Joel Broida UCSD Fall 009 Phys 130B QM II Homework Set 4 1. Consider the particle-in-a-box problem but with a delta function potential H (x) = αδ(x l/) at the center (with α = const): H = αδ(x l/) 0 l/ l x (a) Find the first-order correction E n (1) to the allowed energies. Why doesn t the perturbation affect the states with n even? (b) Find the first three nonzero terms in the first-order correction to the wave function (equation (.10) in the class notes) for the ground state of this system. (c) Find the second-order corrections E n (), and sum the series to obtain a closed expression. Hint: Expand the summand in terms of partial fractions. Then look at the cases n = 1, 3 and maybe 5 to try and see the pattern.. The ground-state hydrogen atom wave function is ψ 100 = 1 π a 3/ 0 e r/a0 where using a 0 = /m e e = 0.59Å is taking the proton to be infinitely heavy. Assume that the charge of the proton is distributed uniformly throughout the volume of a sphere of radius R = 10 13 cm. Use first-order perturbation theory to find the shift in the ground-state hydrogen-atom energy (in ev) due to the finite proton size. Hints: (1) Be careful finding the potential energy of the electron when it has penetrated the nucleus (i.e., the proton). At a distance r from the nuclear center, let the charge enclosed be Q(r). Find an expression for this in terms of r and R. Now use the definition of potential energy F = V along with elementary electrostatics to find V (r) for r R. 1
() Justify assuming that the exponential factor in ψ is essentially equal to 1 within the nucleus. 3. In this problem you will work out the perturbation treatment of the (nondegenerate) helium atom ground-state energy. The He atom consists of two electrons and a nucleus of charge +e. If we treat the nucleus as infinitely heavy, we let the origin of our coordinate system be at the nucleus (treated as a point particle). We may as well let the nucleus have charge +Ze instead of +e, because then our treatment will also include heliumlike ions such as H, Li + and Be +. The Hamiltonian is H = 1 m e m Ze Ze + e e r 1 r r 1 where the subscripts 1 and on refer to the coordinates of electrons 1 and, r 1 and r are the distances of the electrons from the nucleus, and r 1 is the distance from electron 1 to electron : r 1 = [(x 1 x ) + (y 1 y ) + (z 1 z ) ] 1/. By using the usual relation between Cartesian and spherical coordinates, it is possible to write r 1 in terms of r 1, θ 1, φ 1, r, θ, φ. It should be clear that this Hamiltonian represents the separate Hamiltonians of individual electrons in a hydrogenlike atom, plus an electron-electron interaction potential energy. The Schrödinger equation is a function of six variables, three for each electron. But because of the 1/r 1 term, the equation is not separable in any coordinate system, and we must turn to approximation techniques for its solution. To use perturbation theory, we must write H in the form H = H 0 + H where H 0 is the Hamiltonian of an exactly soluble problem. Let us write H 0 = 1 m Ze e r 1 m Ze e r H = e r 1. Then H 0 = H 0 1+H 0 is the sum of two independent hydrogenlike Hamiltonians, one for each electron. As such, the solution is the product of two unperturbed wave functions: ψ (0) (r 1, θ 1, φ 1, r, θ, φ ) = ψ (0) 1 (r 1, θ 1, φ 1 )ψ (0) (r, θ, φ ) with the unperturbed energies defined by E (0) = E (0) 1 + E (0) H (0) 1 ψ(0) 1 = E (0) 1 ψ(0) 1 and H (0) ψ(0) = E (0) ψ(0).
The energies E (0) i are just the usual hydrogenlike-atom energies given by E (0) i = Z n i e a 0 ; n i = 1,,... and hence the zeroth-order energies of our system are given by ) e ( 1 E (0) = Z n + 1 1 n a 0 where n 1 = 1,, 3,... and n = 1,, 3,.... The lowest level is non-degenerate with n 1 = n = 1, and the zeroth-order wave function is ψ (0) = 1 ( ) 3/ ( ) 3/ Z 1 Z e Zr1/a0 e Zr/a0 π 1/ a 0 π 1/ a 0 which is simply the product of two independent ground-state wave functions. Note that what we have is usually written as the 1s(1)1s() state, where the the number in parenthesis denotes which electron is in which state. In this case, both electrons are in the 1s state (or orbital). For our unperturbed system, the ground state (i.e., n 1 = n = 1) has energy E (0) = Z e a 0 where e /a 0 = 13.606 ev is the ground-state energy of the hydrogen atom. The definition of a 0 is /m e e, but if we replace m e by the reduced mass for 4 He (i.e., we no longer regard the nucleus as infinitely heavy), then e /a 0 becomes 13.604 ev, a somewhat more correct number. Using this together with Z = for helium, we take the unperturbed (zeroth-order) ground-state energy to be E (0) = 8(13.604) = 108.83 ev. The experimental first ionization energy of He is 4.59 ev, and the second ionization energy is easily calculated since it is the ionization energy of the hydrogenlike ion He +, and is equal to (13.604) = 54.4 ev. Thus the groundstate energy of the helium atom is (4.59 + 54.4) = 79.01 ev. So we see that our zeroth-order energy is off by 38%. This shouldn t be surprising since we have neglected the term e /r 1 which is not an insignificant energy. Now for the perturbation correction. Evaluate the first-order energy correction E (1) = ψ (0) H ψ (0), and find the percent error in the ground-state energy. Hints: The integral you must evaluate is over the six variables r 1, θ 1, φ 1, r, θ, φ. Use the fact that 1/r 1 3
can be expanded in terms of the spherical harmonics as (see e.g., Jackson s Classical Electrodynamics, 3rd ed. Section 3.6) 1 r 1 = l l=0 m= l 4π l + 1 r< l r> l+1 [Yl m (θ 1, φ 1 )] Yl m (θ, φ ) where r < means the smaller of r 1 and r, and r > is the larger of these. Use this expression in your integral, and then multiply and divide by Y0 0(Y 0 0) = 1/4π. Now use the orthonormality of the spherical harmonics. Next, first integrate r 1 where over the range 0 r 1 r you have r > = r, and over the range r r 1 you have r > = r 1. This will give you some definite integrals that you can explicitly do to arrive at a result that depends on r. Lastly, you will have integrals over r that you can also explicitly perform. The final result is a very simple, clean expression in terms of Z, e and a 0. 4. Suppose we want to know how the energy of a system depends on some parameter λ. For example, the energy levels of the one-dimensional harmonic oscillator are given by ( E n = ω n + 1 ) ( k = n + 1 ) m and this depends on both k and m. In other words, we want to know E n / λ. (a) From the Schrödinger equation H(λ)ψ n (λ) = E n (λ)ψ n (λ) we have E n = n H n. Assuming that ψ n is normalized, i.e., n n = 1, show E n λ = n H n. λ (b) We can use the result of part (a) to derive the first-order perturbation result as follows. As before, write H = H 0 + λv (where 0 λ 1) and E n = E n (0) + λe n (1) +. Now use the result of part (a) to show that E n (1) = n V n in agreement with our perturbation theory treatment. (c) The one-dimensional harmonic oscillator has the Hamiltonian d H = m dx + 1 kx. Letting λ = k, find x for any harmonic oscillator stationary-state without evaluating any integrals. 5. Neglecting spin, the interaction energy of a hydrogen atom in a magnetic field B = B z ẑ is given by H = µ l B = eb z m e c L z. (a) Using first-order perturbation theory, find the effect of H on the n = level of hydrogen. Sketch (i.e., diagram) the effect of H on the n = level. 4
(b) Now work out what happens if B = B xˆx. Compare your result with what you obtained in part (a). Why do you think it came out this way? 5