Greetings, so we will discuss the Foldy Wouthuysen
transformations today, and the whole reason
to do, so is to recognize that.
Now, this is the dirac equation, the operators
of 4 by 4 matrix vector operators, quantum
operators the wave function is a 4 by 1 wave
function it has got 4 components. And the
unit operator or the beta operator has got
0 elements in the off diagonal blocks, but
the alpha operator has got the poly matrices
on the off diagonal block. So, the alphas
are odd operators, the betas and the unit
operators they are called as even operator.
So, the odd operators are the one which makes
the, so called large component and the small
component, they are the one which makes the
particle states and an antiparticle states.
You can really decoupled, because of the presence
of these odd operators, so what one hopes
to do is to transform the dirac equation,
at this h psi equal to i h cross h psi by
del t is the dirac equation that I am now
looking at this is a dirac equation.
You subjected to a transformation to the primed
representation, which is the Foldy Wouthuysen
transform representation, and in this representation.
It is our hope that the operators in the transform
representation will also have the odd operators,
but hopefully they will be less important
than, what they were in the original unprimed
representation. For example, this could happened,
if they get scaled by a factor 1 over m, so
that is a kind of strategy that we are going
to apply following the technique that was
introduced by Foldy Wouthuysen.
So, this would involved transformation of
the wave function psi to psi prime through
a transformation operator e to the i s. And
if we did that the corresponding Hamiltonian
would transform to the sum of these two terms,
which is e to i S h e to the minus s. And
there would be this additional term, because
it cannot be assumed that the operators S
is independent of time, it may be in some
cases, but it does not have to be. So, this
is the general form of the transformed Hamiltonian
under the Foldy Wouthuysen transformation,
this we discussed in our previous class.
So, let us begin with this c alpha dot pi
plus e phi, and to illustrate the method we
first take a very simple case which is the
case of a free electron, so the vector potential
and the scalar potential phi do not appear
in our equation of motion. So, the Hamiltonian
is just c alpha dot pi plus beta m c square,
instead of pi you get p pi is a generalized
momentum, which includes the magnetic vector
potential. So, you get only the alpha p term
instead of alpha dot pi, because there is
no vector potential, likewise this term e
phi is also missing, and we have a much simpler
relation to work with.
And you can already see that even for this
equation even for the free electron, you have
got the odd operator alpha and the even operator
beta. And if you wanted to get freed of the
odd operator, you could give a Lorentz boost
to the electron frame, if you go over to the
electron frame, then the mechanical momentum
p of the electron in that frame would be 0,
and you would get rid of the alpha dot p term.
So, there is some hope that he has this can
actually be achieved, and the residual equation
that you will have will then have a relationship
which will not involve the odd operators.
So, that can be done, but the general transformation
operator, which is e to the i S followed by
that reasoning Foldy Wouthuysen consider this
particular operator, which is minus over i
twice m c beta alpha dot p scaled by some
unknown function omega of p over m.
This is a hitherto unknown function, but it
is some arbitrary function of p over m and
at in an appropriate juncture, we could choose
yet to be whatever we will find it to be the
most convenient one for us, for our purposes.
We have not the objective very well defined,
we want get rid of the odd operators our motives
are you know very clear. So, this is this
does not involve any times independence, so
h prime is e to the i S h e to the minus i
S, and that is what I have written here h
prime is this.
This is the free electron Hamiltonian c alpha
dot p plus beta m c square, this is the free
electron Hamiltonian, and if you moved this
beta outside the bracket move it to the left.
And take it outside the bracket, then you
would need to pre-multiply this by beta inverse
right the first term. So, that is what to
do over here, beta and beta inverse of beta
m c square gives you the appropriate second
term, and to make the first term appropriate
you must include the beta inverse, but beta
inverse is same as the beta.
So, you get e to the i S beta and then you
get c beta alpha dot p plus m c square e to
the minus i S, so that is the term that you
get. And then since the transformation operator
S is already made up of beta alpha dot p,
which is what you have over here which is
beta alpha dot p, this is just plus m c square
multiplied by the unit operator. So, that
is going to commute with e to the minus i
S without any difficulty, but even this one
mode and this e to the minus i S can be written
before this bracket.
So, what you have is h prime operator is e
to the i S beta e to the minus i S, and then
you have c beta alpha do p plus m c square.
Now, let us look at this term over here, beta
e to the minus i S, I am looking at this part
alone, which is beta e to the minus i S. I
write S explicitly there is a minus i here,
and a minus i here, so I get minus 1 over
twice m c the rest of the operator.
Now, let us analyze this expression further,
because essentially beta e to the minus i
S is this operator here, and you can use a
power series expansion as you have for e to
the x. So, that is a well known series, it
is an infinite series, expect that x for us
is an operator, and you must take it appropriate
power of that operator. So, we plug in this
infinite series, and it has got minus 1 over
2 m c to the power n, and then you have got
beta alpha dot p times. This omega function
which is yet to be determined to the power
n, so this is to the power n and then you
have a omega to the power n.
So, let us have a look at this expression,
so you have minus when to the n here, this
is your transformation operator S, now this
is the identity that one can use. You can
easily establish this by simply using the
explicit form of the beta and alpha matrices,
so very simple block diagonal structure. So,
you can you know carry out the matrix multiplication
law, and you will find that this operator
to the power n is a same as the minus 1 to
the n, beta alpha dot p to the n, and then
there is a beta with 6 out.
So, you can substitute this operator by this
term, you get minus 1 to the n which is already
here another minus went to the n over here.
So, that will give you minus 1 to the two
n which is always be plus one no matter what
n is odd or even, and then the remaining terms
take a rather simple form. So, this is your
expression for beta e to the minus i S, and
using the same kind of analysis, you can show
that e to the plus i S has this form, and
you need both of them.
So, you have e to the i S which has this form
you using exactly the same reasoning, and
if you look at these two forms you find that
if you multiply this expression on the right
by beta, then you find that beta e to the
minus i S is equal to e to the plus i S beta.
This is not a commutation relation remind
you, there is a minus sign here, and a plus
sign here, and you have this operator identity.
So, this is the one we can make use of over
here, and this beta e to the minus i S, then
becomes e to the i S beta, and together with
this you get e to the i 2 S beta, and that
you have got the c beta alpha delta p plus
m c square. So, now you have got e to the
i 2 s, but you know how to expand e to the
i S, so e to the i 2 S will just have this
coefficient multiplied by 2. So, instead of
1 over 2 m c you will get the 1 over m c,
in the rest of the term rest of the expansion
would be essentially the same. So, you have
got 1 over m c instead of 1 over 2 m c, and
then the rest of the expansion, and then followed
by c alpha dot p plus beta m c square, now
this is your transformed Hamiltonian.
Now, let us look at this expansion in some
further details, you recognize that this beta
alpha dot p this 4 bracket is rest to the
power n, so each factor 1 over m c is be is
rest to the power n. The beta alpha dot p
operator is rest to the power n, and omega
function is rest to the power n, so you have
this power series expansion now of these operators.
So, for n equal to 0, you get the unit operator,
then this is the term corresponding to n equal
to 1, which is beta alpha dot p over this
m c. And then you have got this omega to the
power 1, then you get the second term which
will have a 1 over factorial 2, which is sitting
over here. Then you have the beta alpha dot
p over this m c to m power 2 and then omega
to the power 2, so you write each term explicitly.
That is what we have, we bring this term to
the top of this slide, and now to analyze
this it is useful to introduce beta alpha
dot p to be equal to gamma, and then look
at the form of this gamma. So, beta alpha
we can find out exactly what it is, beta is
this matrix, alpha is this matrix, so beta
alpha is this matrix operator, all that we
have done is to use the matrix multiplication
property.
And gamma is beta alpha dot p, so this is
beta alpha, this you should take the scalar
product with this p, for some reason this
dot appear nicely as a dot or not as a box,
as I did earlier, so it has decided to be
kind to wash now. And you have this sigma
dot p, and minus sigma dot p in these two
locations, 0 in the diagonal locations, and
now take the square of this, to get the gamma
square.
So, let us take the square of this, and you
can go head and work this out and find that
this is nothing but minus p square times the
4 by 4 unit matrix. You know how to handle
sigma dot p times sigma dot p, so it will
give you the p dot p, and then you will get
a cross product of p with itself which we
will throw, so you get essentially minus p
square.
So, now you have gamma square and this really
allows us to write this expression in a form
which is very familiar to all of us, because
now you write these terms explicitly. And
you had the n to the power 0 term, then n
to the power 1, n to the power 2, n to the
power 3, and so on. Those are the terms which
are written out explicitly, but whenever I
have beta alpha dot p square, I have replaced
that by minus p square, so I have a minus
p square over here.
Here, I will have square of minus p square,
this is the 4th power, so in all the even
powers I will get raising exponent of p square
by 2. So, that is an advantage that you get,
and here you have got all the even powers
in a first row, and all the odd powers in
the second row, but notice that in the all
the odd powers. Since, you get powers of you
know p and then p square and p to the 4 and
so on, so if you take one of those powers
factored out this beta alpha dot p by p. If
you factored out, then you will get odd powers
over here, now I will show you how it is done,
because you can take this beta alpha dot p
as a common factor in each of these terms.
So, let us do that this beta alpha dot p,
I have factored out then you get this term
to the power 1, you have got a term in cube,
but this one comes with a minus sign then
next one is come with a plus sign. And you
recognize that power series or you recognize
the both the power series, so these are the
cosine and the sine power series. So, we have
succeeded in writing this power series in
a very familiar form, so this is a cosine
term, this is a sine term, but the sine term
must be operated upon by beta alpha dot p
by p which we factored out.
So, that we can get these odd powers, and
the remaining powers came in p square p to
the 4 p to the 6 and so on, so that is what
we have got. And now you have got a rather
simple form, these you instead of those infinite
power series and so on, you have got cosine
and sine functions, which are of course, power
series and infinite terms in that. But, then
these are familiar well known easy functions
to work with, we know we used them all time
in geometry trigonometry, so we can do some
very simple mathematics for that.
So, this is our there are two terms over here
which operate further, which pre-multiply
these two terms, so you have a total of 4
terms. This times this, then this times this,
and then this times this, and then the second
term times, the second term over here, so
these are the 4 terms. Now, here you get alpha
dot p and alpha dot p again, and this one
is a dot this one is also a dot, but we know
what it is.
So, we have this beta p square by p coming
from this using the same kind of analysis
as we did earlier, and now if you look at
this 4th term, it has got the operator beta
alpha dot p, and then there is another beta
on this side. So, the essential operator structure,
which is sitting in the last term involves
beta alpha dot p beta in that order, which
is what you have in the bottom here, beta
alpha dot p beta.
That is what we are looking for, and to get
that we first get beta alpha, then get beta
alpha dot p, and then get beta alpha dot p
beta and all you have to do is used the dirac
matrices. And carry out the multiplication
systematically, it is a very simple thing
to do, and you find that this beta alpha dot
p or alpha dot p, when it is bracketed by
beta on either side is nothing but minus of
alpha dot p.
So, you can simplify this operator over here
in the last term, so that is what we have
done here, you have got the beta alpha dot
p beta equal to minus alpha dot p, and that
is what comes over here in the 4th term. Have
good enough, now let us combine these two
terms and these two terms, so this has got
beta, this term has a beta, this term also
has a beta, so these two terms are combined.
So, beta and the remaining part of this term
is m c square cosine of this function, and
the remaining part of this term after having
extracted beta is p c here is a p square by
p and here is a c here. And then you have
got a sine function, so these two terms come
as a first two terms, and the first and the
4th term come as the next two terms.
Now, we can really make a choice that we will
find to be extremely convenient for us, let
us take this cosine function multiplied by
p c outside, because then you will get m c
over p over here times the tangent of this
function. So, let us get this cosine function
now, and now if you choose your omega, because
that is a choice, which we have left free
for us, and we exercise that. Now, it is almost
like you know manthra exercising her ride
whenever she wanted, so we have our freedom
now, but this is going to be for a good cause.
So, what is the choice, you choose omega to
be m c over p tan inverse p over m c, because
what it does to this term is throw it off,
with this choice of omega, this 1 minus m
c over p tangent term goes to 0. What you
are left with, h prime equal to first two
terms, where is the odd operator is gone,
so the choice allows us and this is sometime
called as a Foldy Wouthuysen triangle, because
you will see that this is almost like root
of m square c square plus p square. So, you
are left with only the even operator, where
you find the Hamiltonian and the odd operator
is eliminated, now this is exactly what we
wanted to achieve.
And using this identity for tan inverse function,
you can write this as beta c times root of
m square c square plus p square, there is
beta of course, but beta has got 0es in the
off diagonal locations. So, by choosing this
particular angle, we have succeeded in throwing
of the odd operators, that was the motive
of carrying out this transformation.
So, we know that is possible to do that, we
had restricted ourselves to a free electron
which is not the general case, so we have
to now figure out, how to do it, when you
have an electron in an electromagnetic field.
And now, you cannot expect S to be independent
of time, you will have to take this term into
account, because you have got a vector potential
which easily could be time dependent. So,
now the operator S which will be involve in
the Foldy Wouthuysen transformation will be
will need to have the provision to be a times
dependent operator, so how do we choose it.
So, first of all we have to see how these
operators are determined, because we definitely
have times look at these operators, where
h is whatever be the form of the Hamiltonian,
and now it will include the vector potential
and the scalar potential as well. So, here
are well known techniques and this is another
reason, why I think it is nice to learn the
Foldy Wouthuysen transformations, because
you learn some techniques, which you can find
useful in many other applications in physics.
And these are very powerful techniques simple
techniques, but powerful techniques.
So, omega prime is this, and it is good to
look this expression as the limiting value
of e to the i psi S omega e to the minus i
psi S in the limit psi going to 1, and then
you are taking the limit of a function of
psi. You are taking the limit of a function
of psi this function is this operator, we
know what it is and you can expand this function
of psi in a power series.
This is a well known power series expansion
of a arbitrary function of psi, here notice
that the derivatives are to be taken as psi
equal to 0, and after you take the derivatives,
you complete the process of taking the derivative
and then take the going to 1. So, there is
no contradiction in the derivative begin taken
as psi equal to 0, and then taking the limiting
value of this function psi going to 1. It
is absolutely no contradiction, you do it
step by step one step at the time, so it is
mathematically absolutely correct.
So, let us do this, so let us have this power
series expansion now, take the derivatives
as psi equal to 0, and then after taking the
derivatives take the limit psi going to 1.
So, let us do it term by term, these are the
derivatives as psi equal to 0, and then you
take the limit psi going to 1. So, all these
psi square over factorial 1, becomes 1 over
2, psi cube over factorial 3 becomes 1 over
6, psi to the 4 over factorial 4 becomes 1
over 24, and so on, so that is a kind of you
know set of terms that we get. So, this is
your expression for omega, and in general
you will need the nth partial derivative of
f with respect to psi, at psi equal to 0.
So, that is what you need in general, so this
is the nth derivative that you are going to
look for, and now let us take a look at this
f of psi, this is your f of psi. Now, S may
or may not commute with omega, we will not
make any assumption on that, and you take
the first derivative with respect to psi,
then take the second and find a general expression
for the nth derivative, because that is what
you need.
So, the first derivative is e to the i S e
to the i psi S, because e to the i psi S and
S commute with each other, S may not commute
with omega, but it does commute with e to
the i S, because it is just a power series
n S. And over here from the second term you
get minus i S, because this is e to the minus
i psi S, now these two terms commute. So,
you interchange their positions, and now you
can combine these two terms by sandwiching
this commutator S omega minus omega S in between
these two exponential operators.
So, this is your first derivative with respect
to psi, which involves this commutator, so
this is your first derivative. Now, if you
do the same thing and get the second derivative,
you will get more commutators, and this is
a very nice series that you get, it is a beautiful
series. So, you get the commutator of S comma
omega with S, then you get another commutator,
when you go to the third partial derivative,
you are extending the same technique, you
can develop an expression for the nth partial
derivative.
So, this is the expression for the nth partial
derivative, you will have to raise this i
to the power n, and then you will have so
many you know commutators to work with. Now,
this is a very good result, and then you take
the value as psi equal to 0, so e to the i
psi S becomes 1, e to the minus i psi S also
becomes 1, and you can forget about it, because
you have take the derivative at psi equal
to 0. So, this is the term for n equal to
0, then for n equal to 1, you get i times
S comma omega, then you get i square and then
this commutator. And that is how you get subsequence
terms you will have an infinite series actually.
Now, we have figure out how to handle the
first term and this is the infinite series
that we had, but we also have to work with
this time derivative term. So, you have got
an infinite set of series coming from, how
we handle this term and then we still have
to figure out what to do with this. And over
here we are not going to assume that the time
derivative operator commutes with S, it may
or it may not in general it will not.
So, when you take that into the account very
carefully, you get commutators of S with S
dot whereas, dot is the time derivative of
S, so I will not show you the intermediate
steps, but you know how to work it out. And
you will get commutators of S with S dot,
and then each commutator with S, and subsequently
you can have another infinite series. So,
now, you have got two infinite series, which
are summed over, they are added to each other
in your transformed Hamiltonian.
Your original Hamiltonian has got this odd
operator, and your whole intention is the
expect the right hand side to have some form,
in which the odd operator would become if
not eliminated at least less important. And
our first goal is to look for such transformation,
which will reduce the odd operators at least
by a factor of m, by 1 over m m is a huge
mass, it is half a million electron volts.
So, that is that is the motivation, that is
how we look for this transformation operator.
Notice, that over here I am strike the full
Hamiltonian inclusive of these terms, the
first term, the second term and the third
term, but in the 4th term, if I approximate
h only by beta m c square, and choose S 2
have to be an operator of order 1 over m.
Then I have 4 of these operators in this term,
so I get 1 over m to the power 4, scaling
coming from that. And then if I left the Hamiltonian
to be written only by beta m c square, the
whole importance of this term will be reduce
by 1 over m cube.
And I can develop an approximation scheme
in which we propose that retain only those
terms which are of the order 1 over m cube,
if it is 1 over m to the 4 or smaller throw
it, that does not look like a bad approximate
at all. And to retain to develop an approximation,
which will have the leading term, which and
then all the subsequent terms will become
diminishingly small, because they have higher
power a S, because there are more number of
S operators, which appearing in these commutators.
So, if S is of order 1 over m, the more number
of S'es you have the weaker the commutators.
So, you can make this approximation beta m
c square, for h in this term, but you cannot
do it over here, because this will be 1 over
m cube, this is of order m, and then you will
have only 1 over m square approximation. So,
that is not what we doing, we are doing 1
over m cube, we want to retain terms at least
up to 1 over m cube. So, you can make an approximation
to the Hamiltonian in this commutator, which
have got 4 of these S operators as you see.
And correspondingly you make this term here
also you have 3 of these S operators, in the
times derivative terms which is again of order
1 over m cube. So, you make an approximation
which is consistent, and put a period over
here you are truncating the infinite series,
that is an approximate, which is why in trigs
book on the quantum mechanics he very categorically
states. This the Foldy Wouthuysen transformations
go as far as they do it is not that is exact,
and this is the reason it is not exact, but
it is good.
So, you truncate the series, which is various
approximations involve put a full stop over
here, forget about throughout the rest of
the terms. Now, this is your transformed Hamiltonian
h prime, now let us analyze this, bunch of
commutators you are not choosing S'es yet.
And you choose your Foldy Wouthuysen transformation
operator to be given by minus i beta theta,
theta is this coupling, it includes the generalized
momentum, it includes the dirac operator alpha.
So, let us write it explicitly this is your
Foldy Wouthuysen transformation operator,
it has got the similar form as you found was
used for the free particle transformation
operators.
So, there is some clue available from free
particle transformation, and you can actually
carry out the Foldy Wouthuysen transformation
further more by carrying out these transformations
further, you can reduce the importance of
the odd operators further. So, this is the
first Foldy Wouthuysen transformation as a
matter of fact, it turns out that is most
promising is subjecting the whole system of
equations to 3 transformations. Namely the
first, second, and the third Foldy Wouthuysen
transformations, so we will see how they look.
So, this is the first Foldy Wouthuysen transformation,
through an operator which is minus i beta
theta over twice m c square, and now you can
plug in this explicit form of the dirac operators
beta and theta. And you know the commute,
how to find the commutation of beta theta
with h, and using that you would know how
to find the commutator of S comma h with S,
so it is simply, but laborious. And that is
where your youth will coming handy, because
you can do such things without getting tired,
you have boundless energy, although it is
said that youth is vested in the young people,
so this is the operator i S comma h.
So, this is the Foldy Wouthuysen transformation
operator S, then you have got the dirac Hamiltonian
and find the commutator of the Foldy Wouthuysen
transformation operator with a first term,
with the second term, and with the third term.
Do it term by term simplify, use the properties
of dirac operators, the dirac matrices, you
see how it is developing. Find out how these
dirac operators multiply each other, theta
beta is minus of beta theta.
Beta epsilon is same as epsilon beta, but
these this is different, so you used a correct
signs over here, you also know that beta square
is equal to 1. And when you combine all of
those terms this is what you get for i S comma
h, but that is only the this box over here.
And these things some time to do, especially
after removing the careless mistakes it takes
a little while.
And then you do it, you know what this i S
comma h is, this is what turns out to be,
then using, this you find the second term,
and then using this you get the next term,
by now you have missed your dinner. Lost a
little bit of sleep, and then when you handle
all of these terms very carefully, you get
this S comma S comma h. 
Then you can get the next one, similarly and
I am not going to show you all the terms,
but now you know how to do it.
This is very simple, same thing with the times
dependent, also that is the other infinite
series, which you have truncated happily to
1 over m cube, but at least those terms those
commutators, you have to determine there is
no escape from that. So, you get the commutator
of S with S dot, and what you get is the transformed
Hamiltonian h prime, which can be written
as beta m c square plus 2 new operators epsilon
prime and theta prime, whose leading term
is of the order of 1 over m.
So, all the odd operators are now reduced
by factor of m, that is a significant achievement
in getting this transformed Hamiltonian, which
is to reduce the importance of the odd operator.
So, that you avoid the mixing of the particle,
and the antiparticle states, so that is the
leading term that you get in the odd operator.
And that is where I will stop for today, I
will be happy to take some questions, but
essentially you have seen, how the first Foldy
Wouthuysen transformation works. And then
we will have a second Foldy Wouthuysen transformation,
and then we will have a third Foldy Wouthuysen
transformation. And when you do that, you
will see very happily where the spin orbit
interaction term really comes from, you really
need to subject the dirac Hamiltonian to 3
conjugative Foldy Wouthuysen transformations.
For that term to become manifest in the 2
component form, but for now if there are any
questions, I will be happy to take otherwise
goodbye for now.
