solset1 - PH 216, Problem set 1, Due 4/6 1. In class, we...

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Unformatted text preview: PH 216, Problem set 1, Due 4/6 1. In class, we estimated the ground-state energy of the 3D SHO (see Shankar 12.6.42) using a `guess' of 0 (a; ) = R(a; r)Y00 (, ), r Suppose we try to get the first energy E1 using the trial function 1 = R(a; r)Y10 , Where the Y enforces orthogonality with our 0 . A trial function with zero nodes that has the right asymptotics ( r+1 as r 0 and 0 as r ) is 2,1,0 = N (r/a0 ) exp(-r/2a0 ) cos . what then is E1 , and how does this compare with the first level of the actual 3D SHO? Solution: We have E(a) = (a)|H|(a) = (a)|T |(a) + (a)|V |(a). As in class, we can use the quantum virial theorem, 1 T = - V T + V = En = -2 /2ma2 n2 = -T 2 for an energy state of energy En . So we just have to evaluate 1 2 V = r dr d (r2 /2)(r/a)2 exp(-r/a) cos2 , 32a3 0 where I have taken the normalization of the wavefunction from Shankar 13.1.27. But since V does nothing to angles, the angular integral just falls out as 4, so we get m 2 m 2 15a2 m 2 dr r6 exp(-r/a) = 720a6 = . V = 48a5 0 48a5 Combining our results leads to E(a) written as: E(a) = 2 + 15a2 m 2 8ma2 by varying a and minimizing (a)|H|(a). With this in hand, we minimize the energy with respect to a: dE 2 =- + 30am 2 = 0, da 4ma3 which gives amin = 1 2 120m2 2 1/4 . Thus, we get E1 = E(amin ) = 15 2.74. 2 The exact result for the nth energy level of the 3D SHO is given by En = (n + 3/2), and with n = 1, E1 = 2.5. This gives E1 0.24 which is greater than zero as required by the variational principle. 2. (Shankar 16.2.8) Consider the = 0 radial equation for the three-dimensional Coulomb 3 problem. Since V (r) is singular at the turning point r = 0, we cannot use n + 4 as in eq. (16.2.42) of Shankar. (a) Will the additive constant be more or less than 3 ? (Only heuristic reasoning is 4 required, but you are welcome to derive the result formally if you like.) (b) By analyzing the exact equation near r = 0, it can be shown that the constant equals 1. Using this constant, show that the WKB energy levels agree with the exact results Solution: (a) Heuristically, as discussed in class if the wavefunction must vanish at both ends, the accumulated phase must add up to n half-cycles, so (n + 1). But if the wavefunction need not vanish, then you fit somewhat less, and the WKB formalism reveals that you basically take another /4 cycles past the turning points, so (n + 1/2) or (n + 3/4) for zero or one endpoints fixed at = 0. For the H atom with l = 0, it's unclear to me intuitively what the condition should be. Near the origin we know that r, so it vanishes, and in fact even in an infinite square well in 1D, the wavefunction vanishes linearly near the endpoints. For l = 1, there is an infinite angular momentum barrier and the wavefunction vanishes as r2 near r = 0. This might suggest (n + 3/4) for both cases, but it is not quite clear, since for l = 0 the equation is singular. To get the answer we can also analyze the radial equation more carefully, as follows. We need to determine the asymptotic (r ) behavior of the = 0 hydrogen atom wave function in the semi-classical limit. Since the semi-classical limit corresponds to large quantum numbers, n 1 (where the bound state energy approaches zero), we can solve the Schrodinger equation for the hydrogen atom with = 0 in the limit of E 0. Defining the reduced radial wave function, u(r) rR(r), eq.(13.1.2) on p. 353 of Shankar (with = E = 0) reads: 2 d 2me2 + 2 u(r) = 0 , (1) dr2 r with boundary conditions u(r = 0) = 0 , 2 u(r ) = 0 . (2) It is convenient to define Eq. (1) then reduces to 2me2 r . 2 (3) d2 u u() + = 0. d2 This can be converted to a more recognizable form by defining 1 = 4 z2 and u() = zv(z) . Then, du du dz 2 du = = , d dz d z dz d2 u d 2 du d 2 du dz 4 d2 u 1 du = = = 2 - . d2 d z dz dz z dz d z dz 2 z dz Changing variables from to z, du d dv = (zv) = z + v , dz dz dz d2 u d dv d2 v dv = z +v =z 2 +2 . 2 dz dz dz dz dz It then follows that: 2 d2 u 4 d v dv v = 2 z 2+ - . d2 z dz dz z Inserting this last result into eq. (3) yields: z2 We recognize this as Bessel's equation of order 1.1 The general solution is: v(z) = AJ1 (z) + BN1 (z) , where A and B are coefficients to be determined. The boundary condition at r = 0 [cf. eq. (2)] implies that B = 0. The constant A is determined by the normalization condition. Hence, the reduced radial wave function is given by: u() = 2A J1 (2 ) . The asymptotic form for the Bessel function J1 (z) is given by: 2 3 J1 (z) cos z - 4 , as z . z 1 d2 v dv + z + z 2 - 1 v(z) = 0 . 2 dz dz More generally, Bessel's equation of order p reads: z 2 v + zv + (z 2 - p2 )v = 0, with general solution v(z) = AJp (z) + BNp (z). 3 Hence, To match this to a WKB wave function, we note that u() 2A cos 2 - 3 , 4 as . (4) 1/2 1 e2 k(r) = 2m E + . r In the semi-classical limit where E 0, we simply obtain 2me2 2me2 -1/2 k(r) = , 2 r 2 and r (5) 0 k(r )dr Hence, one can rewrite eq. (4) as: r 2A 3 u(r) cos k(r )dr - , [k(r)]1/2 4 0 0 d = 21/2 . as r . In class, we argued that for a spherically symmetric potential that is regular at the origin, the asymptotic form of the reduced radial wave function in the region 0 < r < rc , where rc is the turning point closest to the origin, is given by r C u(r) cos k(r )dr - , as r . [k(r)]1/2 2 0 Thus, the phase shift is increased by 1 . This means that in the derivation of the 4 3 energy quantization condition [eq. (16.2.42) of Shankar], the factor of (n + 4 ) must be increased by 1 . The correct quantization condition for the = 0 bound 4 states of the Coulomb potential is thus given by: rc k(r)dr = (n + 1) , n = 0, 1, 2, 3, . . . (6) 0 (b) We now use eq. (6) to compute the energies of the = 0 bound states of the Coulomb potential. The turning point rc is obtained by setting k(rc ) = 0, where k(r) is given by eq. (5). Thus, E + e2 /rc = 0, which yields rc = - Using eq. (6), 1 -e2 /E e2 . E 0 1/2 e2 dr = (n + 1) . 2m E + r 4 (7) Keep in mind that E < 0 for the bound state energy levels. Defining the dimensionless real positive variable, rE x- 2 , e eq. (7) can be rewritten as 1/2 1 1/2 2me4 1 - 2 -1 dx = (n + 1) , (8) E x 0 The integral is recognized as a beta function: 1/2 1 1 1 3 ( 1 )( 3 ) 1 -1/2 1/2 2 -1 dx = x (1 - x) dx = B 2 , 2 = 2 = 1 , 2 x (2) 0 0 where we have used ( 3 ) = 1 ( 1 ) and ( 1 ) = . Inserting the above result 2 2 2 2 into eq. (8) and solving for E, we obtain E= -me4 , 22 (n + 1)2 n = 0, 1, 2, 3, . . . which coincide with the exact energy levels of the hydrogen atom! 3. The interactions of heavy quarks are often approximated by a spin-independent nonrelativistic potential that is a linear function r, the separation between the quarks: V (r) = A + Br. Thus the "Charmonium" particles, the and , with rest energies of 3.1 GeV and 3.7 GeV, are believed to be the n = 0 and n = 1 zero-angular momentum bound states of a charm quark and antiquark of mass mc c2 = 1.5 GeV . Similarly, the particles (energy 9.5 GeV) and are believed to be the n = 0 and n = 1 zeroangular-momentum bound states of the `bottom' quark and antiquark, which have rest mass mb c2 = 4.5 GeV . (a) Using dimensional analysis, derive a relationship between the energy splitting between, on the one hand, and and, on the other hand, and . Use this to find the energy of the . (b) Call the n = 2, zero angular-momentum charmonium particle the . Use the WKB approximation to estimate the energy splitting of the and in terms of the energy splitting between the and the , and then use this to give a numerical estimate of the rest energy of the . Solution: In the center-of-mass system of a quark and it antiquark, the equation of relative motion is 2 2 + V (r) (r) = ER (r), = mq /2, - 2 r where ER is the relative motion energy, mq is the mass of the quark. When the angular momentum is zero, the above equation in spherical coordinates can be simplified to 2 1 d 2 d - r + V (r) R(r) = ER R(r). 2 r2 dr dr 5 Let R(r) = 0 (r)/r. Then 0 (r) satisfies d2 0 2 + 2 [ER - V (r)]0 = 0, dr2 i.e., d2 0 2 + 2 (ER - A - Br)0 = 0. dr2 (a) Suppose the energy of a bound state depends on the principal quantum number n, which is a dimensionless quantity, the constant B in V (r), the quark reduced mass , and , namely E = f (n)B x y x . As, [E] = [M ][L]2 [T ]-2 , [B] = [M ][L][T ]-2 , = [M ] = [M ][L]2 [T ]-1 , we have and hence 2 1 x = z = ,y = - 3 3 E = f (n)(B)2/3 ()-1/3 , where f (n) is a function of the principal quantum number n. Then E = E - E = f (1) = and similarly E = Hence E = E As (B)2/3 b 1/3 (B)2/3 c 1/3 - f (0) (B)2/3 c 1/3 (B)2/3 c 1/3 [f (1) - f (0)], [f (1) - f (0)]. 1/3 1 = . 3 c b 1/3 E = E + 0.42 GeV = 9.5 GeV + 0.42 GeV 9.9 GeV. (b) Applying the WKB Approximation to the equation for 0 we obtain the BohrSommerfeld quantization rule ER - A 2 2(Er - A - Br)dr = (n + 3/4) with = B 0 6 E - E 0.42 GeV, which gives, writing En for ER , [3(n + 3 )B/4]2/3 4 En = A + . (2)1/3 Application to the energy splitting gives E - E = E - E = and hence (B)2/3 c 1/3 21 16 2/3 2/3 - - 9 16 2/3 , 2/3 , (B)2/3 c 1/3 33 16 21 16 Thus (33)2/3 - (21)2/3 E - E = 0.81. E - E (21)2/3 - (9)2/3 E - E = 0.81 (E - E ) = 0.81 (3.7 - 3.1) GeV 0.49 GeV, and E = 3.7 GeV + 0.49 GeV 4.2 GeV. 7 ...
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