Applied Nuclear Physics Homework #2

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1 Applied Nuclear Physics Homework #2 Author: Lulu Li Professor: Bilge Yildiz, Paola Cappellaro, Ju Li, Sidney Yip Oct. 7, 2011

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3 1. Answers: Refer to p16-17 on Krane, or 2.35 in Griffith. (a) x < 0: E 0 V 0 /C > 0 2 d 2 ψ 2m dx 2 E 0ψ 1 (x) 0 d 2 ψ dx 2 + k2 1ψ 1 (x) 0 ψ(x) 1 Ae ik1x + Be ik1x in which k1 2 2mE0 2 x > 0: E 0 + V 0 > 0 2mV0 C 2. 2 d 2 ψ 2m dx 2 (E 0 + V 0 )ψ 2 (x) 0 d 2 ψ dx 2 + k2 2ψ 2 (x) 0 ψ(x) 2 Ce ik2x + De ik2x in which k2 2 2m(E0+V0) Boundary Conditions: dψ dx 1 2mV0( 1 +1) C 2 2 2mV0(C+1) C 2. ψ 1 (x 0) ψ 2 (x 0) A + B C + D dψ (x 0) dx 2 (x 0) k 1(A B) k 2 (C D) } { B A 1 k2/k1 1+k 2/k 1 C A 2 1+k 2/k 1 Reflection coefficient is: R B 2 A 2 ( ) 2 ( 1 k2 /k 1 1 C k 2 /k C + 1 (b) The fraction of the particle flux reflected back to the left is the same as the reflection coefficient, which ) 2. is ( 1 C+1 1+ C+1 (c) If this were a classical system, none of the particle flux would be reflected back to the left. ) 2 3

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5 2. Answers: This problem appears to be identical to the Rectangular Barrier Problem (with E < V ) we did in class. (a) If we call the regions from left to right 1,2,3 respectively. Wave Function General Form: We basically start with 2 d 2 ψ 2m dx 2 (E V )ψ d2 ψ dx 2 2m(E V ) 2 ψ The solution to the above Schrodinger Equation in 3 regions look like the following. We ve solved them many times so I am not going to bother going over the details again, besides noting that the e ik1x term in ψ 3 (x) goes away because we assume nothing is traveling to the left in region 3. ψ 1 (x) Ae ik1x + Be ik1x, ψ 2 (x) Ce κx + De κx, ψ 3 (x) F e ik1x 2mE in which k 1, κ 2 Boundary Conditions: 2m(V b E) 2. ψ 1 (x 0) ψ 2 (0), dψ dx dψ 1 x0 dx, ψ 2 (x a) ψ 3 (x a), dψ 2 x0 dx dψ 2 xa dx 3 Solving for the coefficients: We divide all the wave functions by A, so we are left with four unknowns B/A, C/A, D/A, F/A. With the above four boundary conditions, we can solve for them. I am lazy so I used mathematica to get the following: B A (k 2 + κ 2 ) sinh(aκ) 2ikκ cosh(aκ) + (k 2 κ 2 ) sinh(aκ) C A 2k(k iκ) (k iκ) 2 e 2aκ (k + iκ) 2 D A 2e 2aκ k(k + iκ) (k iκ) 2 + e 2aκ (k + iκ) 2 F A 4πe a( ik+κ) kκ (k iκ) 2 + e 2aκ (k + iκ) 2 That is to say, our wave functions are in the form of: ψ 1 (x) Ae ikx + Ae ikx (k 2 + κ 2 ) sinh(aκ) 2ikκ cosh(aκ) + (k 2 κ 2 ) sinh(aκ) ψ 2 (x) Ae κx 2k(k iκ) (k iκ) 2 e 2aκ (k + iκ) 2 + 2e 2aκ k(k + iκ) Ae κx (k iκ) 2 + e 2aκ (k + iκ) 2 ψ 3 (x) Ae ikx 4πe a( ik+κ) kκ (k iκ) 2 + e 2aκ (k + iκ) 2 xa 5

6 (b) The tunneling probability is: T F A 2 8k 2 κ 2 k 4 + 6k 2 κ 2 κ 4 + (k 2 + κ 2 ) 2 cosh(2aκ) 8 6 k2 κ κ2 2 k + ( k 2 κ + ) κ 2 k cosh(2aκ) 8 6 E V E V E E + V 2 E(V E) cosh(2aκ) 8 8 V 2 E(V E) + V 2 E(V E) cosh(2aκ) V 2 8 E(V E) (cosh(2aκ) 1) V 2 4 E(V E) sinh2 (aκ) in which we use that cosh 2x cosh 2 x + sinh 2 x 2 cosh 2 x 1 2 sinh 2 x + 1. (c) The reflection probability is: R B A k2 κ 2 csch(aκ) 2 (k 2 +κ 2 ) csch(aκ)2 V 2 E(V E) csch(aκ) 2 E(V E) V 2 Alternatively, knowing R + T 1, we can do: 1 R 1 T V 2 4 E(V E) sinh2 (aκ) (d) In classical mechanics, R 1, T 0, because none of the particles can pass the potential barrier. The word tunneling is appropriate for this phenomenon because it is suggesting some of the particles in the beam can travel beyond the potential barrier that is larger than the KE of the beam, which appears to be almost as if the particles tunnel through the barrier. 6

7 3. Answers: Take away message: Frequency/wavelength is related to KE which is E V : the smaller KE is, the tighter the wavefunctions are. Amplitude is related to the change in potential: everytime potential changes (whether increase or decrease), amplitude decreases. If the potential does a jump change, amplitude just decreases; if the potential gradually decreases, then amplitude graduate decreases as well. 7

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10 4. Answers: First we justify the solution has to be symmetric, because the probability cannot be zero in the clasically forbidden area. Then we can just solve for the x > 0 region, in which we devide it into three regions, and the solutions are in the form of: The BCs are: ψ 1 (x) Ae κx + Be κx ψ 2 (x) C sin(kx) or C cos(kx) ψ 3 (x) De κx dψ dx 0 A B (1) 1 x0 ψ 1 (x a) ψ 2 (x a) (2) dψ dx dψ 1 xa dx (3) 2 xa ψ 2 (x 3a) ψ 3 (x 3a) (4) dψ dx dψ 2 x3a dx (5) 3 x3a Note here we can combine the continuous requirement for ψ, dψ dx into requiring 1 ψ the boundaries. dψ dx to be continuous at 10

11 (a) Odd Parity Case (assuming ψ 2 (x) C sin(kx)). Apply the boundary conditions for an expression of ka: Simplifying the equations, we get: (2) A exp(κa) + A exp( κa) C sin(ka) (6) (3) κa(exp(κa) exp( κa) kc cos(ka) (7) (4) C sin(3ka) D exp( 3κa) (8) (5) kc cos(3ka) κd exp( 3κa) (9) (6) exp( κa) k cot(ka) κexp(κa) κ tanh(κa) (10) (7) exp(κa) + exp( κa) (8) (9) tan(3ka) k κ k cot(3ka) (11) κ } (10) cot(ka) cot(3ka) tanh( ka cot(3ka)) (12) (11) Graphically solving for ka: The Eqn. 12 we arrive at is an expression of ka only, so we can determine all possible wavenumbers in the classically-allowed regions. We can plot it graphically using WolframAlpha: Figure 1: Plot of solution, odd parity WolframAlpha also tells me the two sides of the equations intersect at x values (x ka) of the following four lowest values (I already got rid of the negative values because ka has to be positive in this case): k 1 a λ 1 2π k a k 2 a λ 1 2π k a k 3 a λ 1 2π k 4 1.6a. It is important to notice that, although mathematica thinks that ka π 2 is a solution too, it is not included because it is a trivial solution that causes both sides of the equation to be zero. 11

12 Lulu Li Homework #2 Oct. 7, 2011 Figure 2: Sketch of the lowest three states, odd parity Sketch of the lowest three eigenfunctions: Write a normalized eigenfunction for the lowest wavenumber: sin(ka) C exp κa + exp κa sin(3ka) C (8) or (9) D C exp 3κa (6) or (7) A C in which we plugged in for the lowest order case ka , and from Equation 11 κ values solve from ka value. Then normalizing the eigenfunction is just a matter of integrating and solving for C: Z 0 a ψ12 (x) dx + Z 3a ψ22 (x) dx + a Z 3a 12 ψ32 (x) dx 1 2

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14 (b) Even Parity Case(assuming ψ 2 (x) C cos(kx)). Apply the boundary conditions for an expression of ka: Simplifying the equations, we get: (2) A exp(κa) + A exp( κa) C cos(ka) (13) (3) κa(exp(κa) exp( κa) kc sin(ka) (14) (4) C cos(3ka) D exp( 3κa) (15) (5) kc sin(3ka) κd exp( 3κa) (16) (13) exp( κa) k tan(ka) κexp(κa) κ tanh(κa) (17) (14) exp(κa) + exp( κa) (15) (16) cot(3ka) k κ k tan(3ka) (18) κ } (17) tan(ka) tan(3ka) tanh(ka tan(3ka)) (19) (18) Graphically solving for ka: The Eqn. 12 we arrive at is an expression of ka only, so we can determine all possible wavenumbers in the classically-allowed regions. We can plot it graphically using Mathematica as seen in Figure 3: Figure 3: Plot of the solution, even parity Reading from Figure 3, we know the approximate range of the three lowest solution, so we can use Findroot to find them as shown in Figure 4: Figure 4: Find Root, even parity 14

Lulu Li 22.101 Homework #2 Oct. 7, 2011 The three lowest values for ka are(i already got rid of the negative values because ka has to be positive in this case): 2π 2.6643a k1 2π 2.02378a k2 a 3.

15 Lulu Li Homework #2 Oct. 7, 2011 The three lowest values for ka are(i already got rid of the negative values because ka has to be positive in this case): 2π a k1 2π a k2 a λ1 k2 2π k3 a λ a k3... k1 a λ1 Sketch of the lowest three eigenfunctions: Figure 5: Sketch of the lowest three states, even parity Write a normalized eigenfunction for the lowest wavenumber: cos(ka) C exp κa + exp κa cos(3ka) (15) or (16) D C C exp 3κa (13) or (14) A C in which we plugged in for the lowest order case ka , and from Equation 18 κ value can be solved to be /k. 15

16 Then normalizing the eigenfunction is just a matter of integrating and solving for C: a ψ 2 1(x) dx + 3a ψ 2 2(x) dx + 0 a 3a ψ 2 3(x) dx 1 2 Hence 16

Physics 218 Quantum Mechanics I Assignment 6

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