We have now a new problem to solve: finding the eigenvalues and eigenvectors of the Hamiltonian H equals ħω times a^† a plus 1/2, knowing that the commutator of a and a^† 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 ω 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\ he 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^† a, still with the same commutator for a^† 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^† returns the next eigenstate phi_(n+1) multiplied by a coefficient root of n+1. In the same vein when applied to phi_n, a returns the eigenstate phi_(n-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^† phi_n or a phi_n. In each equation the coefficient is just the square root of the bigger of the two indices, 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 1/2 times ħω, 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^†, we can generate any state phi_n from the lower state 0. Phi_n is equal to a^† to the power n applied to the ground state with a 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^†. 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 1/2 ħω. 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 a 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^† 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^† 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^† 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^† 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^† is 1, so that we can rewrite a times a^† as 1 plus a^† a. The term a^† 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 p 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 of 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.
