FIG.5:Even parity sotio a/2 FIG.6:Odd parity solution. which gives m(份）-号 ehn ct型y多aytn -m(-)-面 Since values of m,,and Vo are given,it is a transcendental equation of E.Ouly Evalues that satisfy the above equation can be the energy value of that particle. 3.Odd parity case The natural boundary conditions lead to the following equations cm(）)=De即（←） 5co(受）=-号pem(-) which gives mu(份)-号 (3) This is another limitation on wave functions set by natural boundary conditions which is equivalent to the transcen- dental equation m(-到-v-可 (4) It is by energy odd parity states
6 −a/2 a/2 Wave function must be continuous here Derivative of wave function must be continuous here Be β x A cos kx Be−βx x FIG. 5: Even parity solution. −a/2 a/2 Wave function must be continuous here Derivative of wave function must be continuous here β x C sin kx −βx x −De De FIG. 6: Odd parity solution. which gives k 2 tan ( ka 2 ) = β 2 (1) This is the limitation on wave function set by natural boundary conditions. It is actually a transcendental equation of the form 1 2 √ 2mE ~ 2 tan (√ 2mE ~ 2 a 2 ) = 1 2 √ 2m(V0 − E) ~ 2 (2) Since values of m, a, and V0 are given, it is a transcendental equation of E. Only E values that satisfy the above equation can be the energy value of that particle. 3. Odd parity case The natural boundary conditions lead to the following equations C sin ( ka 2 ) = D exp ( −β a 2 ) k 2 C cos ( ka 2 ) = − β 2 D exp ( −β a 2 ) which gives k 2 cot ( ka 2 ) = − β 2 (3) This is another limitation on wave functions set by natural boundary conditions which is equivalent to the transcendental equation 1 2 √ 2mE ~ 2 cot (√ 2mE ~ 2 a 2 ) = − 1 2 √ 2m(V0 − E) ~ 2 (4) It is the equation submitted by energy values E of odd parity states
4.Gruphical solution of the energy defining equation The energy value E must satisfy one of the two equations 左m()-号 ou(份）)=-号 Multiplied by a,they become ()m(侵)=婴 ()(受）)=-() Let -如- --mg-可。 with and nlarger than 0.We notice that and n are not independent each other,rather,they are linked by 2+=那6a2 Thus the energy defining relations can be decomposed into two set simultaneous equations ξtanξ=n 2+=票2 and -Ecot=n +f=6e2 1.For a given potential well (Voa2),numbers of intersection points (bound states)are limited meof(6a2)产：(0，)（侵，x)(m,要）（竖，2x)… number of intersection points:1 lowest state with even parity alwaysexist ≥行e6a2≥ lowest state with odd parity can exist 0d2ar≥2a 1st excited state with even parity will appear 4 Vm)=-6( It is obviously not possible for Voa to satisfy the condition for lowest state with odd parity. Problem 1 Zen Jinyan,Tertbook,page 80-84,Problem 1,.41
7 4. Graphical solution of the energy defining equation The energy value E must satisfy one of the two equations k 2 tan ( ka 2 ) = β 2 k 2 cot ( ka 2 ) = − β 2 Multiplied by a, they become ( ka 2 ) tan ( ka 2 ) = βa 2 ( ka 2 ) cot ( ka 2 ) = − ( βa 2 ) Let ξ = 1 2 ka = 1 2 √ 2mE ~ 2 a η = 1 2 βa = 1 2 √ 2m (V0 − E) ~ 2 a with ξ and η larger than 0. We notice that ξ and η are not independent each other, rather, they are linked by ξ 2 + η 2 = m 2~ 2 V0a 2 Thus the energy defining relations can be decomposed into two set simultaneous equations ξ tan ξ = η ξ 2 + η 2 = m 2~ 2 V0a 2 and −ξ cot ξ = η ξ 2 + η 2 = m 2~ 2 V0a 2 For a given set of values m, V0, a, the energy E can be evaluated by finding out the intersection points of above two sets of simultaneous equations. Group of ξ 2 + η 2 = m 2~2 V0a 2 with different ( V0a 2 ) values are shown in Figure 7. Our main results are: 1. For a given potential well ( V0a 2 ) , numbers of intersection points (bound states) are limited value of ( m 2~2 V0a 2 ) 1 2 : (0, π 2 ) ( π 2 , π) (π, 3π 2 ) ( 3π 2 , 2π ) · · · number of intersection points: 1 2 3 4 · · · 2. Bound states have definite parities. As the energy increases, even parity states and odd parity states appear alternatively with the leading (lowest energy) one to be even parity. lowest state with even parity always exist m 2~ 2 V0a 2 ≥ π 2 4 , i.e.V0a 2 ≥ π 2~ 2 2m , lowest state with odd parity can exist m 2~ 2 V0a 2 ≥ π 2 , i.e.V0a 2 ≥ 2π 2~ 2 m , 1st excited state with even parity will appear 3. No odd parity bound state for δ potential well. In the symmetric finite potential well, let the width a → 0, the depth V0 → +∞ and keep V0a = γ =constant The potential becomes ˙ δ potential V (x) = −γδ (x) It is obviously not possible for V0a 2 to satisfy the condition for lowest state with odd parity. Problem 1 Zen Jinyan, Textbook, page 80-84, Problem 1,3,4,12
1x/22 3π e43新/2 (a) 1r/22 3e43/2 (b FIG.7:Group with differnt (Voa)values IL.GENERIC PROBLEM OF QUANTUM MECHANICS V(time-independent potential)for any sbent time (r,0)starting wave function To do this you must solve h地=-+v回业 The strategy isfirst to solve the stationary(time-independent)Schrodinger equation this yields,in general,an infinite set of solutions (r）2()()… To fit(0)you write down the general linear combination of these solutions z,0)=∑回 =1
8 FIG. 7: Group of ξ 2 + η 2 = m 2~2 V0a 2 with differnt ( V0a 2 ) values II. GENERIC PROBLEM OF QUANTUM MECHANICS A lot has happened in the last several sections, so let us recapitulate, from a somewhat different perspective. Here is the generic problem of Quantum Mechanics. V (x) time-independent potential Ψ (x, 0) starting wave function } ⇒ Ψ (x, t) for any subsequent time t To do this you must solve i~ ∂Ψ ∂t = − ~ 2 2m ∂ 2Ψ ∂x2 + V (x) Ψ Ψ (x, t) = ψ (x) f (t) = ψ (x) exp (−iEt/~) The strategy is first to solve the stationary (time-independent) Schr¨odinger equation − ~ 2 2m d 2ψ dx2 + V ψ = Eψ this yields, in general, an infinite set of solutions ψ1(x) ψ2(x) ψ3(x) · · · E1 E2 E3 · · · To fit Ψ (x, 0) you write down the general linear combination of these solutions Ψ (x, 0) = ∑∞ n=1 cnψn (x)
9 the miracle s that you can match the special initial state by appropriate choice of the To construct (r,t)you simply tack to each term its characteristic time dependence exp(-iEt/h) k到-6国m--a. (⑤) The separable solutions themselves 重n(，)=n()ep(-iEnt/h are station n tha states,and the exponentials do not cancel,when you calculate(r,t). A.Properties of wave functions In the case of the infinite square well,the solutions 回=层(）=128… have some interesting and important propertics: 1.They are alternately even and odd (for symmetric V(r)) 2.has (n-1)nodes (zero-crossing) (universal) 3.Orthogonality (quite general) a()(a)d在=0m≠n Normalization (e)中m(e)dz=1 In fact we can combine the orthogonality and normalization into asingle statement We say that the's are orthonormal 4.Completeness (holds for most potentials) Any function f()can be expressed as a linear combination of (6) We o th Btion()othing ba the ooran he fnct fmction ):theof Etion何好安以 =(国)f)d
9 the miracle is that you can always match the special initial state by appropriate choice of the constants c1, c2, c3,· · · . To construct Ψ (x, t) you simply tack to each term its characteristic time dependence exp (−iEnt/~) Ψ (x, t) = ∑∞ n=1 cnψn (x) exp (−iEnt/~) = ∑∞ n=1 cnΨn (x, t) (5) The separable solutions themselves Ψn (x, t) = ψn (x) exp (−iEnt/~) are stationary states, in the sense that all probabilities and expectation values are independent of time. This property, however, is emphatically NOT shared by the general solution (5); the energies are different, for different stationary states, and the exponentials do not cancel, when you calculate |Ψ (x, t)| 2 . A. Properties of wave functions In the case of the infinite square well, the solutions ψn (x) = √ 2 a sin (nπx a ) , n = 1, 2, 3 · · · have some interesting and important properties: 1. They are alternately even and odd (for symmetric V (x)) 2. ψn has (n − 1) nodes (zero-crossing) (universal) 3. Orthogonality (quite general) ∫ ψ ∗ m (x) ψn (x) dx = 0 m ̸= n Normalization ∫ ψ ∗ m (x) ψm (x) dx = 1 In fact we can combine the orthogonality and normalization into a single statement ∫ ψ ∗ m (x) ψn (x) dx = δmn = { 0 if m ̸= n 1 if m = n We say that the ψ’s are orthonormal. 4. Completeness (holds for most potentials) Any function f(x) can be expressed as a linear combination of ψn f(x) = ∑∞ n=1 cnψn (x) = √ 2 a ∑∞ n=1 cn sin(nπ a x) (6) We recognize that Equation (6) is nothing but the Fourier series for f(x), and the fact that ”any” function can be expanded in this way is sometimes called Dirichlet’s theorem. The expansion coefficients cn can be evaluated–for a given f(x)–by a method I call Fourier’s trick, which beautifully exploits the orthonormality of {ψn (x)}: Multiply both sides of Equation (6) by ψ ∗ n (x), and integrate cn = ∫ ψ ∗ n (x) f(x)dx
10 FIG.8:The starting wave function in the example. B.Construction of the most general solution The stationary states of infinite square well are evidently .e,=V月血()e-awm We claimed that the most general solution to the time dependent Schrodinger equation is a linear combination of stationary states e=立-层m()e-wam D)yprebtial wave function)can be xpanded n(by appropriate 红0=言a回 completeness a=V层厂m(晋k,0t orthonormality Here isa procedure: 重(z,o)→G→(,)→dynamical quantities C.An example A particle in the infinite square well has the initial wave function 亚(c,0)=Ar(a-,(0≤x≤a ouem=nd。 tion:First we determine A by normalization 1=人eoP在=42a-Pk=4蜀
10 Aa2 /4 a x (x,0) FIG. 8: The starting wave function in the example. B. Construction of the most general solution The stationary states of infinite square well are evidently Ψn (x, t) = √ 2 a sin (nπ a x ) e −i(n 2π 2~/2ma2 )t We claimed that the most general solution to the time dependent Schr¨odinger equation is a linear combination of stationary states Ψ (x, t) = ∑∞ n=1 cn √ 2 a sin (nπ a x ) e −i(n 2π 2~/2ma2 )t (Don’t believe? Check it !) Any prescribed initial wave function Ψ (x, 0) can be expanded on {ψn (x)} by appropriate choice of the coefficients cn Ψ (x, 0) = ∑∞ n=1 cnψn (x) completeness cn = √ 2 a ∫ a 0 sin(nπ a x)Ψ (x, 0) dx orthonormality Here is a procedure: Ψ (x, 0) → cn → Ψ (x, t) → dynamical quantities C. An example A particle in the infinite square well has the initial wave function Ψ (x, 0) = Ax(a − x), (0 ≤ x ≤ a) outside the well Ψ = 0. Find Ψ (x, t). Solution: First we determine A by normalization 1 = ∫ a 0 |Ψ (x, 0)| 2 dx = |A| 2 ∫ a 0 x 2 (a − x) 2 dx = |A| 2 a 5 30 ∴ A = √ 30 a 5