Lecture Notes on Quantum Mechanics-Part II Contents -Infinite square potential well 22, .Solution re potential well of finite depth 23455567 891010 ally:find A. IV.The Free Particle:Revisited 2 tion E>V 是 tia wellV cattering from square well -Vo <0< nRBinatiepoteigate s in 6-potential well (E VII.Summary on Part I 39
Lecture Notes on Quantum Mechanics - Part II Yunbo Zhang Instituter of Theoretical Physics, Shanxi University In this chapter we solve the stationary Schr¨odinger equation for various types of one dimensional potentials, including square potential well, harmonic oscillator, scattering of potential barrier, and δ-potential. Contents I. Square Potential Well 2 A. An idealized case - Infinite square potential well 2 1. Solution 2 2. Energy quantization 2 3. Wave function normalization 3 4. Shift of origin of coordinate 4 B. Square potential well of finite depth 5 1. Solution 5 2. Even parity case 5 3. Odd parity case 6 4. Graphical solution of the energy defining equation 7 II. Generic Problem of Quantum Mechanics 8 A. Properties of wave functions 9 B. Construction of the most general solution 10 C. An example 10 III. The Harmonic Oscillator 11 A. Algebraic method: ladder operator 12 1. Commutator 13 2. Ladder operators 14 3. Normalization algebraically: find An 16 B. Analytic method: differential equation 18 1. Discussion on the energy quantization 20 2. Discussion on the wave function 20 IV. The Free Particle: Revisited 24 V. Penetration of Potential Barrier 29 A. Case I: Barrier penetration E < V0 29 1. Boundary conditions 30 B. Case II: Barrier penetration E > V0 32 C. Case III: Bound states in finite potential well −V0 < E < 0 33 D. Case IV: Scattering from square well −V0 < 0 < E 33 VI. The δ-function potential 33 A. Bound states and scattering states 33 B. From square potential to δ-potential 35 C. Bound states in δ potential well (E < 0) 36 D. Scattering states in δ-potential well (E > 0) 38 E. δ-function barrier 39 VII. Summary on Part II 39
L.SQUARE POTENTIAL WELL We consider a particle moving in asquare potential well (Figure 1) va={& and in this case the stationary Schrodinger equation -会票e+veio=Be can be solved exactly. A.An idealized case-Infinite square potential well First we solve the idealized model of above potential,that is when Voa infinite square potential well 1.Solution For the potential v田={x 8。 it is easy to know -_平e=B)0<r<a 2m dr ()=0 for z<0,>a Let k2=2mE/h,we have 票o+=0 →(z)-Asi血(kx+ A and 6 are two constants to be determined shortly. According to the natural boundary conditions,(should be continuous at=0and=a.So (0)=(a)=0 the distinct solutions are with n=1,2.3· and 回=Am()
2 I. SQUARE POTENTIAL WELL We consider a particle moving in a square potential well (Figure 1) V (x) = { 0, 0 < x < a V0, x < 0, x > a and in this case the stationary Schr¨odinger equation − ~ 2 2m d 2 dx2 ψ(x) + V (x)ψ(x) = Eψ(x) can be solved exactly. The physics here is the one-dimensional material-wave propagation in a potential well. We will confine our discussions to cases with energy E less than V0. A. An idealized case - Infinite square potential well First we solve the idealized model of above potential, that is when V0 → +∞, a infinite square potential well 1. Solution For the potential V (x) = { 0, 0 < x < a +∞, x < 0, x > a it is easy to know − ~ 2 2m d 2 dx2 ψ(x) = Eψ(x) for 0 < x < a ψ(x) = 0 for x < 0, x > a Let k 2 = 2mE/~ 2 , we have d 2 dx2 ψ(x) + k 2ψ(x) = 0 ⇒ ψ(x) = A sin(kx + δ) A and δ are two constants to be determined shortly. 2. Energy quantization According to the natural boundary conditions, ψ(x) should be continuous at x = 0 and x = a. So ψ(0) = ψ(a) = 0 which tells us δ = 0, and ka = nπ, with n = 1, 2, 3 · · · (because A = 0 and n = 0 give trivial, non-normalizable solution ψ(x) = 0. Furthermore n = −1, −2, −3 · · · give nothing new since we can absorb the minus sign into A). So the distinct solutions are kn = nπ a , with n = 1, 2, 3 · · · and ψn(x) = A sin (nπ a x )
3 V() >x 0 a FIG.1:One dimensional square potential well ddte te urather the hpthe iceheelcammot have ny old energy- 3.Ware function normaliztion Well,how do we fix the constant A?Answer:We normalize: ar女-2gh=ar号= This ony determines the magnitude of A,but it is simplest to pick the positive realoo (the phase of A carries no physical significance anyway).Inside the well,then,the solutions are 4回)=V月n(g，n=1,23 the dexcited state We know the standard deviation
3 0 a x V(x) V0 FIG. 1: One dimensional square potential well. Curiously, the boundary condition at x = a does not determine the constant A, but rather the constant k, and hence the possible values of E: En = ~ 2k 2 n 2m = n 2 π 2~ 2 2ma2 In sharp contrast to the classical case, a quantum particle in the infinite square well cannot have just any old energy– only these special allowed values. The energy of system (particle) is quantized. Note: For infinitely high potential well, waves can’t penetrate into the forbidden regions. They are fully reflected and standing waves is established. (Figure 2) 3. Wave function normalization Well, how do we fix the constant A? Answer: We normalize ψ: ∫ ∞ −∞ |ψn(x)| 2 dx = ∫ a 0 |A| 2 sin2 ( nπ a x)dx = |A| 2 a 2 = 1 This only determines the magnitude of A, but it is simplest to pick the positive real root: A = √ 2 a (the phase of A carries no physical significance anyway). Inside the well, then, the solutions are ψn(x) = √ 2 a sin(nπ a x), n = 1, 2, 3 · · · As promised, the time-independent Schr¨odinger equation has delivered an infinite set of solutions, one for each integer n. The first few of these are plotted in Figure 2; they look just like the standing waves on a string of length a. ψ1, which carries the lowest energy, is called the ground state: E1 = π 2~ 2 2ma2 ̸= 0 Ψ1(x, t) = √ 2 a sin(πx a ) exp ( −i π 2~ 2ma2 t ) the others, whose energies increase in proportion to n 2 , are called excited states. Heisenberg uncertainly relation can be easily checked on the lowest energy state. We know the standard deviation (uncertainty) in position space is obviously ∆x ∼ a and it is also clear ∆p ∼ ~/a from the ground state energy E = p 2/2m ∼ ~ 2 2ma2 . So ∆x · ∆p ∼ ~
n=3 1=2 个V例 n=1 PIGFinstwave functionsand the correodingylevels of nteonedqe potential wellshifted 4.Shift of origin of coordinate We now shift the origin of coordinate so that the potential takes the form The wave function in the shifted coordinate system are .回=(g- -(层g以面a=15 Vasin(,forn=2,4,6… ··,wave functions are
4 n =3 n =2 n =1 x V(x) 0 a FIG. 2: First 3 wave functions and the corresponding energy levels of infinite one dimensional square potential well. -a/2 n =3 n =2 n =1 x V(x) 0 a/2 FIG. 3: First 3 wave functions and the corresponding energy levels of infinite one dimensional square potential well - shifted coordinate system. 4. Shift of origin of coordinate We now shift the origin of coordinate so that the potential takes the form V (x) = { 0, − a 2 < x < a 2 +∞, |x| > a 2 The wave function in the shifted coordinate system are ψn(x) = √ 2 a sin(nπ a x − nπ a a 2 ) = { √ 2 a cos( nπ a x), for n = 1, 3, 5 · · · √ 2 a sin( nπ a x), for n = 2, 4, 6 · · · which show a regular symmetry about the origin x = 0, called the parity. For n = 1, 3, 5 · · · , wave functions are symmetric (even functions) about x = 0, and are said possessing even parity. For n = 2, 4, 6 · · · , wave functions are
5 anti-symmetric (odd function)about r =0.they have odd parity.There is a minus sign in the wave functions for the B.Square potential well of finite depth Now we consider the situation that the depth of the barrier is finite.We use here the shifted coordinate system 1.Solution The potential is now va={8. -号<x<号 ()for a particle with energy E less than V (E<V)differs from zero for all values V(x) B -a/2 0 a/2 FIG.4:One potential well of finite depth Region A or C Region B @+6Au国=0多回+2二回 +k2=0 Requirements of general rules on determining the correct wave functions .Naturel boundary conditions:wave function must be limited,especially forr The potential V(r)is symmetric about origin,the wave functions must be of definite parity 2.Even parity case From the continuation of wave function and its derivative(Figure 5),we have Acs(g)=Bep(←) 5Asm()=-号Bem(←)
5 anti-symmetric (odd function) about x = 0, they have odd parity. There is a minus sign in the wave functions for the 1st and 2nd excited states compared with the unshifted system, which however, does not lead to big problem because it gives a trivial phase of π. B. Square potential well of finite depth Now we consider the situation that the depth of the barrier is finite. We use here the shifted coordinate system. 1. Solution The potential is now V (x) = { 0, − a 2 < x < a 2 V0, |x| > a 2 ψ(x) for a particle with energy E less than V0 (E < V0) differs from zero for all x values. -a/2 x V(x) 0 a/2 E V0 V0 A B C FIG. 4: One dimensional square potential well of finite depth. Region A or C Region B − ~ 2 2m d 2 dx2 ψ(x) + V0ψ(x) = Eψ(x) − ~ 2 2m d 2 dx2 ψ(x) + 0 · ψ(x) = Eψ(x) d 2 dx2 ψ(x) − 2m ~2 (V0 − E)ψ(x) = 0 d 2 dx2 ψ(x) + 2mE ~2 ψ(x) = 0 2m ~2 (V0 − E) = β 2 2mE ~2 = k 2 d 2 dx2 ψ(x) − β 2ψ(x) = 0 d 2 dx2 ψ(x) + k 2ψ(x) = 0 ψ(x): exp(βx) or exp(−βx) cos(kx) or sin(kx) Requirements of general rules on determining the correct wave functions • Naturel boundary conditions: wave function must be limited, especially for x → ±∞. • The potential V (x) is symmetric about origin, the wave functions must be of definite parity. 2. Even parity case From the continuation of wave function and its derivative (Figure 5), we have A cos ( ka 2 ) = B exp ( −β a 2 ) − k 2 A sin ( ka 2 ) = − β 2 B exp ( −β a 2 )