Chapter 2 The free Electron and state density functions 21. Overview of the free electron 2.1.1. The model As detailed in Chapter 1, the potential(V)(rigorously termed the energy) for a free electron (within the zero order approximation for electronics) is that of a flat-bottomed basin, as shown by the horizontal ID model of Figure 2. 1. It can also be described by V=v=0 For a free electron placed in a potential v =0 with an electronic state described by its proper function with energy and amplitude denoted by E and y respectively, the Schrodinger equation for amplitude △y+2mE"v=0 2. 1. 2. Parameters to be determined: state density functions in k or energy spaces With oUt =zy [2.2]
Chapter 2 The Free Electron and State Density Functions 2.1. Overview of the free electron 2.1.1. The model As detailed in Chapter 1, the potential (V) (rigorously termed the potential energy) for a free electron (within the zero order approximation for solid-state electronics) is that of a flat-bottomed basin, as shown by the horizontal line in the 1D model of Figure 2.1. It can also be described by V = V0 = 0. For a free electron placed in a potential V0 = 0 with an electronic state described by its proper function with energy and amplitude denoted by E0 and 0 \ , respectively, the Schrödinger equation for amplitude is: 0 00 2 0. ² m '\ \ E = [2.1] 2.1.2. Parameters to be determined: state density functions in k or energy spaces With: 0 2 ² = , ² mE k = [2.2]
18 Solid-State Physics for Electronics quation [2.1] can be written as: (x) Figure 2.1.(a) Symmetric wells of infinite depth with the origin taken at the center of the wells); and (b) asymmetric wells with the origin taken at the well's extremity wwhen 0<x<l we have v(x)=0 so that v-x) a ID model and asymmetric forms, the corresponding solutions for the wave function(y%anc We shall now determine for different depth potential wells, with both symmetric the energy (E ) To each wave function there is a corresponding electronic state (characterized by quantum numbers). It is important in physical electronics understand the way in which these states determine how energy levels are filled In solid-state physics, the state density function(also called the density of states) particularly important. It can be calculated in the wave number (h) space or in the energies(E)space. In both cases, it is a function that is directly related to a dimension of space, so that it can be evaluated with respect to L= l(or V=LS for a 3D system). In k space, the state density function [n(h)I is such that in one dimension n(h)dk represents the number of states placed in the interval dk(i.e between k and k+ dk in k wave number space). In 3D, n(k) ck represents the ber of states placed within the elementary volume d k in k space Similarly, in terms of energy space, the state density function [ZE)] is such that Z(e)dE represents the number of states that can be placed in the interval dE (i.e inclusively between E and E+ dE in energy space). The upshot is that if F(E)is the occupation probability of a level denoted E, then the number m(e)dE of electrons distributed in the energy space between E and E+dE is equal to NE)dE=Z() F(ede
18 Solid-State Physics for Electronics equation [2.1] can be written as: 0 0 '\ \ k ² 0. [2.3] Figure 2.1. (a) Symmetric wells of infinite depth (with the origin taken at the center of the wells); and (b) asymmetric wells with the origin taken at the well’s extremity (when 0 < x < L, we have V(x) = 0 so that V(- x) = f for a 1D model) We shall now determine for different depth potential wells, with both symmetric and asymmetric forms, the corresponding solutions for the wave function (Ȍ0 ) and the energy 0 ( ). E To each wave function there is a corresponding electronic state (characterized by quantum numbers). It is important in physical electronics to understand the way in which these states determine how energy levels are filled. In solid-state physics, the state density function (also called the density of states) is particularly important. It can be calculated in the wave number (k) space or in the energies (E) space. In both cases, it is a function that is directly related to a dimension of space, so that it can be evaluated with respect to L = 1 (or V = L3 = 1 for a 3D system). In k space, the state density function [n(k)] is such that in one dimension n(k) dk represents the number of states placed in the interval dk (i.e. between k and k + dk in k wave number space). In 3D, n(k) d3 k represents the number of states placed within the elementary volume d3 k in k space. Similarly, in terms of energy space, the state density function [Z(E)] is such that Z(E) dE represents the number of states that can be placed in the interval dE (i.e. inclusively between E and E + dE in energy space). The upshot is that if F(E) is the occupation probability of a level denoted E, then the number N(E) dE of electrons distributed in the energy space between E and E + dE is equal to N(E) dE = Z(E) F(E) dE. V = V0 = 0 V(x) – L/2 0 L/2 V o f V o f V = V0 = 0 V(x) O L V o f V o f x x
The Free Electron and State Density Functions 19 nodes at extremities)symmetric wells(ID modey enabling the establishment of 2.2. Study of the stationary regime of small scale( 2.2.1. Preliminary remarks For a symmetric well, as shown in Figure 2. la, the Hamiltonian is such that H (x)=Hx), because v(x)=v(-x)and r If/ denotes the inversion operator, which changes x to-x, then IH(x)=H(x)=H(r) H(r) being invariant with respect to 1, the proper functions of l are also the proper functions of H(see Chapter 1 ). The form of the proper functions of I must be such that ly(x)=tv(x ). We can thus write: lv(x)=tv(x)=v(x),andon multiplying the left-hand side by 1, we now have 1[/v(x)=t/v(x)=ty(x) →t-=l,andt=±1 (-x)=v(x The result is that v(x)=ty(x)=±v(x) =y(-x) →(x)=±(-x) In these cases, the form of the solutions are either symmetric, as in u(x)=v(x), or asymmetric, as in v(x)=-v(x) 2.2.2. Form of stationary wave functions for thin symmetric wells with widthL equal to several inter-atomic distances(L =a), associated with fixed boundary conditions(FBC) 0
The Free Electron and State Density Functions 19 2.2. Study of the stationary regime of small scale (enabling the establishment of nodes at extremities) symmetric wells (1D model) 2.2.1. Preliminary remarks For a symmetric well, as shown in Figure 2.1a, the Hamiltonian is such that H (x) = H(– x), because V(x) = V(– x) and ² ² ² ² d d dx d x ª º ¬ ¼ . If I denotes the inversion operator, which changes x to – x, then IH(x) = H(– x) = H(x). H(x) being invariant with respect to I, the proper functions of I are also the proper functions of H (see Chapter 1). The form of the proper functions of I must be such that I \ \ ( ) ( ). x tx We can thus write: I \ \ \ () () ( ) x tx x , and on multiplying the left-hand side by I, we now have: > @ > @ 2 () () ²() 1, and 1 ( ) () I I x tI x t x t t . I x x \ \ \ r \ \ The result is that () () () ( ) ( ). I x t x x x x (– x) \ \ r\ \ r\ \ [2.4] In these cases, the form of the solutions are either symmetric, as in \ \ () ( ) x x , or asymmetric, as in \ \ ( ) ( ). x x 2.2.2. Form of stationary wave functions for thin symmetric wells with width (L) equal to several inter-atomic distances ( L a | ), associated with fixed boundary conditions (FBC) 0. 2 2 § · §· L L \ \ ¨ ¸ ¨¸ © ¹ ©¹ [2.5]
20 Solid-State Physics for Electronics This limiting condition is equivalent to the physical status of an electron that cannot leave the potential well due to it being infinitely high. The result is that between x =- and x=+- there is a zero probability of presence, hence the preceding FBC (2)-(2- The general stationary solution to equation [2.3]is r(x )=A cos kx+B sin kx The use of the boundary conditions of equation [2.5] means that =A coS k-+B sin k-=0 2 = A coS K These last two equations result in the two same conditions either A=0 and B sin==0, so that both y= B sin kx and = nT (n is whole), so that k= 2n -=N- where N is an even integer. The solution for solution y is thus yw= B sint, with N being even, or B=0 and A cosk==0, so that y=A coskx and KL= I+nT(n is an integer),so that k=-(2n+1]=NI where N is an odd integer. The solution for A cos-x hn being odd [2.7
20 Solid-State Physics for Electronics This limiting condition is equivalent to the physical status of an electron that cannot leave the potential well due to it being infinitely high. The result is that between 2 L x and 2 L x = there is a zero probability of presence, hence the preceding FBC: 0 2 2 § · §· L L \ \ ¨ ¸ ¨¸ © ¹ ©¹ . The general stationary solution to equation [2.3] is: 0 \ ( ) cos sin x A kx + B kx. The use of the boundary conditions of equation [2.5] means that: 0 cos + sin 0 2 22 L LL Ak Bk § · \ ¨ ¸ © ¹ or 0 cos sin 0. 2 22 L LL Ak Bk § · \ ¨ ¸ © ¹ These last two equations result in the two same conditions: – either A = 0 and 2 sin 0 L B k , so that both 0 \ B kx sin and 2 kL Sn (n is whole), so that 2 L L kn N S S where N is an even integer. The solution for solution 0 \ is thus 0 sin N L N B x S \ , with N being even; [2.6] – or B = 0 and cos 0 L 2 A k , so that 0 \ A cos kx and 2 2 kL n S S (n is an integer), so that > @ 2 1 L L knN S S where N is an odd integer. The solution for 0 \ is thus: 0 cos N L N A x S \ , with N being odd. [2.7]
The Free Electron and State Density Functions 21 The normalization condition [L/a l af ds =l gives A=8-E two solutions in equations [2.6] and [2.7] can be brought together in w==sin L x +-. where N= 1.2.3. 4etc [2.8] For both symmetric and asymmetric solutions, k is of the form k=kN =N 2.9 where N is an odd integer and the symmetric solution and is an even integer for the asymmetric solution. Thus, N takes on successive whole values i.e. 1, 2, 3, 4,etc The value N=0 is excluded as the corresponding function Vo = B sin kox =0 has no physical significance(zero probability of presence). The integer values N=-l 2,-3=-N yield the same physical result, for the same probabilities as lvA-P-LvewF=lP. Summing up we can say that the only values worth retaining are N =1, 2, 3, 4, etc y. This quantification is restricted to the quantum number N without involving spin As we already know, spin makes it possible to differentiate between two electrons with the same quantum number N. This is due to a projection of kinetic moment on he axis which brings into play a new quantum number, namely ms =t We thus find that each N state can be filled by two electrons, one with a spin m,=+1/2 and wave function vQ+), and the other with a spin m, =-1/2 and wave function 2.2.3. Study of energy From equation [2.2] we deduce that: E0=k. With k given by equation [2.91, we find that the energy is quantified and takes on values given by
The Free Electron and State Density Functions 21 The normalization condition 0 ( ) / 2 2 / 2 x 1 N L L \ dx ³ gives 2 L A B , and the two solutions in equations [2.6] and [2.7] can be brought together in: 0 2 sin 2 N N L x L L S § · \ ¨ ¸ © ¹, where N = 1, 2, 3, 4, etc. [2.8] For both symmetric and asymmetric solutions, k is of the form kk N N , L S [2.9] where N is an odd integer and the symmetric solution and is an even integer for the asymmetric solution. Thus, N takes on successive whole values i.e. 1, 2, 3, 4, etc. The value N = 0 is excluded as the corresponding function 0 0 \ B kx sin 0 has no physical significance (zero probability of presence). The integer values N' = – 1, – 2, – 3 (= – N) yield the same physical result, for the same probabilities as 00 0 2 22 . N' N N \\ \ Summing up, we can say that the only values worth retaining are N 1, 2, 3, 4, etc. This quantification is restricted to the quantum number N without involving spin. As we already know, spin makes it possible to differentiate between two electrons with the same quantum number N. This is due to a projection of kinetic moment on the z axis which brings into play a new quantum number, namely 1 . 2 ms r We thus find that each N state can be filled by two electrons, one with a spin 1/2 ms and wave function 0 , N\ and the other with a spin 1/2 ms and wave function 0 . N\ 2.2.3. Study of energy From equation [2.2] we deduce that: ² ² 2 0 . k m E = With k given by equation [2.9], we find that the energy is quantified and takes on values given by: