Old quantum theory

The simplest system in the old quantum theory is the harmonic oscillator, whose Hamiltonian is:

${\displaystyle H={p^{2} \over 2m}+{m\omega ^{2}q^{2} \over 2}.}$

The old quantum theory yields a recipe for the quantization of the energy levels of the harmonic oscillator, which, when combined with the Boltzmann probability distribution of thermodynamics, yields the correct expression for the stored energy and specific heat of a quantum oscillator both at low and at ordinary temperatures. Applied as a model for the specific heat of solids, this resolved a discrepancy in pre-quantum thermodynamics that had troubled 19th-century scientists. Let us now describe this.

The level sets of H are the orbits, and the quantum condition is that the area enclosed by an orbit in phase space is an integer. It follows that the energy is quantized according to the Planck rule:

${\displaystyle E=n\hbar \omega ,\,}$

a result which was known well before, and used to formulate the old quantum condition. This result differs by ${\displaystyle {\tfrac {1}{2}}\hbar \omega }$ from the results found with the help of quantum mechanics. This constant is neglected in the derivation of the old quantum theory, and its value cannot be determined using it.

The thermal properties of a quantized oscillator may be found by averaging the energy in each of the discrete states assuming that they are occupied with a Boltzmann weight:

${\displaystyle U={\sum _{n}\hbar \omega ne^{-\beta n\hbar \omega } \over \sum _{n}e^{-\beta n\hbar \omega }}={\hbar \omega e^{-\beta \hbar \omega } \over 1-e^{-\beta \hbar \omega }},\;\;\;{\rm {where}}\;\;\beta ={\frac {1}{kT}},}$

kT is Boltzmann constant times the absolute temperature, which is the temperature as measured in more natural units of energy. The quantity ${\displaystyle \beta }$ is more fundamental in thermodynamics than the temperature, because it is the thermodynamic potential associated to the energy.

From this expression, it is easy to see that for large values of ${\displaystyle \beta }$, for very low temperatures, the average energy U in the Harmonic oscillator approaches zero very quickly, exponentially fast. The reason is that kT is the typical energy of random motion at temperature T, and when this is smaller than ${\displaystyle \hbar \omega }$, there is not enough energy to give the oscillator even one quantum of energy. So the oscillator stays in its ground state, storing next to no energy at all.

This means that at very cold temperatures, the change in energy with respect to beta, or equivalently the change in energy with respect to temperature, is also exponentially small. The change in energy with respect to temperature is the specific heat, so the specific heat is exponentially small at low temperatures, going to zero like

${\displaystyle \exp(-\hbar \omega /kT)}$

At small values of ${\displaystyle \beta }$, at high temperatures, the average energy U is equal to ${\displaystyle 1/\beta =kT}$. This reproduces the equipartition theorem of classical thermodynamics: every harmonic oscillator at temperature T has energy kT on average. This means that the specific heat of an oscillator is constant in classical mechanics and equal to k. For a collection of atoms connected by springs, a reasonable model of a solid, the total specific heat is equal to the total number of oscillators times k. There are overall three oscillators for each atom, corresponding to the three possible directions of independent oscillations in three dimensions. So the specific heat of a classical solid is always 3k per atom, or in chemistry units, 3R per mole of atoms.

Monatomic solids at room temperatures have approximately the same specific heat of 3k per atom, but at low temperatures they don't. The specific heat is smaller at colder temperatures, and it goes to zero at absolute zero. This is true for all material systems, and this observation is called the third law of thermodynamics. Classical mechanics cannot explain the third law, because in classical mechanics the specific heat is independent of the temperature.

This contradiction between classical mechanics and the specific heat of cold materials was noted by James Clerk Maxwell in the 19th century, and remained a deep puzzle for those who advocated an atomic theory of matter. Einstein resolved this problem in 1906 by proposing that atomic motion is quantized. This was the first application of quantum theory to mechanical systems. A short while later, Peter Debye gave a quantitative theory of solid specific heats in terms of quantized oscillators with various frequencies (see Einstein solid and Debye model).

One-dimensional problems are easy to solve. At any energy E, the value of the momentum p is found from the conservation equation:

${\displaystyle {\sqrt {2m(E-U(q))}}={\sqrt {2mE}}=p={\text{const.}}}$

which is integrated over all values of q between the classical turning points, the places where the momentum vanishes. The integral is easiest for a particle in a box of length L, where the quantum condition is:

${\displaystyle 2\int _{0}^{L}p\,dq=nh}$

which gives the allowed momenta:

${\displaystyle p={nh \over 2L}}$

and the energy levels

${\displaystyle E_{n}={p^{2} \over 2m}={n^{2}h^{2} \over 8mL^{2}}}$

Another easy case to solve with the old quantum theory is a linear potential on the positive halfline, the constant confining force F binding a particle to an impenetrable wall. This case is much more difficult in the full quantum mechanical treatment, and unlike the other examples, the semiclassical answer here is not exact but approximate, becoming more accurate at large quantum numbers.

${\displaystyle 2\int _{0}^{\frac {E}{F}}{\sqrt {2m(E-Fx)}}\ dx=nh}$

so that the quantum condition is

${\displaystyle {4 \over 3}{\sqrt {2m}}{E^{3/2} \over F}=nh}$

which determines the energy levels,

${\displaystyle E_{n}=\left({3nhF \over 4{\sqrt {2m}}}\right)^{2/3}}$

In the specific case F=mg, the particle is confined by the gravitational potential of the earth and the "wall" here is the surface of the earth.

This case is also easy to solve, and the semiclassical answer here agrees with the quantum one to within the ground-state energy. Its quantization-condition integral is

${\displaystyle 2\int _{-{\sqrt {\frac {2E}{k}}}}^{\sqrt {\frac {2E}{k}}}{\sqrt {2m\left(E-{\frac {1}{2}}kx^{2}\right)}}\ dx=nh}$

with solution

${\displaystyle E=n{\frac {h}{2\pi }}{\sqrt {\frac {k}{m}}}=n\hbar \omega }$

for oscillation angular frequency ${\displaystyle \omega }$, as before.

Another simple system is the rotator. A rotator consists of a mass M at the end of a massless rigid rod of length R and in two dimensions has the Lagrangian:

${\displaystyle L={MR^{2} \over 2}{\dot {\theta }}^{2}}$

which determines that the angular momentum J conjugate to ${\displaystyle \theta }$, the polar angle, ${\displaystyle J=MR^{2}{\dot {\theta }}}$. The old quantum condition requires that J multiplied by the period of ${\displaystyle \theta }$ is an integer multiple of the Planck constant:

${\displaystyle 2\pi J=nh\,}$

the angular momentum to be an integer multiple of ${\displaystyle \hbar }$. In the Bohr model, this restriction imposed on circular orbits was enough to determine the energy levels.

In three dimensions, a rigid rotator can be described by two angles — ${\displaystyle \theta }$ and ${\displaystyle \phi }$, where ${\displaystyle \theta }$ is the inclination relative to an arbitrarily chosen z-axis while ${\displaystyle \phi }$ is the rotator angle in the projection to the x–y plane. The kinetic energy is again the only contribution to the Lagrangian:

${\displaystyle L={MR^{2} \over 2}{\dot {\theta }}^{2}+{MR^{2} \over 2}(\sin(\theta ){\dot {\phi }})^{2}\,}$

And the conjugate momenta are ${\displaystyle p_{\theta }={\dot {\theta }}}$ and ${\displaystyle p_{\phi }=\sin(\theta )^{2}{\dot {\phi }}}$. The equation of motion for ${\displaystyle \phi }$ is trivial: ${\displaystyle p_{\phi }}$ is a constant:

${\displaystyle p_{\phi }=l_{\phi }\,}$

which is the z-component of the angular momentum. The quantum condition demands that the integral of the constant ${\displaystyle l_{\phi }}$ as ${\displaystyle \phi }$ varies from 0 to ${\displaystyle 2\pi }$ is an integer multiple of h:

${\displaystyle l_{\phi }=m\hbar \,}$

And m is called the magnetic quantum number, because the z component of the angular momentum is the magnetic moment of the rotator along the z direction in the case where the particle at the end of the rotator is charged.

Since the three-dimensional rotator is rotating about an axis, the total angular momentum should be restricted in the same way as the two-dimensional rotator. The two quantum conditions restrict the total angular momentum and the z-component of the angular momentum to be the integers l,m. This condition is reproduced in modern quantum mechanics, but in the era of the old quantum theory it led to a paradox: how can the orientation of the angular momentum relative to the arbitrarily chosen z-axis be quantized? This seems to pick out a direction in space.

This phenomenon, the quantization of angular momentum about an axis, was given the name space quantization, because it seemed incompatible with rotational invariance. In modern quantum mechanics, the angular momentum is quantized the same way, but the discrete states of definite angular momentum in any one orientation are quantum superpositions of the states in other orientations, so that the process of quantization does not pick out a preferred axis. For this reason, the name "space quantization" fell out of favor, and the same phenomenon is now called the quantization of angular momentum.

The angular part of the hydrogen atom is just the rotator, and gives the quantum numbers l and m. The only remaining variable is the radial coordinate, which executes a periodic one-dimensional potential motion, which can be solved.

For a fixed value of the total angular momentum L, the Hamiltonian for a classical Kepler problem is (the unit of mass and unit of energy redefined to absorb two constants):

${\displaystyle H={p_{r}^{2} \over 2}+{l^{2} \over 2r^{2}}-{1 \over r}.}$

Fixing the energy to be (a negative) constant and solving for the radial momentum ${\displaystyle p_{r}}$, the quantum condition integral is:

${\displaystyle \oint {\sqrt {2E-{l^{2} \over r^{2}}+{2 \over r}}}\ dr=kh}$

which can be solved with the method of residues,[12] and gives a new quantum number ${\displaystyle k}$ which determines the energy in combination with ${\displaystyle l}$. The energy is:

${\displaystyle E=-{1 \over 2(k+l)^{2}}}$

and it only depends on the sum of k and l, which is the principal quantum number n. Since k is positive, the allowed values of l for any given n are no bigger than n. The energies reproduce those in the Bohr model, except with the correct quantum mechanical multiplicities, with some ambiguity at the extreme values.