Okay so just towards the end of the last lecture,
I just talked about the measurement which
is probabilistic in the sense that you take
a large number of identically prepared systems
and then if you want to measure a specific
energy En then the value will be mod Cn squared
which is what is the probability for finding
the energy En right, everybody is there, that
is where these kind of stopped.
So recall the definition of psi, you just
insert an identity operator which is in parentheses
here on the same side and then the coefficient
Cn is nothing but the product of phi n with
psi. Everybody with me, we did this okay.
So then if you want to find the expectation
value of a system prepared in the state psi,
then Hamiltonian expectation value of the
energy then it will be weighted average where
mod Cn squared is the probability for this
energy En to be measured and so on.
So you sum it over and that will give you
the average energy. So what have you learnt
in classical mechanics?
You have a phase space. Phase space is defined
by position and momentum right and then you
have dynamical variables which I call it as
capital A, capital B which are functions of
this example. Then, if you want to know the
time evolution, Hamiltonian helps in finding
the time evolution in classical physics. What
are those equations? First you need to write
the Poisson bracket of A and B which is governed
by the Poisson bracket of x and p.
You all know Poisson bracket right okay. The
definition of a Poisson bracket of two variables
in a phase space defined by x and p involves
these partial derivatives in this fashion.
So that is the definition. If you want to
find what is the Poisson bracket of x then
p here, so del x/del x is 1 del p/del p is
1 del x/del p is 0 del p/del x is 0.
So it is just 1, so I am just trying to bring
in some kind of a correspondence between a
Poisson bracket which is learnt in classical
mechanics with commutator in quantum mechanics.
That is the main motivation why I am going
and comparing and contrasting classical mechanics
with quantum mechanics okay. So this is the
formal definition of this Poisson bracket
and evolution equation depends on the Hamiltonian.
And you can write x dot dx/dt involving the
partial derivative. The Hamiltonian is also
a dynamical variable which is dependent on
x and p. So you will have an equation which
gives you the time evolution of the position
coordinate, another equation which gives you
the evolution of the momentum right. What
does these equations called? Hamilton's equations
okay, so these two equations are classical
mechanics, this one is called Poisson bracket.
And these two equations which I have written
which gives you the evolution in phase space
time evolution, they are called Hamilton’s
equation.
So now we want to see an analogy to quantum
physics. So you clearly see that here the
phase space time evolution is governed by
some equation. We also want to write the state
vector which we write in quantum mechanics
to be governed by some equation that is 1
and what will be the analog of Poisson bracket
when you go to quantum mechanics. So those
are the questions you can ask.
And so the first thing is all the dynamical
variables, example your Hamiltonian and other
operators which are functions of x and p they
get promoted to become operators in Hilbert
space. To be more precise, you can take linear
operators on Hilbert space. Poisson bracket
which we had you can replace that by a commutator
bracket. What is a commutator bracket?
A commutator of A, B is AB-BA okay. A and
B are Hermitian operators. Let us take them
to corresponding to observables, its commutator.
When you do the commutator what happens? So
suppose I want to find a dagger of this. Someone
what is this? So B dagger sorry BA-AB. Is
that right? This is nothing but minus of commutator
of A with B. Something bothering us. If you
have two observables which are Hermitian operators,
the commutator is not a Hermitian operator
right.
If you want to make it Hermitian, what do
we do? Suppose I call this to be C operator
then it is not Hermitian. What do I do? Yes,
multiply by an i and then this dagger that
i will also get in here. So i so you will
get this to be again ic hat. So these are
the reasons why you will have an i here because
your left hand side is not going to be Hermitian.
If you want to make both left hand side and
right hand side to match, you put an i factor.
Why ih cross? Why do you need a h cross? By
dimensional arguments, the Planck constant
is the relevant one for the product of x and
p which you can introduce okay. So this is
one naïve way of arguing why it is postulated
that the Poisson bracket goes to commutator
bracket and the commutator bracket is ih cross
delta j. You are done for a finite dimensional
Hilbert space, is that right?
You proved this for a finite dimensional vector
but if you have a commutator of A with B as
i times c and what happens? So c in fact here
is an identity operator right. So if you take
x operator with p operator, let us do one-dimensional
that will be ih cross identity operator. In
general, it could be an infinite dimensional
space. If I take the trace on both sides,
what will the left hand side be? Left hand
side is 0 by this property. Is that correct?
Then, what is the problem, tell me? Is this
allowed? This contradicts; left hand side
is not equal to right hand side right. So
which means this identity is valid only for
finite dimensional linear vector space that
is the only way you can try and say that this
is in the infinite dimensional vector space
you cannot have this. In general, if you have
an AB commutated as ic, you have to impose
what, even in the finite dimensional Hilbert
space if this side is 0, this should be traceless
matrices okay right.
If I want this, take trace, if this is 0,
this implies c is traceless in finite dimensional
Hilbert space but in infinite dimensional
Hilbert space, this property cannot be satisfied
because you cannot give a matrix representation
even though we can write a Dirac braket notation,
you cannot write a matrix representation to
argue the trace AB is=trace BA in an infinite
dimensional Hilbert space, so that is just.
There is also something else we could prove;
delta A is your standard deviation of an operator
A. What is a formal definition? We saw this
in one of the lectures. What is delta A?
It is the expectation value of A squared-expectation
value of A the whole squared and then you
take the square root. So for some dynamical
variable of some operator, the standard deviation
is given this way. Product of two, this is
what we call it as an uncertainty, standard
deviation is what we call it as an uncertainty
the observable corresponding to the operator
A, time is uncertainty in B can be proven
that this is going to satisfy.
It is going to be related to the commutator
bracket. The commutator was 0, then what does
it mean? What does Heisenberg's uncertainty
principle tell you? You always write delta
A delta B is greater than or equal to h cross
by 2 or something right. What does that mean?
Delta x delta p when you write, you say that
if you can precisely look for the position
which is delta x. Your delta p will be very
large right.
So this is what is you cannot simultaneously
measure x and p. If the commutator of A and
B is 0, then delta A delta B is 0. So you
can simultaneously measure both of them with
precision if the commutator is 0. So x, p
commutator is what? X, p commutator is nonzero
okay. So because x, p commutator is nonzero,
you do not have, you have an uncertainty principle
which is your familiar Heisenberg's uncertainty
principle.
In general, if two operators are commuting
that is this one is 0 then you can show that
those two can be simultaneously measured,
is that clear? There is no uncertainty that
measuring 1 will not give you precise value
for other one. This we will prove in one of
the tutorials okay. So hence nonzero commutator
of two operators will imply uncertainty principle
and we can say one is precisely measurable,
the other one may not be precisely measurable.
So these are some things which you can play
around. This I already said. If you take the
conjugate and transpose of two operators,
so what we showed that the commutator of A,
B is same as negative of commutator of A,
B dagger right. So it is not Hermitian, to
make it Hermitian we introduced an i. For
x, p commutator, you need a dimensionless
constant.
X, p commutator whatever is the dimension
you have to introduce Planck constant, so
that the both sides dimensions are matched
and some of the interesting properties of
the commutators which one of the things which
you should remember if you remember when I
was saying the properties of operators on
the linear vector space, linear operators
the order matters. You cannot say whether
I call A, B and B, A are one in the same.
A, B is different from B, A in general. When
will it be same? When the commutator of A
with B is 0, then the order will not matter
or you can measure this for that in whichever
order right. So you need to keep track that
you cannot violate the order, so if A, B,
C, D is the order I first do the expansion
of this side. I take the A out and B commutator
with CD and then this one I keep the A.
But since this one should be order is AB,
the B should come this side. B cannot be here
which you can do in your classical mechanics
but in quantum mechanics the A will be to
the left, B will be to the right okay and
what else can we do, we can further do the
C, D also the same way. So C, D also if you
do it what happens? C comes to the left and
D goes to the right okay, so it will become
AC with B and D.
And you can also write A with B, C commutator
at D this side. These are the two terms which
you will get from this term and similarly
the other two terms you will get from. One
way in which you could do is you can expand
this as ABCD- CDAB right.
You could do that and you can also verify
that I can write this as A times B with C,
D+A times C, D sorry A times C, D and then
put the B here right this one. and you can
explicitly write this further as AC times
BD. This term will allow one more term, which
is A with BC times D. Similarly, this term
will be A with C D B+A with D B where will
the C go, in the left. Is This clear?, is
the pattern clear to you?
This is the way to play around with the commutator
brackets when you have products of operators
okay. Suppose I give you x squared commutator
with p squared, how will you do that? x square
commutator with p square.
There is one more thing you can use here x
with x where there is no order right. You
can use that fact here. The commutator of
x with x is 0. Say this is not AB; it is not
two different operators, the same operators.
How will you write this? We can write this
as x with x, p squared+x with p squared x
and then what do we do? So that is x hat with
x, p hat with p hat+x hat p hat x hat p hat
x hat with p hat okay.
That is the first term, the second term will
be x with p hat with p hat x hat+p hat x hat
p hat x hat. Now you use xp to be ih cross
right. I am doing it in one dimension. What
is the simplification finally? Twice x, p
and then twice x, p with this one will be
p x hat twice p x ih cross that is it. So
something which you see here the beauty what
is happening? This x, p with p, x should have
the property that it is Hermitian right.
That day I was telling you that the order
matters and whenever I have to write a Hermitian
operator, so this if you have taken A with
B as C, the C should be same as C dagger help
and you can verify that if you take this operator,
two factor anyway will let us not worry. If
you take x, p with p, x and take the dagger
of it, it will be same as this will become
px+xp which is same.
And we did not do anything. We just used the
commutative property and we see that it always
comes as a symmetric combination which I was
trying to insist last time that in quantum
mechanics because the order matters and you
have a dot product of two operators, you first
write that as a AB+BA remember and I said
that is what is the Hermitian combination.
If you had a cross product of two operators,
so we find x squared+p squared to be 2ih cross
sorry x p+p x, this is Hermitian. So 
I am just trying to say that if you have A
dot B as operators even in three dimension,
you have to it is not same as B dot A which
you are familiar in your classical physics.
This when you go to quantum physics, this
is classical, quantum you have to make it
Hermitian which means A dot B+B dot A/2.
Is this Hermitian? If you just took A dot
B is it Hermitian? Not Hermitian right because
A dot B dagger is B dot A. A is Hermitian
and B is Hermitian but A dot B is not Hermitian.
It goes to a new operator whereas this one
A dot B goes to B dot A and B dot A goes to
A dot B that is a Hermitian. That comes naturally
in the commutative brackets, I did not do
anything. Using the properties of commutator,
I end up getting combination of xp+px naturally.
I did not get just xp okay. So the order really
matters and you have to be careful with the
ordering.
So just to flash that slide, so the last where
the ordering is done is the one which will
help us to find the commutators of any products
of operators. Let me stop here.
