Question

Consider an electron moving in a spherically symmetric potential V = kr, where k>0. (a) Use the uncertainty principle to estimate the ground state energy. (b) Use the Bohr-Sommerfeld quantization rule to calculate the ground state energy. (c) Do the same using the variational principle and a trial wave function of your own choice. (d) Solve for the energy eigenvalue and eigenfunction exactly for the ground state. (Hint: Use Fourier transforms.) (e) Write down the effective potential for nonzero angular momentum states.

Answer 1

(a) The uncertainty principle states that Aparat where Ap= [[p - 7211/2 = [(02–2pp + f2)]1/2 = (p2 - p2,1/2,

Ar= (p2 – 72,1/2 The potential is spherically symmetric, so we can take p = 0, i.e. Ap Vp2. For an estimate of the energy E = + ki, we shall also take Ar~. Then Em compet kar > (97)2 For the ground state energy E, we have ae aap m kħ 2(Ap)2 = 0, giving and Ap= (makin)." D=} ( )" (b) The Bohr-Sommerfeld quantization rule gives 4 P, dr = n(h, 4 P, do = nại. Choose polar coordinates such that the particle is moving in the plane Q = n/2. The ground state is given by nr = 0, ng = 1, and the orbit is circular with radius a. The second integral gives Po = Iw = maʼw = ħ. The central force is Pr = Y = -k=-mw?a.

Combining we have a = (h2/mk)1/3, and hence Eo = P}/2ma? + ka = ž (k?M2/m)1/3. (c) The notion in the ground state does not depend on 6 and 6. Take a trial wave function = exp (-Ir) and evaluate À = (4 Ñ ) (414) where À = 12 + kr. AS wo) a - em L. * * ( -v) pedre + k 1° re-2Ar de " (3 - ) c-28-? der (vln a ſº z-20 ? dr = Tapijt (41) « e-211 r2 dr = we have For stable motion, H is a minimum. Then taking M=0, we find

and hence 9 k 379 k22 1/3 H = a1 =ž (ā m) . (d) The Schrödinger equation for the radial motion can be written as h2 d2 2m dr2 X + (kr - Ex = 0, where x = TR, R being the radial wave function. For the ground state, the angular wave function is constant. By the transformation v = (art" (--) the Schrödinger equation becomes the Airy equation 2,– yx(y) = 0, whose solutions are Ai(--) and Ai(x), where x = -lyl, for y < 0 and y > 0 respectively. The boundary conditions that R(r) and R r) be continuous at r = , i.e. y = 0, are satisfied automatically as Ai(x) = Ai(-2), Ai'(x) = Ai'(-x) for 3 + 0. The condition that R(r) is finite at s + 0 requires that Ai(-x) = R(T) + 0 as s + 0. The first zero of Ai(-x) occurs at x = to ~ 2.35. Hence the ground state energy is / 23 and the ground state eigenfunction is R(r) = – 4:1-2) with a = (prek) * (. - -). (e) The effective potential for nonzero angular momentum is Veft = kr + h? 1(1 + 1)/2mr2.