In the last lecture, we looked at the Heisenberg
equation of motion.
This was an equation for the operators and
we had d A of t by d t, where a is an operator
plus commutator of A of t with H by i H cross
plus delta a by delta t. Where, this was a
term which survived only if there was an explicit
time dependence. So, this derivative means
differentiation of the part of the operator
which had explicit time dependence. So we
also realized that this was very close to
the classical physics equations, where you
had d A classical by d t. Where, this is some
dynamical variable it is the Poisson bracket
of A classical with the Hamiltonian plus delta
A classical by delta t.
And therefore, we realized that the Poisson
bracket here is simply the commutator bracket
by i h cross and you went from quantum physics
to classical physics in the limit h cross
going to 0. We explicitly saw using the commutator
of x with p x what exactly happened and how
you approached the Poisson bracket.
So, let us take a simple example; to look
at the Heisenberg equation of motion. Let
us take the Hamiltonian to be h cross omega
a dagger a of course, plus half this is the
harmonic oscillator Hamiltonian. And, so what
is d a of t by d t? This would be the commutator
of a with a dagger a plus delta a by delta
t and that is 0 because, a does not have an
explicit time dependence. And therefore, that
term drops out so this is simply there is
an h cross omega here; so it is simply omega
by i a. Similarly, d a dagger by d t can be
got. Let me take, a more non trivial example.
Let me consider, the operator A of t to be
p of t this, is a linear momentum operator
sin omega t minus m omega x of t cos omega
t. So in contrast to this case we have put
in an explicit time dependence in the operator
A.
Now, basically, this is simply a recapitulation
of what we did towards the end of the last
lecture. The A simply differs in sign from
what we considered in the last class. But,
I am repeating it so that I may reiterate
an important fact. So, what is d a of t by
d t? So first of all this Hamiltonian can
well be written as p squared by 2 m plus half
m omega squared x squared. So, the first part
involves the commutator of p sin omega t,
I will remove the argument t in the momentum
with the Hamiltonian. And, that would just
be half m omega squared x squared 
minus m omega cos omega t the commutator of
x with p squared by 2 m. So, this is the explicit
commutator term plus delta a by delta t and
of course, there is an i h cross. So, let
me start with i h cross this is this quantity
plus i h cross delta a by delta t.
So, this is simply given by half m omega squared
sin omega t commutator of p with x squared
minus m omega cos omega t by 2 m commutator
of x with p squared plus i h cross delta a
by delta t.
So, this quantity can easily be obtained and
I have 
d A by d t, is half m omega squared sin omega
t times 2 i h cross. That comes from this
term with the minus sign because, it is a
commutator of p with x that is involved and
I use the fact that commutator x p is i h
cross. The next term is minus m omega cos
omega t by 2 m times 2 i h cross p. I have
used the a b c rule for commutators, so that
x p is i h cross plus i h cross delta a by
delta t. And, since this is A the 1st term
gives me omega p cos omega t 
and the 2nd term gives me plus m omega squared
x sin omega t.
So therefore, d A by d t 
you can see that the 2 cancels here and I
just have a minus i h cross m omega squared
x sin omega t from the 1st term. The 2nd term
gives me a minus i h cross omega cos omega
t p and from the explicit time dependence
I have a plus i h cross omega cos omega t
p. And, then I have a plus i h cross m omega
squared x sin omega t. So, plus i h cross
m omega squared x sin omega t plus i h cross
omega cos omega t p and therefore this is
0.
So, I have a situation where the explicit
time dependence of a is 0 which means, that
this is a constant of the motion having said
that the important point to note is the following.
It is not as if a commutes with the Hamiltonian
this is a very say tailored operator. I have
chosen this to demonstrate the fact that,
if A is a constant of the motion because d
by d t of a is 0, the total time derivative
is 0. It does not mean that a commutes with
the Hamiltonian in general it means that the
commutator of a with h apart from the i h
cross suitably cancels the explicit time evolution
of A.
So, here is an example where d A by d t is
0 because commutator of A H by i h cross is
equal to minus delta A by delta t. So that
is a thing worth remembering and you cannot
naively imagine that always the explicit time
dependence would not be there or that the
commutator of a with h is 0 because, a is
a constant of the motion. That need not necessarily
be true.
Now, proceeding on these lines now that we
know the Schrodinger picture and we also know
the Heisenberg picture. We can choose our
favourite picture depending on the context
to work out the time evolution of quantum
expectation values. As I have emphasized again
and again expectation values are the experimentally
measured quantities and quite independent
of the picture you use answer should match
for the expectation values and their dynamics.
Whether, you work in the Heisenberg or in
the Schrodinger picture.
So, let me work with the Schrodinger equation,
I have i h cross d by d t of psi of t is H
psi of t. In general, the Hamiltonian could
also have a time dependence and therefore,
if I take the Hermitian conjugate equation
minus i h cross d by d t bra psi of t is psi
of t H. My aim is to see how expectation values
evolve in time under appropriate Hamiltonians
that I will select later on. So, I wish to
find d by d t of expectation value of say
an operator A which, could have an explicit
time dependence that is allowed, this is what
I want to find out. That means this is identical
to d by d t of A of t, this is my notation.
So, I can use this equation clearly this involves
d by d t psi of t and d by d t of ket psi
and d by d t of bra psi. So this will give
me i h cross let me bring the i h cross to
the other side.
So, the 1st term is d by d t Bra psi and that
is out here, so I have psi of t H minus 1
by i h cross, A of t psi of t that is my 1st
term. Then of course, plus psi of t, any explicit
time dependence that differentiation has to
be there and the 3rd part is plus psi of t
a d by d t of ket psi of t is h by i h cross.
So I can put the 1 by i h cross here, psi
of t. So this is what this expectation value
is about and you can see already that there
is a commutator it is 1 by i h cross, commutator
of A with H here, between these 2 terms.
So, I can write this down as 1 by i h cross
expectation value of the commutator of A with
H so this is what I have. Of course, there
is a psi of t here and this is my equation,
so d by d t of expectation value of A of t,
is expectation value of the commutator of
A with H by h cross i h cross plus expectation
value of delta A by delta t. So this is what
I have, I have done this in the Schrodinger
picture. Now, I could have done this in the
Heisenberg picture as well.
Now, in the Heisenberg picture I would have
an equation that says, that d A by d t is
commutator of A with H by i h cross plus delta
A by delta t. So, if I take expectation values
and I realize that in the Heisenberg picture
expectation value of d by d t of A is expectation
value of d by d t; could well be written as
d by d t of expectation value of A. Because,
this expectation value amounts to writing
psi A H of t psi and since psi does not evolve
in time in the Heisenberg picture, it amounts
to simply differentiating A with respect to
time. So, as you can see it gives me the same
equation. it gives me d by d t of expectation
value of A is expectation value of the commutator
of A with H by i h cross plus expectation
value of the explicit time dependent term
delta A by delta t.
This is the Ehrenfest relation and this gains
a lot of importance in the understanding of
how to go from classical physics to quantum
physics. The contribution of Ehrenfest becomes
very significant, because in the early days
of quantum mechanics there was this very grey
area. Even now, there are some very formidable
problems to be tackled in the area, when you
pass from classical to quantum physics called
semi classical physics. How do you go smoothly
from the quantum to the classical world? This
relation given here by Ehrenfest helps us
significantly in this process, because of
the following reason: Expectation values are
in numbers and the idea is this if you want
to go from quantum physics to classical physics.
You could perhaps, replace all operators by
their expectation values because, this is
precisely the kind of equation that you will
have in classical physics, for a dynamical
observable; for a dynamical variable. So,
in classical physics you will have d by d
t of any dynamical variable. It is the Poisson
bracket of that variable with the Hamiltonian;
the classical Hamiltonian plus delta by delta
t of that dynamical variable. Where, this
is the explicit time dependent part.
But, we need to understand this Ehrenfest
relation in better detail its ramifications
and its precise interpretation. Look at this
example; a simple example let us again consider
the harmonic oscillator of Hamiltonian that
should do for our purpose. So you consider
H is let us say the particle has unit mass
m is 1 and so you have a Hamiltonian which
is p squared by 2 plus half omega squared
x squared. This is the Hamiltonian and I want
to find d by d t of expectation value of X.
The Ehrenfest relation there is of course,
no explicit time dependence so we can forget
that term.
So, this is simply the statement that d by
d t of expectation value of x is 1 by i h
cross commutator of x with H which is 1 by
i h cross. So the commutator essentially is
the commutator of x with p there is a 2 out
here and I have 2 i h cross p. So, this is
the same as expectation value of p, so I have
d by d t of expectation X is expectation p.
Let us look at d by d t expectation p once
more, this is 1 by i h cross expectation value
of the commutator of P with H and that is
1 by i h cross. The commutator of P with H
just gives me a half omega squared minus 2
i h cross x expectation value and then that
is just minus omega squared expectation x.
But, this is the same as the derivative of
the potential minus V prime of x its expectation
value.
So, look at what I have I have d by d t expectation
x is expectation p d by d t expectation p
is minus omega squared expectation x. But,
these equations are familiar these are precisely
the classical equations, if you replace the
operator by expectation values and if you
identify these to be the analogues of the
classical dynamical variables. These are precisely
the equations that I have for a particle,
which is executing simple harmonic oscillation.
And therefore, it is in this sense that the
Ehrenfest theorem is naively stated as quantum
mechanical expectation values follow classical
equations of motion.
But, you should understand that this is the
general statement of the ehrenfest relation
this is merely an example. Because, if I did
not have half omega squared x squared as my
potential but, I had a Hamiltonian p squared
by 2 m plus V of x an arbitrary potential.
Then, d by d t expectation p is minus expectation
V prime of x where V prime of x is simply
the derivative of v with respect to x.
I would like to call this object therefore,
as d by d t expectation p is expectation F
of x. I use the notation F because at least,
in classical physics for a conservative system
I know that the force is the negative gradient
of the potential and just by way of notation
I use F here. It is just a notation as far
as this is concerned. But, you see the harmonic
oscillator example is a very interesting example
because, it is quadratic in x. And therefore,
I could have well replaced the operator by
its expectation value.
In other words, expectation V prime of x in
this case can be just written as V prime of
expectation x. Because, V had an x square
in it and therefore, V prime was linear in
x and therefore, expectation V prime is the
same, as V prime expectation x. You cannot
in general, replace all x to the power of
n expectation by expectation x to the n. It
worked very nicely in the harmonic oscillator
case simply because, it was a quadratic potential.
But, in general as you know this replacement
is not possible because there are higher moments
that contribute. It is not as if you can write
the expectation value of x to any power, in
terms of just the power of the mean there
are higher moments that make a contribution.
All of quantum physics is about understanding
the wave function reconstructing the wave
function from expectation values. And, that
means that you need to know not just the mean
but, all the higher moments; the infinite
set of higher moments before you can reconstruct
the full wave function.
Now, in practice in the laboratory, you would
not be able to get an infinite set of moments,
expectation values. So, you have a finite
set of them but, surely it will involve the
variance perhaps the higher moments, the skewness
the kurtosis. And you try to reconstruct the
wave function to some level of approximation.
So, if V of x for instance is some a x to
the 4 V prime of x is 4 a x cubed and expectation
V prime of x is 4 a expectation x cubed, which
is not the same as 4 a expectation x the whole
cubed.
So, when can you simply replace? When can
you do this replacement? Well, you can do
this certainly it is an approximation in general.
It can be done if the position probability
density is very close to expectation x of
t for all times. Then this can be done, but,
if that is not true obviously the higher moments
will contribute and you will not be able to
make such a replacement. And that is why,
if you want to find for instance d by d t
of expectation x squared. It will not simply
involve expectation p squared.
Let’s calculate d by d t expectation x squared.
Now, this object is 1 by i h cross commutator
of x squared with h, I am using the harmonic
oscillator Hamiltonian. So this involves commutator
of x squared with p squared by 2 and therefore,
that is 1 by 2 i h cross x squared with p
p plus p commutator of x squared with p. And
this is 1 by 2 i h cross i h cross x p twice
and this is plus 2 i h cross p x therefore,
this is essentially x p plus p x.
So, d by d t of expectation x squared involves
x p plus p x and it is not as if the algebra
simply closes between x squared and p squared,
other moments are also involved. Similarly,
if I find d by d t of p squared that too will
involve the expectation value of x p plus
p x. So this 2 is proportional to the expectation
value of x p plus p x, can be checked out.
So, inevitably, the algebra does not simply
close between x squared and p squared or x
to the m and p to the m it involves many other
moments. So this is the other important point
about the Ehrenfest relation.
Now, 1 thing is clear clearly, this relation
and what it predicts depends crucially upon
the state of the system that we consider.
Of course, it depends crucially upon the Hamiltonian
of the system I have used just a Hamiltonian
of the oscillator form. It is evident that,
if I use something like a dagger squared a
squared, it is going to create more complications.
But, for the moment I restrict myself to the
oscillator Hamiltonian and then I find that,
depending upon the state; depending upon the
Hamiltonian the dynamics of the expectation
value is going to change tremendously.
So, the way in which I would like to demonstrate
the Ehrenfest relation in the context of expectation
values of course, is to look at states of
the radiation field. Since there are very
many interesting states of the radiation field,
some of which you already know like: coherent
state that is laser light, the photon added
coherent state and the squeezed state of light.
It would be interesting to use these 3 states
of the radiation field and see what happens.
Let us recall, that the Hamiltonian was essentially,
this is the Hamiltonian in quantum optics
was essentially E squared plus B squared.
It is clear that I have set I have ignored
mu naught and epsilon naught for the moment
and we did appropriately right E and B in
terms of a and a dagger. The photon destruction
and creation operators and we could get the
Hamiltonian in the form of the oscillator
Hamiltonian. This is just a formal mapping
this is the Hamiltonian for the Hamiltonian
density for a free electromagnetic field.
So, if there is a radiation field propagating
in free space, then, instead of looking at
the behaviour of the radiation field propagating
in free space. I could well, look at a particle
of unit mass subject to an oscillator potential,
so that is all that this needs.
So, if I look at various states of the radiation
field propagating. I could mimic that or I
could have an equivalent formalism where,
I take a particle of unit mass subjected to
a parabolic potential omega squared x squared
by 2. And, look at its expectation values
expectation x expectation p and so on and
that should tell me the behaviour of expectation
x and expectation P in the case of the radiation
field. I call these the x and p are the quadrature
variables. Clearly, in the case of optics
these expectation values would be related
to the intensity of the electric field. And,
I would just formally define x as root of
h cross by omega a plus a dagger by root 2
and p as root of omega h cross a minus a dagger
by root 2 i.
So while a and dagger have physical meaning
that they are the photon destruction and creation
operators. x and p are the hermitian counterparts
and would have in principle relations with
would be related in principle to the intensity
of the electric and the magnetic fields.
So given that, I would now look at 3 types
of states of the radiation field the 1st step
that I want to look at it is the coherent
state; the standard coherent state ket alpha.
And as you know in terms of the photon number
states, this is e to the minus mod alpha squared
by 2 summation n equals 0 to infinity alpha
to the n by root n factorial ket n. Yesterday,
we looked at this state in the position representation
it is a gaussian. So, now I am interested
in finding out what happens at time t. So
what is psi of t if the initial state is alpha?
Yesterday, we showed that a gaussian continues
in the harmonic oscillator a Hamiltonian when
subject to the harmonic oscillator Hamiltonian
the gaussian continues to be a gaussian merely,
oscillating. The cheap way of seeing it is
by saying that in this context psi of t is
e to the minus mod alpha squared by 2 summation
n alpha to the n by root n factorial e to
the minus i H t by h cross ket n. And, this
is simply e to the minus i omega n t ket n,
I have used the Hamiltonian h cross omega
a dagger a. Otherwise, you would have to put
an n plus half there.
So, this can be written as e to the minus
mod alpha squared by 2 summation over n alpha
e to the minus i omega t to the power of n
by root n factorial ket n. And therefore,
psi of t is simply alpha e to the minus i
omega t, executes harmonic motion simple harmonic
motion and continues to be a coherent state.
Therefore, in the position representation
it is a gaussian, so this is the way in which
the state changes in time.
Now, let us look at expectation value of X.
Let us now calculate expectation value of
x of t in the initial state alpha. So alpha
evolves and at any instant of time alpha is
just alpha e to the minus i omega t so that
is the state of the system at time t. And
therefore, this expectation we’ve set n
equals 1. So, there is a root of h cross by
omega and then there is an a plus a dagger
sandwiched between the ket and the Bra. So
there’s an a plus a dagger here and of course,
there is a root 2. So this is the object that
we need to calculate this is certainly true,
this is alpha e to the minus i omega t times
the ket itself; it is an Eigen value equation.
Similarly, I can act a dagger on the Bra and
pull out alpha star e to the minus e to the
plus i omega t as an eigen value when a dagger
acts on this bra the eigen value is alpha
start e to the plus i omega t. Therefore,
x of t is root of h cross by 2 omega alpha
e to the minus i omega t plus alpha star e
to the i omega t. I need to take the complex
conjugate of this number which is what I have
put down there and that is all, that is the
expectation value of X. Of course, I can write
alpha as alpha 1 plus i alpha 2, write e to
the minus i omega t in terms of sins and cos
and then expectation x of t, is apart from
an overall constant which you put down here.
That involves h cross omega the number and
so on it is just alpha 1 cos omega t plus
alpha 2 sin omega t. So, this is what we have
for a expectation value of X.
So, here I have it in animation, this is the
oscillator potential the black parabola and
what we have is an initial state which is
the coherent state. And, what we have here
in pink is the gaussian form as I have just
now told you. We demonstrated it explicitly
the gaussian continues to remain a gaussian
as time proceeds.
So you can see that once it comes to the extreme
right or left it is simply a gaussian, so
is it at the centre no change in amplitude,
no change in height, no change in width, no
change in height. Execute simple harmonic
motion in the following sense that as we have
seen expectation X apart from constants is
alpha 1 cos omega t plus alpha 2 sin omega
t and expectation P you can easily verify
is alpha 2 cos omega t minus alpha 1 sin omega
t. The blue dot here which keeps moving is
the expectation value; the red horizontal
line is merely the energy value that I have
selected. So, there is a constant energy which
is given and that is a constant of the motion.
And you can see that expectation x of t the
blue dot for an initial coherent state, simply
moves like a particle of unit mass subject
to the parabolic potential, it simply executes
simple harmonic motion.
Now, look at the inset and that will tell
you how far, the blue ball moves. Moves all
the way till it touches the potential the
parabolic potential that is what you see from
the inset. So those would be the classical
turning points, so you see what we have shown
you is this the expectation value expectation
X and P behave exactly like the position and
momentum of a classical linear harmonic oscillator.
The correspondence is actually even more exact
because, the amplitude of oscillation you
can easily check in position space is root
of 2 h cross by omega times mod alpha. But,
the mean energy in this state is h cross omega
mod alpha squared because it is simply the
expectation value of a dagger a in the state
ket alpha. Notice that this is a conservative
system so energy does not change in time.
So the position of the classical turning point
is root of 2 e divided by omega.
So, the quantum mechanical expectation values
in the coherent state behave exactly, like
the classical position and momentum of a linear
harmonic oscillator. And the state remains
a minimum uncertainly state at all times.
So you can see as it goes from 1 end to the
other that this really behaves like a classical
harmonic oscillator. The dynamics of the higher
powers of expectation X and expectation P
and their combinations can be established
similarly, with the help of the Ehrenfest
theorem.
So, now, let us consider the case when the
initial state is a photon added coherent state.
You will recall, that a photon added coherent
state; the n photon added coherent state for
instance adds m photons to the standard coherent
state. That means, you repeatedly apply a
dagger m times 2 ket alpha and then you normalize
it to get the normalized photon added coherent
state, alpha comma m. So, this is the initial
state that we will now consider because, we
would now, look like to look at a situation
where the state does not have coherence.
So, here in this figure, I show the periodic
motion of the position probability density
that, is the pink curve as you can see that
is the pink curve. And of course, the blue
dot is expectation x the initial state is
a 1 photon added coherent state. You will
recall that the I have already mentioned that
this was produced in the laboratory a few
years ago. And therefore, understanding the
nature and behaviour the dynamics of the single
photon added coherent state becomes that much
more interesting and important. As you can
see this is in sharp contrast to the coherent
state; the state is not a gaussian function
of x. It is not a minimum uncertainty state
at any time and moreover the violet curve;
the pink curve which is the position probability
density, it does not retain its shape as it
evolves under the influence of the oscillator
potential.
So, the dotted red line again corresponds
to the mean energy in the photon added coherent
state. That means the expectation value of
h cross omega a dagger a. But, remember a
dagger a is no longer equal to mod alpha squared,
which was its value in the case of a coherent
state. Now, if you look at the probability
density in a position space it has 2 maxima
rather than just 1. Moreover, the curve changes
shape in a rather drastic but, regular fashion
as it oscillates back and forth. The expectation
values x of t and p of t continue to vary
sinusoidally, as in simple harmonic motion.
Now, this feature is guaranteed by the Ehrenfest
theorem and it is really implicit in the equations
that we wrote down that is: d by d t of expectation
x is expectation p and d by d t expectation
p is minus omega squared expectation x. These
equations remain valid for all states and
all times since, the Hamiltonian is quadratic
in x and p. However, there is an important
difference between the present case and that
of the coherent state. Notice that the amplitude
of oscillation falls short of the classical
turning point the classical turning point
was, root of 2 e by omega. But, then if you
look at the inset you will see that the blue
dot does not touch the edge of the potential
at all; in fact does not come up to the classical
turning point.
So, this is a new feature that we have seen
and in a sense there is a simple way of understanding
why the amplitude of oscillation of expectation
value of x falls below the classical turning
point. In the case of the fock state you will
recall that expectation x is 0 and expectation
x remains equal to 0 for all time. In the
case of the coherent state it periodically
reaches the classical turning point that is
what we saw. So, the photon added coherent
state interpolates between the fock state
and the coherent state and therefore, it is
possible when the amplitude of oscillation
of expectation x for the photon added coherent
state lies somewhere between 0 and the classical
turning point given by root of 2 e by omega.
Finally, I want to consider the squeezed state
we have examined this state in 1 of the earlier
lectures. So the state is given as a super
position of the 0 photon state and the single
photon state and this is an example of a squeezed
state; it is not the squeezed vacuum it is
a super position, which certainly shows squeezing
it is normalized as you can see, the mean
value of energy in the state is quarter h
cross omega. Because, E is h cross omega;
the Hamiltonian is h cross omega a dagger
a and then you can easily check that this
term contributes whereas, that gives me a
0. And this is the expectation value of E.
Now, if you calculate for this state the variance
if you calculate delta x which is given as
root of 3 h cross by 8 omega and delta p is
root of 3 h cross omega by 2. It is clear,
that there is squeezing in the x quadrature
because, if you set h cross and if you set
omega equals 1. Then this object is certainly
less than what you would have for the ground
state of the oscillator or the standard coherent
state. So, there is squeezing in the x quadrature.
Now, if you let this state evolve subject
to this Hamiltonian you can trivially check
that this, is the state at a later time. Apart
from a factor half it is root 3 ket 0 plus
e to the minus i omega t ket 1.
I will leave it to you as an exercise to calculate
delta x at time t as a function of time and
delta p as a function of time. These are the
expressions, so these would be exercises for
you to do. Apart from a coefficient 3 h cross
by eight omega delta x is essentially, 2 minus
cos squared omega t. So, the variance would
be remove the half and get the rest of it.
Similarly, delta p whole squared which is
the variance is 3 h cross omega by eight times
2 minus sin squared omega t. Just by way of
information and to complete the picture, if
you look at the manner in which this initial
state evolves in time when subject to the
Hamiltonian h cross omega a dagger a. If you
find the probability density this is interesting
because, apart from constants it is e to the
minus omega x squared by h cross. Times there
is a function of x squared here, there is
x squared there; a constant and then there
is a cos omega t as well x cos omega t.
If you look at expectation value x of t as
you know, ket zero is expectation value at
all times is 0. Ket 1 has an expectation value
the power of quantum super position is seemed
here because, despite the fact that the expectation
value is 0 in this state; fock state. You
find that expectation x of t turns out to
be non-zero and in fact varies as cos omega
t.
So, I have an animation here for you to show
the behaviour of this initial squeezed state.
So I have for you the periodic motion of the
position probability density as before, this
is the pink curve and then of course, expectation
x of t is the blue dot. And, this is for the
initial squeezed state that I was talking
about, the horizontal red line here is simply
the mean value of the energy; that is a constant.
The black is the parabolic potential the oscillator
potential. And then, you can see various features
you can see for instance that in the inset
expectation x of t very clearly, does not
get anywhere near the classical turning point;
it does not touch the classical turning point.
It stops there and it stops a little short
of the classical turning point you can see
that.
The squeezed state changes its shape and this
is the position probability density and that
keeps changing its shape as time elapses.
And of course, in all the 3 cases that I have
shown you it is pretty clear that there is
a tunnelling, because, as you can see, the
position probability density is non-zero even
outside the parabolic wall. So there is a
nonzero thing out here and here, so there
is certainly this quantum mechanical the leakage,
leakage beyond the containing potential.
So, there is a leakage into the classically
forbidden regions, which we can see very clearly
here. So, these are the 3 cases; the 3 cases
of non classical states of 2 of them are certainly
non classical states. The coherent state is
considered to be a classical state for reasons
that I have explained to you. And once again
in the case of the squeezed state you can
find out the time evolution of expectation
x squared and expectation p squared by just
using the ehrenfest’s relation in its general
content.
So, in any faux state n the expectation value
of x is 0 but, looking at this it is clear
that quantum super position produces various
changes. There are more stringent quantitative
measures of non classicality of quantum mechanical
states in general. You see even among states
of radiation there are many other interesting
non classical states for instance, there is
a so called cat state ket alpha plus ket minus
alpha or the Yurke Stoler state which is ket
alpha plus i ket minus alpha; the squeezed
vacuum state and so on. In all these cases,
the Ehrenfest theorem provides a convenient
way to analyze the dynamics and to understand
why are the behaviour of the higher moments
of the quadrature concerned, the role played
by quantum fluctuations.
So, you see we have looked at 3 states of
the radiation field ket alpha and then, the
so called non classical states the P A C S
which was the photon added coherent state
and the squeezed state. And, we have demonstrated
how expectation values behave in time, there
are other interesting non classical states
for instance, there is the Cat state: this
is ket alpha plus ket minus alpha or ket alpha
minus ket minus alpha. Then the Yurke stoler
state, this would be: ket alpha plus i times
ket minus alpha and so on In all these cases,
you can look at the Ehrenfest’s relations
and see how exactly expectation values evolve
in time. Now, what we have used is the Hamiltonian
h cross omega a dagger a and then you find
that for all these states considered, there
is just periodic motion and nothing more.
Now, you look at a Hamiltonian like the Kerr
Hamiltonian a dagger squared a squared, very
many interesting things happen for instance
even a coherent state; an initial coherent
state does not continue to be a coherent state
for all times. Almost immediately, since this
is a wave packet; that means it is a super
position of plane wave states for instance.
And therefore, it does not have a sharp momentum
or a sharp position but, it is more or less
there it is a gaussian. Approximate momenta
and approximate positions can be given, it
is somewhat localized that is what I mean.
You will find that almost immediately after
being subject to this Hamiltonian, the coherent
state puts out it is no longer a gaussian.
Changes in all sorts of fashions very very
interesting fashion.
But, then because of some very interesting
peculiar properties, it could revive that
means at periodical instance of time it could
exhibit this feature that, it comes back to
its initial state apart from a phase factor.
And then, expectation values return to themselves
to their initial values at those instance
of time and so on. These are called wave packet
revivals and we will not be considering that
here, it is a more advanced topic. But, I
hope I have given you a flavour for what kind
of complicated dynamics can be there, what
sort of interesting dynamics can be there
in the behaviour of expectation values depending
upon the states initial state and depending
upon the Hamiltonian.
