Physics 218 Quantum Mechanics I Assignment 6

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1 Physics 218 Quantum Mechanics I Assignment 6 Logan A. Morrison February 17, 2016 Problem 1 A non-relativistic beam of particles each with mass, m, and energy, E, which you can treat as a plane wave, is incident on a potential barrier of the form U 0 0 x L U(x) (1) 0 otherwise The incident beam energy is adjusted so that the special condition E U 0 is obtained. Note that U 0 > 0. Part (a) Find the solutions to the Schrodinger equation in all three regions (x < 0, 0 x L, andx > L). Call the regions x < 0, 0 x L and L < x regions I, II and III respectively. In regions I, III the schrodinger equation in the position representation is 2 d 2 ψ Eψ (2) 2m dx2 This differential equation is simple to solve (just guess a solution of the form ψ Ae ikx ). The solutions consist of right-moving and left-moving plane waves. We know that in region I we have an incoming plane wave and that there will be a reflected wave from the boundary. In region III, we will only have an outgoing wave (plane wave moving to the right). We find that the solutions for the wave function in regions I and III are ψ I (x) Ae ikx + Be ikx (3) ψ III (x) Fe ikx (4) where k 2mE/ 2 2mU 0 / 2. In region II, we have that the schrodinger equation is 2 d 2 ψ 2m dx + U 0ψ U 2 0 ψ (5)

2 Logan A. Morrison Assignment 6 Problem 1 Page 2 / 14 Thus, we can see that our differential equation is d 2 ψ dx 2 0 (6) The solution to this equation is also easy to determine. Here, we can simply integrate both sides twice. Doing so, we find that ψ(x) Cx + D (7) Now, we need to determine the constants, A G. To do so, we need to satisfy our boundary conditions. In other words, we need to inforce continuity of the wave function at x 0 and x L: ψ I (0) ψ II (0) and ψ II (L) ψ III (L) and continuity of the derivative at x 0 and x L: ψ I (0) ψ II (0) and ψ II (L) ψ III (L). Continuity of the wave function at these two points gives us and continuity of the derivative of the wave function at x 0 and L gives us A + B D (8) Fe ikl CL + D (9) ika ikb C (10) ikfe ikl C (11) The amplitude of A is fixed by the incoming beam. Thus, all we can solve for is the ratio of amplitudes to A. Define B B/A, C C/A, D D/A and F F/A. Our new set of equations looks like Now, in matrix form, this looks like B D 1 (12) LC D + e ikl F 0 (13) ikb C ik (14) C + ike ikl F 0 (15) B 1 0 L 1 e ikl C 0 ik D ik ike ikl F 0 To solve this set of algebraic equation, we need to augment the matrix with the column solution vector and row reduce the matrix. We could do this by hand but it is rather tedious and unilluminating. Mathematica is much better at doing it. Mathematica gives the following solutions: Part(b) B kl 2i + kl C 2L 2i + kl D 2(i + kl) 2i + kl F 2ie ikl 2i + kl In terms of the given quantities and any physical constants you need, what fraction of the beam is transmitted to the other side of the barrier? (16) (17) (18) (19) (20)

3 Logan A. Morrison Assignment 6 Problem 1 Page 3 / 14 We have already done the heavy lifting. We know that the faction of the beam transmitted will simply be F/A 2 : ( ) ( ) F F 2ie ikl 2ie ikl A A 2i + kl 2i + kl (21) k 2 L 2 (22) mU 0 L 2 / 2 (23) mu 0 L 2 /2 2 (24) (25) Part (c) What is your answer to part (b) in the limit L? Part (c) We can see that as L that the transmission amplitude goes to 0.

4 Logan A. Morrison Assignment 6 Problem 2 Page 4 / 14 Problem 2 Consider a particle of mass m in a one-dimensional infinite square well with a small delta-function barrier added in the middle. Take the width of the well to be 2a, with the well centered at x 0, and the amplitude of the delta function barrier to be α 2 /10ma. Answer the first two parts before doing any calculations. Part (a) Sketch the wavefunction of the ground state. The wave function will have the following form Figure 1: Plot of the wave function with the delta function added in the middle. Not this is highly exaggerated. This is because we know that the ground state wave function (for one dimensional problems) has no nodes. It is symmetric about, x 0 and must be continuous. It should look approximately like the old ground state, but with a perturbation at x 0. This perturbation will look like a slight dip. We can see this by looking at the finite barrier case, in which we have that the solution decays in the barrier. Part (b) Based on your sketch, should its energy be higher or lower than that of the ground-state energy of the same particle in a plain infinite square well of the same width? Explain. The energy will be higher. This is because the solution is almost the old groundstate, but we need higher energy eigenfunction to reconstruct this new function with the dip in the peak.

5 Logan A. Morrison Assignment 6 Problem 2 Page 5 / 14 Part (c) Set an upper limit on the energy of the ground state with the variational principle, using the wavefunction of the ground state of the plain well. By what percentage is this limit higher than the energy of the ground state of the plain well? The variational principle states that E gs ψ H ψ (26) where ψ is any wave function and E gs is the ground state energy. Let x ψ 0 1 ( πx ) a cos 2a (27) and in the position representation, our Hamilitonian for a x a is H ˆp2 2m ma δ(x) H 0 + H (28) where H 0 is the hamilitonian of the infinite square well and H is our new addition to the square well, the delta function. Recall that the ground state energy of the infinite square well is E 0 2 π 2 /2m(2a) 2 ψ 0 H 0 ψ 0 where ψ 0 is the ground state wavefunction of the infinite square well. Now, we have that E gs ψ 0 (H 0 + H ) ψ 0 (29) ψ 0 H 0 ψ 0 + ψ 0 H ψ 0 (30) 2 π 2 8ma 2 + L/2 L/2 2 π 2 8ma ma 2 π 2 8ma ma a ψ 0 x x H ψ 0 dx (31) L/2 L/2 a a δ(x) ψ 0 x x ψ 0 dx (32) δ(x) cos ( πx ) dx (33) L 2 π 2 8ma + 2 (34) 2 10ma 2 The percent difference of this new energy and the old ground state energy is ( 2 π 2 8ma ma 2 2 π 2 8ma 2 ) 2 π 2 8ma % (35) 10π2 Part (d) Solve for the energy of the ground state exactly. What is the percent change relative to the plain well in this case?

6 Logan A. Morrison Assignment 6 Problem 2 Page 6 / 14 We need to solve the following differential equation 2 d 2 ψ 2m dx (x) + 2 δ(x)ψ(x) Eψ(x) (36) 2 10ma The issue here is the delta function. However, we know that the delta function is zero in the regions a x < 0 and 0 < x a. So let s solve for the wave function in these regions and patch the solutions together. Note that the wave function must be zero at x a, 0 and a. So all we need to know is how the derivatives match up at x 0. Let s integrate the Schrodinger equation around x 0 to determine this: [ ɛ lim 2 ɛ 0 2m ɛ 2 2m d 2 ψ dx (x)dx ma ( dψ dx dψ ) xɛ dx x ɛ ɛ ɛ δ(x)ψ(x)dx E ɛ ɛ ] ψ(x)dx (37) + 2 ψ(0)dx 0 (38) 10ma Therefore, if we call ψ II the wave function in the region 0 < x < a and ψ I the wave function in the region a < x < 0, we have dψ II dx dψ I x0 dx 1 ψ(0) (39) x0 5a and hence, the derivative must be continuous. Okay, now in the regions where a < x < 0 (REGION I) and 0 < x < a (REGION II), the we just have the free Hamilitonian. The solution for the free schrodinger equation will be ψ I (x) A sin (kx) + B cos (kx) (40) ψ I (x) C sin (kx) + D cos (kx) (41) where k 2mE/ 2 Now we have that the wave function is continuous at x 0. This results in B D. We has have that the wave function is zero at the endpoints. This results in 0 A sin ( ka) + B cos ( ka) (42) 0 C sin (ka) + B cos (ka) (43) Lastly, we have to enforce the discontinuity of the derivative at x 0. The derivatives at x 0 are Enforcing the discontinuity, we find that kc ka B 5a Now, if we use this information and subtract equations 42 and 43, find that ψ II(0) kc (44) ψ I(0) ka (45) (46) C B 5ka + A (47) 0 (A + C) sin(ka) (48) ( 2A + B ) sin(ka) (49) 5ka

7 Logan A. Morrison Assignment 6 Problem 2 Page 7 / 14 Now here is where w come to a branching point. There are two solutions to this equation. One is that ka nπ. Then second is that 2A B. If we take the first solution, we find that (from equations 42 and 43) that the 5ka wave function is ( nπx ) SOLUTION # 1: ψ(x) A sin (50) a Now, we already know the ground state energy of this solution: E 2 π 2 /2ma 2 and this is clearly larger than what the variationaly principle gave us. Thus, the other set of solutions must contain the ground state. If we enforce 2A B 5ka we have that following B 10kaA (51) C A (52) D 10kaA (53) and A is fixed by normalization. If we plug these into equ. (42), we find that 0 A sin(ka) 10kaA cos(ka) tan(ka) 10(ka) (54) Figure 2 displays the tan(ka) and 10ka functions. Solving for the lowest intercept point (using Mathematica), Figure 2: Plot of 10ka(orange) and tan(ka)(blue). The first interct point gives us the ground state energy. we find that ka E ( ) (55) 2ma 2 This energy is approximately 7.94% above the ground state of the plain square well. Thus, we can see that the variational principle did pretty well in estimating the ground state! 2 Part (e) It might have seemed odd at first to call the delta function amplitude small when there is no scale to compare it to the potential is zero or infinite everywhere. In retrospect, having solved the problem, what does small mean

8 Logan A. Morrison Assignment 6 Problem 3 Page 8 / 14 Calling the amplitude small is justified since the effects of the delta function on the wave function is small. We found that there is only a small shift in the energy eigenvalues. In fact, if we have kept α in our solution, as alpha gets larger, the shift of the energy get s large and larger. Problem 3 The one-dimensional quantum mechanical potential energy of a particle of mass m is given by V 0 δ(x) a < x < V(x) x < a (56) with a, V 0 positive, real constants. At time t 0, the wave function of the particle is completely confined to the region a < x < 0. Part (a) Write down the normalized lowest-energy wave function of the particle at time t 0. If the particle is completely confined to the region a < x < 0, then the wave function must be zero outside that region. In this region, the Hamiltonian is that of a free particle. In this region the position rep. schrodinger equation is The solution to this is 2 d 2 ψ Eψ(x) (57) 2m dx2 ψ(x) A sin(kx) + B cos(kx) (58) where k 2mE/ 2. Since the wave function must be zero at x 0, B 0. And since it must be zero at x a, we have ka nπ where n is an integer. Thus, ψ n (x) 2 ( nπx ) a sin a (59) with E 2 π 2 n 2 /2ma 2. So the lowest energy solution is thus E 2 π 2 /2ma 2 with ψ(x, t 0) 2 ( πx ) a sin a (60)

9 Logan A. Morrison Assignment 6 Problem 3 Page 9 / 14 Part (b) Give the boundary conditions which the energy eigenfunctions must satisfy, where region I is a < x < 0 and region II x 0. ψ k (x) ψ I k (x) and ψ k(x) ψ II k (x) (61) Since the potential in infinite for x < a, the wavefunction must be zero in that region. Thus, The wave function must be continuous at x 0 which implies ψ I k ψ I k ( a) 0 (62) (0) ψii k (0) (63) At x 0, the derivative will be discontinous due to the delta function. Let s derive the discontinuity. First, the schrodinger equation is If we integrate this equation around x 0 we find [ lim ɛ 0 2 2m 2 ɛ 2 d 2 ψ 2m dx + V 0δ(x)ψ Eψ (64) 2 ɛ d 2 ψ ] 2m ɛ dx dx + V 2 0 δ(x)ψ(x)dx E ψ(x)dx (65) ɛ ɛ ) dψi k x0 dx + V 0 δ(x)ψ(0)dx 0 (66) x0 ( dψ II k dx Therefore, the discontinuity of the derivative is (ψ II k ) (0) (ψ I k ) (0) 2mV 0 ψ I 2 k (0) 2mV 0 2 These are all of our boundary conditions. Our boundary conditions are Part (c) ɛ ψ II k (0) (67) ψ I k ( a) 0 (68) ψ I k (0) ψii k (0) ψ k(0) (69) (ψ II k ) (0) (ψ I k ) (0) 2mV 0 2 ψ k (0) (70) Find the (real) solutions for the energy eigenfunctions in the two regions (up to an overall constant) which satisfy the boundary conditions.

10 Logan A. Morrison Assignment 6 Problem 3 Page 10 / 14 We need to solve the followin equation in the two regions 0 < x < and a < x < 0: 2 d 2 ψ Eψ (71) 2m dx2 First, let s focus on region 1: a < x < 0. Since the potential is positive everywhere, the energy will be positive. Thus, we have sinusoidal solutions: where k 2mE/ 2. The boundary condition at x a give us that Now let s take a look at region two. Here we have the same solution ψ I k (x) A sin(kx) + B cos(kx) (72) A sin(ka) + B cos(ka) 0 (73) B Atan(ka) (74) ψ II k (x) C sin(kx) + D cos(kx) (75) Now, we need to patch these solutions together. Our boundary conditions state that the wavefunction must be continuous at x 0. This means that ψ I k (0) ψii k (0) (76) B D (77) Now, we have one more boundary condition to satisfy: the discontinuity of the derivative. Recall that we require (ψ II k ) (0) (ψ I k ) (0) 2mV 0 ψ k (0). This requirement gives us hence 2 (ψ II k ) (0) (ψ I k ) (0) 2mV 0 2 ψ k (0) (78) kc ka B 2mV 0 2 (79) C A + A tan(ka) 2mV 0 k 2 (80) (81) C A + A(1 + tan(ka)) 2mV 0 k 2 (82) B A tan(ka) (83) D A tan(ka) (84) (85) Now, our energy eigenfunctions are sin(kx) + tan(ka) cos(kx) a < x < 0 ψ(x) A [ 1 + tan(ka) 2mV ] 0 sin(kx) + tan(ka) cos(kx) 0 < x < k 2 (86)

11 Logan A. Morrison Assignment 6 Problem 3 Page 11 / 14 Part (d) The t 0 wave function can be expressed as an integral over energy eigenfunctions: ψ(x) f (k)ψ k (x)dk (87) Show how f (k) can be determined from the solutions ψ k (x). We know that ψ(x, t) x ψ (88) dk x k, t k, t ψ (89) dkψ k (x, t) f (k) (90) dkψ k (x, 0) f (k)e iω kt where k, t are the energy eigen functions at time t. Note the the time dependence for the energy eigenfunctions is trivial. It is just e iω kt, where ω k E/ k2 2m. Thus, ψ(x, 0) (91) (92) dkψ k (x, 0) f (k) (93) Now, if we want to determine f (k) k ψ, then we can insert in a complete set of position eigenstates: (94) f (k) k, 0 ψ (95) dx k, 0 x x ψ, 0 (96) dx ψ k (x, 0)ψ(x, 0) (97) (98) Part (e) Give an expression for the time development of the wave function in terms of f (k). What values of k are expected to govern the time behavior at large times?

12 Logan A. Morrison Assignment 6 Problem 3 Page 12 / 14 As shown above, since ψ k (x, t) are energy eigenfunctions, the time dependence is trivial, just the e iω kt where ω k 2 k 2m. Thus, ψ(x, t) ψ k (x) f (k)e i 2 kt/2m dk (99) Now, as t, the dominating terms will will be the terms with the smallest ω k. This is due to the Riemann- Lebesgue-Lemma. This lemma states that lim f (z)e itz dz 0 (100) t where f (x) is an L 1 integrable function (essentially just square integrable function over R). Thus, as t only the smallest ω k survive. ω k is small when k is small, so only k around zero will survive.

13 Logan A. Morrison Assignment 6 Problem 3 Page 13 / 14 Problem 4 Find the wave function for a free particle of energy E moving in one dimension in a constant imaginary potential iv, where V E. Calculate the probability current and show that an imaginary potential represents absorption of particles. Find an expression for the absorption coefficient in terms of V. We wish to solve the following differentail equation: Rearranging, we find 2 d 2 ψ ivψ Eψ (101) 2m dx2 d 2 ψ 2m(E + iv) ψ (102) dx2 2 The solution to this is [ 2m ] ψ(x) A exp ± i E + iv x (103) 2 Since V E, we can binomially expand the complex square root: E + iv E 1 + i V (104) E ( E 1 + iv ) (105) 2E Thus, our solution is [ 2mE ( ψ(x) A exp ± i 1 + iv ) ] x (106) 2 2E Here we have two solutions. We can think of each as left and right-moving plane waves. Let s just take one solution, say, the right-moving plane wave for our calculation of the probability current and the absorption current. Recall the the probability current in one-dimension is defined by J(x, t) ( ψ dψ ) 2mi dx ψdψ (107) dx 2mE ( For convenience, let α 1 + iv ). Then 2 2E J(x) ( (Ae ) iα x d ( ) Ae iαx ( Ae iαx) d ( )) Ae iα x (108) 2mi dx dx ( iαa 2 e i(α α )x + iα A 2 e ) i(α α )x (109) 2mi 2m (α + α )A 2 e i(α α )x (110) 2mE A 2 2mE 2 m exp V 2 E x (111) 2E 2m A 2 m exp V E x (112) (113)

14 Logan A. Morrison Assignment 6 Problem 3 Page 14 / 14 Thus, the absorbtion coefficient is just the term in the exponent, Absorbtion Coeff. 2m V E (114)

Applied Nuclear Physics Homework #2

22.101 Applied Nuclear Physics Homework #2 Author: Lulu Li Professor: Bilge Yildiz, Paola Cappellaro, Ju Li, Sidney Yip Oct. 7, 2011 2 1. Answers: Refer to p16-17 on Krane, or 2.35 in Griffith. (a) x