum number”.
18.9. Fermions and bosons. Spin of the electron
There are two basic types of identical (indistinguishable) particles in quantum mechanics. Fermions,
and bosons. Bosons are particles that can exist in the same place of space at given time instant,
fermions cannot. A wavefunction that describes identical particles must give identical probability
distribution after the exchange of the particles, i.e.:
2 1,2= 2 2,1
This means that we have two options for such particles. Either:
1,2= 2,1
which is true for bosons (particles that can exist in the same point of space); or
1,2=− 2,1
which is true for fermions (particles that cannot occupy the same position in space). Why can't the
fermions exist in a single place? Because the wavefunction is antisymmetrical. Such wavefunction
must be equal to zero in the symmetry point (the origin of coordinate system where the symmetry
holds).
A phenomenon which is related to being a boson or a fermion is the spin of a particle. The spin is an
internal angular momentum of a particle that was discovered experimentally in the subtle structure
of atomic spectra and predicted by the relativistic Dirac equation (it appears that a new operator
introduced by the Dirac equation behaves just like the angular momentum operator [the same
commutation rules, relation for its square], although it has a completely different structure). It can
be classically imagined as „spinning” of the particle, but in reality it is a purely quantum property.
Just try to imagine how could an electron spin if it is a point particle? What is the velocity of
spinning if radius limits to zero? Is it in accord with special relativity? These are just a few
questions worth realizing.
Few yers ago I have proposed a heuristics that helps to understand the relation between symmetry
or anti-symmetry of the wave function and the value of a related angular momentum.
Consider two electrons that are in some distance with respect to each other. Now interchange their
position by rotating them about the symmetry axis. Their angular momentum wave functions should
be similar to the component of the wave function for the orbital momentum characterized by the
m quantum number, so
1~eis 
1
2 ~e is
2
Now consider a joint wavefunction 1,2= 1 2 . To interchange the electrons we can rotate
them about the symmetry axis. Then:
 2,1=e is   e is   =e is  e is  e is 2 = 1,2 e is2 
1
1
1
2
If we want to have the joint wave function to be anti-symmetric, we must have the exponential term
multiplied by 2n1 . This can be done for non-integer values of
5 3
1 1 3 5
s=...− ,− ,− , , , ... . This is characteristic for fermions. Bosons have an integer spin,
3
2 2 2 2 2
so the wavefunction is symmetric.
In case of an electron,
s=±
1
. The total angular momentum related to spin is calculated
2
similarly as in the orbital angular momentum,
L S = s s1 (we know this by the Dirac
equation). Of course, the „spinning” electron (let us remain with the classical analogy) resembles a
circulating charge, so it should provide a magnetic dipole, just like the orbiting electron. By similar
reasoning that lead to (***) in 18.8., taking into account the magneto-mechanical anomaly
(which states that the magnetic moment of an electron is twice larger than predicted by classical
analogy), we obtain:
e
L
2m S
p m=−2 B  s  s1
pm=−2
the projection of this angular momentum to the z axis is:
p m=−2 B s=±B
The magneto mechanical anomaly is not possible to explain based upon classical mechanics. It is
even impossible to explain by including the relativistic effects to classical interpretation. One must
consider relativistic quantum mechanics by Dirac equation... This is beyond our possibilities.
However, again, I have figured out some heuristic hint: if spin is s=1/2, then the wavefunction
e is needs an angle of 4  to return to its starting position. Thus, we could say that the electron
makes a „double loop” in his circulation and instead of the factor  r 2 in the magnetic moment
we should use 2  r 2 . This is however rather a mnemotechnical calculation to encourage some
imagination.
18.10. The total angular momentum
The total angular momentum

J =
L  LS of an electron orbiting around nucleus is quantized in
similar manner to L and LS:
J =ℏ  j j1
Similarly, the total magnetic moment is a vector sum of the orbital and spin magnetic moments.
pm= pmo pms
However, because the spin magnetic moment is twice larger than classically expected from the
length of spin angular momentum, the direction of
pm is not parallel to the direction of
Therefore to calculate the total magnetic moment of the electron we cannot simply multiply
B  j  j1 .We must find the contributions of L and LS and add them separately:

J .
p m=−B [  l l1 cos 2  s  s1cos  ] (*)
To find the cosines we make use of the cosine theorem:
where: c 2=a 2b 2−2ab cos  . Applying this to the diagram of angular momentum, we obtain the
cosines:
L2 J 2 −L2s l l1 j  j1−s s1
=
2 LJ
2  l l1  j j1
2
2
2
L J − L s  s1 j  j1−l l1
cos = s
=
2 Ls J
2  s  s1  j  j1
cos =
substituting the cosines into (*) the contributions of
 l l1
and
 s s1
in the
denominator vanish, and we obtain an expression to calculate the magnetic moment based upon the
total angular momentum j(j+1) as:
p m=−b g  j j1
where the correcting factor that accounts for the magneto-mechanical anomaly reads:
g=
3j  j1s  s1−l l1
2j  j1
18.11. The uncertainty principle
I hope that you remember a technique of Fourier series expansion from the math. For example, the
expansion of a function:
f  x =
{
0 : x∈− ,−0.1∪0.1 , 
1:
x ∈−0.1,0 .1
The corresponding Fourier series can be written as:
f  x =
∞
1
2
n
∑n=1
sin cos nx
10 
n
10
The quality of the approximation can be seen in the following figure:
It is clearly seen that the more terms in the summation we include, the more „localized” is the
aproximated function.
The same happens with quantum particles. To obtain a quantum particle that is localized in space
(i.e. corresponds to a  x  function similar to the above f(x)), we need to superpose together
many elementary waves, which form such called wave packet. But! The more elementary waves
we introduce, the more different wavelengths are in action, and because of the de'Broglie formula,
p=
h

the quantum particle „at once” has many different momenta! This „localization” introduces the
uncertainty in the value of momentum. Of course if we relax the localization of the particle, and
allow it to be anywhere from minus infinity, to plus infinity, a single wave would be sufficient to
describe the state. But to narrow the existence of the particle to some specific region, we must
sacrifice the certainty about its momentum. This is expressed by the Heisenberg's uncertainty
principle:
 p x≥
ℏ
2
This principle can be derived by considering the average of a quantum operator which we will show
for the interested readers just in the following.
18.12. The operators and orthogonality in quantum mechanics
In quantum mechanics the state of a quantum particle is described in terms of its wavefunction
 x  . To assess some measurable quantities, we need to extract them from the wavefunction by
means of the, such called, operators. For example, to extract a measurable quantity
act on  x  by means of an operator
A , we would
A . The hat notation is a typical one for operators.
As an example, consider the momentum operator. Given a wavefunction of the typical form like
 x =Ae ikx , we can act on it with an operator like
−i ℏ
p =−i ℏ
d
:
dx
d
d
ikx
ikx
 x =−i ℏ
A e =−i ℏ ik A e =ℏ k  x 
dx
dx
But we know that k equals to k =
h 2 h
2
= = p . So, we have obtained
so ℏ k =

2  
momentum on the right hand side! We have constructed a differential operator that acts on the
wavefunction, and as a result, it returns the wavefunction multiplied by the measurable quantity.
This is called the Sturm-Liouville problem, and can be formalized by:
p   x= p  x 
where for a pair of the wavefunction  x  and the corresponding measurable p, the
wavefunction is called the eigenfunction (the eigenvector) while the measurable is called the
eigenvalue. Sometimes  x  is not an elementary wave, but is a superposition of elementary
waves:
 x =∑n a n n  x
∑n a n a ∗n =1
The second condition conserves the normalization of the wavefunction provided the elementary
waves are normalized. Compared to the normalization condition of a wavefunction by itself, it also
gives the orhogonality condition for the elementary waves:
∞
∑n a n a∗n =1=∫−∞  x ∗  x dx
∞
∑n a n a∗n =∑n a∗n ∫−∞  x ∗n  x dx
∞
a n=∫−∞   x ∗n  x dx
We have expanded the conjugated wave into a coniugated series of elementary waves on the right
hand side. Repeating this to the remaining composed  x  , we obtain
∞
a n =∑ m a m ∫−∞  m  x n  x dx
∗
Since the integral on the right hand side is „1” for m=n, and am equals to an for m=n, the remaining
integrals must be zero to assure the general validity of this formula:
∞
∫−∞  m  x ∗n  x dx= mn
In general, the eigenvalue problem becomes a superposition of elementary problems:
p   x=∑ n p n a n  n  x 
Multiplying this by ∗ x (note! To multiply, we must put this term on the left on left hand side
to prevent entering under the action of an operator):
 ∗ x p  x=∑n ,m p n a n a∗n  n  x  ∗m  x
If we now integrate the expression over the whole space, the mixed  terms on the right hand
side vanish for m≠n, and constants can be thrown in front of the intgeral:
∞
∞
∫−∞ ∗  x  p  x =∑n pn a n a ∗n ∫−∞ n  x ∗n  x  dx
∞
∫−∞ ∗  x  p  x =∑n pn a n a ∗n
∞
∫−∞ ∗ x p  x =∑n p n P  n=〈 p 〉
where P(n) is the probability of the n-th wavefunction, and <p> is the average measurable quantity
p. The last formula is commonly known as the sandwich formula, and is shortly written for any
possible operator
A
as:
∞
〈 A〉=∫−∞ ∗  x  A  x dx
or, using the Dirac notation (a shorter notation):
 ∣〉
〈 A〉=〈∣A
Quite handful? There is no need to write the integral, its boundaries and dx...
18.13. Derivation of the Heisenberg uncertainty principle.
Consider two operators
X and Y that act on the wavefunction to give observables X and Y. We
will be interested in determining the standard deviation of the observables, which is a measure of
the uncertainty. To make the calculations easier, we will consider operators
 X −〈 X 〉 and
A=
 Y −〈Y 〉 , i.e. operators that have zero mean, but the same uncertainty.
B=
To proceed, we construct an artificial operator that „measures” two things simultaneously:
 Ai
  B
Q=
If we act with this operator on the wavefunction, the operator and obtain some new function. We
can calculate the square of this function, which should be positive:

=Q
∫−∞   x ∗  x dx≥0
〈∣∣〉≥0
 〉≥0
〈∣Q ∗ Q∣
 Y  X i  Y ∣〉≥0
〈∣ X −i
∞
Multiplying the operator terms we obtain:
〈∣ X 2∣〉〈 ∣i[ X Y −Y X ]∣〉2 〈∣Y 2∣〉≥0
Noticing that the average of squares of operators is just their variance (the sandwich formula), we
have:
2
2 2
 X  〈 ∣i [ X Y −Y X ]∣〉  Y ≥0
This inequality is satisfied if the discriminant of this quadratic expression (with respect to  ) is
less or equal to zero:
−〈 ∣[ X Y −Y X ]∣〉2 −4  2X  2Y ≤0
so:
−〈 ∣[ X , Y ]∣〉 2≤4  2X  Y2
∣〈∣[ X , Y ]∣〉∣
≤ X  Y
2
where [ X , Y ]= X Y −Y X is called the commutator. It can be easily shown that the commutator
 , B]=[

[A
X , Y ] and  X = A ,  Y = B . Thus, the uncertainty relation can be generalized to
the operators with nonzero mean by:
∣〈 ∣[ A , B ]∣〉∣
2
The commutator in case of the momentum operator
≤ A  B
p =−i ℏ
d
and position operator
dx
x = x
reads [ p , x ]=−i ℏ . Substituting to the uncertainty relation above, this yields:
 p  x≥
ℏ
2
i.e. the momentum-position uncertainty principle that we have learned about previously.
It is also possible to apply the above formula to the angular momentum operator. It equals to:
 r × p= y p z −z p y  i z p x − x p z  j x p y − y p x  k
L=
We see the components of L standing next to the corresponding unit vectors.
Let us now calculate the commutator between
L x and
L y :
[ y p z− z p y , z p x −x p z ]=.
 y p z z p x − y p z x p x −z p y z p x z p y x p z 
− z p x y p z −x p x y p z −z p x z p y  x p z z p y =.
 y p x p z z − yx p z p x −z 2 p y p x x p y z p z 
− y p x z p z − yx p z p x −z 2 p x p y x p y p z z =.
y p x  p z z −z p z  x p y  z p z− p z z =.
−i ℏ y p x i ℏ x p y =i ℏ  x p y − y p x =i ℏ L z
it is also worth calculating the commutator between
L 2 and
L z , which turns out to be:
 2 2  2
[ L x  L y  L z , L z ]=.
L x L x L z − L z L x L x  L y L y L z − L z L y L y =.
L x [ L x , L z ][ L x , L z ] L x  L y [ L y , L z ][ L y , L z ] L y =.
i ℏ L x L y  L y L x − L y L x − L x L y =0
where the commutators between [Lx,Lz] and [Ly,Lz] are calculated by analogy to [Lx, Ly], and the
sign difference comes from the different orientation of the perpendicular axis in the coordinate
system to the two commuting axes.
Reference: K. Zalewski, Wykłady z nierelatywistycznej mechaniki kwantowej, PWN 1997.
Jim Branson, University of California, San Diego (commutation of the angular momentum).
18.14. Interesting implications of the uncertainty principle
One of the implications is the idea of virtual particles. To consider them, we write the uncertainty
principle for the energy:
 E t ≥
ℏ
2
which follows heuristically from the momentum-time uncertainty:
 E  t=
2
Because
E=
p
and
2m
dt 1
= , we have:
dx v
dE
dt
 p x
dp
dx
 E  t=
p
1
p
m
ℏ
 p  x=  p  x= p  x≥
m
v
m
p
2
The conclusion is that for a period of time  t there are possible energy fluctuations of  E .
This holds also for zero mean energy, i.e. for particles that do not exist. So a particle can occur from
nothing and then it can disappear.
The virtual particles allow an interesting phenomenon in astronomy, i.e. the black hole radiation. If
a pair of virtual particle-antiparticle is being created on the edge of the horizon, and one of the
particles falls into black hole, the second one ceases to be virtual because it cannot recombine. It is
emitted out of the black hole as a real particle.
The virtual particles also are the basis of the modern theory of interactions, where the interaction is
being reduced to an exchange of a virtual particle between two interaction centers. This explains
well the range of the nuclear physics interactions. We will investigate this later.
Another example from astronomy is a phenomenon of a „degenerate electron gas”. Normal stars do
not collapse beacuse they conduct nuclear reactions and release heat that accelerates its particles
creating a pressure that overcomes the gravitational collaps. However, after the nuclear fuel is
burned, there is nothing (classical) that could stop the collaps. However, there are white dwarf stars,
that cool down to black dwarfs, and they do not collapse. The explanation is based upon the
Heisenberg's uncertainty principle. As the star collapses, its electrons become closer and closer to
each other, reducing the uncertainty in the position. The reduced uncertainty in the position enlarges
the momentum uncertainty. The particles start to move very fast and to exert a specific type of
pressure which is created the degenerate electron gas.
If this pressure is not sufficient to prevent the collapse of a massive star, it can stop as a neutron star
(held by the denegerate pressure of neutrons) or it can collapse to a black hole.
Questions:
1. Derive the Schroedinger equation from the de Broglie hypothesis and a wave equation.
2. What is the Born interpretation of quantum mechanics?
3. Derive the Bohr radius in the Bohr model of hydrogen atom.
4. Derive the energy of hydrogen atom in the Bohr model.
5. Explain the problem of potential barrier with finite length and the tunnel effect. Write down
the final formula for the transmission probability.
6. What is the reason for primary quantum number in hydrogen atom? What happens during
the power series solution? [Form of the answer: ...assuming a form of solution ... we obtain
a formula for power series coefficients ... where ....].
7. What is the form of the
F  solution for the hydrogen atom? How does it imply integer
values for m quantum number?
8. How is the angular momentum quantization to the power series solution of P?
9. Write down the Heisenberg's uncertainty and discuss it in frame of view of the wave
packets.
10. *Derive the general uncertainty relation.
11. Derive the energy-time uncertainty.
12. *Derive the sandwich formula for averages in Quantum Mechanics.
20. Relativistic quantum mechanics
20.1. The kinetic energy in relativistic case
Let's make use of the famous Einstein's equation for the energy (which we will derive in the
following sections):
E=mc2
In this equation m stands for the, such called, relativistic mass,
m=
m0

2
1−
v
c2
This term contains a dependency on velocity: the higher the velocity, the larger the „relativistic
mass”, and the larger the energy of the body. We can now calculate:
  
2
v
2
1
c
2
2 2
2 4
2 4
2 4
2 4
E −m 0 c  =m c −m 0 c =m0 c
−1
=m
c
=
0
2
2
v
v
1− 2
1− 2
c
c
 
m0 v
2
1−
v
2
c

2
2
2 2
c =p c
So the energy of the particle reads:
E=  m20 c 4 p 2 c 2
Introducing a possible loss due to a potential energy, we alter the equation as:
E−U = m20 c 4 p2 c 2
20.2. The Klein-Gordon equation
Now we could try to substitute the energy expression with operators,
E i ℏ ∂ ;
∂t
p −i ℏ ∇ .
But how can we take a square root of a derivative?! It complicates the problem. However this
singularity can be removed if we consider the formula for a square of the energy:
 E−U 2=c 2  m 0 c 2 p2 c 2 
Multiplying both sides by the wavefunction, and changing the energy and momentum into the
corresponding operators, we end up in the Klein-Gordon equation1:

2
i ℏ ∂ −U =c 2  −ℏ ∇ 2m0 c 2  
∂t

This is an equation that describes the particle with an integer spin. We are more interested in an
equation that describes a particle with spin one half, because that equation predicsts such interesting
properties of the matter as the existence of spin, magneto-mechanical anomaly, antimatter.
20.3. The Dirac equation
The equation that we are interested in is the Dirac equation. Dirac considered the Klein-Gordon
equation and tried to separate it into two non-squared equations:
i ℏ ∂ −U
∂t
 0=
c
=−i ℏ ∂
∂
then the Klein-Gordon equation (without the multiplication by a wavefunction) reads:
20=∑= x , y , z 2m 0 c 2
or:
[
]
20− ∑= x , y , z 2m0 c 2 =0
Dirac has proposed that this equation could be written as:
[   ∑
0
= x , y , z
 ][
]

 0 m0 c 0 − ∑= x , y , z  0 m0 c =0
if he only finds proper coefficients  . Utilizing the formula ab a−b=a 2 −b2 the above
equation results in:
20−
∑
= x , y , z

2
 0 m0 c =0

20− ∑  ,= x , y , z    m0 c ∑= x , y , z [   0  0 ]20 m20 c 2

1 I skip the introduction of a vector potential to express the effect of potential fields. Such discussion can be
found in the book by Lucjan Piela on the Ideas of Quantum Chemistry.
It is clear that to satisfy this equation, the mixed terms in  must vanish. Because the derivatives
with respect to position and time can be interchanged in order, we can write the above equation in a
form that will give the conditions for parameters  :


20− ∑ ,= x , y , z    m0 c ∑= x , y , z [ 0  0 ]20 m20 c 2 =0
We can rearrange the terms of the first summation even further to obtain even more structure in this
equation:


20− ∑ 2  2∑ , [   ]  m0 c ∑ [ 0  0 ]20 m20 c 2 =0
It is now clear that if   =0 and 2=1 , we recover the original Klein-Gordon
equation:


20− ∑ =x , y , z 220 m20 c2 =0
Clearly,  cannot be a number, since for normal numbers ab+ba=2ab... But they can be
matrices.The choice for such matrices is not unique, however their dimension must be larger than
or equal to 4x4. The typical choice is:
 
0
0 x
0
 x=
=
0
x 0
1


0
0
1
0
0
1
0
0
1
0
0
0




0 0 0 −i
0 y
 y=
=0 0 i 0
0 −i 0 0
y 0
i 0 0 0

 z=

0
z



0 0
z
0 0
=
1 0
0
0 −1

1
= 1 0 = 0
0 −1
0
0

1 0
0 −1
0 0
0 0
0 0
0
1 0
0
0 −1 0
0 0 −1
Where  i are the Pauli matrices, introduced earlier by Pauli, that meet certain requirements
needed by the Dirac matrices:
 x=
  
 
0 1
0 −i
1 0
,  y=
, z =
1 0
i 0
0 −1

Concluding, we have two equations:

 − ∑

   m c =0
0 ∑ =x , y , z  0 m0 c =0
0

=x , y , z

0
0
Applying a wavefunction to this opeator equation, we have:
∑
 − ∑
0 
0

  m c  =0
= x , y , z
 0 m0 c =0
= x , y , z


0
0
It is clear that since 0 is a 4x4 matrix, the wavefunction must be a 4-component vector.
Otherwise it is impossible to preform a matrix multiplication:
[ ][]
1

= 2 = 

1
2
Such „vector wavefunction” is called a spinor. Assuming a stationary wavefunction of the form
i
 x , y , z , t=  x , y , z  e
E
t
ℏ
, with  =
0
i ℏ ∂ −U
∂t
c

 E−U −m c − ∑
we obtain:

  =0
 E−U m0 c 2 ∑= x , y , z c    =0
2
0
= x , y , z
c 

Which is the stationary Dirac equation for an positron (above, E is negative to compensate for the
m0 c 2 ) and for the electron (below, positive energies). Please note that we have shifted from the
4x4 Dirac matrices to the 2x2 Pauli matrices. Shifting the energy scale we typically rewrite these
equations as:


 E−U 2 m 0 c 2  ∑ =x , y , z c    =0
 E−U −
∑
=x , y , z

c    =0
20.4. The big and small component of the Dirac equation
We can decompose the two 4-component equations into dwo 2-component equations by evaluating
the matrix multiplications for the first two, and the remaining two rows of equation's matrices:


 E−U 2m 0 c 2  ∑= x , y , z c    =0


 E −U − ∑ =x , y , z c     =0
(*)
(please note the difference between  and  ).We can divide the first of these equations by
2m 0 c 2 and calculate  :

= 1
E−U
2
2m0 c
−1

1
1
c    ≈
∑=x , y , z    (**)
2 ∑= x , y , z
2m
2m 0 c
0c
∂
recall now that =−i ℏ ∂  is the momentum operator in direction  . Because Pauli
matrices contain ones and zeros in their entries, the product   ~m0 v . Therefore we can see
that there is a relation between the spinor components like ~
v
 , so for normal velocities
2c
≪ and therefore  is called the small component, and  is called the big
component.
20.5. Schroedinger equation out of the Dirac equation
Because the Dirac equation is supposed to be a relativistic generalization of the quantum
mechanics, it must reproduce the Schroedinger equation in the limit of small velocities. This time
we need to calculate the sum

z
∑= x , y , z  =  i
x
y

 x −i y
=⋅
− z
Now substitute the approximation (**) of the previous section 20.4. to the second eq. Of (*) in 20.4:
 E−U =
Expanding this yields:
1
⋅2  (*)
2m 0


2z x −i  y x i  y 
z  x −i y − x −i y  z
1
 E−U =

2m 0  x i  y  z− z x i  y   x i y  x −i  y 2z
 E −U  =


1 2 0

2m 0 0 2
Which is a Schroedinger equation for two conjugated wavefunctions. This means, that such a
version of the Schroedinger equation includes the notion of spin. We will discuss spin in a moment
in greater depth.
20.6. Electromagnetic Hamiltonian
The magnetic Lorentz force depends upon velocity and requires different treatment than the ussual
potential forces that we can introduce into the Schroedinger equation. Since the energy operator of
the Schroedinger equation resembles an analogy tothe Hamiltonian of classical mechanics, we can
discuss the topic from the classical concept (recall the ideas from previous semester).
We will propose the Lagrangian to take the form of:
1 2
L= mv −qV q v⋅A
2
 = A is the vector potential of a magnetic field.
where: V is the potential, v is the velocity, rot H
The Lagrange functional fullfills the equation of motion (recall the previous semester):
 
d ∂L
∂L
−
=0
dt ∂ ẋ i ∂ xi
For the x component we can write:
mdv x
∂ Ax
∂ Ax
∂ Ax
∂ Ax
∂V
q
v x q
v y q
v z q
=−q
dt
∂x
∂y
∂z
∂t
∂x
The last line can be rearranged into:
m

 [
 
dv x
∂ A y ∂ Ax
∂ A z ∂ Ax
∂V ∂ Ax
=−q

q v y
−
v z
−
dt
∂x
∂t
∂x
∂y
∂x
∂z
which is the x component of the 2nd law of dynamics:
]
m
where
∂
A

and
E=−∇ V −
∂t
d v
=q 
E q 
v ×
B
dt

B =∇ × A .
We have proven the Lagrangian is correct. But in order to apply it to the quantum mechanics, we
need the energy operator, the Hamiltonian. Hamiltonian in classical mechanics is given by:
mv 2
mv2
H =E k V =
V =
V
2
2m
where normally we would substitute momentum in the square bracket. But momentum does not
have to be equal mv, as in the case of a generalized momentum, which is defined in the Lagrange
mechanics as:
p i=
∂L
=mv iqAi
∂ vi
Substituting, we have:
1
H = mm v −q 
A 2 qV
2
Concluding, in quantum mechanical Dirac equation, in presence of an external magnetic field we
will substitute
= p−q 
A
=−i ℏ ∂ −q 
A (*)
∂
This part is written after the lecture notes of Peter Hadley „The Hamiltonian of a charged particle
in a magnetic field”, TU Graz, Austria.
20.7. The spin in magnetic field
We are ready now to investigate the behaviour of an electron in the magnetic field. For this sake we
consider the non-relativistic limit of the Dirac equation (*) in section 20.5 and subsitute the
momentum defined by (*) in the previous section 20.6. On the way we make use of a mathematical
identity that specifies:
⋅⋅=⋅i ×
Now let us work with the Dirac equation:
 E−V  =
1
1
i
⋅⋅=
⋅
⋅×
2m 0
2m 0
2m 0
The dot product on the right hand side gives 2~ p2 . The cross product
∣
∣
i
j
k
∂
∂
∂
−i ℏ
−q A x −i ℏ
−q A y −i ℏ
−q Az
×=
∂x
∂y
∂z
∂
∂
∂
−i ℏ
−q A x −i ℏ
−q A y −i ℏ
−q Az
∂x
∂y
∂z


2
2
∂ Az
∂ Ay
∂
∂
−ℏ2
i ℏ q
−i ℏ q
q 2 A y Az −q 2 Az A y ...
∂ y∂ z
∂z∂y
∂y
∂z
∂ Az ∂ Ay 
∂ Ax ∂ Az 
∂ A y ∂ Ax
×= i i ℏ q
−
 jiℏq
−
k i ℏ q
−
∂y
∂z
∂z
∂x
∂x
∂y
×=i ℏ q rot 
A=i ℏ q 
B
×=i −ℏ2





Substituting the expressions for the dot and cross product, we have:
 E−V  =
p2
ℏq
−
⋅
B
2m 0
2m0
Expanding the dot product:
 E−V  =
[
]
p2
ℏq
Bz
B x −iB y
−

2m 0
2m0 B x iB y
−B z
For a field that is aligned in the z axis:
 E−V  =
[ ]
Since =
1
2
[
]
p2
ℏ q Bz
0
−

2m 0
2m 0 0 −B z
, the above equation splits into two:

p2
ℏq B
 E −V 1=
1−

2m0
2m 0 1
p2
ℏq B
 E−V  2=
2 

2m 0
2m 0 2
or:
2
ℏqB
p
 E−V 
 1=

2m0
2m 0 1
ℏq B
p2
 E−V −
 2=

2m0
2m 0 2
or:
2
p
 E−V MB 1=
1
2m 0
p2
 E−V −MB 2=

2m 0 2
with magnetization
M=
quantum number equal to
ℏq
=B . On the other hand
2m 0
s=
M ~ B Lsz =B s so for the spin
1
we can conclude the magneto-mechanical anomaly, because
2
M =2 B s .
19. The nuclear physics
19.1. The origin of the E=mc2
The E=mc2 equation is one of the most famous in the history of science. One of the first derivations
that were privded by Einstein (it later appeared incomplete to be precise), considered the emission
of two photons from a particle of mass m.
In the upper line we see the situation in the stationary point of view. The particle of mass M releases
two photons of frequency f (the wavelength =cf ). Each of them carry the same amount of
momentum to the left and to the right—the net momentum is zero and nothing happens. There is no
need to shake any fundamentals of the science.
However, when looking from the moving reference frame, where the particle of mass M moves with
velocity v, the two photons, that were identical, are not identical anymore! They undergo the
Doppler shift in their frequencies so
f 1≠ f 2 . Consequently, the momenta
h hf
=
 c
differ and
the momentum of the system, i.e.
p=Mv
changes into
p=Mv−
hf 1 hf 2

c
c
in absence of external forces! This violates the momentum conservation principle!
To rescue this principle, we must allow mass M to change so that the value of p is conserved. We
have no other parameters that we can manipulate. To find the correction for M, we first need to
understand the Doppler effect in special relativity.
19.2. The relativistic Doppler effect
The speed ofa wave c relates the wavelength and the period of the wave. I.e. the wavefront moves
forward by one wavelength in one full period:

c= = f
T
If an observer oves away from the light wave, the second peak of the wave must pass additional
distance of  S =v T (where T is the period of the wave), so evidently, the period must be larger
and deserves a new symbol. If T =

then:
c
v T
v T
T =
=T 
c
c
c−v
T
=T
c
c
T
T =
T=
c−v
v
1−
c
 
Furthermore, the time in the moving coordinate system is dilated, so T' should be corrected by
dividing through

1−
v2
. This gives the final formula:
c2

v2
c2
T ' =T
v
1−
c
v
v
1− 1
c
c
T '=T
v
v
1− 1−
c
c
c−v
T ' =T
cv
1−
 
 

So in terms of the frequencies we have:
f '=f

cv
c−v
19.3. Deriving the mass defect.
Recall now the change in the momentum due to the photon emission considered in 19.1.
p=
Mv

2
1−
v
c2
−
hf 1 hf 2

c
c
Substituting the frequencies, we have:
Mv


2
=
mv

2
−
hf
c

cv hf

c−v c
v
v
1− 2
2
c
c
 Mv hf −cvc−v −hf
=
=
c
c2 −v 2

v2 c
1− 2
c
M
−2hf
=
=E v
2
v
v2
2
1− 2 c 1− 2
c
c
2
 M c =E 0
1−


c−v
cv
2v
 c 2−v 2

And that's it!
19.4. The consequences of the mass-energy equivalence
Due to the discovery of the mass-energy equivalence it was possible to construct a nuclear bomb
and nuclear power plants. We are also able to understand the interactions between matter and
antimatter (e.g. of electrons and positrons).
In case of the nuclear fission bomb (i.e. where the nucleus is divided into smaller pieces), the
uranium U235 subject to an incoming neutron becomes split into barium Ba141 and krypton Kr92 and
three new neutrons which sustain the chain reaction. The mass defect per single atom is
˙
 m=235.043930u1.008664u−140.91441191.9261563 1.008664u=0.186035u
This corresponds to the following mass in kg:
0.186035⋅0.001[ kg / g ] /N A=3.1⋅10−28 kg
Thus, the energy per atom is:
˙m 2
−28
8
˙
˙
E=3.1 10 kg 3 10
=28 pJ
s
[
]
Consider now a mole of U235, i.e. a quarter of kilogram undergoing nuclear fission:
E=28 p J⋅N A=1.69 10˙ 13 J =16.9TJ
Now, because a ton of TNT delivers after explosion 4.184 GJ , we can say, that the above energy
relates to:
16.9TJ
=4kt
4.184GJ
In fact this calculation is oversimplified for a nuclear bomb, as not whole fissible material
undergoes fission durring explosion. The Little Boy contained 64kg of material, of which only less
than 1kg underwent fission, giving rise to „only” 16kt. The reason for this is that the bomb
undergoing explossion does not hold the exploding parts of material in single place, and that the
mass of the explosive approaches the critical mass in progress of the chain reaction, and also that
part of the material does not undergo fission (14%) but only absorps a neutron and releases a
gamma photon of radiation.
19.5. The critical mass
The critical mass to initiate a chain reaction relates to the problem that neutrons generated by fision
should excite secondary atoms, and not leak away to space.
On the left one can see a small amount of uranium, where the generated neutrons „miss” the atoms
and leak to space. On the other hand, when there is more mass of the material, the neutrons have
larger probability to react and the chain reaction becomes sustained.
We can roughly estimate the critical mass of the uranium to sustain a chain reaction. The density of
uranium is equal to =19.1
neutrons equals to
g
4 kg
=1.91⋅10 3 (wikipedia), the collision cross section for fast
3
cm
m
s=1.27b=1.27⋅10−24 0.01m2=1.27⋅10−28 m2
lattice spacing for uranium can be roughly estimated by  x=

(Diven, Phys. Rev. 105). The
0.235kg
=2.7⋅10−10 m . Now, the
N A
number of layers for a neutron that covers the possible directions equals to:
 x 2
n=
=5.7⋅108
s
This gives a radius for a spherical mass of uranium, which can be actually divided by two, as
typically in fission reactions two neutrons are generated (three in the example reaction quoted
above):
r =n  x /2=8cm
4
3
Rescaling this to mass, we obtain m= 0.08m  =38.7kg .
3
Reference:
R. L. Murray, Nuclear Energy—an introduction to the concepts, systems and applications of nuclear
processes. Butterworth-Heinemann, 2000.
TODO: rozpad alfa,beta,gamma,reakcja bomby atomowej
Questions
1. Derive the relativistic Doppler effect
2. Derive the E=mc2 formula
3. Explain the problem of a critical mass in a nuclear bomb
4. Derive the Klein-Gordon equation
5. Comment on the derivation of the Dirac equation. How is it different compared to KleinGordon equation?
6. What are the suggestions to treat spin matrices as operators of an internal angular
momentum?