Chapter 2 The Particle in a Box The stationary-state wave functions and energy levels of a one-particle,one-dimensional system are found by solving the time-independent Schrodinger equation(1.19).In this chapter,we solve the time-independent Schrodinger equation for a very simple system, a particle in a one-dimensional box(Section 2.2).Because the Schrodinger equation is a differential equation.we first discuss differential equations. 2.1 Differential Equations This section considers only ordinary differential equations,which are those with only one independent variable.[A partial differential equation has more than one independent variable.An example is the time-dependent Schrodinger equation(1.16).in whicht andx are the independent variables.]An ordinary differential equation is a relation involving an independent variable x.a dependent variable y(x).and the first,second,.nth deriva- tives of y(y'.y".y).An example is y"+2x(y')2+y2 sinx 3e 2.10 The order of a differential equation is the order of the highest derivative in the equation. Thus.(21)is of third order. A special kind of differential equation is the linear differential equation,which has the form Ancy+An-1ya-)+.+Acy'+Aoy=g) 2.2 where the A's and g(some of which may be zero)are functions ofx only.In the nth-order linear differential equation (2.2).y and its derivatives appear to the first power.A differ 含C0 that canno puto22 s nonlnear.g二0n22.e Schreo(near momogeneecn-oner diferen 10m eous.Ihe one g by the c efficient of y",we can put every linear homogeneous second- ial equation into the form y"+P(x)y'+e(x)y=0 2.3) Suppose y and y2 re two independent functions,each of which satisfies (23).By inde ple of y.Then the general olution of the 2.4 21
21 Chapter 2 The Particle in a Box The stationary-state wave functions and energy levels of a one-particle, one-dimensional system are found by solving the time-independent Schrödinger equation (1.19). In this chapter, we solve the time-independent Schrödinger equation for a very simple system, a particle in a one-dimensional box (Section 2.2). Because the Schrödinger equation is a differential equation, we first discuss differential equations. 2.1 Differential Equations This section considers only ordinary differential equations, which are those with only one independent variable. [A partial differential equation has more than one independent variable. An example is the time-dependent Schrödinger equation (1.16), in which t and x are the independent variables.] An ordinary differential equation is a relation involving an independent variable x, a dependent variable y1x2, and the first, second, c, nth derivatives of y (y , y , c, y(n) ). An example is y + 2x1y 22 + y 2 sin x = 3ex (2.1) The order of a differential equation is the order of the highest derivative in the equation. Thus, (2.1) is of third order. A special kind of differential equation is the linear differential equation, which has the form An1x2y1n2 + An-11x2y1n-12 + g+ A11x2y + A01x2y = g1x2 (2.2) where the A’s and g (some of which may be zero) are functions of x only. In the nth-order linear differential equation (2.2), y and its derivatives appear to the first power. A differential equation that cannot be put in the form (2.2) is nonlinear. If g1x2 = 0 in (2.2), the linear differential equation is homogeneous; otherwise it is inhomogeneous. The onedimensional Schrödinger equation (1.19) is a linear homogeneous second-order differential equation. By dividing by the coefficient of y , we can put every linear homogeneous secondorder differential equation into the form y + P1x2y + Q1x2y = 0 (2.3) Suppose y1 and y2 are two independent functions, each of which satisfies (2.3). By independent, we mean that y2 is not simply a multiple of y1. Then the general solution of the linear homogeneous differential equation (2.3) is y = c1y1 + c2y2 (2.4)
22 Chapter2The Particle in a Box where c and c2 are arbitrary constants.This is readily verified by substituting (2.4)into the left side of (2.3): cLyi+c2y?+P(x)ciyi P(x)czy2 Q(x)cLy Q(x)c2y2 cilyi+P(x)yi +e(x)y]+cly+P(x)y:+Q(x)y2] =c10+920=0 2.5) lution of a equation c fnthorder usually hasn arbitrary se const we may have b undary conditi are condi int or po t he z roat the An important special case is a linear homogeneous second-order differential equation with constant coefficients: y"+py'+gy=0 (2.6 where p and g are constants.To solve (2.6).let us tentatively assume a solution of the form y =e".We are looking for a function whose derivatives when multiplied by constants will cancel the original function.The exponential function repeats itself when differenti- ated and is thus the correct choice.Substitution in (2.6)gives se+pse+ge=0 2+p四+9=0 (2.7) soluti 1f2.6is e not equal,give two 2.8 For example,for y"+6y'-7y =0.the auxiliary equation is s2 6s-7 =0.The quadratic formula gives s=1.2=-7,so the general solution is ce+cze7 2.2 Particle in a One-Dimensional Box a po is infinite e al the nt of l l.wh ential ene is 2 tem nreal hut this model car origin at the left end of the line segment (Fig.2.1). FIGURE 2.1 Potential to energy function V(x) for the particle in a v()t one-dimensional box =0 x=I
22 Chapter 2 | The Particle in a Box where c1 and c2 are arbitrary constants. This is readily verified by substituting (2.4) into the left side of (2.3): c1y 1 + c2y 2 + P1x2c1y 1 + P1x2c2y 2 + Q1x2c1y1 + Q1x2c2y2 = c13y 1 + P1x2y 1 + Q1x2y14 + c23y 2 + P1x2y 2 + Q1x2y24 = c1 # 0 + c2 # 0 = 0 (2.5) where the fact that y1 and y2 satisfy (2.3) has been used. The general solution of a differential equation of nth order usually has n arbitrary constants. To fix these constants, we may have boundary conditions, which are conditions that specify the value of y or various of its derivatives at a point or points. For example, if y is the displacement of a vibrating string held fixed at two points, we know y must be zero at these points. An important special case is a linear homogeneous second-order differential equation with constant coefficients: y + py + qy = 0 (2.6) where p and q are constants. To solve (2.6), let us tentatively assume a solution of the form y = esx . We are looking for a function whose derivatives when multiplied by constants will cancel the original function. The exponential function repeats itself when differentiated and is thus the correct choice. Substitution in (2.6) gives s2esx + psesx + qesx = 0 s2 + ps + q = 0 (2.7) Equation (2.7) is called the auxiliary equation. It is a quadratic equation with two roots s1 and s2 that, provided s1 and s2 are not equal, give two independent solutions to (2.6). Thus, the general solution of (2.6) is y = c1es1x + c2es2x (2.8) For example, for y + 6y - 7y = 0, the auxiliary equation is s2 + 6s - 7 = 0. The quadratic formula gives s1 = 1, s2 = -7, so the general solution is c1ex + c2e-7x . 2.2 Particle in a One-Dimensional Box This section solves the time-independent Schrödinger equation for a particle in a onedimensional box. By this we mean a particle subjected to a potential-energy function that is infinite everywhere along the x axis except for a line segment of length l, where the potential energy is zero. Such a system may seem physically unreal, but this model can be applied with some success to certain conjugated molecules; see Prob. 2.17. We put the origin at the left end of the line segment (Fig. 2.1). I x 5 0 x 5 l II III to ` to ` x V (x) Figure 2.1 Potential energy function V(x) for the particle in a one-dimensional box
2.2 Particle inaOne-Dimensional Box23 We have three regions to consider.In regions I and III.the potential energy Vequals infinity and the time-independent Schrodinger equation(1.19)is 尝- Neglecting Ein comparison with we have 女 and we conclude thatis zero outside the box 1=0,m=0 2.9 For region II.between zero and I,the potential energy Vis zero.and the Schrodinger equation (1.19)becomes +=0 2.10) dr2 equation(2.7)gives s2+2mEh2=0 s=±(-2mE)/Ph 2.11 s=±i(2mE)p/h 2.12 wherei=V-1.Using (2.8).we have =(2mE)+Cze-i(2mE)x 213) Temporarily.let 0=(2mE)1/2x/h 如=c1e0+c2e0 We have e=cos0+isin 0 [Eq.(1.28)]and ei=cos(-0)+isin(-0)=cos0- isin 0,since cos(-0)=cos0 and sin(-0)=-sin0 2.14 Therefore, =(c +c2)cos+(ic1 ic2)sin0 where A and Bare new arbitrary constants.Hence. n A cos[h(2mE)x]+B sin[h(2mE)Px] 2.15) conditions It seen tulate that the wave onable to pos
2.2 Particle in a One-Dimensional Box | 23 We have three regions to consider. In regions I and III, the potential energy V equals infinity and the time-independent Schrödinger equation (1.19) is - U2 2m d 2c dx 2 = 1E - 2c Neglecting E in comparison with , we have d 2c dx 2 = c, c = 1 d 2c dx 2 and we conclude that c is zero outside the box: cI = 0, cIII = 0 (2.9) For region II, x between zero and l, the potential energy V is zero, and the Schrödinger equation (1.19) becomes d 2 cII dx 2 + 2m U2 EcII = 0 (2.10) where m is the mass of the particle and E is its energy. We recognize (2.10) as a linear homogeneous second-order differential equation with constant coefficients. The auxiliary equation (2.7) gives s2 + 2mEU-2 = 0 s = { 1-2mE21>2U-1 (2.11) s = {i12mE21>2>U (2.12) where i = 2-1. Using (2.8), we have cII = c1ei12mE21>2 x>U + c2e-i12mE21>2 x>U (2.13) Temporarily, let u K 12mE21>2x>U cII = c1eiu + c2e-iu We have eiu = cos u + i sin u [Eq. (1.28)] and e-iu = cos1-u2 + i sin1-u2 = cos u - i sin u, since cos1-u2 = cos u and sin1-u2 = -sin u (2.14) Therefore, cII = c1 cos u + ic1 sin u + c2 cos u - ic2 sin u = 1c1 + c22cos u + 1ic1 - ic22sin u = A cos u + B sin u where A and B are new arbitrary constants. Hence, cII = A cos3U-112mE21>2x4 + B sin3U-112mE21>2x4 (2.15) Now we find A and B by applying boundary conditions. It seems reasonable to postulate that the wave function will be continuous; that is, it will make no sudden jumps in
24 Chapter 2 The Particle in a Box value (see Fig.3.4).If is to be continuous at the pointx=0,then and u must ap- proach the same value atx=0: li细吻=im如 0=limA cos(2mE)]+Bsin[(2mE)]) 0=A since sin0=0 and cos0=1 (2.16 With A =0.Eq.(2.15)becomes u =Bsin[(2/h)(2mE)x] 2.17) Applying the continuity condition,we get Bsim[(2m/h)(2mE)21=0 (2.18) o because this would make the wave function zero everywhere we would an empty box.Therefore sin[(2π/h)(2mE)'P=0 The zeros of the sine function occur at0(,±m,±2r,±3m,.=±nr.Hence, (2π/h)(2mE)1/21-±nm 2.19 The value n=0 is a sp ecial case From (2.19)n=0 co onds to E=0.Fo E=0,the ro oots 12)of the auxilia uation are equal and (2.13)is not the complete solution of the Schrodinger equation.To find the complete solution.wereun to10 which for E=0 reads d2uu/dx2=0.Integration gives divu/dx c and u =cx +d, where c and d are constants.The boundary condition that=0 atx =0 gives d=0. and the condition that =0 at x=/then gives c =0.Thus,u =0 for E=0.and therefore E=0 is not an allowed energy value.Hence,n =0 is not allowed. 一n=4 Solving (2.19)for E,we have 2h2 E n=1,23. (2.20 -月=3 Only the energy values (2.20)allowto satisfy the boundary condition of continuity atx= I.Application of a boundary condition has forced us to the conclusion that the val ues of the energy are quantized(Fig.2.2).This is in striking contrast to the classical result that the particle in the box can have any nonnegative energy.Note that there is a minimum 0上=1 e,gre tne energy ol e part st energy Is he state.St n the ground-state e clas the article cal particle sits motion inside the box with zero kinetic energy and ero po n a one-dimensio nergy. EXAMPLE A particle of mass 2.00 x 10-26g is in a one-dimensional box of length 4.00 nm.Find the frequency and wavelength of the photon emitted when this particle goes from the n 3 to the n 2 level
24 Chapter 2 | The Particle in a Box value (see Fig. 3.4). If c is to be continuous at the point x = 0, then cI and cII must approach the same value at x = 0: lim xS0 cI = lim xS0 cII 0 = lim xS0 5A cos3U-112mE21/2x4 + B sin3U-112mE21/2x4 6 0 = A since sin 0 = 0 and cos 0 = 1 (2.16) With A = 0, Eq. (2.15) becomes cII = B sin312p>h212mE21>2x4 (2.17) Applying the continuity condition at x = l, we get B sin312p>h212mE21>2l4 = 0 (2.18) B cannot be zero because this would make the wave function zero everywhere—we would have an empty box. Therefore, sin312p>h212mE21>2l4 = 0 The zeros of the sine function occur at 0, {p, {2p, {3p, c = {np. Hence, 12p>h212mE21>2l = {np (2.19) The value n = 0 is a special case. From (2.19), n = 0 corresponds to E = 0. For E = 0, the roots (2.12) of the auxiliary equation are equal and (2.13) is not the complete solution of the Schrödinger equation. To find the complete solution, we return to (2.10), which for E = 0 reads d 2cII>dx 2 = 0. Integration gives dcII>dx = c and cII = cx + d, where c and d are constants. The boundary condition that cII = 0 at x = 0 gives d = 0, and the condition that cII = 0 at x = l then gives c = 0. Thus, cII = 0 for E = 0, and therefore E = 0 is not an allowed energy value. Hence, n = 0 is not allowed. Solving (2.19) for E, we have E = n2h2 8ml 2 n = 1, 2, 3, c (2.20) Only the energy values (2.20) allow c to satisfy the boundary condition of continuity at x = l. Application of a boundary condition has forced us to the conclusion that the values of the energy are quantized (Fig. 2.2). This is in striking contrast to the classical result that the particle in the box can have any nonnegative energy. Note that there is a minimum value, greater than zero, for the energy of the particle. The state of lowest energy is called the ground state. States with energies higher than the ground-state energy are excited states. (In classical mechanics, the lowest possible energy of a particle in a box is zero. The classical particle sits motionless inside the box with zero kinetic energy and zero potential energy.) Ex a m p l e A particle of mass 2.00 * 10-26 g is in a one-dimensional box of length 4.00 nm. Find the frequency and wavelength of the photon emitted when this particle goes from the n = 3 to the n = 2 level. E n 5 4 n 5 3 n 5 2 n 5 1 0 Figure 2.2 Lowest four energy levels for the particle in a one-dimensional box
2.2 Particle inaOne-Dimensional Box25 By conservation of energy,the energy of the emitted photon equals the energy difference between the two stationary states (1.4);sce also Section =-h (32-2)(6.626×104Js) 8ml2 8(2.00×10-29kg)(4.00×10°m)月 =1.29×102s where u and stand for upper and lower.Use of r=cgives A 2.32 x 10m.(A common student error is to set equal to the energy of one of the states instead of the energy difference between states.) EXERCISE For an electron in a certain one-dimensional box,the longest-wavelength transition occurs at 400 nm.Find the length of the box.(Answer:0.603 nm.) Substitution of (2.19)into (2.17)gives for the wave function =B(） n=1,2,3 2.21) in(-) uld another indep simply get a constant,- constant B in Eq.(2.21)is still arbitrary.To fix its value,we use the normaliza- tion require nent.Eqs.(1.24)and (1.22): P==1 r(受k=1=1时 2.22) where the integral was evaluated by using Eq.(A.2)in the Appendix.We have B=(2/1)2 0soa0-20 not be a real numbe and co e 0 to 2 tion 17ch0 for the particle in n=1,2.3. (2.23 Graphs of the wave functions and the probability densities are shown in Figs.2.3 and 2.4. energies(20)and the wave functions(2)is calledaquant h腿ot twe and a different state
2.2 Particle in a One-Dimensional Box | 25 By conservation of energy, the energy hn of the emitted photon equals the energy difference between the two stationary states [Eq. (1.4); see also Section 9.9]: hn = Eupper - Elower = n2 uh2 8ml 2 - n2 l h2 8ml 2 n = 1n2 u - n2 l 2h 8ml 2 = 132 - 22216.626 * 10-34 J s2 812.00 * 10-29 kg214.00 * 10-9 m22 = 1.29 * 1012 s -1 where u and l stand for upper and lower. Use of ln = c gives l = 2.32 * 10-4 m. (A common student error is to set hn equal to the energy of one of the states instead of the energy difference between states.) Exercise For an electron in a certain one-dimensional box, the longest-wavelength transition occurs at 400 nm. Find the length of the box. (Answer: 0.603 nm.) Substitution of (2.19) into (2.17) gives for the wave function cII = B sina npx l b, n = 1, 2, 3, c (2.21) The use of the negative sign in front of np does not give us another independent solution. Since sin1-u2 = -sin u, we would simply get a constant, -1, times the solution with the plus sign. The constant B in Eq. (2.21) is still arbitrary. To fix its value, we use the normalization requirement, Eqs. (1.24) and (1.22): L - 0 0 2 dx = L - 0 c0 2 dx = 1 L 0 - 0 cI 0 2 dx + L l 0 0 cII 0 2 dx + L l 0 cIII 0 2 dx = 1 0 B0 2 L l 0 sin2 a npx l bdx = 1 = 0 B0 2 l 2 (2.22) where the integral was evaluated by using Eq. (A.2) in the Appendix. We have 0 B0 = 12>l21>2 Note that only the absolute value of B has been found. B could be - 12>l21>2 as well as 12>l21>2 . Moreover, B need not be a real number. We could use any complex number with absolute value 12>l21>2 . All we can say is that B = 12>l21>2eia, where a is the phase of B and could be any value in the range 0 to 2p (Section 1.7). Choosing the phase to be zero, we write as the stationary-state wave functions for the particle in a box cII = a 2 l b 1>2 sina npx l b, n = 1, 2, 3, c (2.23) Graphs of the wave functions and the probability densities are shown in Figs. 2.3 and 2.4. The number n in the energies (2.20) and the wave functions (2.23) is called a quantum number. Each different value of the quantum number n gives a different wave function and a different state