In the last lecture we introduced the smallest
unit of a quantum state. And we said that
this is called a qubit which is essentially
corresponds to our classical Cbits of 0 and
1, the difference between the quantum bit
and the classical bit was while a classical
bit stay in either value zero, take either
the value zero, or the value one. A quantum
bit which is a qubit can be simultaneously
in a linear combination of the state 0 and
1 we also saw that while that is true now,
if you are making a measurement let us suppose
that you make the measurement in a computational
basis,
then you would get the either this state 0
or the state 1with the probability which is
given by mod a2/ mod a2+b2 and mod b2/ mod
a2+b2 respectively. I also gave you some physical
realizations of Qubits saying that they could
be representing the spin half particle, it
is spin projection along some arbitrary chosen
direction like Z direction or it could be
the polarization direction associated with
a photon or even the ground state and the
excited states of an atomic system.
Now we talked about the Pauli matrices and
we are, what we are going to do is this that
the well-known Pauli matrices are σx, σy
and σz are σx =0 1 1 0, σy = 0 -i i 0 and
σz=1 0 0 -1 each one of this has an eigen
value + 1 or -1 for example σz has an eigen
value +1 corresponding to the state 1, 0 and
an eigen value -1 corresponding to state 0
,1 you can yourself find out what are the
eigen states of σx and σy.
Let me find out an arbitrary direction, the
Pauli matrix in an arbitrary direction let
us just call it σn . As you are aware, that
an arbitrator direction in space the unit
vector has components which are x component
is equal to rsin θ cos Φ , y component is
equal to r sin θ sin Φ and z component is
r cos θ or if I take r=1 that makes the σn
that is the Pauli matrix along the arbitrary
direction n as  σz times cosθ which makes
these diagonal elements cos θ, - cos θ +
σx times sin θ cos Φ which makes it sinθ
cosΦ let me just write down here and then
I will write it more compactly here.
So I have got sin θcos Φ + σy times sin
θ sin Φ but you notice there is an –i
in this so therefore it will be - i times
sin θ sin Φ but you can use cos Φ - i sin
Φ equal to e-i Φ and then write this as
e-i Φ times sin θ so this term is e-i Φ
times sin θ and correspondingly this one
is ei Φ times sin θ.
Now it is very trivial to find out what are
the eigen values and the eigen vectors of
this matrix eigen value as you know you, have
to simply find out the secular matrix which
is λ - cos θ here λ+ cos θ there and of
course these two elements just become negative
and then you find out the determinant put
it to equal to zero if you do that you get
the value of λ that is the eigen value to
be equal to + or - 1. Now which is not surprising
actually the eigen values λ.
The reason why it is not surprising is because
we have said that the eigen value along the
z direction, which is an arbitrary direction,
there is no God given direction called z,
is +1 or - 1 so, if I take in any n-th direction
that is also should be + 1or - now you can
use the same matrix in σn to find out what
are the eigen states, the eigen states associated
with either of these eigen values but I am
going to be interested only in the eigen states
corresponding to λ equal to + 1 because that
is what is the standard convention for the
pictorial representation that we talked about.
And, if you just do a little bit of an algebra
I would urge to do that for getting some practice
in to eigen values and eigen vectors you will
find that this eigen vector
is given by cosine θ/2 and ei Φ sin θ/2.
If you look at the slide here now, see what
I have done now is this, I now take this state
which I wrote down that is the column vector
cos θ/2 and eiɸsinθ.
So this is my state which I will write as
state Θ, Φ. Pictorial representation that
I am talking about is known as a Bloch sphere
representation returning back to the slide.
Let us look at the directions I said that
now this is a sphere of does not quite look
it on the on the slide it looks a little overlised
but on the other hand it is a sphere of unit
radius. So it is a unit sphere and what I
am going to do is to associate every point
on the unit sphere with a unique state with
a unique state having a value θ ,Φ, now
let us look at what it means. Now supposing
I take the point, take a position well remember
r is equal to 1so I take let us say θ=0,
Φ=0. Now what is θ=0, Φ=0 ? Okay, let me
let me rewrite this state θ, Φ in a slightly
different way. So this is cos θ/2 times 1
0 + eiɸ sin θ/2 0 1 or in our vector notation
ket notation this is cos θ/2 0 state + eiɸ
sin θ/2the state 1.
Now supposing I take the state which is θ
=0, ɸ=0 put it here you notice because θ=0
this term becomes equal to 0. And since cos
0 = 1 so θ=0, Φ=0 which is nothing but the
North Pole in the slide that I showed you.
So North Pole corresponds to the state 0 look
at that picture there so this is the picture
on the slide.
This is the point which stands for the point
|0> that is the North Pole. Now let us go
to the South Pole, now if you go to South
Pole.
So I have got to remember that North Pole
corresponds to the state 0, now let us look
at South pole, the South pole has θ =π and
ɸ =0. Now that obviously stands for the state
1, remember that I had written in this form
so if I have θ=π, cos π/2 = 0 so this term
goes away. ɸ=0 so e0=1 and this is sin ɸ/2
= 1 so the state will be 1.
So if you go back to the slide you find that
this is your state 1 ,okay. Let us go to a
slightly different point. Let us consider
the point where the positive x-axis meets
the equator, the point where the positive
x-axis meets the equator.
Now since it is on the equator θ=π/2 and
Φ=0.
Now you can immediately go back there θ=π/2
means cos π/4 which is 1 over square root
of 2, eo=1, sin π/4 which is again 1 over
square root of 2.
So this state is 1 over square root of 2 0
+ 1 remember that I had called this the diagonal
state one of the basis of the diagonals. And
likewise you can convince yourself that the
point where the negative x-axis meets the
equator that is simply the other bases in
the diagonal states that is 0-1. Like this
every point on the Bloch sphere stands for
a unique quantum state and the θΦ value
can be obtained.
I will later on talk to you these states which
lie on a bloch sphere they all are what are
known as a pure state what is pure about it,
we will be talking about as we go along.
The next questions is that how much of information
is there in a qubit. Now remember that in
principle, a qubit contains infinite amount
of information. Now what is the reason ? The
reason is that remember my θ and Φ are the
parameters and θ and Φ expansion in terms
of binary digits may be essentially non terminating.
But we have told you that in spite of the
fact there is huge amount of information contained
in that state much of this information is
not really available to you because when you
measure it you either get a state 0 or you
get a state 1.
You can calculate what is the value of σz
in this context you can easily see it by finding
out the way the expectation value of a quantity
is calculated we have seen that the state
is given by cos θ/2 e1ɸsinθ/2 so, my σz
operator is 1, 0, 0, -1 so the expectation
value of σz is given by the row vector cos
θ/2 e1ɸsinθ/2 multiplied with this. Trivial
matrix calculation which will give you cosθ
as the expectation value of σz.
And likewise we can determine σx. Now remember
that I have mentioned to you earlier though
I have not yet proved it a quantum state cannot
be exactly copied there is a theorem which
will be proving as we go along that there
is a no cloning theorem of quantum state.
But then you
imagine somehow we have prepared identical
copies of quantum states and we are able to
make a large number of mathematics.
So I can do the calculate the expectation
value of σx by that I can find out what are
the directions nz and nx and you remember
that I have nx2+ny2+nz2=1 so by these calculations
I can find the values of nz and nx but the
signature or the sign of ny will remain undetermined.
So this is what we talked about.
Now the point then is this, that we have so
far said that when you make a measurement,
you get either a state 0 or a state 1 with
a probability. Now this probability will allow
you to determine so the.
For example, if I have a state α 0 + β 1
now, if I make a repeated measurement on identically
prepared copies I will get state 0 with a
probability α2/α2+β2 and this state 1 with
the probability β2/ α2+β2. Now these are
the probabilities which will get which will
enable us to determine the magnitudes of α
and β, is there a way of finding out the
relative phase, supposing I consider a state
for convenience ψ = 0 + 1 but I have a eiθ
and I normalize this with a factor 1 over
square root of 2.
Now I know even if this state if I measure
the states in a computational basis I would
get the state 0 or state 1, each with the
probability of ½ this is what comes from
here, but how about the information on θ
can I get it ? Now if you are making a measurement
in the computational basis you cannot, but
I will just not show you that if you make
a measurement in a diagonal basis okay, you
will be in a position to find out the information
about phase.
So let us suppose ψ I have said is 1 over
square root of 2 0+ eiθ θ is the relative
phase by square root of 2. And let us measure
it, not in a computational basis, but in a
diagonal basis. Now in order to make the measurement
in a diagonal basis I must first express the
state 0 and 1 in the diagonal basis. Now recall
that my diagonal basis is + = 1 over square
root of 2 0 + 1 and - = 1 over square root
of 2 0-1.
If you do a little bit of algebra this will
tell you that 0 can be written as + + - by
square root of 2 and 1 can be written as +
-, - by square root of 2. So this ψ that
I wrote down just now can be now expressed
as this 0 there is another 1 by square root
of 2 here so I get 1 over 2 times + + - +
eiθ/2 + - -. Now if you rearrange these terms
you find that this is given by here, I will
write it as ½ of this + ½ of this ,so let
us look at this, let me write it down that
becomes clear.
This is fairly straight forward algebra but
let me write it down. See basically what I
have done is this I have taken out an overall
eiθ/2 out so that this term now has a factor
e-iθ/2 and this term has e+iθ/2. Now if
I now collect the + then I get e-iθ/2 + e+iθ/2
by 2 which is cos θ/2 and likewise the cefficient
in term with – becomes -sinθ/2. Remember
that this is the overall phase factor and
in a quantum state as we have stated earlier
and overall phase factor does not matter because
a state ψ and another state which is C times
ψ they have no physical difference. So therefore,
this is not going to matter.
But suppose I am now making a measurement
in the +, - basis. The probability with which
in the diagonal basis I will get the value
to be + is cosine2 θ/2 and the probability
with which I will get the – will be sin2
θ/2. So therefore, this measurement in the
diagonal basis gives us some information about
the phase. So so far we have been talking
about single qubit, but notice that we all
realize that a single qubit is just one bit
of a information.
It may be with linear combination it may be
lot more informative or may contain lot more
information that make classical bit, but nevertheless
we are still talking about 1 qubit. Now in
order to do computation we need to go to more
and more qubits as you are aware in classical
bits we talk about situations like where let
us say 00 now suppose I talk about two bits
what do I have there, I have 00, 01, 10, and
11.
Likewise in, if you take three qubits, three
classical bits and you have eight different
combinations. Now in that case how do we handle
this situation ? Now what we are looking for
is the qubits which come from a composite
system looking at this slide.
Suppose I have a composite system let me call
it A and B they, and let us suppose α is
a basis in the subsystem A and β is represents
a basis in the subsystem B. Now I define my
composite system like this that is αβ AB
and if I take vectors in the subsystem A and
subsystem B as α׳, β׳ then the corresponding
bra multiplied with this ket gives me. Now
analogously I can now define an operator in
this composite space.
Suppose I talk about this operator this is
a direct product sign MA x NB now it acts
on state ψ,ɸ so what happens actually is
the state MA acts on the state ψ, NB acts
on the state Φ and I know that an operator
acting on the state gives me a linear combination
of vectors in that subspace and likewise.
So I would get a situation like this it is
getting more and more complicated.
So as I said that the classical bits referring
to classical bits are 00, 01, 10 and 11 the
quantum two Qubit state will then be a linear
superposition of states α 00+ etc. . The
same normalization holds good, α00 absolute
square up to α11 absolute square is equal
to 1. Now we will talk about two qubit system
and its expansion to n cubit system in the
next lecture.
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