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values of the total angular momentum quantum number of the electron are 3/2ís and 5/2ís.
I havenít proved this but I made the statement that if you took 2 angular momentum; J 1 and
J 2 and defined that to be the total angular momentum J, then this J runs from modulus
j 1 - j 2 up to j 1 + j 2. i didnít prove the statement but its easy to see that its
plausible from the rules of what happens when you take the raising operator and act on states
and so on.
But in the present instance, for an electron L + S is the total angular momentum, J the
quantum number of L is l and that of S is always a Ω. The spin of the electron is a
1/2 and therefore if this l = 2, then the only allowed values are 3/2ís and 5/2ís.
So this statement is true. What happens is l was 0? If you were in the S state of the
electron in an atom, what happens then? The only allowed values are 1/2 because you have
to take modulus j1 - j2. So the only allowed value is Ω. Then the total angular momentum
is in fact just the spin angular momentum of the electron.
As you know in the hydrogen atom which I have assumed you studied in the chemistry class,
the states of the electron for a hydrogen atom are specified by the principle quantum
number n which takes on the values 1, 2, 3, etc and then an orbital angular momentum quantum,
l and this runs over the values 0, 1 up to n ñ 1. So when n = 1, you have only the possibility
of l = 0. So you have the 1s state and then the 2s 2 p states and so on.
The magnetic quantum number m which is the quantum number corresponding to any component
of the orbital angular momentum takes on values from - l up to + l in steps. And then there
is a spin quantum number and of course we know that the spin quantum number S is always
a 1/2 and the projection, denoted as mS is = - 1/2 and a Ω. These are the only 2 possibilities.
and now you could ask what the degeneracy of the electron is when a given state with
n, l, m etc. the general state with principle quantum number n and that degeneracy factor
gn so happens that for the 1 over r potential, the Coulomb potential energy levels depend
only on n and not on any of these quantum numbers unless you break degeneracy by including
some other interaction.
For instance you put a magnetic field, that will couple to the spin of the electron and
there is magnetic potential energy - mu dot v and that will distinguish between + 1/2
and - 1/2 for ms and so on. There are other factors which I will come to which will also
tell you that the quantum numbers could depend on l, m etc. But if you donít have any of
that then this gn, since En,l,m and the other quantum numbers is a function of n alone for
the hydrogen atom, its actually - 1 over n squared in Rydberg units, therefore this gn
is = the number of possibilities over ms = - 1/2 up to + 1/2 times 1 because thatís the number
of possibilities you are counting, the number of possibilities for m = 0 - l up to + l and
then you have to count the number of possible values of l from 0 to n - 1 and this is = gn.
This straight away gives you factor 2. So this is = twice this and this gives you 2
l + 1. So you have to sum l = 0 to n ñ 1, 2 l + 1 and this is = 2 n squared. So that
was asked as a separate question and that was in fact in one of the fill in the blanks.
It said the degeneracy of the electron in the hydrogen atom is gn taking into account
its spin is 2 n squared. If you didnít take the spin into account, if you ignore that
internal degree of freedom, then itís just n squared. Now if you switch on a magnetic
field, the following is going to happen. there will definitely be a coupling between the
intrinsic magnetic moment of the electron and the external field that will break this
degeneracy in mS. And then if you have any potential which is not central then there
is also a dependence on m because if itís a central potential, no axis in space is distinguished
from anything else and therefore there is degeneracy over m.
In general, if you had a central potential alone V (r), which was not a coulomb potential
or some very special potential like the harmonic oscillator potential but an arbitrary general
central potential with no special symmetries whatsoever, then what would the degeneracy
be for a particle of spin Ω? There is a factor 2 which comes from the spin. But over and
above that, there is a degeneracy in m. so in such a case, the energy levels would be
a function of n as well as l and what would be the degeneracy of each of these states
for a given n and given l? It would be 2 l + 1. all the m values would give you exactly
the same energy. If you included spin, its twice 2 l + 1 in that case. Now what happens
in practice even in the hydrogen atom is that there is an extra contribution called the
spin orbit coupling.
There is a coupling between the spin of the electron and its orbital angular momentum
in the following sense. Itís actually a more complicated effect. But itís as if, if you
pretend for a minute that you are on the rest frame of the electron, you still have an intrinsic
angular momentum S or a magnetic moment. you see the proton going around you which is a
current loop and therefore there is magnetic dipole moment associated with the orbital
motion and that will be proportional to the orbital angular momentum the magnetic moment
which was l and therefore there is going to be a term in the Hamiltonian proportional
to l dot S. thatís called lS coupling and the moment you have this or spin orbit coupling
and the moment you have an l dot S coupling, you have broken the symmetry which the coulomb
potential has. therefore the energy levels become dependent on l. in other words, this
accidental degeneracy that these energy levels donít depend on l is gone out and there is
in fact a difference in energy between the different l states for a given n.
And thatís the reason why the spectrum of the hydrogen atom looks like 1S. This corresponds
to n = 1, l = 0. I am ignoring now. So there is no external field and therefore no dependence
on the spin at all. And then there is a 2S state and 2p state. This corresponds to n
= 2, l = 0, n = 2, l = 1. These states are split. They would not be split if you didnít
have a spin orbit coupling. Both these would have exactly the same energy. And then of
course there is a 3S, 3p and a 3d state and so on. They are all split because of spin
orbit coupling. As the effective split is extremely small compared to the actual energy
value of the eigenvalue of either the 2 S or the 2 p state, there is a small splitting.
But as you go up in the periodic table and as the atomic number z value of the nucleus
increases, it turns out that the spin orbit coupling becomes more and more prominent and
there comes a stage when itís quite significant.
In fact, it is so significant and there are other effects as well, there is an effect
called the hyperfine splitting which arises due to the coupling between the magnetic moment
of the electron and the magnetic moment of the nucleus itself. And thatís an even smaller
effect but as z increases, this becomes sufficiently large so that you have significant amount
of splitting. And then this neat ordering of these energy levels; the S state, the p
state and then the d state and so on is gone. There could be the f state, maybe lower in
energy, then the d state and so on. This happens when you reach the transition metals and is
responsible for a lot of interesting effects. but the important point I want to make is
that both the spin orbit coupling as well as the effect of an external magnetic field
serve to break the symmetry that the original hydrogen atom has the coulomb problem has.
And so the general lesson is the application of a field breaks symmetry in some sense.
After all, this space in this room is isotropic but all the directions are not equivalent
because there is a vertical direction in which there is a gravitational field that breaks
the translational symmetry. And thatís the reason if you have a particle moving around
in this room under the effect of this gravity, the x and y components of its momentum are
conserved in the classical picture because there is no force in that direction but the
z component certainly is not. So linear momentum consolation is broken in one direction because
of the symmetry breaking field.
The next question was the particle mass moving in 1 dimension is incident upon a delta function
barrier, not an attractive delta function but a barrier and the statement was the reflection
coefficient of the particle is identically = 0. This is certainly not true. We saw explicitly
that there is a definite effect of the barrier even though itís an infinitely thin barrier.
The fact is the potential goes to infinity at that point and therefore there is a finite
reflection coefficient. we compute at this reflection coefficient as a limiting case
of what happens in a rectangular barrier. What you have to do is to take the product
of the width of this barrier times the height, retain that to be a finite constant and then
you can compute what it is in the limit. The second statement is false. The only normalizable
eigen state of the raising operator a dagger of the linear harmonic oscillator is the ground
state of the Hamiltonian. Is this true or false? Itís false because certainly, the
ground state of the Hamiltonian is not an eigenstate of a dagger. And in fact, a dagger
has no normalizable eigenstates whatsoever in the space of square integrable functions.
What about a? Does that have eigen states? The ground state of the Hamiltonian is also
is an eigenstate of a. but then, all coherent states are eigen states of a. now there are
several reasons why you call it a coherent state. I will mention some of these a little
later. I am going to talk about coherent states specifically. But right now, by coherent state
I mean an eigen state of the lowering operator a for the harmonic oscillator. So, one should
really call them harmonic oscillator coherent states if you like. So you have been careful
to do that every now and then. They are also minimum uncertainty states. As you saw that
in suitable units, delta x and delta p are each = 1 over square root of 2. They are Gaussian
wave packets in the position representation or in the momentum representation. But they
are not the ground state of the harmonic oscillator itself but displaced with respect to the ground
state. There are several ways of defining coherent states and they all happen to coincide
in this particular instance.
The next question was the angular momentum commutation relations together with a requirement
that the eigenstates of the angular momentum be normalizable suffice to determine the possible
eigen values of j squared and j dot n where n is a unit vector along any arbitrary direction.
This is true. We saw that this was so but the important thing is to put in the normalization
condition and that was true even for the harmonic oscillator because i required normalization
on the states and thatís how I got a discrete spectrum. If you donít require normalizable
eigen states, then there is no guarantee that the spectrum is discrete. In a more general
family of wave function or eigen states, you could have a continuous spectrum for the same
operator. So again in these cases, given an operator its spectrum is decided by what class
of eigenstates you would like to look at.
You would like to have for physical reasons. We want normalizability for the conservation
of probability and the interpretation of quantum mechanics. But conceivably mathematically
they could be more. There could be a continuous spectrum with non-normalizable states. So
that statement was true. A particle moves in 1 dimension in a symmetric potential V
(x) is V (- x) .all the energy levels of the particle are discrete. So you are given that
you only have bound states all the wave functions die down exponentially fast at infinity on
both sides. The position space wave function of the particle in the ground state phi 0
of x is an even of function of x. True. In this problem, given that the potential is
V (x) = V (- x) in 1 dimension, there is no degeneracy everything is a discrete spectrum
and there is nothing no degeneracy. There are no other functionally independent constants
of the motion. Nothing other than the Hamiltonian. Because classically you know that saying the
Hamiltonian H of qp = constant fixes the phase trajectories. You canít have a second independent
constant of the motion functionally.
Quantum mechanically it means you cannot have another operator other then the Hamiltonian
or some function of the Hamiltonian which commutes to the Hamiltonian. So the levels
are non-degenerative. And then if the potential is symmetric the Hamiltonian commutes with
the parity operator and therefore you can find a complete set of common Eigen states
of the 2. Now the parity operator has Eigen states all even functions and all odd functions.
But the Hamiltonian has special Eigen functions. They form a complete set in the space. So
the conclusion is every eigenstate of the Hamiltonian is also a parity Eigen state.
Which means every Eigen state of the Hamiltonian is either an even function of x or an odd
function of x in the positions space representation. In the ground state there are no nodes. If
you write down the Schrodinger equation, every time the function crosses the axis, it has
to go up come down again go to 0 on both sides. It costs energy because there is a second
derivative term in this Schrodinger equation. So the lowest energy state would just start
of from 0 at - infinity it goes up and it comes down and itís normalizable. It has
no nodes. by the requirement that it should be either in even or odd state, since it never
crosses the axis it is mostly in even state. So that is a true statement.
Consider the set of functions phi n of x = the oscillator Eigen functions. This is the normalization
constant e to power - f squared Hn of x where Hn is the Hermite polynomial. Any arbitrary
square integral function of x in - infinity infinity is any l2 function can be written
as a linear combination of the elements of the set in a unique manner. True or false?
This is true because this set forms an orthonormal basis. Once itís an orthonormal basis and
any function in that function space can be written uniquely. So the coefficients determine
the function completely. What happens if you didnít have, suppose for a minute I looked
at l, 2 from 0 to infinity instead of - infinity to infinity. Letís say we looked at the solutions
of the 3 dimensional Schrodinger equations and I want the radial part of the wave function.
The radial coordinate r and from 0 to infinity you would still require square integrability.
So you would now want from 0 to infinity the function mod square etcetera is finite. What
would happen in that case? Would be Hermite polynomials form a complete set still? Would
they be orthogonal? Do you think so? Would they be a good basis to expand things in?
They may not be unique. Do you think so?
First of all we have this weight factor e to the power - 1/2 x squared. Hn of x is a
polynomial. That grows at infinity. But e to the power - 1/2 x squared dies down faster
than any power and therefore you guaranteed that all these polynomial would be integrable
when you square it. But if you function if you are going to variable is going to run
from 0 to infinity in x for example then you donít need e to the power - x squared e to
the power - x would do. You put the x squared because it should be it - infinity too. And
I donít want a mod x because that would mean there is cusp at the origin and so on. So
thatís one reason why we have e to the power - x squared appearing there naturally. But
from 0 to infinity e to the power - x would do. And what do you call that set of functions
with the weight factor e to the power - x integrable? They are the Laguerre polynomials.
Thatís why that appears in the hydrogen atom problem in the natural way. Consider a particle
with spin quantum number Ω. The particle can never be in a spin state in which the
uncertainty product delta Sx delta Sy is 0. There were mixed results from this question.
So let me explain whatís happening.
We would like to find out whatís delta Sx delta Sy in any state. In any arbitrary state,
remember that the uncertainty principle says that this is greater than or = 1/2 the modulus
of the commutator of Sx with Sy. But this is = h cross over 2. This is ih cross S z
times the expectation value of Sz in an arbitrary state. How big can Sz be? No matter what state
you are in, that canít be bigger than Ω. Because the Eigen values Sz would be either
+ 1/2 or - 1/2 in units of h cross. So this is finite. Can it be 0? Yes of course. Classically,
you can see that if this is your axes of quantization and you put your spin along the x axis, itís
an Eigen state of Sx for example. Then the expectation value of Sz is 0. Just as when
you put it in an Eigen state of Sz, The expectation value of Sx and Sy are 0. In exactly the same
way, you could have it in a state where the expectation value of this is 0. Classically,
this would mean the spin quote and quote lies along the x in the xy plane. There is no z
projection. So the least value of this quantity in fact is 0. So the statement is actually
false. It can definitely be in such a state. So you see the significance of the generalized
uncertainty principles. Really you should go back to this algebra and see look at here
and this could be 0. It canít be negative because the product of uncertainties can be
never being negative. It could be 0 in some situations. Again the harmonic oscillator
functions but I will put a tilde and put a p there. Otherwise itís exactly the same
function as on the position basis. Are Eigen functions of the Fourier transform operator?
The answer is yes. They definitely are because this is the momentum space Eigen function
of the harmonic oscillator. The position space Eigen function had exactly the same functional
form but we know the momentum space Eigen function is a Fourier transform of the position
space function. Therefore here is a set of functions whose Fourier transforms are essentially
constants time the original functions. What can those Eigen value is be? There are the
fourth roots of unity. Because we know that the fourth power of the Fourier transform
operator is the identity operator in this function space. Those Eigen values can only
be 1 i - 1 and ñ i. the ground state of the oscillator corresponds to Eigen value 1. That
we know follows also from the fact that the Fourier transform of Gaussian is a Gaussian
and with suitable normalization coefficients, itís exactly the same function.
Consider the 1 dimensional potential barriers V (x). Let the reflection and transmission
coefficients be R and T. so I have in arbitrary potential barrier and i would like to find
out for incidence from the left, there is reflection R and there is transmission T.
and now you are asked suppose you started from the right, there would be a reflection
R prime and a transmission T prime and the question is: is R = R prime and T = T prime?
Now the way we derived this expression was simply to say e to the power ikx was incidence
from the right e to the power - ikx should been incident) from the right and e to the
power - + ikx should mean going rightwards from the left. The way we derived this entire
expression had nothing to do with the directions in which we moved. Itís completely arbitrary
all we did was to say that itís a plane wave asymptotically on either side. So in fact
the reflection and transmission coefficients are identical. it doesnít matter which direction
you have been incidence. It doesnít require the symmetry property of the potential because
we never use that. So do it for a non-symmetric potential and you will discover that this
is exactly the same answer.
So here is a simple potential which would look like this. Do it for a potential of this
shape. It would remain exactly the same thing. We want to make sure the potential vanishes
at infinity. So i said there is a finite barrier. This is just a barrier at some point and it
definitely doesnít mean that the potential extends. If it does, then the energy has to
be greater than that value. Otherwise you canít even have such a state.
So his point is if the potential were to be a barrier like this and this was 0 of energy
you canít have an energy state below this value. To start with it, it has to be above
that. And of course itís easy to see what would happen if you had a lack of symmetry.
Suppose you have V (x) looking like this. Now we computed things because they are analytically
computable. We computed things where the potential barrier was actually compact. Beyond a certain
stage on either side there was no potential at all. But thatís not absolutely essential.
You could have a situation like this. This asymptomatically goes to 0 and these asymptotes
to some finite value here. One could look at this barrier as well. Itís just that itís
only at - infinity and + infinity that you apply these plane wave conditions otherwise
itís not necessary a plane wave at all the function Eigen function can be quite complicated.
But for computing R and T you donít need to know the exact potential except the formal
expressions that you have when you compute the numerical values of the coefficients which
would depend on the potential. You need to solve the Schrodinger equation here.
Let j be the total angular momentum quantum number of a system. Then in any arbitrary
state psi of the system the operator J squared must necessarily have the expectation value
h cross squared J times J + 1. What we are actually saying is if it has definite value
then itís an eigen state of J squared. So we are really saying that an arbitrary state
of the system is an Eigen state of the total angular momentum. True or false? Itís true.
You havenít given any other conditions. So itís certainly true. So the idea is that
you have maybe an orbital part. You have other in degrees of freedom. So the total wave function
is a product of the spin wave function and the spatial wave function for instance and
itís always an Eigen state. Just like an electron no matter what you do to this electron,
its spin quantum number is Ω. Therefore you are guaranteed that if you measure S squared,
the answer is going to be 3 quarters h cross squared. So that is certainly true always.
Thatís a good point he has raised.
In the normal course of events, if you took a Cartesian coordinate delta p, this is certainly
greater than or = 1 1/2 the modulus of the expectation value of x with p. but it so happens
that this is = ih cross the times unit operator and therefore, this thing here becomes h cross
over 2 the modulus of the expectation value of the unit operator. But the expectation
value of the unit operator is by definition, 1 no matter what you do. Thatís the reason
you get the usual commutation relation saying it can never be 0. its greater than or = 1/2
h cross. But in the spin case, this commutator is Sz. so what you have to deal with is this
quantity and this could be 0. So thatís the reason its different from what it is for
the usual pair of Cartesian coordinates. So I will give a problem where you work this
out explicitly and you will see this. The next 1 was fill in the blanks and the first
question was a simple question but was a little bit of trick.
It said a particle moves in a 1/2 oscillator potential.V(x) is given to you in any case-
1/2 m omega squared x square and infinity. For x greater than 0, itís the oscillator.
For x less than 0, there is a wall. so its like attaching a spring to 1 end to the origin,
the other end to this particle and letting this particle vibrate but not fully along
the x axis but whenever it hits the x = 0, it bounces back. So in this 1/2 oscillator
potential the question asked is: whatís the ground state of this problem.
So the potential is infinite here. This is 1/2 m omega squared and squared and its infinite
for x less than 0. Now, you are asked what the ground state energy Eigen value is and
the eigen functions. Now you see the earlier problem which you solved the oscillator problem
at the Hermite polynomials etcetera and you used to boundary conditions at - and + infinity
the wave function went to 0. Now itís exactly the same differential equation but it applies
in the region x greater than 0. so in this region, you have the equation - h cross squared
over 2 m d 2 phi over dx 2 + 1/2 m omega squared x squared phi = E phi and this is for x greater
than 0. Since the potential is infinite, the wave function is 0 in this region including
at this point. So when you solve this and you impose the condition of square integrability
that the function be finite at infinity the energy levels turn out to be discrete as before
the solutions of a Hermite polynomials but they would be those Hermite polynomials which
vanish at the origin. They are the odd ones that vanish at the origin. The even ones donít
vanish at the origin. Therefore they canít be solutions to this problem. Whatís the
lowest energy value? Its 3 1/2 h cross omega. So this spectrum has all the odd eigenvalues
only. So the lowest energy value E ground state = 3 1/2ís h cross omega purely from
the boundary condition. So the even ones are not allowed in this case and it picks out
a way other one. So by the way that now is the set that forms a complete set. So only
the odd ones are sufficient they form a complete set once again.
So, if you insist on the weight function e to the power - x squared instead of e to the
power ñ x, then you donít get the full set of Hermite polynomials provided you impose
the boundary condition. You look at a class of function which vanishes at the origin.
Then e to the power - x squared times the odd Hermite polynomials form a complete set.
What
happens if the spring constant becomes negative? What happens if the potential is - 1/2 constant
times x squared?
This is your potential. Can you have bound states in this problem? There are normalizable
Eigen functions of the Hamiltonian. No bound states at all. Classically the origin is an
unstable equilibrium point. It just falls off from the origin. So corresponding to that
in quantum mechanics there are no bound states. The general lesson is that bound states would
correspond to periodic solutions in the classical case. Anything that is periodic becomes a
bound states in quantum mechanics. No periodic orbits are allowed here. The next was the
Hamiltonian of a perturbed oscillator is given by the usual h cross omega a dagger a + 1/2
+ lambda times a + a dagger and the question is: whatís the exact value of the ground
state? This is just the shifted oscillator. All you have to do is to recognize that you
have shifted the oscillator because a + a dagger is x. what you have d1 is to take p
square + x square + put an x term so complete squares. Itís like shifting the oscillator
and changing the 0 of the energy. All you are asked to find is: whatís the change in
the 0 of the energy.
So thatís easily found because you have h cross omega a dagger + 1/2 + lambda a + a
dagger. Notice lambda is real and this is a Hermitian quantity. So the Hamiltonian is
Hermitian and the Eigen values are guaranteed to be real. So you could write this as h cross
omega times a dagger a + lambda over h cross omega a + a dagger + Ω h cross omega. This
is the 0 point energy level. So all you have to do is to define b = a + lambda over h cross
omega. So b dagger is a dagger + lambda by h cross omega. The important thing is the
commutator of b with b dagger is still = 1. Thatís all that decides the spectrum. So
this whole thing becomes = h cross omega b dagger b + a 1/2 - lambda squared over h cross
omega. If I complete squares, there is an extra term i have added which is lambda squared
over h cross square omega squared but there is 1 sitting outside here and thatís it.
So whatís the ground state energy?
This portion has a ground state energy which is 1/2 h cross omega. So E ground state = h
cross omega by 2 - lambda square over h cross omega. Thatís the exact answer. You just
subtract that portion out. Thatís it.
So whatís been done by adding this thing is to start with this potential and move to
a potential which looks like this. Thatís all thatís happened. Itís still a discrete
set of eigenvalues. All we are saying is that what was 0 here has been shifted to - lambda
squared over h cross. The next one said: let S be the spin operator for spin 1/2 particle
and n and n prime 2 arbitrary directions and you are asked to find the commutator S dot
n S dot n prime.
Remember that we have relations for the Pauli matrices. So S dot n S dot n prime = h cross
over four because S is h cross over 2 times a Pauli matrix. Then the commutator of sigma
dot n with sigma dot n prime. So this is = h cross square over 2 i n cross n prime dot
sigma. Itís got to be operator then at the end. So there has to be a sigma. You can re-express
it in terms of S if you like. Now up and down are the usual Eigen states and you are asked:
whatís the operator Sx Sy in this basis. Itís just a simple trivial exercise, understanding
whatís meant by the basis.
So we know that Sx Sy = h cross squared over 4 sigma x sigma y but sigma x sigma y is i
sigma z which is = ih cross over 2 Sz. so we can retain it in either of the forms. It
doesnít matter really. Then, what is this matrix? As a matrix this is = h cross squared
over 4 i and then a 1 0 0 ñ 1. Thatís it. So as an operator, remember this is in the
basis in which you diagonalized along Sz itself.
So as an operator it says Sx Sy = ih cross squared over 4 [|up> = p over m. thatís the equation of motion. Of course
if you solve it, then you need to know what is the expectation of r at z = 0. So this
is the equation of motion. It follows from Heisenbergís equation of motion. for a free
particle the Hamiltonian is p squared over 2 m. so dp over dt is 1 over ih cross the
commutator of r with H. the expectation value of a dagger a in a coherence state alpha with a linear harmonic oscillator
S tells us 1 of the meanings of this complex number alpha which is the eigen state of a
and the expectation value is not hard to find.
Remember, that the coherent state alpha was defined as e to the - mod alpha squared summation
n = 0 to infinity alpha to the n over square root of n factorial acting on n. You could
now ask whatís a dagger a acting on this and find with an alpha on the left hand side.
So alpha a dagger a alpha = e to the - mod alpha squared because you are going multiply
2 of these fellows here. And then a summation n = 0 to infinity alpha to the n alpha star
to the n because this would have been m on the left but I am going to use the orthonormality
condition.
So it becomes alpha star on this side. Divided by n factorial because you canít do square
root of n factorial. You canít sum but n factorial in the denominator is easy. and
then put an alpha a dagger a. when it acts on n, it just produces an n and n and of course
this is mod alpha squared to the power n. this sum is easy to do. You start from n = 1
onwards. n = 0 is 0 by definition. This series is = mod alpha squared. So if you put beta
= a 1 alpha which incidentally is alpha on alpha, then this quantity here is nothing
but norm of beta on beta = mod alpha squared. So this is the meaning of mod alpha squared.
Itís in fact the expectation value of the number of quanta. If we regard this as state
of the radiation field, itís a number of protons in a coherent state. The average value
of the number photon because you are not in the number state. You are not an Eigen state
of a dagger a but in the superposition these states. Therefore the sum average value and
thatís mod alpha square in this problem. Now what is the allowed value of mod alpha
squared? It is 0 to infinity. Its 0 if and only if alpha is 0. When alpha is 0, then
the state that you get is in fact the ground state of the harmonic oscillator in which
the expectation value of n is 0 anyway. Itís the vacuum state for the radiation field.
Otherwise you get a finite number.
Now of course you can also generate this. You could write this as = a dagger to the
n over square root of n factorial on 0 in which case you get e to the power alpha a
dagger on 0. The advantage of this is there is no summation. It tells you there is some
operator acting on the ground state produces this coherence state but an even better way
of writing this is by using Lemma combinations.
You can also write this as e to the alpha a dagger - alpha star a acting on 0. Thatís
a very important way of writing it. This is called the displacement operator D of alpha.
Itís a function and alpha star but I will just omit the alpha star there. itís a unitary
operator. So in a sense a coherent state is a unitary transformation on the ground state
of the harmonic oscillator. And this unitary transformation is parameterized by a complex
number here. Therefore, the natural question to ask is: whatís this really telling you?
Well, this set of unitary transformation forms a group and there is a group multiplication
law which says D (alpha) D (beta) is D(alpha + beta) times a phase factor. Itís called
wild group because after all, a and a dagger form a a dagger and the unit operator form
the Heisenberg algebra. We will come back to this because they play a significant role
in quantum optics. Now let me stop here today.