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 potential: U = 0 One-dimensional problems are easy to solve. At any energy E, the value of the momentum p is found from the conservation equation: 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: which gives the allowed momenta: and the energy levels One-dimensional potential: U = Fx 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. so that the quantum condition is which determines the energy levels, 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. One-dimensional potential: U = kx2 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 with solution for oscillation angular frequency , as before. Rotator 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: which determines that the angular momentum J conjugate to , the polar angle, . The old quantum condition requires that J multiplied by the period of is an integer multiple of the Planck constant: the angular momentum to be an integer multiple of . 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 — and , where is the inclination relative to an arbitrarily chosen z-axis while is the rotator angle in the projection to the x–y plane. The kinetic energy is again the only contribution to the Lagrangian: And the conjugate momenta are and . The equation of motion