This course gives you access to basic tools and concepts to understand research articles and books on modern quantum optics. You will learn about quantization of light, formalism to describe quantum states of light without any classical analogue, and observables allowing one to demonstrate typical quantum properties of these states. These tools will be applied to the emblematic case of a one-photon wave packet, which behaves both as a particle and a wave. Wave-particle duality is a great quantum mystery in the words of Richard Feynman. You will be able to fully appreciate real experiments demonstrating wave-particle duality for a single photon, and applications to quantum technologies based on single photon sources, which are now commercially available. The tools presented in this course will be widely used in our second quantum optics course, which will present more advanced topics such as entanglement, interaction of quantized light with matter, squeezed light, etc... So if you have a good knowledge in basic quantum mechanics and classical electromagnetism, but always wanted to know: • how to go from classical electromagnetism to quantized radiation, • how the concept of photon emerges, • how a unified formalism is able to describe apparently contradictory behaviors observed in quantum optics labs, • how creative physicists and engineers have invented totally new technologies based on quantum properties of light, then this course is for you.
1.2.2 Material harmonic oscillator /2
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来自 École Polytechnique 的课程
Quantum Optics 1 : Single Photons
80 个评分
从本节课中
Quantization of light: one mode
In this first lesson, you will discover what is canonical quantization, apply it to the quantization of a single mode of the electromagnetic field, and find that it behaves as a quantum harmonic oscillator. The notion of photon will then naturally emerge, as well as the weird but fundamental notion of vacuum fluctuations.
与讲师见面
Alain AspectProfessor
Physics
Michel BruneProfessor
Physics
We have now a new problem to solve, finding the eigenvalues and
eigenvectors of the Hamiltonian H equals hbar omega times a_dagger a plus one-half,
knowing that the commutator of a and a_dagger is 1.
Note that we have no x or p in these expressions.
In fact, this is one of the merits of Dirac approach.
It is independent of the kind of harmonic oscillator we want to quantize.
It will be applicable to any harmonic oscillator oscillating at frequency omega
of which we want to know the quantum behavior.
We can thus forget about the initial problem.
What we are going to establish now will be valid for any harmonic oscillator.
It turns out that this problem can be solved by purely algebraic reasoning
in the Hilbert space of the state vectors.
Taking into account the constraint that the eigenvectors must have a finite norm,
which is taken equal to 1.
The reasoning is detailed in standard textbooks on quantum mechanics.
For instance the one of Cohen-Tannoudji, Diu, and Laoe.
This is my preferred textbook because I studied quantum
mechanics in it by myself at a time when there was no MOOC.
The book of Dalibard and Basdevant associated with quantum mechanics course
of Ecole Polytechnique is also an excellent reference.
I will now give the results of the beautiful reasoning of Dirac.
These results will be used all the time in this course.
Be sure to remember them.
One first looks for a slightly simpler problem,
finding the eigenvalues and eigenvectors of the operator
N_hat = a_dagger a, still with the same commutator for a_dagger and a.
This operator is called the number operator.
One finds that its eigenvalues are non-negative integers.
That is, n equals 0,1, 2, etc.
The corresponding eigenvectors, phi_n obey remarkable properties.
Firstly, when applied to phi_n, a_dagger returns the next eigenstate phi
n plus 1 multiplied by a coefficient root of n plus 1.
In the same vein when applied to phi_n a returns the eigenstate phi_n minus 1,
multiplied by a coefficient root of n.
For the lower value of n, that is to say 0,
one can show that this formula returns the number 0.
Be careful, this means in fact that the right hand side of
this last equation is a vector whose norm is null.
Clearly, phi_0 plays a particular role,
and we will denote it ket 0.
It is a state associated to the eigenvalue 0 of the number operator,
as any state vector in quantum mechanics, it has a norm of 1.
So be careful. The state 0 has norm 1,
it should not be confused with the null vector which has norm 0.
Let me tell you now a small trick of the trade to remember
the coefficient in the equations giving a_dagger phi_n or a phi_n.
In each equation the coefficient is just the square root
of the bigger of the two and these is as you can check.
With this you have no excuse not to remember these relations
which we will use all the time.
We can now come back to our initial problem, finding the eigenvalues and
eigenvectors of the Hamiltonian of the harmonic oscillator.
[MUSIC]
It is clear that the eigenvectors phi_n of big N are also
eigenvectors of the Hamiltonian H.
The eigenvalues of the Hamiltonian are then deduced from
the eigenvalues of capital N.
We find the well known expression for the energy levels of the harmonic oscillator
n plus one-half times hbar omega with n,
a non-negative integer.
Let us come back to the fundamental properties of the eigenstates phi_n,
which are the quantum states associated to the energy levels.
Using the relation with the creation operator a_dagger
we can generate any state phi_n from the lower state 0.
Phi_n is equal to a_dagger to the power n applied to the ground state
with normalization factor 1 over root factorial n.
In your basic course on quantum mechanics you have learned that the ensemble of
the eigenstate vectors, of the Hamiltonian constitutes a complete
basis of the space of states of the system described by that Hamiltonian.
We have thus
a means of generating such a basis as soon as we know the lowest energy state.
ket 0 and the creation operator a_dagger.
This state 0 clearly plays an important role.
[MUSIC]
To finish this section on quantum oscillators in general,
let us see interesting properties of that state zero corresponding to
the smallest energy also called the ground state.
In fact, its energy is not null,
it has a well known value of one-half hbar omega.
Is there some physics associated with that non-null energy?
You could tell me that the reference level for
the energy being arbitrary, we better not waste time on that question.
But wait for a moment.
I have not chosen an arbitrary reference for the energy.
My reference is the bottom of the harmonic potential.
It means that the non zero energy is either potential energy above that bottom or
kinetic energy or both.
To be specific, I come back for
a while to the particular case of mechanical oscillator.
Let us look for the position X.
or momentum P in the lower energy state.
You can think of a pendulum close to its lower position of equilibrium.
We calculate first the average position in the ground state
using the dimensionless position capital X.
The value is 0 because we know that a applied to
the state 0 on the right yields 0 and
similarly a_dagger applied to 0 on the left also gives 0.
Since it is nothing else then the hermitian conjugate of the previous relation.
So the average position is 0 but it does not mean that the pendulum is at rest.
The position could fluctuate around 0.
I should rather say it could have a spread, a dispersion around 0.
Let us investigate that question by looking into the average of
the square of X.
I expand carefully the square keeping the order because we know that a and
a_dagger do not commute.
The first and third terms give 0, because of a applied to state 0 on the right.
And similarly, for
the second term, because of a_dagger applied to the state 0 on the left.
But what about the last term?
I will now teach you another trick of the trade.
Always manage to have a on the right and a_dagger on the left.
This is not the case here.
To obtain the desired form, we use the fact that the commutator of a and
a_dagger is 1, so that we can rewrite a times a_dagger as 1 plus a_dagger a.
The term a_dagger a is now ordered as we like and
it gives 0 and we are left with a finite term one-half.
So there is a spread of x.
The wave function of the pendulum,
has finite width, it extends symmetrically around 0.
So the potential energy is not null in the ground state.
A similar calculation with the momentum P gives a null average and
a non-null average of the square.
There is also a spread around 0 for the momentum, so
the kinetic energy is not null in the ground state.
We can speak of fluctuations.
Doesn't this ring a bell in your mind?
In fact, you know that quantum property that neither x nor be can be exactly 0.
You remember Heisenberg dispersion relation between x and p?
I know that it is often called Heisenberg uncertainty relation,
but I prefer to keep the word uncertainty for experimental inaccuracy and
speak here of a spread, a dispersion, in the position or the momentum.
So let us see the connection to Heisenberg relation dispersion relation.
From the calculation above, we can immediately obtain the dispersion of
the real values of small x and small p,
not the dimensionless ones and we obtain the value of
delta x squared, also called its variance.
Go back to the definition of the reduced position to check this result.
Similarly, we can obtain the variance of p.
From this, we obtain the product of the rms values of the dispersions.
That is of the square root of the variances, wow!
We are just at the lowest value permitted by the Heisenberg relations!
It means that the ground state has the smallest possible fluctuations
permitted by Heisenberg relations.
It is called a minimum dispersion state.
We will come back to this property in the case of quantized light.
The ground state for radiation is called the vacuum.
We will see that even in the vacuum, the quantized field is not 0.
It has fluctuations.
After all these preliminaries it is time to turn towards the case of light or,
more generally, of the electromagnetic field.
