THE QUANTUM PHYSICS OF SOLIDS 681 



forms for sodium is really very small — so small that we should not 

 expect the present theory to evaluate it. Some indication that the 

 energy of caesium is very nearly the same in the body-centered form 

 and face-centered form (or possibly the hexagonal close-packed form) 

 is furnished by a transformation at high pressures observed by Bridg- 

 man. In the next paper we shall meet a case where the theory does 

 seem able to differentiate between the energy of face-centered and 

 body-centered structures. In general, however, the procedure is to 

 use the crystal structure found by x-rays and to calculate the energy 

 for a series of values of the lattice constant as was done for sodium. 

 We must now return to a discussion of the curves Eo and Ep of 

 Fig. 15. About the curve £o we shall only say that it is obtained by 

 solving Schroedinger's equation for and finding the energy and wave 

 function of an electron in the lowest state in the energy band. The 

 wave function for this state, however, possesses the interesting feature 

 of being very nearly the same as the wave function for a free electron 

 having zero energy of motion. From this fact it is possible to draw 

 the conclusion that the distribution in energy of motion of the valence 

 electrons in sodium is the same as the distribution in energy of free 

 electrons in an electron gas. Accepting this conclusion, we can then 

 use the formulas given for the distribution of states for free electrons 

 in order to calculate the mean energy of motion of the valence electrons 

 in sodium. The results of this calculation, which we give in a foot- 

 note,^* lead to the energy curve Ef- This energy curve is, from its 



^3 We shall first derive a general expression for Ep without specifying the particular 

 form of N{E). Since in this footnote all energies are measured from £o, we shall 

 omit the subscript M from Em and use simply the symbol E in the equations. The 

 total number, denoted by n, of atoms in the crystal is equal to the total number of 

 valence electrons. Let the volume of the crystal be V. Because of the duplicity 

 due to the spin there are 2n states in the band, and half of them will accommodate 

 the n electrons so that the band will be filled only up to a certain energy £max. We 

 must therefore have 



r^""' N{E)dE. (i) 



Once the distribution function N{E) is known, this equation serves to determine 

 -Emax. The average energy of motion of an electron in these occupied states is, from 

 the definition of an average, the total energy of motion divided by the total number 

 of electrons: 



Ef = - r^""' EN{E)dE. . (ii) 



Substituting the value of N{E) for free electrons into the first equation gives 



rt = I VilmE^^^Jh-^yi^ (iii) 



The quantity Vj-n is the volume per electron which in the case of a monovalent metal 



