In the previous lecture, I described the Klein-Gordon
equation and how its degrees of freedom can
be separated in the two component language,
which were identified with particle and antiparticle
modes. Now, this is a generic feature of relativistic
quantum mechanics that with every particle
mode, an antiparticle mode necessarily appears.
But historically, it was a lesson learnt over
many years, the simple reason being that when
these equations were developed, none of the
antiparticles were experimentally observed.
And...So, what I describe is the modern
understanding of what we call particles and antiparticles.And they have an important relation
between the properties, which can be just 
summarized as these objects have the same value of the
mass, but all charge values are opposite.
And for that reason, which one we call particle
and which one we call antiparticle is sometimes
a matter of convention. Most often the objects,
which are observed quite frequently in our
world – they are called particles; and everything
which is rare is put in the antiparticle label.
And for example, an electron is a particle
and a positron is antiparticle or a proton
will be called a particle and antiproton will
be called antiparticle.
Now, it should be noted that, though the masses
are the same for both these objects, all the
charges are opposite; and the charges may
come in many different types. For example,
I was discussing electromagnetism; and so
the electric charges will have opposite values.
If I am discussing weak interactions, the
weak charges will have opposite values. If
I am discussing strong interactions, the color
charge will have opposite values. Sometimes
the charges are not associated with any interactions,
but they will still have opposite values like
baryon number or a lepton number. And this
is always associated with a symmetry between
particle and antiparticles that, you flip
the signs of all possible charges, which may
come. If some charges are flipped, the other
ones are not flipped; that will not be called
an antiparticle. And this is our modern understanding
or the modern referring system of how do we
associate these objects with each other.
And, the only way we can have objects, which
do not have this kind of association is when
all charges are exactly 0. And in that particular
case, there is no possibility of creating
of another object, which has opposite charges.
And in such cases, the objects will be necessarily
all neutral with respect to all the charges.
And in such cases, we may as well refer to
them as particle and antiparticle referred
to the same object. Or, in other words, the
object – it happens to be its own antiparticle.
And there are few example of this category.
For example, photon is completely neutral
with respect to all the charges and we can
refer to photon being its own antiparticle.
Now, this emergence of antiparticle solution
is a very important lesson of reactivity.
But, in the early days, that was not what
people studied heavily. The important point
of development of quantum mechanics was that,
it has to match with observed phenomenon in
atomic physics. And that was the essential
point of developing the subject. And we have
to have a theory, which can explain all the
experimental observations described in various
atomic phenomena, energy levels, quantization
of radiations and transitions, how the levels
change in presence of external fields, etcetera,
etcetera. And what we learn in quantum mechanics
today at the beginning like a particle in
a box or harmonic oscillator, which are much
simpler problems; they were not the ones,
which were historically analyzed first, because
the theory had its meaning only if it could
match with experiments. So, everything was
tested against observations in atomic physics.
And the bench mark of the test was the solution
of the hydrogen atom problem. If the theory
could explain the hydrogen atom spectrum and
its various properties, then it was worthwhile
studying.
So, now, I move on to discussing the solution
of the hydrogen atom problem with the Klein-Gordon
equation. And that is what Schrodinger tried
at the beginning after writing down the equation;
it was rather unfortunate that, he obtained
the solution, but he had to discard it, because
it did not match with experiment as well as
he had expected, but that is a little bit
of twist of history. And I will give the explanation
little more after describing the solution.
So, let us take now the Klein-Gordon equation and  couple it to electromagnetic fields. And that equation
when we solve it, we will get a solution of
the hydrogen atom problem in a fully relativistic
frame work. So, the standard prescription
for coupling quantum theory to electromagnetism
is so-called minimal prescription; which says
that, you take the momentum operator and replaced
it by p mu minus A mu; where, A mu is the
gauge potential for the electromagnetic field.
And if you do that, then the modified equation
will describe the particle in the presence
of electromagnetic field specified by the
potential A mu. And so we have the Klein-Gordon
equation now written in terms of a derivative
operator or gradient in the particular form
– del mu minus A mu whole square minus m
square c square multiplied by psi equal to
0. Now, we have to solve this equation. And
in various cases, depending on the field,
there will be various kind of solutions. It
can be easily seen that, the particular situation,
where this differential operator 
is different compared to Schrodinger equation,is in the time component. The space part of
this produces the Laplacian. And that is exactly
the same structure as what is happens in the
Schrodinger equation. And so one can look
at the situation in electric and magnetic
field separately.
In case of just a magnetic field, where this
vector potential is not 0 and B is defined
as curl of A; and we will put this scalar
potential equal to 0. In that particular case,
what we have in this equation is nothing but
the operator, which 
happens to be the same. And so it produces
the same solution and the same spectrum as
happens in the case of Schrodinger equation.
In particularly, it gives the usual dipole
coupling of the electron in presence of a
magnetic field, where the dipole is proportional
to the orbital angular momentum. Or, in other
words, the Gyromagnetic ratio 
happens to be equal to 1 for this particular
interaction. So, this is a standard. And it
has been observed in certain situations; but
it fails when the electron spin also contributes
to the magnetic moment. And that is something,
which does not come out from the Klein-Gordon
equation. However, this was not the first
concern; the magnetic field was known to have
a smaller effect than the electric field itself.
And, the particular solution of the hydrogen
atom problem corresponds to solving the equation
with the coulomb field, where the potential
A 0 is essentially Z e by r with presence
of a nucleus, which has atomic charge Z e.
So, it is not necessarily just the proton,
but for any nucleus a single electron going
around it. And now, the point is now to solve
this equation in presence of this coulomb
field; and we will put the magnetic field
equal to 0. And this case is a simple enough,
so that one can directly compare the solution
with respect to the non-relativistic case.
And we will see term by term what are the
changes, which occur. The first thing to observe
that, in this particular case, this is a central
potential, which implies that, the angular
momentum is conserved. And so the equation
can be separated into radial and angular parts
essentially, because the angular parts, which
are associated with the spherical angles theta
and phi – they only occur as a part of the
Laplacian; and the potential does not produce
any term, which depends on those angles.
So, now, let us write the solution, parameterize
the same way as in case of a hydrogen atom.
So, I will decompose this general wave function
psi into its radial part and an angular part.
And since the angular operator only comes
as a part of the Laplacian operator, the solutions
are the well-known spherical harmonics. And
the potential does not do anything to the
angular part and we can just use the solutions.
The part, which depends on the potential,
is the radial wave function.
And now, we can write down the reduced equation
for the radial part assuming that, the angular
part basically just gives the same value as
is well-known in case of quantum angular momentum;
it will produce a contribution, which are
the eigenvalues of the spherical harmonic
and a term, which appears as l times l plus
1 divided by r square. So, the equation, which
we have after this separation of variable
looks like minus 1 by r square d by dr r square
d by dr plus l into l plus 1 by r square into
R. This is differential operator, which comes
from the radial part of the Laplacian acting
on the wave function.
And, the result has to be the other terms
in the equation, which are essentially the
energy and the radial potential. And then,
can be very easily written as 
the combination E minus A 0 whole square minus
the rest mass contribution. And this is the
equation, which we have to solve. It helps
to compare with the corresponding equation
in the non-relativistic case, where the left-hand
side is identical; it comes from the same
Laplacian operator. But, the right-hand side
is different the energy part of the equation
is linear in the non-relativistic case.
Now, the comparison can be illustrated by
writing down the coefficient of each power
of r in these two equations. The differential
operator is identical while the other part,
which are basically coefficients of 1 over
r square, 1 over r and the constant term – they
are different in these two particular equations.
And by knowing the solution of one, we can
actually write the solution for the other
by just replacing these appropriate changes
in the various terms, which are basically
just changes in certain coefficients. So,
let me write a little table, which compares
this particular coefficient.
So, let me denote the two cases here: relativistic
and the non relativistic cases. The coefficients
of 1 over r square in these two situations
happen to be this – the non-relativistic
case of layered l into l plus 1. But, here
you got an extra term, which comes from A
0 square. The coefficient of linear term in
1 over r; in one case, it comes from the cross
product of this expansion of the square; in
another case, it is purely the contribution
of A 0. So, this is…
And then, there is a constant piece, which
does not depend on r and which essentially
is a function of the eigenvalue energy. And
all we have to do is write the general solution
of the non-relativistic equation in terms
of these coefficients and then replace those
coefficients by the corresponding relativistic
answers. And we automatically have a solution
for the relativistic case. So, the solution
of these equations 
are the so-called Associated 
Laguerre functions, which are essentially
an exponential fall off multiplied by a finite
order polynomial. And the order of the polynomial
is related to the radial quantum number.
Now, one can look at these various quantum
numbers in an explicit case by writing down
a general solution. But, many behaviors are
kind of well-known and I will just point them
out rather easily. So, one feature is that,
the solutions behave r is to l prime near
the origin, where l prime is determined by
solving this particular coefficient in the
same way as this particular structure. And
this is a quadratic equation, which can be
easily solved. And the answer is easily written
in terms of this 
solution involving a square root. There are
things, which I have implicitly assumed in
this solution. One of them is that, the negative
sign in front of the square root is discarded
for this solution of the quadratic equation,
because otherwise, the solution will not be
normalizable; it will blow up as r goes to
0 with a large negative power.
So, we are forced to take the plus sign to
make the function well-behaved as r goes to
0. The other feature happens if the second
correction, which is 4 z square alpha square
exceeds the first term. In this particular
case, the number will become imaginary; and
that applies or indicates to an unstable problem.
And in that case, we must have to do some
thinking and reinterpreting what the solution
means, because as long as this number is a
positive number, things are fine. But, if
it either turns negative or becomes imaginary,
something is not physical in the solution,
which it produces.
So, let me state the first condition, which
is already included in by discarding one of
the signs of the square root. But, the other
condition is at… There is an instability
expected when the atomic charge is larger
than 137 by 2. This is a problem in the sense
that, there are lots of nuclei, which we know
which are experimentally observed, which satisfy
this particular condition. So, something goes
wrong with this solution; and clearly, this
is one case, where it does not match with
experimental observation. But, let me just
continue and construct the complete solution
and we will come back to this situation after
that. And that remaining part – the angular
quantum number now has gotten fixed and we
have to find a radial quantum number.
And, the radial quantum number is 
related to the degree of the Laguerre polynomial.
And that degree is defined or rather constrained
to be a non-negative integer. This is a convention.
And it is dictated by the fact that, unless
this radial quantum number is not an integer,
the polynomial will become an infinite series
and the function will then not have a good
behavior as r goes to infinity; you must truncate
the polynomial and that truncation is enforced
by making this number and integer; otherwise,
the series of the recurrence relation defining
Laguerre functions does not terminate. And
then, the final solution is return in terms
of a total quantum number, which is a combination
of the radial 1, the angular 1 and an extra
1, which is related by all the conventions
of where these numbers start and either from
0 or 1. So, this total quantum number is the
one which fixes the energy.
And, the relation is that, this number is
essentially the ratio of 2 of the terms, which
I wrote down on the previous slides. These
are coefficients of 1 over r and r-independent
term. The l part we have already taken care
of and it does not appear anymore. But, these
2 coefficients dictate the value of the energy.
And it is essentially is the ratio of those
2 terms; and it can be expressed as 2 E Z
e square by h cross square c square into half
h cross c divided by square root of mod E
square minus m square c to the power 4. This
particular expression, which can be simplified
after canceling all the constants; and expressing
the result in terms of the so-called fine
structure constant, which is nothing but e
square by h cross c. So, this is a final result.
Just to compare it directly, we can construct
the same quantum number in the non-relativistic
case, where the corresponding terms and the
ratios evaluate to Z e square into 2m by h
cross square into half h cross divided by
square root of 2m into mod E is equal to Z
e square by h cross into square root of m
by 2 mod E. The result that, it is n prime
plus l plus 1 in this particular case; which
makes up the total quantum number and it has
an expression, which is again related to the
energy. And the energy levels, which we are
familiar with in case of the hydrogen atom
problem, are now obtained by inverting this
expression. So, E is written as the function
of the total quantum number. And that gives
the well-known formulae, which are very easily
to remember.
And so I will write down those results. So,
this energy is a… This is a relativistic
case. Energy is m c square 1 plus z square
alpha square by n R square is to minus half.
And the corresponding version for the non-relativistic
case gives the Rydberg formula, where energy
is proportional to 1 over the non-relativistic
total quantum number square. The difference
between the 2 formulae is that, in the relativistic
case, we have the rest mass energy m c square,
is always part of the solution.
And, if you want to compare the non-relativistic
case, you have to subtract out the m c square
part and look at the remaining part, which
follows from it. Other than that, the results
are all identical. So, the non-relativistic
case is an energy can be written as m c square
into minus Z square alpha square by 2 n NR
2. It is essentially the first term in the
expansion of this quantity, but with total
quantum number appropriately replaced by its
analog; and the difference being that, the
value of l, which appears here is different
than the value of l prime, which appears there.
So, these are the kind of standard results.
And, the question as I said is compare this
again with experiments. So, this is the so-called
Rydberg formula. And this includes relativistic
kinetic energy. And this is exactly what Schrodinger
arrived at. He had the Rydberg formula, which
actually worked very well in fitting the experimental
data. And he thought now he has a more powerful
version, which includes the corrections due
to relativistic kinetic energy; and so it
should be able to explain the observed linear
energy levels in particularly the fine structure
splitting more accurately than the Rydberg
formula.
And, he could just work out the fine structure
corrections as a next term in the particular
expansions. And surprise was that, it did
not work as well as he thought it would; and
he was disappointed; he had to put aside his
solution, because it did not explain what
he thought it would explain. And after some
time, he decided to just publish the non-relativistic
limit of his particular formulation. And that
is the equation, which now carries his name;
that is the Schrodinger’s equation.
That was not the first equation, which he
wrote down, but it was his compromise that,
the relativistic version did not work. So,
he took the non-relativistic limit. And that
produced what was expected in the non-relativistic
limit. And that then became successful in
dealing with lots of problems in atomic physics
and chemistry. So, where did this fine structure
description go wrong? Let us just work out
the details by expanding this object in a
Taylor series. The leading part is m c square,
which is a rest mass energy. The next term
is the Rydberg formula, which is also OK.
And we have to go to the still further order
in expansion and simultaneously relate the
value of the relativistic total quantum number
and the non-relativistic one to match the
various expressions.
So, that relation between the various quantum
numbers can be easily worked out by rewriting
l prime in terms of l, which turns out to
be the non-relativistic quantum number corrected
by a small term with denominator 2l plus 1.
And now, one can take this expansion again;
plug it back into the energy levels and find
out what the fine structure splitting is.
So, the energy expansion becomes m c square
minus m z square e to the power 4 by 2 h cross
square n R 2 plus 3 by r m z to the power
4 e to the power 8 by h cross to the power
4 into c square n R 4 minus… And then, you
substitute for n R in terms of the non-relativistic
constant number. These two terms will give
rise to the rest mass and the Rydberg energy,
but something will be leftover.
And, that gives a fine structure 
of these energy levels. And the term is this
particular coefficient. But, now, everything
rewritten in terms of the non-relativistic
quantum numbers. And this is the place, where
the detail matching with experiment failed.
Schrodinger was not just disappointed, because
this thing did not work as well, because there
could have been lots of other sources of error.
But, he was more disappointed, because the
formula, which has matched with the experiment,
also existed; and that was already derived
in the semi-classical quantization formulation,
which Bohr started and Sommerfeld generalized.
And that formula 
essentially gave the very small modification
of this particular quantity and it had this
number replaced by 2 times k, where k – positive
integer.
And, in this particular case, l happens to
be an integer and 2l plus 1 is an odd positive
integer. And the only difference is that,
the odd one does not work; the even one works
in matching with the experiment. And nowadays,
we know the reason for this particular shift;
it is just that, what is appearing here is
the value of the total angular momentum and
not just the orbital part. And the relativistic
theory, which includes the new contribution
to the angular momentum – essentially, the
spin of the particle adds to the value of
l; l is an integer; s is half. And once you
combine this formula, instead of having an
odd integer over here, you will obtain an
even integer over here. And this formula now
works.
So, Sommerfeld had already derived this particular
formula, which met with experiment. And Schrodinger
got this slightly different answer; it did
not work and he had to put it aside saying
that, this is not the correct description
of the hydrogen atom. The change from this
odd number to even number followed a few years
later when Dirac obtained his equation and
solved it for the hydrogen atom. That equation
automatically included the spin of the electron;
and for that reason, shifted this number exactly
by this particular amount and arrived at the
correct result. So, this is the summary of
what all goes on in case of a solution of
the hydrogen atom. One can also derive this
expression in our modern language. So, the
fine structure of the energy level can be
obtained by treating. So, this can be obtained
also by including the correction to kinetic
energy, which happens to be the term in perturbation
theory. And this term is nothing but the whatever
expansion of square root of p square c square
plus m square; the leading term is m c square,
which is here.
The next term is p square by 2m, which is
already included in the solution, which produces
the Rydberg formula. And the third term is
this. And if you calculate it in perturbation
theory, it exactly produces this fine structure
relations. So, we now understand what exactly
this thing is coming from; it is the relativistic
correction to the kinetic energy. The part,
which is missing in this particular formulation,
is the spin. And once you combine the spin
as well as the relativistic corrections, you
get the correct number, which matches with
experiment. So, that is the story of a Klein-Gordon
equation, its solution for the hydrogen atom
problem; and why that was put aside. And Dirac
had to look for some other equation, which
will describe the hydrogen atom spectrum properly
including the fine structure case.
But, let me make a Digration here in the arguments,
which led Sommerfeld to derive the correct
formula. And it is a little bit unusual that,
he could get the correct formula without knowing
anything about the spin, because when he derived
his formula, spin was not discovered by experiments;
and still his answer, which does not include
the spin value of the electron at any stage
in the derivation matches with experiment.
And that was a little bit of a luck in the
sense that, there were 2 corrections to the
derivation, which he followed as we understand
today: one was the spin was left out and another
was the correction to the semi-classical quantization,
which also was left out. It just happened
that, both these effects were equal and they
canceled out each other in the formula and
he got the final answer correct. And that
is a topic of semi-classical quantization.
And, I will give an outline of this quantization;
it is often not given in much detail in text
books. But, it is worthwhile to understand
the principles of it, because it is useful
in connecting classical and quantum mechanics
at various different levels and arguments.
So, this so-called semi classical or Bohr
Sommerfeld quantization follows from the so-called
action-angle variable formulation and description
of the system in phase space. And what it
expresses is that, for canonical formulation,
you take the canonically conjugate variables
and look at the revolution. And if the evolution
is periodic, then the period is quantized
in units of Planck’s constant. And this
was a formulation, which was set up by Bohr.
And Sommerfeld expanded it to include many
other situations. And the reason that, these
kind of objects are quantized, has to do with
the behavior of these objects in classical
mechanics. And they are the so-called adiabatic
invariants. And one can prove that, if you
change the parameters of a classical system
very slowly, that is where the word adiabatic
comes from. This particular quantity do not
change. And that is the statement that, integral
of p d q over a closed orbit is invariant.
In quantum theory, the rule is that, the area
enclosed by the orbit in phase space is an
integral multiple of the Planck’s constant.
This semi-classical rule has two obvious limitations.
One of them is that, you can apply it only
when the variables go through a periodic behavior.
And another one is that, this kind of quantization
requires that you will be able to apply it
to individual canonically conjugate pair only
when the theory or the equations of dynamics
can be separated in terms of these variables.
And 
so it can be applied only to, where limited
number of possibilities. But, still the formulation
was successful, because many of the interesting
motions fell into that particular category
particularly when the motions are periodic
and the variables can be separated. And this
is the principle, which Bohr emphasized.
And also, it allowed him to connect classical
and quantum mechanics by imposing the condition
that, classical behavior 
is recovered when these quantum numbers become
large. And this is named after Bohr as his
so-called correspondence principle. So, Bohr
applied these arguments to a very simple situation
and to get the Rydberg formula. And that was
to look at the so-called circular orbits of
an electron going around a nucleus. In this
particular case, the equations are rather
simple; the dynamics just is given by the
centripetal force matched to the electromagnetic
interaction. And in that case, this particular
integral is basically the integral around
the circle phi going from 0 to 2pi. And this
is the value of r does not change for a circular
orbit. One can easily work out what that angular
momentum is. And quantization of that in terms
of this quantum number immediately produced
the result that, the energy is given by the
well-known Rydberg formula with the 
quantization number being an integer. So,
this was the first application of a quantization
role in atomic system.
And, since it was successful, Sommerfeld tried
to generalize to new situations; and that
is where he got inclusion of orbital angular
momentum explicitly into this formula; as
well as he could include relativity also in
this particular formula, because classical
dynamics and its relativistic extensions were
very well-known. And these two things were
included by the well-known results in classical
dynamics. One was that, you do not have to
stick to only circular orbits; but one can
look at elliptic orbits as well, which are
again exact solutions of Kepler's problem
in classical dynamics. And those elliptic
orbiters basically allow the periodices integral
to be applied to the radial variable as well,
instead of just the angular variable.
And, that basically gives a new quantum number,
which now can be written as the one corresponding to the integral of the radial momentum. And it will have values again
and starting with a set of integers. The 0
actually corresponds to the object being in
a circular orbit, where radial momentum vanishes
identically; and after that, there is an integer
quantization. And this generalization provided
a new rule to Sommerfeld for including contributions
to quantization of certain things. And that
is where he could get extra levels of the
hydrogen atom, which were not there in Bohr’s
formula. And once the relativistic corrections
were included in these integrals, he could
obtain the fine structure as well. And that
is the formula, which I quoted earlier in
relation to the Klein-Gordon equation solution
and which matches with experiment. The details
of this we will work out next time.
