Let 
us get started with few more problems on translation
operator for a finite dimension that is the
finite distance special finite distance.
And this problem 4 will involve a problem
on the commutator of the position operator
and the translation operator with a finite
displacement.
So in the question we have T a is exponential
-ip dot a/h cross where p is the momentum
operator in three dimension. So we can write
this as e raise to -ipx ax okay. I can write
this hat as the operator px, ax is a position
fixed position, py ay+pz az/h cross. So we
are given that p is a momentum operator in
three dimension and ri is a position operator
in some certain direction. So rx or r1 will
give me x direction. Then, I can have y direction
and z direction.
So let us start by writing a specific case
like you can you have to first evaluate ri
Ti a okay and then we generalize the result.
We will generalize the result later. So as
we know that the momentum operator px, py
and pz commute with each other. So will be
Tx, Ty and Tz. So this T cap a I can write
it as T cap x ax sorry there will not be a
cap Ty ay and Tz az. So these translation
operators will also commute with each other.
So when we are taking the commutation relation,
we have to use this hint. So these operators
x, y, z are in 3 directions, 3 translation
directions that is the 3 axes x, y and z and
these Tx Ty and Tz commute with each other
right. Now since we have this, let us evaluate
ri T ai okay. So ri Tai you remember we have
done such exercise in tutorial 5. In that,
we have evaluated this quantity and this came
out to be just try to recollect ai times T
ai.
This is what we had obtained in tutorial 5
and remember we will use these relations while
evaluating the general result. So ri commutator
of ri and pi will be 0 okay. This we should
recollect and we know that ri pi is nothing
but ih cross. We have seen this commutator
relation. These are the general relations
which we have seen. In general, we have seen
the expression of ri and j, ri and pj okay
that is delta ij when i and j are equal the
will not commute.
And when i and j are unequal they would commute
in this particular case. So this is what we
have here. So will be ri, Tj aj, this would
imply this remember because Ti depends on
pi so if I have a pj then I have a Tj which
will be equal to 0 okay. So using 1 2 these
two things.
Using equation 1 and 2 what we obtaind is
let us right now ri this is position operator,
T a okay. So ri, T a, T a can be written as
Tx, Ty, Tz okay and ri will commute only with
Ti, so I will have ri, Ti ai times that is
Tx if I take and here if I take let me keep
it i only okay. So I have a Tj aj Tk ak. Remember
these are operators and we need to denote
it by cap. So this is a ri Ti aj which is
nothing but ai okay I will have ai and this
quantity that is Ti ai Tj aj and Tk ak which
will give me back this term okay.
In the question, we have this. In the question,
we have that we have to determine how the
expectation value of r will change under this
translation. So let us now consider a system
in state psi. So let psi be the state of the
system and now we have to remember this. It
is important to note here that translational
operator is unitary, okay. This is one thing
you must remember.
And now when we suppose we apply a translation
operator that is T a what do we obtain? So
consider this T dagger a this is the operator
T a ri. So we are finding out the expectation
value of ri. This is what we are asked to
find but on this state we are applying the
translation operator. So T on psi on the right
and T dagger on psi on the left because T
will become T dagger.
Psi T dagger ai ri T ai, so this we will evaluate.
Here what I said is I will use ri T ai commutator+T
ai ri. I will substitute for this here. We
have seen what is the commutator of this;
commutator of this was ai T ai correct. So
commutator will give me this and the second
term is this term okay. So what do I get next
from here? This will be nothing but T cap
ai I have got ri which I can actually take
outside.
In the next step, I will do that. T cap ai
this is the first term and the second term
is T dagger ai and I have a T ai ri times
psi okay. Now these two terms I will simplify,
ai is a number, ai can be taken out. It is
a number not an operator sorry. So ai I can
take outside very well and I have a T dagger
T which will give me just 1 okay. This is
what I have correct and when ri is operated
on a and this term will give me 1.
So let me write here okay, so this is the
expectation value of ri on the state psi and
when you operate the translation operator,
it is getting shifted by amount ai. So in
the end what we see is that there is a shift
in the position by ai. So the position is
shifted by ai, thus the expectation value
of ri is shifted by ai when one operates a
translation operator okay. So there is a shift
by amount r ai okay. So this is the fourth
problem we did. Let us go to the fifth problem.
Fifth problem is dividing 3 parts and it is
slightly lengthy but it is interesting to
check these relations commutation relations
and what will be the displacement operator
and all things.
And this was discussed in the last part of
the lecture and you are asked to prove some
of the relations. So we have put some problems
as the tutorial problems so that you actually
sit and solve them. So first part, in the
first part you are given a coherent state
which is defined as the eigenstate of the
ladder operator and you are given a relation.
So for any operator or any complex number
alpha you are given alpha is e raise to -alpha
square/2 e raise to alpha a dagger, so small
a dagger okay and 0.
Just recollect that and this operator or the
normalized eigen ket of a as eigenvalue a
alpha. So when A operated A dagger or A, operator
A is operated on alpha you get eigenvalue
A alpha okay. This you have to remember. Now
the next step would be that we have to define,
we have to calculate x x square P P square
in this coherent state basis. So for that
you need to do some algebra.
Few hints I will give and you can maybe use
those hints to work further. So let us define
dimensionless position and 
momentum operators as A+A dagger/2 and p as
A-A dagger/2 times i. So this is a position
operator x and the momentum operator P which
are defined in terms of the operator A which
has eigenvalue alpha when you operate it on
the eigen ket alpha okay. So now let us rewrite
x and P in terms of A and A dagger, okay.
A and A dagger. So what will be in terms of
A and A dagger?
A will be=x+i P cap/2 and A dagger is simply
the complex conjugate, you can see here correct.
So we know that for simple harmonic oscillator,
we have this relation capital x is nothing
but m omega/2 h cross inside the square root
x cap and p is nothing but 1/2 h cross omega
m inside the square root p cap. So capital
X and small x are related by this relation
okay. We have defined capital X in terms of
dimensionless variable.
So we have to now and remember x and p are
our regular or usual position and momentum
operators right. So it is now easy to evaluate
what is X. This is X is what we are going
to evaluate. X we had defined as A+A dagger/2.
So first we will evaluate X, then we will
evaluate X square, then P and then P square.
So expectation value of X on ket alpha will
be 1/2 I have A dagger A.
So I will write A alpha+alpha A dagger alpha
correct and when you operate this, this will
give me an eigenvalue alpha, this will give
me an eigenvalue alpha square alpha star.
So I will have alpha+alpha star and these
are normalized eigen ket. So these are of
the form R, I can write it as real part of
alpha okay.
Now in the same manner you evaluate X square
will be you will find when you solve this
you will have the real part of alpha which
we evaluated will be square+1/4. So please
check this. Check yourself okay. Then, I will
write what is my P. We have seen that operator
P was i A-A dagger okay and I have skipped
this but you must remember that when there
is a hat it is an operator notation.
So I have i/2 and then I will have alpha-alpha
star kind of a term which is equivalent to
or which can be written as imaginary part
of alpha okay. Similarly, you evaluate this
and you will find that imaginary part of alpha
square+1/4 is what you obtain. Now the next
step would be to calculate delta X dot delta
capital P okay. Delta X dot delta P will be
square root of you know the definition, X
square-expectation value of X square*similarly
P-P square.
We have evaluated all these and when you substitute
this, you will have something like this. Now
you will substitute for capital X and capital
P in terms of small x and small p. This is
what we have right. So when I substitute for
these, I get delta x dot delta p as in the
operator notation let me be very explicit.
So this relation is nothing but Heisenberg
uncertainty principle, delta x dot delta p
should be<or=h cross/2 okay right.
So this coherent state, note coherent state
alpha actually achieves the minima in Heisenberg
uncertainty principle relation which is equal
to the value of the uncertainty principle
because of capital X and capital P, we have
seen that.
The second part or the b part, b part you
are asked to calculate the exponential eB
if commutator of A and B is equal to a constant
then you have to show that e raise to alpha
A dagger e raise to –alpha star A dagger
is having some relation like e raise to alpha
square/2 e raise to alpha A dagger-alpha star
A. So we have this relation and we are going
to use Baker–Hausdorff formula.
So what we have let me write here as B A e
raise to A, B okay and you have this equal
to e raise to A+B times e raise to A, B commutator
of A, B/2. So using this relation and applying
Baker–Hausdorff formula, we have e raise
to alpha a dagger and another term is e raise
to alpha star a okay. This expression I can
rewrite using this result. I can rewrite this
as e raise to I can write this as alpha a
dagger, okay.
This term -alpha star a, so I have used this
times e raise to alpha a dagger-alpha star
a/2 and this will 
be a dagger alpha a dagger-alpha star a and
I can have alpha square outside and commutator
of a dagger and a and a dagger is 1, so I
have alpha square/2. So this is what we have
proved. It is just one step, okay. These relations
will be useful in proving other results also.
So now the c part is that we can rewrite the
coherent state in terms of the displacement
operator. Can we do that?
Now let us write the displacement operator
we have to write in terms of the coherent
state. So the coherent state given to us is
e raise to -alpha square/2 e raise to alpha
a dagger. This is what is given to us and
remember we know that when you operate annihilation
operator on any inner state 0, you will obtain
0, okay. It will not go further. This is the
end of the ladder. So I can write this again
as alpha square/2 times e raise to alpha a
dagger.
I can write this as e star a, I can just multiply
by some coefficient. It will not matter; we
have used this okay. Now remember we had this
relation over here, okay. I will use this
relation in this expression, so when I use
this relation I will have this term times
this and I have 1 exponential-alpha square/2
in the expression already. So I will just
end up having e raise to a dagger-a star a
on alpha.
And this is the definition of displacement
operator and hence we can prove that this
is the coherent state can be written in terms
of displacement operator and you have to show
in this part that the displacement operator
is unitary. This is explicitly seen, it is
very simple, e raise to alpha a dagger-alpha
star a and D dagger alpha will be e raise
to alpha star a-alpha a dagger, okay.
So when I take D dagger alpha D alpha, so
U dagger U is 1 that is it is unitary operator.
It is very simple. You can see it without
even solving ,okay. So these exercises and
you had in your lecture few more exercises,
try and solve it and see if you can do solve
those exercises which were discussed.
