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Physics 215 Solution Set 3 Winter 2018 1. Consider the One Physics 215 Solution Set 3 Winter 2018 1. Consider the one-dimensional problem of a particle moving in a delta-function potential: V (x)= Aδ(x) . − (a) Solve for the bound state energies and wave functions. Consider the cases A> 0 and A< 0 separately. The time-independent Schr¨odinger equation for this problem is ~2 d2ψ Aδ(x)ψ(x)= Eψ(x) . (1) −2m dx2 − Integrating this equation from x = ǫ to x = ǫ, where ǫ > 0 is a positive infinitesimal − quantity, ~2 ǫ ψ′(ǫ) ψ′( ǫ) Aψ(0) = E ψ(x) dx . −2m − − − Z−ǫ where ψ′ dψ/dx. We require that ψ(x) is continuous at x = 0. If this were not true ≡ then dψ/dx would behave like a delta function near x = 0 and d2ψ/dx2 would behave lie the derivative of a delta function.1 Such behavior is not compatible with eq. (1). Hence, it follows that ǫ ǫ ψ(x) dx ψ(0) dx =2ǫψ(0) , ≃ Z−ǫ Z−ǫ which vanishes in the limit of ǫ 0. We conclude that → 2mA lim ψ′(ǫ) ψ′( ǫ) = ψ(0) . (2) ǫ→0 − − − ~2 That is, dψ/dx is not continuous at x = 0; its discontinuity is determined by eq. (2). To solve for the bound state energies, we solve eq. (1) for x = 0. In this case, we can set δ(x) = 0. Normalizable energy eigenstates exist if and only if E6 < 0 and A> 0. That is, we solve the differential equation, ~2 d2ψ + E ψ(x)=0 , for x = 0, −2m dx2 | | 6 where E = E , under the assumption that limx→±∞ ψ(x) = 0. The solution to this equation after−| imposing| the boundary conditions at x = is ±∞ Ne−αx , for x> 0, ψ(x)= αx (Ne , for x< 0, 1Recall that dΘ(x)/dx = δ(x), where Θ(x) is the Heavyside step function. 1 where N is a normalization constant, and 2m E α | | . (3) ~2 ≡ r Note that the sign of α (which is positive) is determined since limx→±∞ ψ(x) = 0. The energy eigenvalues are determined by imposing eq. (2). Noting that N = ψ(0), eq. (2) yields 2mAN 2αN = , − − ~2 which implies that mA α = . (4) ~2 Note that we must require that A> 0 in order that bound state solutions exist, since A< 0 is incompatible with the requirement that α> 0. Finally, inserting eq. (3) into eq. (4) yields mA2 E = E = . −| | − 2~2 That is, there is precisely one bound state. The corresponding energy eigenfunction is 1/2 mA 2 ψ (x)= e−mA|x|/~ , (5) E ~2 where we have determined that the normalization constant is N = mA/~2 by requiring that ∞ p ψ (x) 2dx =1 . | E | Z−∞ The normalization condition yields 1 ∞ ∞ 1 ∞ 1 ~2 = e−2α|x| dx =2 e−2α|x| dx = e−2α|x| = = , N 2 −α α mA Z−∞ Z0 0 | | after using eq. (4). The phase of N is conventionally taken to be unity, in which case N =(mA/~2)1/2, as indicated in eq. (5). (b) In the case of A > 0, consider a scattering process where the incident wave enters from the left with E = ~2k2/(2m) > 0 (where E is the energy eigenvalue of the Hamiltonian). Determine the corresponding reflection coefficient R and the transmission coefficient T as a function of k. Write the coefficients in terms of the dimensionless parameter b E/Eg where E is the ground state energy obtained in part (a), in the case of A > 0.≡ What is − g the behavior of T (b) as b 1? → − 2 Consider a scattering process where the incident wave enters from the left. Then, the solution to ~2 d2ψ Eψ(x)=0 , for x = 0, −2m dx2 − 6 where E > 0, is given by Beikx + Ce−ikx , for x< 0. ψ(x)= ikx (6) (De , for x> 0, and 2mE k = , (7) ~2 r Note that k > 0, under the assumption that no incident wave enters from the right. As discussed in part (a), we demand that the wave function ψ(x) is continuous at x = 0, whereas dψ/dx is discontinuous at x = 0 according to eq. (2). Noting that ψ(0) = B + C = D and lim ψ′( ǫ)= ik(B C) , lim ψ′(ǫ)= ikD , ǫ→0 − − ǫ→0 which we make use of in eq. (2), we end up with two equations, B + C = D, 2mA ik D B + C = D, − − ~2 which can be rewritten in the following form, D C =1 , (8) B − B D 2imA C 1 + =1 . (9) B − ~2k B Adding eqs. (8) and (9), we immediately solve for D/B, D ~2k = . (10) B ~2k imA − Plugging back in to eq. (8) yields, C imA = . (11) B ~2k imA − The corresponding reflection and transmission coefficients are C 2 m2A2 R = = , (12) B ~4k2 + m2A2 2 4 2 D ~ k T = = . (13) B ~4k2 + m2A2 3 It is convenient to rewrite eqs. (12) and (13) in terms of the the energy eigenvalue for the scattering problem, E = ~2k2/(2m) [cf. eq. (7)], and the parameter E mA2/(2~2), 0 ≡ − which we recognize from part (a) is the ground state energy of the attractive delta function potential. In terms of E and E0, eqs. (12) and (13) yield E E R = 0 , T = . −E E E E − 0 − 0 Note that the transmission coefficient T (E) exhibits a pole at the ground state energy E0 of the attractive delta function potential. (c) In the case of A< 0, consider a scattering process where the incident wave enters from the left with E = ~2k2/(2m) > 0. Investigate the location of any poles of the transmission amplitude in the complex k plane and the complex E plane, respectively. Explain your result in light of the fact that the repulsive delta function potential possesses no bound states. In part (b), the transmission amplitude for one-dimensional scattering was obtained in eq. (10). In the case under consideration here, the only difference from part (b) is that the sign of A is negative. From eq. (10), we see that the poles of the transmission amplitude correspond to ~2k imA = 0. Solving for k yields − imA k = . (14) ~2 where k (2mE/~2)1/2 > 0. Since A < 0, it follows that Im k < 0, which means that ≡ in the complex k-plane, the pole is located on the negative imaginary axis. Hence, in the complex E-plane, the pole is located on the negative Re E-axis on the second Riemann sheet. In contrast, poles corresponding to bound states reside on the negative Re E-axis on the first Riemann sheet of the complex E-plane. We conclude that the repulsive delta function potential supports no bound states, as expected. Indeed, if we rewrite eq. (6) in the form, 1 R eikx + e−ikx , for x< 0. ψ(x) T T (15) ∼ eikx , for x> 0, and substitute eq. (14) into eq. (15), we see that the resulting wave function diverges expo- nentially as x . This implies that the resulting wave function cannot correspond to a bound state. → ±∞ 4 2. A one-dimensional potential has the following form: + , for x< 0, ∞ V (x)= V , for 0 < x < b, − 0 0 , for x > b, where V0 and b are positive constants. (a) Find V0 as a function of b such that there is just one bound state, of about zero binding energy, for a particle of mass M. To solve the bound state problem, we require that V <E< 0. The relevant boundary − 0 conditions are ψ(0) = 0, due to the infinitely high barrier at x = 0 and limx→∞ ψ(x) = 0. The solution to the time-independent Schr¨odinger equation is A sin(kx) , for 0 x < b , ψ(x)= −κx ≤ (Be for x > b, where E = E and −| | 2M(V E ) 2M E k = 0 −| | , κ = | | , (16) ~2 ~2 r r with k > 0 and κ> 0.2 Continuity of ψ and dψ/dx at x = b yields two equations, A sin(kb)= Be−κb , Ak cos(kb)= Bκe−κb . − Diving these two equations, we obtain 1 1 tan(kb)= . (17) k −κ Note that eq. (16) implies that 2MV k2 + κ2 = 0 . (18) ~2 We can also rewrite eqs. (17) and (18) as kb cot(kb)= κb , (19) − 2MV b2 (kb)2 +(κb)2 = 0 . (20) ~2 2We require κ> 0 so that ψ(x) 0 as x . The requirement of k> 0 is conventional, since the sign of k can be absorbed into the definition→ of the→ constant ∞ A. 5 Eqs. (19) and (20) can be solved graphically by looking for intersection of the function y = x cot x with the circle of radius r = (2MV b2/~2)1/2 in the first quadrant of the x–y − 0 plane (where x = kb and y = κb). κb bound state kb π 2 It is clear from the figure above that if V0 is less than some critical value, then the radius 1 of the circle, r < 2 π, in which case there are no intersections in the first quadrant, and 1 therefore no bound states. The critical value of of V0 is obtained by setting r = 2 π. That is, 2MV b2 π2 0 = , ~2 4 which yields π2~2 V = . (21) 0 8Mb2 (b) Applying this crude model to the deuteron (a bound state of a proton and a neutron), −13 1 evaluate V0 in MeV, assuming b = 1.3 10 cm and M = 2 mp (where mp is the proton mass).
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