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In the quantum analysis of particles quantization means the properties of particles, such as energy and momentum, can take on only discrete values. In quantum field theory quantization has an entirely different meaning. It means that certain constructs function to keep track of the numbers of things. This is referred to as the second quantization in quantum theory.
In Schrödinger's wave mechanics the Hamiltonian operator for a system is constructed from its Hamiltonian function
by replacing the momentum p with −ih∇, where i is the square root of negative one and
h is Planck's constant divided by 2π. (An operator is just a function that takes a function over space as an input and returns
a function over space as an output. Usually the operator for a variable is denoted by a hat mark over the symbol for the variable. Such notation
is not availabe so the operator corresponding to a variable is indicated by the capitalized form of its symbol. In the case of a variable that is
a capital the operator is indicated by the use of "^" as a superscript. )
For example, the Hamiltonian function for a particle attached to a spring (a harmonic oscillator) is ½p²/m + ½kx², where m is the mass of the particle, k is the stiffness coefficient of the spring and x is deviation from equilibrium. For this one dimensional system ∇ is (∂/∂x) and ∇² is (∂²/∂x²). Thus the Hamiltonian operator for a haromonic oscillator is
A harmonic oscillator oscillates with a frequency ω equal to (k/m)^{½}. Thus the Hamiltonian operator for a harmonic oscillator may also be represented as
In wave mechanics the time independent wave function φ(z) satisfies the equation
where z is the vector of the coordinates of spatial location, and E is energy. This means that φ(z) is an eigenfunction of the operator H^ and E is an eigenvalue. The wave function φ(z) is such that φ(z)² is the probability density of finding the particle at point z.
In a system of units such that the speed of light in a vacuum is unity and h (hbar) (Planck's constant
divided by 2π) is also equal to unity the eigenvalues are integers plus ½. P.A.M. Dirac proposed a
brilliant bit of notation. He suggested that the eigenfunctions be denoted in the form n> and
are thus labeled by values closely related to their eigenvalues. This means that the time independent Schrödinger
equation
For a harmonic oscillator the Hamiltonian operator takes the form
The operators X and P are required to be such that their commutator [X, P]=X(P)−P(X) be equal to iI, the imaginary unit times the identity operator. This is called canonical quantization and is the same as the condition in Heisenberg's Matrix Mechanics of
where P and Q stand for momentum and location, respectively, and I is the infinite identity matrix. (h is Planck's constant divided
by 2π.)
Now define the following two operators
where β is equal to (mω/2)^{½} and γ is equal to i/(mω).
Consider now the commutator of A and A*.
But [X , X]=0 and [P , P]=0 so the above equation reduces to
Since [P, X] is equal to −[X, P],
Upon the substitution of the values for β and γ
where I is the identity operator.
Now consider the operator A*A
But (XP − PX) is the commutator [X, P] and is equal to iI. Therefore
Upon the substitution of the values for β and γ
Then
Note that
Therefore
If L is an eigenfunction of H^ with an eigenvalue of λ then the energy of the state L is λ and
Thus L is also an eigenfunction of A*A but its eigen value is (λ/ω−½).
In the Appendix it is estabished that the eigenvalues of the eigenfunctions of A*A are nonnegative integers. For this reason the operator A*A is called the number operator and is denoted as N^.
If the eigenvalue of N^ is the integer n then the eigenvalues for H^ is equal to ω(n+½) and thus
Thus the quantization of a quantum field involves a count. The question is what is it a count of. The quick answer is that it is a count of the number of particles associated with the field. It might possibly be the number of charged particles creating the field. But this is not the case. The count is of the number of perturbations of the field; i.e., the number of photons.
When an electron or other charged particle creates a photon the casual perception is that the photon goes off on its own at the speed of light. But this leaves out the nature of the photon as a perturbation in the electromagnetic field of the charged particle. Thus no matter how far the photon travels it forever belongs to the electromagnetic field of the charged particle that created it until it is absorbed and extinguished by another charged particle. If later another photon is created by the original charged particle it too belongs to the field.
Let A, B and C be operators. Then for the commutator [AB, C]
Proof:
Now let A be any operator and A* its conjugate. Applying the above indentity to [A*A, A] gives
Likewise
If A satisfies the canonical quantization condition that [A, A*] is equal to the identity operation the above two relationships reduce to
These can also be expressed as
where I is the identity operation, which is the same as multiplication by 1.
Let an eigenvalue of A*A be denoted as α. Expressed in the Dirac notation this means
Not only is α> an eigenfunction of A*A, but Aα> is also an eigenfunction of A*A because
A reapplication of the above would show that A^{n}α> for n over some range is an eigenfunction of A*A with (α−n) as its eigenvalue. Therefore the eigenfunction of A^{n}α> is denoted as (α−n)>.
Similarly A*α> is an eigenfunction of A*A with an eigenvalue of (α+1) and therefore (A*)^{n}α> is an eigenfunction of A*A with an eigenvalue of (α+n). It is represented as (α+n)>.
The eigenfunction of an operator cannot be the zero function. Therefore there must be an integer n such that A^{n}α> is an eigenfunction of A*A but A^{n+1}α> is not. This implies that α is equal to that integer and hence the eigenvalues of A*A are necessarily integers from 0 to some maximum integer. The eigenfunctions can be generated from 0> by repeated applications of A*; i.e.,
where the constant is to be evaluated.
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