CHAPTER 1
The Origin of the Quantum Theory
The Rayleigh-Jeans Law
1. Blackbody Radiation in Equilibrium. Historically, the quantum theory began with the attempt to account for the equilibrium distribution of electromagnetic radiation in a hollow cavity. We shall, therefore, begin with a brief description of the characteristics of this distribution of radiation. The radiant energy originates in the walls of the cavity, which continually emit waves of every possible frequency and direction, at a rate which increases very rapidly with the temperature. The amount of radiant energy in the cavity does not, however, continue to increase indefinitely with time, because the process of emission is opposed by the process of absorption that takes place at a rate proportional to the intensity of radiation already present in the cavity. In the state of thermodynamic equilibrium, the amount of energy U(v)dv, in the frequency range between v and v + dv, will be determined by the condition that the rate at which the walls emit this frequency shall be balanced by the rate at which they absorb this frequency. It has been demonstrated both experimentally and theoretically, that after equilibrium has been reached, U(v) depends only on the temperature of the walls, and not on the material of which the walls are made nor on their structure.
To observe this radiation, we make a hole in the wall. If the hole is very small compared with the size of the cavity, it produces a negligible change in the distribution of radiant energy inside the cavity. The intensity of radiation per unit solid angle coming through the hole is then readily shown to be I(v) = c/4π U(v), where c is the velocity of light.
Measurements disclose that, at a particular temperature, the function U(v) follows a curve resembling the solid curve of Fig. 1. At low frequencies the energy is proportional to v2, while at high frequencies it drops off exponentially. As the temperature is raised, the maximum is shifted in the direction of higher frequencies; this accounts for the change in the color of the radiation emitted by a body as it gets hotter.
By thermodynamic arguments Wien showed that the distribution must be of the form U(v) = v3f(v/T). The function f, however, cannot be determined from thermodynamics alone. Wien obtained a fairly good, but not perfect, fit to the empirical curve with the formula
U(v) dv ~ v3e-hv/κT dv (Wien's law) (1)
Here κ is Boltzmann's constant, and h is an experimentally determined constant (which later turned out to be the famous quantum of action).
Classical electrodynamics, on the other hand, leads to a perfectly definite and quite incorrect form for U(v). This theoretical distribution, which will be derived in subsequent sections, is given by
U(v) dv ~ κTv2 dv (Rayleigh-Jeans law) (2)
Reference to Fig. 1 shows that the Rayleigh-Jeans law is in agreement with experiment at low frequencies, but gives too much radiation for high frequencies. In fact, if we attempt to integrate over all frequencies to find the total energy, the result diverges, and we are led to the absurd conclusion that the cavity contains an infinite amount of energy. Experimentally, the correct curve begins to deviate appreciably from the Rayleigh-Jeans law where hv becomes of the order of κT. Hence, we must try to develop a theory that leads to the classical results for hv< κT, but which deviates from classical theory at higher frequencies.
Before we proceed to discuss the way in which the classical theory must be modified, however, we shall find it instructive to examine in some detail the derivation of the Rayleigh-Jeans law. In the course of this deviation we shall not only gain insight into the ways in which classical physics fails, but we shall also be led to introduce certain classical physical concepts that are very helpful in the understanding of the quantum theory. In addition, the introduction of Fourier analysis to deal with this classical problem will also constitute some preparation for its later use in the problems of quantum theory.
2. Electromagnetic Energy. According to classical electrodynamics, empty space containing electromagnetic radiation possesses energy. In fact, this radiant energy is responsible for the ability of a hollow cavity to absorb heat. In terms of the electric field, ε(x, y, z, t), and the magnetic field, H(x, y, z, t), the energy can be shown to be
E = 1/8π [integrity] (E2 + H2) dτ (3)
where dτ signifies integration over all the space available to the fields.
Our problem, then, is to determine the way in which this energy is distributed among the various frequencies present in the cavity when the walls are at a given temperature. The first step will be to use Fourier analysis for the fields and to express the energy as a sum of contributions from each frequency. In so doing, we shall see that the radiation field behaves, in every respect, like a collection of simple harmonic oscillators, the so-called "radiation oscillators." We shall then apply statistical mechanics to these oscillators and determine the mean energy of each oscillator when it is in equilibrium with the walls at the temperature T. Finally, we shall determine the number of oscillators in a given frequency range and, by multiplying this number by the mean energy of an oscillator, we shall obtain the equilibrium energy corresponding to this frequency, i.e., the Rayleigh-Jeans law.
3. Electromagnetic Potentials. We begin with a brief review of electrodynamics. The partial differential equations of the electromagnetic field, according to Maxwell, are given by
[nabla] X ε = - 1/c [partial derivative]H/[partial derivative]t (4)
[nabla] x H = (5)
[nabla] X H = 1/c [partial derivative]ε/[partial derivative]t + 4πj (6)
[nabla] x ε = 5πρ (7)
where j is the current density and ρ is the charge density. We can show from (4) and (5) that the most general electric and magnetic field can be expressed in terms of the vector and scalar potentials, a and I, in the following way:
H = [nabla] x a (8)
and
ε = - 1/c [partial derivative]a/[partial derivative]t - [nabla]φ (9)
When ε and H are expressed in this form, (4) and (5) are satisfied identically, and the equations for a and I are then obtained by the substitution of relations (8) and (9) into (6) and (7).
Now, eqs. (8) and (9) do not define the potentials uniquely in terms of the fields. If, for example, we add an arbitrary vector, -[nabla]ψ, to the vector potential, the magnetic field is not changed because [nabla] × [nabla]ψ = 0. If we simultaneously add the quantity 1/c [partial derivative]/[partial derivative]t ψto the scalar potential, the electric field is also unchanged. Thus, we find that the electric and magnetic fields remain invariant under the following transformation of the potentials:*
[MATHEMATICAL EXPRESSION OMITTED] (10)
The above is called a "gauge transformation."
We can utilize the invariance of the fields to a gauge transformation for the purpose of simplifying the expressions for ε and H. A common choice is to make div a = 0. To show that this is always possible, suppose that we start with an arbitrary set of potentials, a(x, y, z, t) and ψ(x, y, z, t). We then make the gauge transformation of eq. (10) to a new set of potentials, A' and ψ'. In order to obtain div a' = 0, we must choose ψ such that
div a - [nabla]2ψ = 0
But the above is just Poisson's equation defining ψ in terms of the specified function, div a. Its solution can always be obtained and is, in fact, equal to
[MATHEMATICAL EXPRESSION OMITTED]
Thus, we prove that a gauge transformation that yields div a' = 0 can always be carried out.
We now show that in empty space the choice div a = 0 also leads to I = 0 and, therefore, to a considerable simplification in the representation of the electric field. To do this, we substitute eq. (9)into (7), setting ρ = 0 since, by hypothesis, there are no charges in empty space. The result is
div ε = - 1/c div [partial derivative]a/[partial derivative]t - [nabla]2φ = 0
But since div a = 0, we obtain
[nabla]2φ = 0
This is, however, simply Laplace's equation. It is well known that the only solution of this equation that is regular over all of space is I = 0. (All other solutions imply the existence of charge at some points in space and, therefore, a failure of Laplace's equation at these points.) We should note, however, that the condition ψ = 0 follows only in empty space because, in the presence of charge, eq. (7) leads to [nabla]2ψ = -4πρ, which is Poisson's equation. This equation has nonzero regular solutions, provided that ρ is not everywhere zero.
We conclude, then, that in empty space we obtain the following expressions for the fields:
H = [nabla] X a (11)
ε = - 1/c [partial derivative]a/[partial derivative]t (12)
subject to the condition that
div a = 0
Finally, we obtain the partial differential equation defining a in empty space by inserting (11), (12), and (13) into (6), provided that we also assume that j = 0, as is necessary in the absence of matter. We obtain
[nabla]2a - 1/c2 [partial derivative]2a/[partial derivative]t2 = 0 (14)
Equations (11), (12), (13), and (14), together with the boundary conditions, completely determine the electromagnetic fields in a cavity that contains no charges or currents.
4. Boundary Conditions. As pointed out in Sec. 1, it has been demonstrated both experimentally and theoretically* that the equilibrium distribution of energy density in a hollow cavity does not depend on the shape of the container or on the material in the walls. Hence, we are at liberty to choose the simplest possible boundary conditions consistent with equilibrium. We shall choose a set of boundary conditions that are somewhat artificial from an experimental point of view, but that greatly simplify the mathematical treatment.
Let us imagine a cube of side L with very thin walls of some material that is not an electrical conductor. We then imagine that this structure is repeated periodically through space in all directions, so that space is filled up with cubes of side L. Let us suppose, further, that the fields are the same at corresponding points of every cube.
We now assert that these boundary conditions will yield the same equilibrium radiation density as will any other boundary conditions at the walls.† To prove this, we need only ask why the equilibrium conditions are independent of the type of boundary. The answer is that, from the thermodynamic viewpoint, the wall merely serves to prevent the system from gaining or losing energy. Making the fields periodic must have, the same effect because each cube can neither gain energy from the other cubes nor lose it to them; if this were not so, the system would cease to be periodic. Thus, we have a boundary condition that serves the essential function of keeping the energy in any individual cube constant. Although artificial, it must give the right answer, and it will make the calculations easier by simplifying the Fourier analysis of the fields.
5. Fourier Analysis. Now, a(x, y, z, t) may be any conceivable solution of Maxwell's equations, with the sole restriction, imposed by our boundary conditions, that it must be periodic in space with period L/n, where n is an integer. It is a well-known mathematical theorem that an arbitrary periodic function, f(x, y, z, t), can be represented by means of a Fourier series in the following manner:
[MATHEMATICAL EXPRESSION OMITTED] (15)
where l, m, n are integers running from -∞ to ∞, including zero. Any choice of a's and b's leading to a convergent series defines a function, f(x, y, z, t), which is periodic in the sense that it takes on the same value each time x, y, or z changes by L. For a given function, f(x, y, z, t) it can be shown that the al,m,n(t) and the bl,m,n(t) are given by the following formulas:
[MATHEMATICAL EXPRESSION OMITTED] (16)
These formulas illustrate the fact that only the sum of the a's and the difference of the b's are determined by the function f.
From the above, we conclude that f may be specified completely in terms of the quantities al,m,n + a and bl,m,n + b but we prefer to retain the specification in terms of the al,m,n and bl,m,n because of the simpler mathematical expressions to which they lead.
Equations (16) are obtained with the aid of the following orthogonality relations:
[MATHEMATICAL EXPRESSION OMITTED] (17a)
unless
[MATHEMATICAL EXPRESSION OMITTED]
in which case it is L3/2, except when l = m = n = 0, in which case it is L3.
[MATHEMATICAL EXPRESSION OMITTED]
unless
[MATHEMATICAL EXPRESSION OMITTED]
in which case it is L/2. [It is suggested that the reader prove (17a and b) as an exercise, and use the results to obtain (16).]
Fourier analysis, in the preceding form, enables us to represent an arbitrary function as a sum of standing plane waves of all possible wavelengths and amplitudes. The entire treatment is essentially the same as that used with waves in strings and organ pipes, except that it is three-dimensional.
Let us now expand the vector potential in a Fourier series. Because a is a vector, involving three components, each al,m,n and bl,m,n also has three components and, hence, must be represented as a vector:
[MATHEMATICAL EXPRESSION OMITTED]
We assume that a0,0,0 is zero in the above series.
We now introduce the propagation vector k, defined as follows:
[MATHEMATICAL EXPRESSION OMITTED] (1)
By orienting our co-ordinate axes in such a way that the z axis is directed along the k vector, we obtain l = m = 0, and k = 2p/L. From the definition of k, it follows that k/2p is the number of waves in the distance L; hence the wavelength is λ = 2p/k, or
k = 2π/λ
In this co-ordinate system a typical wave takes the form cos 2πnz/L. Thus, the vector k is in the direction in which the phase of the wave changes. Going back to arbitrary co-ordinate axes, we conclude that k is a vector in the direction of propagation of the wave. Its magnitude is 2π/ λ, and it is allowed to take on only the values permitted by integral l, m, and n in eq. (18). With this simplification of notation, we obtain
where the summation extends over all permissible values of k.
[MATHEMATICAL EXPRESSION OMITTED] (20)
6. Polarization of Waves. Let us now apply the condition div a = 0 to (20). We have
[MATHEMATICAL EXPRESSION OMITTED]
It is a well-known theorem that if a Fourier series is identically zero, then all of the coefficients, ak and bk, must vanish.
Problem 1: Prove the above theorem, using the orthogonality relations (17). From the above it follows that k · ak(t) = k · bk(t) = 0. Thus, ak(t) and bk(t) are perpendicular to k, as are also the electric and magnetic fields belonging to the kth wave. Since the vibrations are normal to the direction of propagation, the waves are transverse. The direction of the electric field is also called the direction of polarization.
To describe the orientation of ak let us return to the set of coordinate axes in which the z axis is in the direction of k. The vector ak can have only x and y components, and if we specify the values of these, we shall have specified both the magnitude and the direction of ak.
