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QUANTUM MECHANICS (PHYSICS 137A) 137A HWO03 SOLUTIONS - HW3 SOLN, SP_ “w=” 2025 EXAM REVIEW Problem Set 3 Solutions Problem 1: Infinite Square Well from Scratch a) We substitute the wavefunction W(x, t) = f(t)y(x) into the time-dependent Schrodinger equation a koe “eee = “3 gee OVO) + V (xIF (t)W(x) (1) k= h(x) = ~-—— f(t) + V (x)F(t)b(x) (2) dt 2m dx? We divide by f(t) W(x) to find Lé eo. &y ik — =- f(t) dt 2m W(x) dx? The LHS is only a function of t and the RHS is only a function of x, so both sides must be equal to a constant, which we label E. V (x) (3) b) The equation for f(t) and its solution are aki Fee a) f(t) dt df o£ ae Kl) (5) => fit) = foe” (6) c) The equation for yy(x) is kK 1 dy amb de * V(x)=E (7) RK fy Sm axe 1 MUX) = Eb) (8) d) The potential V(x) > co atx = 0, L, so the boundary conditions (BCs) are (x) =O at x = 0,1. The first BC implies 0=(0)=A+B (9) B=-A (10) Applying this result, the wavefunction becomes w(x) = A(e™ = e™”), The second BC implies 0 = Wil) = A(e™ —e-™) (11) er 4 (12) => 2kL=2nn, where n=1,2,3,... (13) k= mn where n = 1, 2,3,... (14) i Therefore, the energy eigenstates are W(x) =A(e”™ — e”™), Therefore, the energy eigenstates are Acos "* , n=MNag = 1,3,5,..- Dlx) = in (30) Bsin "® , n= neen=2,46,... b) We normalize the even and odd solutions separately. The normalization for the even solutions is Jo “2 1 Suz 2. 1= [WaQ)P dx =|Al? cos’ PE gx =~ yay? a+cos ™ ax (31) -co -172 L 2 -/2 L 1 L v2 1 = =p x+ sin 207% == |APe (32) 2 2nn L yr 2 2 IA] = (33) rl =3A= 7e% (34) ya where a € R is an arbitrary phase. We choose A to be real, so we take A= 2. Therefore, the even L q energy eigenstates are ,(x) = + cos on , where n = noug = 1, 3,5,... q Similarly, the normalization for the odd solutions isB= 2. Therefore, the odd energy eigenstates are “ brlx) = Asin ™, where n= Neven = 24,6... F Fi enter = - c) The energy eigenvalues for the even solutions are £,, = 3? = Smre- where n= Nou = 1,3, 5,....... The RR nytk? zs eigenvalues for the odd solutions areE,, = yt = uke where 1 = Neven = 2, 4, 6....... Combining these, we see that the energy eigenvalues are £,, = ene where n = 1, 2,3,.... These are the same energy eigenvalues as the unshifted infinite square well. d) We are interested in the potential V(x) given by , Vo , ~b a= 3 (41) The + sign in the expression for a is irrelevant since W(x) is parameterized by a’, so we take a= be The normalization condition yields Je fr, 1= |W(x, 0)f dx = C2 (x? = a*)Pdx (42) The integrand is even and the integration limits are about an even interval, so we can rewrite this as Je 5 ys ee 4s 223.4 1=2C ° (x —a’)*dx =2C ° (x" - 2a°x’ +a")dx=2C Ps gre +ax ° (43) . 3 s 8 2 6 ?_ 29 £2 ig £ =2e8 & (44) 5 2 3 2 2 5 2 30 Crs (45) The + sign in the expression for C isi rrejevant since it simply gives W(x, 0) an overall phase which has no physical significance,sowetakeC= #8. b) The probability distribution is P(x, 0) =|W(x, 0)? = C(x? — a°)?. To simplify the plot, we define the dimensionless position, normalization constant, and probability distribution, denoted by a tilde (46) (47) P (%, 0)=aP *0 (48) The boundaries of the well are then located at X= +1. The dimensionless expression for the probability distribution is P (%, 0) = 2(%* -1)°. Figure 1 shows a plot of the probability distribution. Pix) +N \ “5 c os 10 Figure 1: The dimensionless probability distribution P (X, 0) = BUR? -1). c) The probability distribution at t = 0 is symmetric about x= 0, so the probability is P The c, for the odd solutions are r_f ip Ch =C 2 sin 7X Lup =0 (62) (x? — a’)dx, where n= Neven = 2,4,6,--- (61) because the integrand is odd and the integration limits are about an even interval. In summary, the coefficients in the expansion are 1, n=1,5,9,... B54 7, 1=3,7,11... (63) cn= x. , n=2,4,6,... 373 Ten 0 g) The probability that a measurement of energy yields the value &,, is P(E,) = EF, 4. Therefore, the probabilities are Vi.2 -8 b& 960 =lal= o— P(Ex) = lel 7 7 (64) P(E2) = Ical (65) h) The energy eigenstates evolve in time as YalX) > Walxye"* (66) Therefore, the time evolution of the wavefunction is = WO t)= ~ cntbalxde Ene (67) a Problem 4: Gaussian Wave Packet a) The normalization condition yields It oa 2 r 1= [Wx O)Pdx=C ee?” dx= CC? = (68) 00: een: 2a 2q v4 c=. (69) T where we chose C >0. b) The probability distribution is P(x, 0) = |W(x, 0)|? = Ce-?*. The probability distribution is even, so the average value of x is (x) = 0. 2. c) The average value of x’ is. s Ss r x = xP (x, O)dx = C xe? dx = et = 7 (70) co = 2 (a> 4a The uncertainty o, is q Tu a2 A. x= (*)- 0) = WE (74) d) The function p(k) is Jo i J elk) = we W(x, eM gy = Joe erat mtlex dye (72) 2m -co an -o 4 "nt (-18)'74a —1_ aa =V5c g® = onayv*® (73) which is a Gaussian as well. To simplify the plot, we define the dimensionless wavenumber and function ~ k k= ve (74) 31k) =a Mo VE (75) The dimensionless expression for the function is then o(k oe eA, Figure 2 shows a plot of it. Figure 2: The dimensionless function o(k ds: e) The probability distribution is P(k,O) = |@(k)|? = Vane MRe, The probability distribution is even, so the average value of k is (k) = 0, so the average value of p is (9) =k (k) =0. f) The average value of kK is f J co 1 °° 1 av. = KP (k, O)dk = Vi Ke" Pegk = Vo_ = Zap =a (76) -oo 2ma -oo 2na2 The average value of p’is p? =k’ k’ =k’a. The uncertainty in the momentum is q____ _v Op = (k2)-(k)* =k a (77) g) The uncertainty product is 30, = 2
