So, today we will learn, we will continue
our discussion on the relativistic quantum
mechanics. How do we describe particles and
their interactions in that.
So, in the case of a non-relativistic quantum
mechanics, we had the Schrodinger equation
minus h cross square over 2 m grad square
psi plus V the potential interacting potential
psi equal to i h cross doh by doh t the energy
operator acting on the wave virtual psi. And
as we said this is not adequate to describe
the relativistic dynamics. So, for the relativistic
case we had discussed what is called the Klein
Gordon equation. This is basically based on
the energy momentum relation E square is equal
to P square plus m square where we are now
working in units of h cross is equal to 1,
also equal to c.
In that case we said we could write down for
a wave function, particle described by a wave
function phi, doh mu doh mu phi, plus m square
phi plus V phi equal to 0 where doh mu is
now, doh by doh t. And we discussed some the
discuss the interaction of charged particles,
and to some extent found out the expression
for transition amplitude in this case.
Another approach to get to the relativistic
equations is due to Dirac, and we will see
that we can get some other dynamical equation,
describing the wave functions evolution which
is known now as the Dirac equation. In the
case of Klein Gordon equation, we used a square
is equal to P square plus m square; in the
case of Dirac equation what we will do is
to linearize this thing.
Remember the problem in Schrodinger equation
was that it is quadratic in the coordinate
in the position coordinate, or the second
order derivative in the position coordinate,
but first order derivative of time parameter.
And then in Klein Gordon equation we took
both of them to second order.
And in Dirac equation we go the other way,
and then try to get the first order. So, it
is now written as some coefficient alpha dot
P. Alpha is taken given a symbol of a vector,
and dotted with P the meaning is essentially
that alpha 1 some coefficient multiplying
the first coordinate of momentum another coefficient
alpha 2, P 2 plus alpha 3, P 3 plus beta m
psi. So, this is actual meaning of this.
But in a compact way we can write it as alpha
vector dot P, whether alpha is actually a
vector or not we will see later. Now what
are this alphas and betas for that we will
actually gain rely on the energy momentum
relation. We know that H square psi should
be P square plus m square psi. Since energy
operator they; Hamiltonian operator square
of that when it acts on psi, it will give
you the square of energy as the value if psi
is the Eigenstate of energy, and initially
we take psi as the Eigenstate of energy, which
is a physical state.
So, E square is equal to P square plus m square
is what V have. But from eq. 1 if you take
H square, and if you take the square of the
operator on the right hand side, you will
have alpha 1 dot P 1 dot P 1, plus alpha 2
dot node or actually in this ordinary multiplication,
alpha 3 P 3 plus beta m square psi. That will
give you.
Let me call this equation then energy equation
as 2. H square psi then is equal to so, I
have alpha 1 square P 1 square, plus alpha
2 squared P 2 square, plus alpha 3 square
P 3 square, plus beta square m square psi.
Plus, other terms alpha 1, alpha 2, P 1, P
2 plus alpha 1, alpha 3, P 1, P 3 plus, alpha
1 beta m P 1 acting on psi. Plus similarly,
alpha 2 alpha 1, P 2, P 1 plus alpha 2, alpha
3, P 2, P 3 plus alpha 2 beta mP 3 acting
on psi. Alpha 3 alpha 1 P 1 actually P 3 does
not matter here, P 3 P 1 alpha 3 alpha 2,
P 3 P 2 plus alpha 3 beta mP 3 psi.
Earlier 1 is actually P 2. Then 1 more time
beta alpha 1 P m P 1 beta alpha 2 mP 2 plus
beta alpha 3 mP 3 psi. So, these are the terms
when you expand it. I deliberately expanded
it in explicit fashion. So, that we do not
confuse ourselves. And we also did not assume
that alpha 1 alpha 2 is equal to alpha 2 alpha
1, P 1 and P 2 are momentum operators which
we know commute. P 1 and m also commute.
Similarly, other components of P 1, P also
commute with each other and also can meet
with other commute with mass m. But the other
coefficients alphas and beta, we are not assuming
to be commuting. And we will see what are
the relations that it should follow. Say for
example, if you take it to be commuting what
happens. If you take they are commuting, and
then it will be a problem.
So, let us look at it H square psi equal to
alpha 1 square, P 1 square plus alpha 2 square
P 2 square plus alpha 3 square P 3 square
plus beta square m square psi, plus P 1 and
P 2 and P 2 and P P 2 P 1 and P 2 P 1 P 2
are the same they commute. So, I combine alpha
1 alpha 2 plus alpha 2 alpha 1, P 1 P 2 acting
on psi. plus alpha 1 alpha 3 alpha 3 alpha
1 P 1 P 3 psi.
Alpha 1 beta plus beta alpha 1 mP 1 psi plus
alpha 2 alpha 3 plus alpha 3 alpha 2 m; no
m; P 2 P 3 psi. Plus alpha 2 beta plus beta
alpha 2 mP 2 sin plus alpha 3 beta plus beta
alpha 3 mP 3 acting on psi, these are the
only terms.
Now, let us look at the relation energy momentum
relation that let me write it here, again
situation to that we had written earlier H
square psi is equal to P square plus m square
psi, that is equal to P 1 square plus P 2
square plus P 3 square plus m square psi.
But now the first line here, in a square psi
what we have is alpha 1 square P 1 square
plus alpha 2 square P 2 square plus alpha
3 square P 3 square plus beta square, and
m square psi. And there is no other term with
P 1 square P 2 square P 3 square or m square.
So, the coefficients of P 1 square P 2 square
P 3 square, and m square should all be equal
to 1 that is one condition. Let us put that
there.
So, alpha 1 square equal to alpha 2 square
alpha 3 square equal to beta square equal
to 1, all of them are the same and equal to
1. Then other thing is alpha 1 alpha 2 plus
alpha 2 alpha 1. Suppose we put and there
is no such term, in the H square, because
H square is P square plus m square and that
is already accommodated now.
Now, the rest of it all the terms should either
cancel with each other, or should vanish identically.
So, the first term is proportional to P 1
P 2, second one is proportional to P 1 P 3,
then m P 1; P 2 P 3, mP 2; mP 3. Either some
of these components are 0s. So, if P 1 is
equal to 0 P 2 equal to 0 P 3 equal to 0 m
equal to 0 then everything is fine, but that
anyway we will give you a null result; H square
is equal to 0.
So, that cannot be. But then the other thing
that can happen is coefficients of P 1 P 2
identically goes to 0. It should be valid
for again general momentum any momentum it
should be valid it does not matter, what particular
case P 1 or P 2 equal to 0 1 particular term
will vanish, but in general if you want this
to be so, and then each of these coefficients
alpha 1 alpha 2 plus alpha 2 alpha 1 should
vanish. And they will vanish like that only
if either alpha 1 alpha 2 r or either of this
is equal to 0 that will also not give you
a general case.
Because otherwise there would not be either
a P 1 square or a P 2 square term there, in
H square which is not allowed, which is not
a general case. So, now, alpha 1 alpha 2 plus
alpha 2 alpha 1 is equal to 0 means they are
objects which do not commute. So, these are
no ordinary numbers, or ordinary functions.
That tells us that alpha i alpha j plus alpha
j alpha i equal to 0, if i is not equal to
j, i equal to 1 j equal to 2 it is alpha 1
alpha 2, alpha plus alpha 2 alpha 1 which
is equal to 0.
And similarly, you have alpha i beta plus
beta alpha i equal to 0, for all i equal to
1 2 3. These objects, therefore, are called
anti commuting objects, because if you switch
the order in which they are multiplied that
will give you a change of sign. So, that is
one thing. Now so, that is 1 thing that we
notice.
Now, let us say we said alpha 1 square equal
to 1, it is not half an ordinary function
we know we are familiar with one, but one
type of this anti commuting objects; matrices
in general do not commute. If you take a matrix
A and the matrix B multiply it together; so
that you have first element A, and then B
matrices A or M 1 and M 2 in general is not
equal to M 2 M 1 that is clear to you, because
you are familiar with these matrices.
So, it could be a matrix one of the ways this
can be represented is through matrix, and
we will certainly be doing that. So, at least
some mathematical entity we know, which actually
are not which can be non-commute. Now if you
have a matrix form say, and then suppose you
have diagonalize this matrix. If you diagonalize
this matrix, then you have only diagonal elements
for this lambda 1 lambda 2 etcetera, then
alpha 1 square is essentially lambda 1 square,
lambda 2 square, lambda 3 square etcetera.
So, the other elements are all 0s only the
diagonal elements exist. Depends on how many
what is the dimension of the alpha. So, when
I write it as lambda square equal to 1 i actually
mean a unit matrix. So, this is a unit matrix,
and a unit matrix is a diagonal matrix with
diagonal entries equal to 1 etcetera. Which
means that lambda 1 square is equal to lambda
2 square equal to lambda 3 square equal to
1 so, etcetera all of this.
Diagonal elements should be equal to 1, this
1 is actually just 1, the number 1 not a matrix,
and that then tells you that the eigenvalues
lambda, lambdas 1 lambda 2 lambda 3 etcetera
of alpha 1 is either plus 1 or minus one.
So, the lambda eigenvalues can be either plus
1 or minus 1 nothing else. So, it cannot be
0 it cannot be plus 1, it cannot be minus
1, or any other number it can be either plus
or minus 1, that is one consequence.
Then let us see, let us look at the property
of the anti-commuting property of alpha i.
Let us specifically take alpha 1, and then
or I can write as alpha beta is equal to minus
beta alpha 1. Now let me multiply this by
beta 1 from the left side. So, that will give
you beta square alpha 1, but beta square is
equal to 1 therefore, it is equal to alpha
1.
Let me now take the trace of this, beta alpha
1 be down trace off or minus the trace of
alpha 1, but now we have a property of the
matrices trace of a chain of products of matrices
say ABC is equal to, unchanged, if you cyclically
rotate this CAB that is an exercise, which
you can prove later on. That will tell you
that trace of beta alpha 1 beta is also equal
to trace of beta square alpha 1, that is equal
to beta square is equal to 1. So, that will
give me trace of alpha 1 is equal to minus
trace of alpha 1. This is possible only if
trace of alpha 1 is equal to 0.
The quantity is negative of the same quantity;
only if that is identically equal to 0. So,
there are 2 such properties that we have.
Similarly for other alpha values and even
for beta. So, we can replace alpha 1 by alpha
2 or alpha 3. So, that will immediately tell
you alpha 2 trace is also equal to 0, and
instead of multiplying the second line here,
by beta if I multiply by alpha 1, then again
I will get alpha 1 square equal to 1 will
give me trace of beta is equal to 0. So, trace
of all of this is equal to 0.
Now, what does that tell us? One is Eigen
value either plus 1 or minus 1, trace equal
to zero. So, let us say let us take an example,
if I take 2 by 2 case dimensional matrix,
then say alpha 1 can be in the diagonal form
1 say minus 1 this is an example, or let me
take alpha 1. This is now trace less eigenvalue
is either 1 or plus 1 or minus 1.
But now let me take a 3 by 3 case. So, let
me take alpha 1, 1 possibility is to take
1 1 1, because then that is a unit matrix
so; obviously, the trace of that is not equal
to 1, but can I make one of them minus 1 other
is plus 1, and then also trace is minus plus
1. This is not possible to have a trace less
you cannot have eigenvalues either plus or
minus 1, and at the same time trace is equal
to 0 for an odd dimensional case. That, actually
tells you that. In fact, these 2 tells you
that it is an even dimensional matrix.
Now alpha the other property of alpha i and
beta is that they are Hermitian. And how do
we see that? H psi is equal to alpha 1 P 1
plus alpha 2 P 2 plus alpha 3 P 3 plus m beta
psi. Since H is Hermitian, P is Hermitian,
and M is anyway constant Hermitian that will
tell you that alpha i is Hermitian, and beta
is Hermitian.
So, this is another property. So, alphas and
betas are Hermitians objects dimensionality
is; it should be an even dimensional racing
either, 2 by 2, 4 by 4, 8 by 8, 6 by 6 or
whatever cannot be odd dimension 3 by 3 is
not possible. Eigenvalues are either plus
1 or minus 1 and they are traceless.
Let us take 2 by 2 case lowest even dimension
2 by 2. Let us take a general matrix a b c
d, Hermitian will tell you that a b c d e
a dagger is equal to a b c d itself, and that
will tell you that. So, the Hermitian thing
is left hand side is a star, b star first
the conjugate transpose. So, LHS is equal
to equal to a star, c star, b star, d star.
So, that should be equal to RHS tells you
that a is a real number b is equal to c star,
and d is equal to d star a real number.
So, the diagonal elements are real; off diagonal
elements are complex conjugates of each other,
symmetrically. That is what will happen if
you have higher a larger than our dimensions.
So, let us take 2 by 2 and then this is how
it.
So, any Hermitian matrix you can write as
a b, b star d fine. So, now let me see if
I can write it in terms of 0 1 1 0 plus 0
minus i i 0 plus 1 0 0 minus 1 plus some c
0 for whatever reasons I am denoting it by
0.
So, the last 1 here is a unit matrix 2 by
2, and you recognize the first 3 has the pouli
matrices, if I do this first element only
c 3, and c 0. So, let me write it in that
order c 0 and c 3 c 0 plus c 3. Then second
c 1 minus ic 2 c 1 plus ic 2 and c 0 minus
c 3, indeed now off diagonal elements are
complex conjugates of each other diagonal
elements are real.
Now, provided the c s are diagonal I mean
c s are real numbers. What does this mean?
Otherwise this is arbitrary c 1 can be any
number, c 2 can be any number, c 0 can be
in number, c 3 can be any number, which says
that a 2 by 2 in general a 2 by 2 Hermitian
matrix can be written as a combination of
these 4 matrices. They are called the Pauli
matrices.
Pauli matrices: sigma 1 is 0 1 1 0 off diagonal,
sigma 2 is 0 minus i i 0 off diagonal, sigma
3 is diagonal 1 minus 1. So, the Pauli matrices
along with the unit 2 by 2 unit matrix is
enough to describe any arbitrary Hermitian
matrix of dimension 2 by 2. And you can also
see that sigma 1, sigma 2, sigma 3 do not
depend on each other or they are linearly
independent of each other.
You cannot get one in terms of another one.
By observation, you can see that also I mean
just multiply 1 by some number you will not
get the other 1. Then, and the unit matrix
is also linearly independent of this. So,
we have 4 uni linearly independent matrices,
in terms of which we can write any other matrix.
This means that if we have to consider alpha
1 alpha 2 alpha 3 and beta other direct matrices.
In 2 by 2 matrix form it is enough to actually
consider sigma 1 sigma 2 sigma 3, and beta
and the unit matrix, because any other matrix
you consider would be can be written in terms
of this anyway. And now if you look at the
anti-commuting properties, and alpha I, alpha
j, plus alpha j alpha i equal to 0 for i not
equal to 0. All of them anti community with
all the others.
But in the set, we have along with the Pauli
matrices, we have a unit matrix and that commutes
with everything. So, we cannot have 4 linearly
independent matrices, which anti commute with
each other, in case of 2 by 2 matrices. So,
2 by 2 cannot is not enough.
So, the next thing to consider is the 4 by
4. 3 by 3 is an odd dimension. So, we will
not be able to look at that. Now looking at
the 4 by 4 matrices let us look at; that is
the minimum dimension 
for Dirac matrices. There is one particular
representation called Dirac-Pauli representation.
This is alpha 1 is 0 0 0 1 1 0 0 1 1 0 0 0
0 0, alpha 2 0 0 0 minus I, i 0. Alpha 3 equal
to 0 0 1 0, 0 minus 1, 1 0 0 0, 0 minus 1
0 0. And beta equal to 1 0 0 1 0 0 0 0 0 minus
1 minus 1.
This particular representation I would like
you to check that, they satisfy all the anti-commutation
relation, and also that their square is equal
to unit matrix.
In a compact way you can write alpha as 0
sigma sigma 0, sigma is a Pauli matrix, where
now you had to well. Now when you consider
alpha 1, sigma 1 is considered. Alpha 2 sigma
2; alpha 3 sigma 3, and beta is a unit matrix,
and minus of the unit matrix. Each of these
entries here are 2 by 2 matrices. So, it is
written in block form, and you can check this
thing. This is only 1 representation. And
there are other possible representations too,
which will give the same physics.
So, there is no unique representation. It
can be this the rack coordinate representation
it could be some other also. We might mention
some other representation, which is interesting
in particle physics.
So, now, let us consider the Dirac equation
again. H psi is equal to alpha dot P psi plus
beta m psi, where now alphas and beta are
4 by 4 matrices and therefore, psi should
be a 4 by 1 matrix. So, H I will have actually
psi 1 psi 2 psi 3 psi 4; a 4 component object.
Alpha is a 4 by 4 matrix dotted with P, and
then you have a so, alpha dot P itself is
a then a 4 by 4 matrix, then you have psi
1 psi 2 psi 3 psi 4 by 1 matrix. So, that
they can be multiplied plus similarly beta
m psi 4 by 4 matrix, and you have psi 4 by
1 matrix. This should be kept in mind.
Let me, but, now take this first equation
here, 1 and multiplied by or first let me
take the so, let me take this equation.
The Dirac equation and write it in the operator
form i doh by doh t, the Hamiltonian energy
operator acting on psi is equal to alpha dot
minus i gradient, which is the momentum operator
acting on psi plus beta m psi. Multiply this
by beta from the left will give me, i beta
doh by doh t of psi minus i beta alpha dot
grad psi plus beta square m psi beta square
is equal to 1.
Now, I can bring this to one side. So, I will
have i beta doh by doh t plus i beta alpha
dot del acting on psi, minus m psi equal to
0. Beta square is equal to 1. Let me introduce
the notation, gamma mu equal to beta, and
beta alpha. So, gamma 0 is beta gamma 1 is
beta alpha 1 gamma 2 is beta alpha 2 gamma
3 is beta alpha 3.
In this notation, now I can write the Dirac
equation as i gamma 0 doh by doh t minus i
gamma i doh i doh psi minus m psi equal to
0 here, doh mu is equal to doh by doh t, and
grad minus sorry this is plus. So, doh mu;
so, this will be then plus.
Together I can write it as gamma mu doh mu
minus m psi is equal to 0, just to remind
you that it is a matrix equation.
We will write this matrix indices explicitly
a b element of i gamma mu, multiplied doh
mu psi b element of this, is mu psi, and subtract
so, that will give you a element of gamma
mu doh mu psi, and multiplied by m psi a will
give you 0, where a and b are Dirac indices
1 2 3 4. So, this a and b runs through the
Dirac components. The components of gammas
and components of psi.
So, this is not; this has nothing to do with
the Lorenz index mu or the components of vector
in that sense. So, this is what we have here
for this. And now let me look at the properties
of alpha, and beta written in terms of gamma
now.
Beta alpha i plus alpha i beta equal to 0,
let me multiply it by beta beta alpha plus
beta alpha is equal to beta alpha i beta equal
to 0. So, that is beta is gamma 0 beta alpha
i is gamma i plus, gamma i beta alpha i let
me club them together in this fashion, gamma
i gamma 0 equal to 0. So, this is one thing.
So, gamma 0 and gamma i for any i commute
anti commute. Take alpha i and alpha j for
alpha not equal to j sorry i not equal to
j anti commute. Multiply again by beta alpha
i alpha j beta from both sides, alpha j alpha
i beta equals. We knew because of the first
line here beta alpha is equal to minus alpha
i beta, I switch this beta and alpha j in
the first term, as beta and alpha i in the
second term, that will give you minus i in
both cases.
So, I should have added a minus sign here,
and then put this as a minus sign equal to
zero, but because of an overall way that will
give you only an overall minus sign, and the
first one is gamma i is gamma j, plus gamma
j gamma i equal to 0. So, now that is so we
have 2 relations here, one is let me highlight
those. One is this that is gamma 0 gamma i
plus gamma i gamma 0 is equal to 0, and we
have another relation gamma i gamma j, plus
gamma j gamma i for i not equal to j equal
to 0.
So, gammas anti commute as well. So, this
is the same as the relation that we had in
that earlier case.
And now some other properties of gammas. Alpha
i square equal to unity, which will give you
let me write it as beta alpha i alpha i beta
equal to beta square I just multiplied by
beta from both sides, and then again and that
will give me I have to the first 1 is beta
alpha I, second when i switch this beta and
alpha. So, I have beta alpha i become minus
sign is equal to 1, which is basically gamma
i gamma i for gamma i square equal to minus
1, because of the minus sign here.
And beta square is, obviously, equal to 1.
So, gamma 0 square is equal to 1. So, beta
square gamma square is equal to 1 gamma 0
square is equal to plus 1, and gamma i square
is equal to 1. Putting together or these the
anti-commutation relation, these 2 relations,
and this property of the square of the gammas,
we can write gamma mu, gamma nu, plus gamma
nu, gamma mu equal to 2 g mu nu. I would like
you to verify that this in components are
exactly the same as what we had a few we have
now written here.
So, that is essentially the thing that we
want to discuss. There is one exercise that
I would like you to have. So, what I will
write the exercise here. You just show that
gamma 0 dagger is equal to gamma 0, which
is obvious, because it is beta itself. But
more importantly, gamma i dagger is equal
to minus gamma i. One thing that I had not
mentioned, but it is somewhat obvious the
way, we had written is that gamma i runs from
1 to 3 whereas, the Greek indices run from
0 1 2 3 4 dimensions. So, this is another
property the Hermicity.
So, I will leave it as an exercise for you.
Please check that. So, we will consider other
aspects of the Dirac equation in the coming
lectures.
