We are going to spend some more time now on
the properties of spinning particles specifically
with regard to the spin of the electron because
this is an extremely valuable probe not just
in particle physics but also in condensed
matter physics and many other areas of physics.
So its worthwhile spending some time on this
and telling you how spin of electron behaves
under applied electric & magnetic fields.
If you recall, the spin operator for an electron
S has 3 components, S1,S2, S3 or Sx, S y,Sz,
which obey the angular momentum algebra and
they are represented by h cross over 2 sigma,
where sigma are the Pauli matrices. Each Pauli
matrix has eigenvalue + or ñ 1. So this guarantees
that any component of the spin of an electron,
the intrinsic angular momentum has eigenvalues
+ or - 1/2 h cross. So the spin quantum number
is a 1/2 for the electron. As you know, if
you have a quantum number j for the angular
momentum, then the subspace in which the states
of this angular momentum live is 2 j +1dimension.
There are 2 j +1 values for the eigenvalues
of any component of S or the angular momentum.
So in this case, it becomes 2 dimensional
and therefore the states that you could talk
about are | 1/2 , 1/2 > and |1/2, - Ω>. This
corresponds to the quantum number S which
is always 1/2 for an electron and this corresponds
to m where m times h cross is the eigenvalue
of any one component of the angular momentum
And we will conventionally take it to be the
z component. Then a very convenient representation
for the spin operator is this thing here.
And recall the definition of the Pauli matrices.
We have written this down several times. They
are sigma 1= (0 110), sigma 2 = (0 - i i 0)
and sigma 3 is already diagonal and itís
(10 0 -1). There are numerous properties of
the sigma matrices which i have written out
and which we have come across earlier when
we studied linear vector spaces. The square
of every sigma matrix 2 by 2 unit matrix and
the eigenvalue of any sigma matrix is + or
-1. They donít commute with each other. Sigma
3 is already diagonal. So once you diagonalize
that, the other 2 cannot be diagonalized simultaneously
but any of them could be diagonalized and
the eigenvalues would be 1 and ñ 1. this
state of the electron corresponds to the total
angular momentum quantum number 1/2 which
we have already known and Sz eigenvalue is
+1/2 h cross and this corresponds to - 1/2
h cross. Since we will we have in the back
of our minds, what happens when you apply
a magnetic field, in which case you have either
an up state or a down state for the magnetic
moment. its conventional to call this in short
hand, a spin up and this a spin down. the
understanding being that the basis that you
choose corresponds to states in which Sz is
diagonalized. So eigenstates of Sz. then the
orthonormality relations are of course up
up =1=down down and the scalar product down
top is 0. So this is going to be very a convenient
notation for us. We use this notation i am
going to talk about several spins then this
would become very helpful. Now what is the
actual representative of this term here? Remember
in the basis natural basis, you could also
write this as 1 0 and this could be written
as 0 1.
In any 2 dimensional linear vector space,
you could choose 10 and 0 1 as the orthogonal
independent vectors which form a basis in
this space. Now why are we making such a big
first about it? the answer is even though
the quantum number 1/2 is extremely small
compared to classical quantum numbers, for
instance, we know that the angular momentum
component has value + or - 1/2 h cross. Thatís
actually true and angular momentum is quantized.
So if you if you took a stone weighing 1kg
and you whirled it around in a circle of 1m
radius, mr squared is 1 in standard international
units, thatís the moment of inertia. If you
chose the time period to be 1 second for example,
then the angular frequency is 2 pi.
Therefore the angular momentum is i omega
which is 2 pi joule second which is enormous.
But that 2 pi joule second must also be = orbital
angular momentum multiplied by h cross. This
must be of the order 1or 2 pi but this is
of the order 10 to the - 34 in these units
and therefore this quantum number is 10 to
the +34. That is so enormous that the discreteness
of the angular momentum completely gets washed
out absolutely. Whether you add ten or hundred
or whether you add million or billion itís
still doesnít matter. 10 to the 34 is an
enormous number thatís the reason why you
donít see a angular momentum quantization
in daily life in macroscopic objects. But
itís very different story when you have a
quantum number like a Ω. This indeed tiny
so this whole thing this has eigenvalues +
or ñ 1. So you can see numerically this is
extremely small in the normal daily units
that we are used to. So how do you find it?
How do we detect this spin at all? the answer
is that this spin also implies a magnetic
moment for the electron and that magnetic
moment would couple to applied magnetic fields
and then you can manipulate the spin or probe
the spin of the electron using the magnetic
field or using the various protocols for the
magnetic fields like switching them on and
off in different directions etc. Letís see
how this is done.
The magnetic moment of the electron mu e as
i pointed out last time is = the g factor
of the electron ge multiplied by the charge
of the electron divided by twice the mass
of the electron times the spin operator, the
intrinsic angular momentum operator for the
electron. But this is =ge which is 2, and
i will explain why this is 2. This turns out
to be 2 for the electron which is modulus
e with the - sign because electron has a negative
charge. Itís divided by twice mass for the
electron and the spin of the electron is h
cross over 2 times sigma. The reason it is
convenient to do this is because this sigma
is dimensionless and has a eigenvalues + or
ñ 1 for every component. One of the 2 is
cancelled and you end up with - mod e h cross
over twice m e sigma. And of course you are
familiar with this object. This is the Bohr
magneton.
So you could write this as = - mu bohr times
sigma and mub here is the Bohr magneton. Itís
the natural unit of the magnetic moment which
you get for a particle of charge e and mass
me. When you put this thing in a magnetic
field, there is an extra contribution to the
energy of the system or the Hamiltonian of
the system. The electron would have some Kinetic
energy of some kind. So there is a Hamiltonian
but when you switch on a magnetic field, there
is an extra contribution which is the magnetic
Hamiltonian if you like. So let me call this
H magnetic which is = - mu dot B.
The magnetic dipole moment dotted with B is
the potential energy of a magnetic dipole
moment in an applied magnetic field B. and
if you put this in, this becomes = mu B which
is a number and then sigma dot B. thatís
is the Hamiltonian. And now if you start with
the system in some state, you could change
the state by switching on this Hamiltonian
and then letting the system evolve as time
goes along and you would have to see what
happens. Classically I expect that if i have
a magnetic dipole moment placed in an external
magnetic field pointing upwards, then this
dipole moment precissess around the component
of the magnetic moment along the field. It
doesnít change in the magnitude. and the
other 2 components undergo harmonic motion.
So, you have a precessional motion and the
tip of this magnetic moment vector precess
a circle and forms a 1/2 cone about the direction
of the magnetic field. And the magnetic precession
frequency is easily computed. It is the Larmor
frequency of precession which is trivially
computed once you write down the equation
of motion for the dipole moment. so classically,
this mu is different from all these other
muís that I have talked about, if you have
a magnetic moment dipole moment mu, then d
mu over dt is = the torque on this term.
The torque is mu cross B and itís a constant.
Actually itís an angular momentum because
mu is proportional to the angular momentum.
The rate of change of angular momentum is
= the torque. Since the magnetic moment and
the angular momentum are connected through
some gyromagnetic ratio, itís essentially
d mu over dt times some constant. so there
is some constant sitting here and then a mu
cross B. the moment you have an equation of
motion of this kind, you know immediately
that if this is the direction of B and that
is the initial direction of mu then, all that
this does is to trace a path of this kind.
The tip of this vector traces a cone whose
direction along the magnetic field doesnít
change. and thatís easily found because all
you have to do is to dot both sides with B.
the magnitude of this doesnít change either
because all you have to do is to dot both
sides with mu.
So the magnitude doesnít change and this
component doesnít change so the only thing
it can do so move around a circle. And the
precession frequency is determined by this
constant here which will involve the gyromagnetic
ratio and so on and so forth. Quantum mechanically
what do you think will happen? Well by Ehrenfest's
theorem, I expect that the expectation value
of the magnetic moment would obey a classical
equation of this kind. so with the expectation
values put around this mu, I still expect
the same sort of equation. But what does a
system actually do? That depends on what this
Hamiltonian is, what the direction of the
magnetic field is, etc. and we really should
write down now is the state of the system
at any given time. So lets try to compute
it and see what happens but before i do that,
I also point out that if the field is sufficiently
non-uniform then you know that a magnetic
dipole moment not only undergoes a torque
but also a force. so its possible if the field
is sufficiently non-uniform and strong enough
to actually separate out different magnetic
moment states, and this was what was done
in Stern-Gerlach experiment which is a classic
experiment which i will describe a little
later because itís at the root of all thatís
being said today about quantum computing and
quantum information and so on.
So very often in books on quantum computation,
you would see an sg apparatus which means
the Stern-Gerlach apparatus. In different
directions, you have magnetic fields and so
on. we will come back to that we keep that
are the backup our mind but right now we will
take this B to be a constant magnetic field
and ask what happens to an arbitrary state
of the system. You start with an arbitrary
spin state of this electron, what will happen
as a function of time? Well, the first thing
to note is that since these 2 states; the
up and the down actually form a basis in this
space, itís clear that any spin state of
the system denoted by psi( t), can actually
be written uniquely as a superposition of
these 2 basis states. So this can always be
written as a( t) up +b( t) down, where a and
b are 2 coefficients which will depend on
time. And if you normalize this state, since
these are already normalized, we are going
to assume that a( t) whole squared +b( t)
whole squared is 1for all time.
So we will talk about normalized states of
the system. Now lets do the simplest problem
and that is i prepare the system in the up
state say, so i tell you to start with that
psi(0) is just up that means a(0) is unity
and b(0) is 0 to start with, and then I switch
on this Hamiltonian. Now if the magnetic field
is in the z direction too, then nothing is
going to happen. because if B is B times ez
for example, then the H magnetic the Hamiltonian
becomes mu and then sigma 3 B. sigma dot B
is just sigma 3 B. this quantity is an eigenstate
of sigma 3 with eigenvalue +1/2 h cross or
+1 because its sigma 3 alone. So, nothing
is going to happen and psi (t) is e to the
- i Ht over h cross. This is = e to the - i
mu B over h cross sigma 3 t acting on the
up state. So what happens? Itís just phase
factor here because sigma 3 is a diagonal
matrix. So we can compute what this quantity
is and it remains something which is proportional
to the unit matrix and sigma 3. So what happens
finally? This is an eigenstate of sigma 3
with eigenvalue +1. So you can replace the
sigma 3 by 1out there.
So this is = e to the - iB mu B by h cross
t on up. So nothing happens. so if it starts
in this initial eigenstate and you switch
on the Hamiltonian you are already in an eigenstate
of the Hamiltonian and it just remains there
apart from a phase factor e to the i et over
h cross. It doesnít do anything interesting.
Similarly if I started with down and applied
field in the z direction, nothing is going
to happen. But if i start with arbitrary state;
a superposition of up and down then of course
interesting things are going to happen. Because
whenever that Hamiltonian acts on the up state,
you are going to get +1 and when it acts on
the down state you, are going to get ñ 1.
And you will have a superposition of 2 exponentials.
They could be interference and so on. so this
is already telling us, how in the simplest
of instances, quantum interference would play
a role. Now classically, if you got a dipole
lined up upwards and you want change its alignment,
what would you do? There is no use applying
a field in the upward direction. You have
to apply a field in the transverse direction
or at least have a component in the transverse
direction. Letís start with the dipole in
this fashion. This is the electron and the
upstate and letís apply a traverse field
in the x direction. Now letís see what happens.
So this H is mu B. so the B, i am going to
apply is B times ex in the x direction. So
this is mu B times B sigma 1, itís the x
direction, so the first Pauli matrix comes
in. and then what does this do? psi( t) = A
to the -, let me put this factors and so thatís
e to the - i over h cross Bt. and then mu
B mu was modulus e h cross over twice the
mass and then a sigma 1. This acts on the
up state; the initial state. So the h cross
cancels out. And that is =e to the - i mod
e eB over 2 m times sigma 1 acting on this.
What is eB over m? If you have a charge e
of magnitude e in a magnetic field of magnitude
B and the mass of this charge is m, what is
this combination eB over m? Itís a cyclotron
frequency. So this is = e to the - omega cyclotron
t over 2 and then there is a sigma 1 acting
on the up state in this fashion. Now this
is not a pure phase factor by any means because
there is a sigma 1 here and that is not an
eigenstate state of sigma of 1. This is an
eigenstate of sigma 3. So you really have
to write down what this quantity is explicitly.
Letís use general formula which we already
wrote down. e to 
the power - i a dot sigma. This is = cos a
times the unit vector - i a dot sigma over
a. that is the magnitude of this vector (sin
a). That was the generalization of this a.
Eulerís formula e to the i theta is cos theta
+ i sin theta. How do we get this formula?
We use the fact that the square of each sigma
matrix is 1 + the commutation properties of
the various sigma matrices. So letís use
this formula explicitly.
In our present instance a is = omega c t over
2 a unit vector in the x direction, ex. so
psi( t) is = cos omega c t over 2 I - i dot
sigma 1 sin omega c t over 2. This whole thing
acting on the up state. And thatís explicit.
Now so this is what it is exactly at later
instant of time. But letís compute what it
is.
Remember that sigma 1 is 0 110. psi (t) = it
is a ket vector of some kind which going to
be written as column matrix and this is = cos
omega c t/2, cos omega c t/2, and then - i
sin omega c t over 2, - i sin omega c t/2.
Thatís this matrix acting on the up state.
But that is 1 0. And that gives you cos omega
c t /2 and - i sin omega c t/ 2.
Which is =cos omega c t, that is the up matrix
and - i sin omega c over 2, that is the down
state. So thatís my a( t) and my b( t) and
its clear that mod a squared + mod b squares
is 1 at all times. Itís normalized as you
would expect. a (0) is 1 as it should be we
started with the state of and b(0) is 0. Now
what is the significance of these coefficients?
Since I expanded this in the basis up and
down, whatís the significance of these coefficients?
What is the interpretation of their mod squares?
Itís the probability. I start with the electron
in the spin of state. i switch on a transverse
magnetic field at t =0. Then mod a square
(t) gives me the probability that at time
t, the state is up and mod b squared is the
probability that the state is down. The sum
of the 2 must always be 1. So itís immediately
clear that we could write this down and plot
as a function of t.
The probability up (t) =mod a (t) whole squared
and thatís = cos squared omega c t/ 2 and
what does that graph look like? It goes up
and down like this. On the other hand, the
one starts like this and this is = p down
time t = sin squared omega c over 2. And the
frequency with which this happens is precisely
omega c because cos squared is 1. You can
write this in terms of cos twice the angle.
So the periodicity is that of cos or sin omega
c t. so quantum mechanically, this Hamiltonian
which is in the transverse direction plays
the role of a spin flip operator. It flips
the spin from up to down and down to up repeatedly.
And at any time t, the system is actually
in a superposition of the 2 states expected.
So that happens quantum mechanically. You
canít say that the system is definitely in
the upstate of the downstate unlike what you
could do classically. This is because the
Hamiltonian doesnít commute. The up and down
states are not eigenstates of the Hamiltonian
but we interested on using that as 
a basis.
Quantum mechanics is a way of computing probabilities.
Certainly thatís true but the actual rule
for computing these probabilities; expectation
values and so on are quite deterministic.
There is nothing arbitrary about it totally.
What i want to illustrate by this example
is the basic nature of superposition. You
start with an eigenstate and you switch on
a Hamiltonian. The state you start with is
not an eigenstate Hamiltonian. And then the
Hamiltonian takes you through all the other
states of the system which in this case, is
all the other states of the basis. There are
only 2 in this particular instance but it
causes these spin flips. Therefore one sometimes
says that a transverse magnetic field causes
spin flips. This the way to flip the quantum
mechanical spin. And this is with a constant
field in the z direction. I am going to leave
it to you as a problem to work out happens
if i start with an arbitrary state a(0) and
b(0) normalized to unity and the magnetic
field in an arbitrary direction.
Itís not very much harder. The algebra gets
a little more complicated but you can see
easily that in this 2 dimensional space, you
always have a superposition of the upstate
and downstate. Suppose I have an eigenstate
of sigma x or sigma 1, I could pictorially
write it like this. This can also be written
as a superposition of up and down. Any spin
state in this Hilbert space can be written
as a superposition of an up and down. Thatís
important to remember.
Now all this was with a constant time independent
Hamiltonian. If I start now switching the
Hamiltonian on and off, switching a field
on and off as a function of time, matters
will get considerably more complicated. Then
we need a new formulism. We need to know what
to do when you have time dependent Hamiltonian
because e to the power - iht over h cross
is no longer the time development operator.
We have to be a little careful. We will do
that too. He what we have at hand is a 2 level
system in some sense. Just 2 states possible;
an up and a down.
So if you assign energies, remember that we
wrote H is =mu B sigma dot B here. And if
I now say that the magnetic field is along
the z direction for example, then this Hamiltonian
becomes mu B times sigma 3. And this has eigenvalues
+/- 1. So the energy levels of the system
would correspond to the following for the
down state and the upstate state. And there
is a gap between them. This energy is - B
mu B and this energy is +B mu B and there
is a gap of 2 B mu B. this corresponds to
2 level system but itís not an ordinary 2
level system. This is a quantum mechanical
2 level system. This is a general state that
you have and the superposition gives you enormous
freedom because you have 2 coefficients a
and b such that mod a squared + mod b squared
is = 1. Therefore an infinite number of superpositions
are possible.
So in some sense, the fact that you start
with the 2 level system but really you have
an infinite number of possibilities. This
is why itís called the cubit, as opposed
to classical bit which would be a 0 or 1.but
you starting with a 0 and 1 but all possible
superpositionís are allowed. Therefore things
are little more complicated. It is the superposition
thatís going to lead to all the interesting
features.
So this tells you what happens in the simplest
2 level system when i switch on a transverse
magnetic field. Let me also introduce another
concept here then we will come back to this.
And this is the concept of entanglement which
is something which is very common these days.
Suppose you have two 2 level systems, 2 electrons
and look at the spin states of the 2 electrons
together. I am not interested in the other
degrees of freedom. I am not interested in
the column interaction, the potential energy
due to the coulomb interaction, the kinetic
energy due to that motion and so on. I am
looking only at the spin states of the system.
What does the Hilbert space look like? For
the first electron, you have a Hilbert space
H 1 and it has 2 states which would correspond
to either up or down for the first electron.
Thatís this Hilbert space.
For the second one similarly, I have an up
2 and a down 2. I have these 2 states. Now
I put these 2 together and i look at the possible
states of the full system. How many states
would you say there are? There are four possible
states to start with. A naive guess would
be say (up, up), (up, down) or (down, up)
and (down, down). But say there are these
four states but now the matter gets a little
more complicated. We need to know how to add
angular momenta in quantum mechanics. We are
going to do this but let me state the result
which you have already learnt in spectroscopy.
if you have 2 angular momenta j1 and j2 which
are independent of each other, the quantum
mechanical rule for adding these angular momenta
is that the components of j 1commute with
the components of j but among themselves the
components of j 1satisfy the angular momentum
algebra and similarly for j 2. then the question
is if i add these 2 and i call this is the
total angular momentum here, what are the
possible values of this capital j or j square
what are the eigenvalues of j square given
the eigenvalues of j 1square and j 2 squared.
This is the theory of the addition of angular
momenta and i will make short excursion to
that a little later.
And answer is that if j squared has eigenvalues
j times( j +1) h cross squared and J¨1 has
a spin quantum number j1. so its j1 times
(j1+1)h cross squared and J2 is j2 into (j2
+1)h cross squared. Then the rule of addition
of angular momenta tells you that little j
can take on a large number of possible values
ranging from |j1- j2| up to |j1+j2|. We will
establish this is a very well known rule 
in 
addition of angular momenta. But you see already
that quantum mechanical angular momenta are
not going to add up like classical vectors.
Not surprising because these are operators
the vector valued operators.
So this is an operator and thatís an operator
and therefore they have their own peculiarities
and this quantum number can take on all values
in the range. any component adds up linearly
for example if this a j1 z component of it,
then this is = m1 h cross because it must
have eigenvalues which are some multiple of
h cross and this is magnetic quantum number
m1 and this m1 can run from - j 1 to +j1in
steps of unity. Similarly j2 z is m 2 h cross,
then jz is (m1+m2 )h cross. They add up completely
linearly because these 2 operators commute
with each other. and if this there is 1 eigenstate
of j1z and j2 z with these eigenvalues, then
you are also in an eigenstate of capital Jz
with eigenvalue but the allowed values for
the total quantum number run from here up
to there. Take this as a given rule for the
moment then the question is what happens if
i add the spins of 2 electrons?
Each of them is a 2 level system if you like
and what if I add the spins of the 2 electrons.
What are the possible eigenvalues of S 1+S
2? This is the total spin. What are the possible
values of the total spin quantum number S?
Letís use this rule. Whatís j1 in this case?
Itís just S1 and whatís that? spin Ω. No
h cross it just the quantum number. This quantum
number is 1/2 for the electron. It can be
negative with some number 0, Ω, 1, etc. and
i have already said that the electron spin
quantum number is Ω. So it is completely
clear that in this case, this is a 1/2 and
that is a Ω . So this total spin quantum
number can be = 0 or 1/2 +1/2 which is 1.
So you see you have 2 components each of which
has angular momentum quantum number 1/2 but
they can add up to 0 and they can also add
up to 1. They clearly cannot add up to Ω.
They canít add up to 1/2 because it runs
in steps of unity.
But I have also said that something with a
1/2 odd integer quantum number is a fermion
and something with an integer quantum number
is a boson. i will explain what these things
mean they have to do with statistics. So it
implies that you take 2 fermion and you put
them together we can create a boson. This
has profound implications. This is where superconductivity
occurs for instance. So there are many implications
to this but the fact right now i want to emphasize
that the total spin quantum number can be
either 0 or 1. Then the next question is what
are the possible states? What are these actual
states and what do they look like? We have
these four states here 
and now that actual space is the direct sum
of H1 and H2.
The states that we are talking about are four
in number. They could be either 1direct product
2 and so on. There are four of these direct
product states and I would like some slightly
less complicated notation for this. So let
me do exactly as i did when i took 2 simple
harmonic oscillators and constructed the angular
momentum algebra. Let me just write this as
up up, with the understanding that the first
arrow always talks about the state of the
first electron and the second arrow about
the state for the second electron. So I have
four of these states up up up, down down,
up down & down down. They obviously form a
basis in the full Hilbert space. So every
spin state of both the electrons can be written
as superposition of these four states.
Now the question is what are the these states
and what do they correspond to? You see the
moment i say S =0, then it also implies that
Sz is =0. So these are eigenvalues that i
am writing down here on the right hand side
because if this is 0 in units of h cross always,
this must be 0 because a single component
cannot have a non-zero value while the total
quantum number is 0. The square of this angular
momentum has eigenvalue 0. So every component
definitely has a eigenvalue 0 also. Even if
they donít commute with each other, thatís
a state of 0 total angular momentum. This
is a state in which every component of the
angular momentum has eigenvalue 0.
So here is an example where 2 operators which
donít commute with each other can have a
common eigenstate. The trivial case of 0 total
angular momentum. Then every component of
the angular momentum also has 0 as the eigenvalue.
This is the only possibility here but the
moment i say it is 1in units of h cross, then
what are the possible values of Sz? So corresponding
to this you have this and corresponding to
this, what are the possible values? It can
go from - the total quantum number to +the
total quantum number in steps of 1. - little
j to +little j in steps of unity and now that
S is 1, what are the possible values? It can
be ñ 1,0 +1. So we have 3 possible sub states
here if you like or states in this space corresponding
to S =1 and in this space you have a 0 here.
There are 4 possible states. They must be
linear combinations of these 4 possible states.
And it is these states that would like to
describe in terms of those direct product
states or the basis set. So what would they
be and how would you write this down? And
now letís write the states down as the total
S and Sz. recall this is my j m notation.
Thatís exactly the notation I am going to
use. So I would like to write 0 0, that is
this state.
Can it be this ? Because remember this corresponds
to remember in any component the angular momenta
just add up. So this corresponds to S 1z =+1/2
h cross, S 2 z is +1/2 h cross. So Sz must
be = h cross. So that fellow actually corresponds
to this. Similarly this fellow corresponds
to this because both are down and those are
easily written down. So |1,1> can be written
down and that is just up up. |1,- 1>can also
be written down and you leave a gap and write
it 1- 1can be written down and thatís the
down down.
What we are left with is to answer what is
the 10 and what is the 0 0? We need to know
these 2. Which is which this also has Sz =0
because 1is up 1is down. So does this. This
is down & that is up. How will you say this
corresponds to that or this corresponds to
this or any superposition of the 2 how would
you how would you decide? You decide by what
is the called symmetric property of the states.
This thing here is called a singlet state.
S =0 is called a singlet state.
The reason is there is only 1 state. There
is 1only Sz possible. On the other hand S
=1 has 3 possible states associated with it
and therefore what would you call this? This
is called the triplet state. So |0,0> is the
singlet and these 3 together form the triplet.
Now you notice that the singlet state sets
alone out here and the triplet state here
and this state here have certain symmetric
property. Suppose I exchange the 2 electrons,
I interchange all the degrees of freedom of
this electron with that, does this state change
at all? It doesnít change at all. Itís symmetric
under the exchange of the 2 electrons. Itís
completely symmetric. So really under the
exchange of these electrons, this state has
remained exactly the same. It is indistinguishable.
And so this now turns out that all the members
of a multiplet must have exactly the same
symmetric. This is the whole idea of putting
them together and one bin. Therefore this
state must have total Sz =0 that means a 1arrow
is up the other arrow must be down but in
such a way that when you exchange the 2, the
state doesnít change at all.
What would do you? you have this and this
as the 2 possibilities but if i exchange the
2, i go from here to here and this goes from
here to there. So what should i do there?
You have to superpose the 2. So this is up
down + down up divided by square root of 2
because I have to normalize this. Each of
this is normalize to unity. And therefore
if i divide by square root of 2, I normalize
this state. And then of course the other state
is an orthogonal superposition and would this
be? It must be orthogonal. Every state here
must be orthogonal to all the states. So what
is this got to be? Itís this. Now you could
ask why did i choose up down - down up? Why
didnít I choose down up - up down? It just
multiplies it by a phase factor. It doesnít
do anything to it multiplying this by any
complex number of unit modulus doesnít anything
to this state.
So this is antisymmetric and under the exchange
of the 2 electrons. And this is symmetric.
This is the spin part of the state but you
know it is something called the Pauli Exclusion
Principle for these particles which says that
if you have 2 electrons they are absolutely
identical then and no matter what state these
electrons are in, the total state must be
anti symmetric under the exchange of the 2
electrons. But this is only the spin part
of the wave function.
There is also a spatial part. If these are
moving about in the position representation,
there is a psi (r 1, r 2, t), a wave function
and product of that with the spin wave function
is a total wave function that must be anti
symmetric. So this immediately says that when
the 2 electrons are in the singlet state,
the spatial part of the wave function must
be symmetric under the exchange so that this
antisymmetric gives you an overall - sign
and in the triplet state the special part
must be antisymmetric and the spin part is
symmetric. This plays fundamental role in
all of physics. This idea that when you have
these 2 electrons and the Pauli Exclusion
Principle is implemented in this fashion.
So what you have to remember is a that spin
singlet state is anti symmetric and spin triplet
state is symmetric. And notice that we got
these linear combinations by imposing that
all the members of this multiplet have the
same symmetric.
Thatís why their part for the same multiplet
here. Now the question is what I mean by entanglement.
The answer is these states cannot be written
as a direct product state. These are direct
product states but a superposition of these
states. So the fate of 1 electron is inextricably
linked with that of the other electron in
these states. This is called entanglement
and it has very deep consequences. I will
subsequently show how this also leads to the
phenomenon of magnetism because this is a
very simple observation. It is actually going
to lead us to a so called exchange interaction.
This was Heisenbergís discovery. It explains
the origin of ferromagnetism. So we will talk
about that but right now, I brought this up
in order to tell you what entanglement is
but we are going to come to this concept and
talk.
You could derive this from the rules of addition
of angular momenta but what i gave was a very
compelling reason. The free components of
an ordinary vector are not 3 random numbers
or anything like that. They go into each other
under rotations. In an arbitrary rotation,
a given vector becomes linear combination
of the 3 components here. so together those
3 quantities transform in a specific manner
under rotations. In exactly the same way,
this is the fact that these members are all
symmetric under the exchange, is part and
parcel of they being a multiplet. This is
essential, itís necessary to this. But I
want you to appreciate that these are not
writeable. These are not writeable as direct
product states. Thatís the crucial point,
although this one happens to be and this one
happens to be writeable, the other states
are not. We will come back to these concepts.
Let me do that tomorrow.
