## Assignment 10, solutions

Problem 1:

Consider a hydrogenic atom with nuclear charge Zqe in a strong magnetic field B = Bk.  Assume that the Zeeman slitting is much larger than the spin-orbit splitting of the energy levels, so that to first order the spin-orbit interaction can be ignored.
(a)  Find the energies of the 2s and 2p energy levels in the strong magnetic field.  Is the degeneracy completely removed by the Zeeman interaction?
(b)  Estimate the magnitude of the magnetic field, B, required to give a Zeeman splitting in the hydrogenic atom comparable to the binding energy of the ground state of the hydrogen atom.  Can such a magnetic field be created in the laboratory?

Solution:

 Concepts: A magnetic moment in a magnetic field Reasoning: We can use first order stationary perturbation theory to find the energy levels of the perturbed hydrogenic atom. Details of the calculation: (a)  H = H0 + Hz,  H0 = p2/(2μ) – Ze2/r,  e2 = (4πε0),  μ = me = reduced mass of the electron.The eigenstates of H0 are {|n, l, m, ms>}, characterized by the 4 quantum numbers n, l, m, ms, with eigenvalues are En  = -Z2*13.6 eV/n2.  All n = 2 levels have energy E2 = -Z2*3.4 eV, the level is 8-fold degenerate.  The 2s level is two-fold degenerate and the 2p level is 6-fold degenerate. HZ = -μ∙B = -μzB.  μz = (-qe/(2me))(Lz + 2Sz) = -μB(Lz + 2Sz)/ħ. H0 commutes with L2 an Lz, so H0 commutes wit HZ.  The eigenstates of H are {|n, l, m, ms>} and the matrix of Hz in that basis is diagonal.The first order corrections to the energy eigenvalues therefore is ΔEn,l,m,ms = (qeħB/(2me))(m + 2ms) = μBB(m + 2ms),  where μB is the Bohr magneton.The eigenvalues are En  = -Z2*13,6 eV/n2 + μBB(m + 2ms).The 2s level splits in into two and the 2p levels into five sublevels.  The energy difference between the sublevels is ΔE = μBB.For the 2s level we have E2 = -Z2*3.4 eV ± μBB, and for the 2p level we have E2 = -Z2*3.4 eV ± mtμBB, where mt = 0, ±1, ±2 The degeneracy of the 2s level is completely removed, the degeneracy of the 2p level is only partially lifted.  The m = -1, ms = ½ and the m = -1, ms = -½ are not shifted and still degenerate.If we neglect all other interactions, then the n = 2 level splits into 5 sublevels, of which 3 are still two-fold degenerate. (b)  For μBB = 13.6 eV, we need B = 13.6 eV/ 5.79*10-5 eV/T ~ 2*105 T.The largest magnetic field created in the laboratory is less than 100 T.

Problem 2:

In the WKB approximation, find the allowed energies that a ball of mass m, bouncing due to gravity on a perfectly reflecting surface, can have.
You can use the fact that for this problem the WKB approximation gives
∮p dq  = (n - ¼)h.
where p(q) is the momentum of the ball at the height q and the integral is over a full periodic path.  You may leave your answer as an integral equation which could be solved to yield the energy levels.

Solution:

 Concepts: The WKB approximation Reasoning: We are instructed to use the WKB approximation.  For this problem the WKB approximation requires that  ∫q1q2 pdq = (n - ¼)(h/2) or ∮pdq = ∮ħkdq = (n - ¼)h.  Here ∮denote an integral over one complete cycle of the classical motion and  k2 = (2m/ħ2)(E - V(q)). Details of the calculation: For stationary bound states we want ∫qminqmax ħkdq = (n - ¼)(πħ), n = 1, 2, ... . In our problem k2 = (2m/ħ2)(E - mgq).  (Here q is the vertical coordinate.  The q-axis points upward and the origin lies at the surface.) We need (2m)½∫0qmax (E - mgq)½dq = (n - ¼)(πħ). Here mgqmax = E,  qmax = E/(mg). Therefore (2m)½∫0E/(mg) (E - mgq)½dq = (n - ¼)(πħ).

Problem 3:

In one dimension, the potential energy of an electron as a function of x is given by
U(x) = -30 eV exp(-x2/(4 Å2)).
Use the variational method to find the energy of the ground state in units of eV.

Solution:

 Concepts: The variational method Reasoning: The variational method often yields a very good estimate for the ground state energy of a system. Details of the calculation: The Hamiltonian is H = (-ħ2/2μ)(∂2/∂x2) - V0exp(-x2/a2)). Here V0 = 30 eV and a = 2 Å and μ = me. Choose ψ(x) = Aexp(-x2/b2).  Here b is an adjustable parameter. ψ(x) = 0 as x --> ±∞, and dψ/dx must be continuous at all x. A2∫-∞∞exp(-2x2/b2)dx = 1,  A2 = (2/πb2)½. = A2∫-∞∞ exp(-x2/b2)((-ħ2/2μ)(∂2/∂x2) - V0exp(-x2/a2))exp(-x2/b2)dx = 2A2(-ħ2/2μ)∫0∞ [-2/b2 + 4x2/b4]exp(-2x2/b2)dz - 2A2V0∫0∞ exp(-x2(2/b2 - 1/a2))dx = 2A2[(-ħ2/2μ)(2/b2)½ - V0(a2b2/(2a2 + b2)½]∫0∞ exp(-x2)dx - 2A2[(2ħ2/μb4)(b2/2)3/2∫0∞ x2exp(-x2)dx. ∫0∞ exp(-r2)dr = π½/2.  ∫0∞  r2exp(-r2)dr = π½/4. = ħ2/(2μb2) - 2½aV0/(2a2 + b2)½. d/db = 0  -->  -ħ2/(μb3) + 2½abV0/(2a2+b2)3/2 = 0. Let V0' = V0μa2/ħ2. Then a2/b2 - 2½ab2V0'/(2a2 + b2)3/2 = 0, a4/b4 = 2(a2b4)V0'2/(2a2 + b2)3, a4/b4 = 2(a2/b2)V0'2/(2a2/b2 + 1)3. Let y = a2/b2. Then y2 = 2yV0'2/(2y + 1)3.  y(2y + 1)3 = 2V0'2. Here V0' = 30 eV me a2/ħ2 = 17.5. y = 2.6,  b2 = a2/2.6. E = 2.6ħ2/(2μa2) - V0(0.916) = -25.2 eV.

Problem 4:

A proton and a neutron are confined by a three-dimensional potential.  For this problem assume that the proton and neutron do not interact with each other, and neglect spin-orbit interactions.  Both particles have spin ½.  Including the spins, the ground state is four-fold degenerate.  To this system we now add the interaction between the magnetic dipole moments of the particles described by the interaction Hamiltonian
H' = k Sp∙Sn,
where SpSn are the spin operators of the proton and neutron, respectively, and k is a positive constant.
(a)  Consider the following operators:
Sp2, Sn2, Spz, Snz, S2, and Sz,
where S = Sp + Sn.  State which of these operators commute with H'.
(b)  Into how many distinct energy levels does the original ground state split in the presence of H'?   Calculate the corresponding energies and state their degeneracy.

(c)  We now place the system into a uniform external magnetic field, which points in the pos­itive z-direction, B = B k.  The spin-spin interaction described by H' continues to be present and the additional interaction Hamiltonian is
H'B = b(Spz + Snz)Bz,
where b is a positive constant.
Calculate the corrections to the energies of the states identified in part (b) due to the presence of the magnetic field.
(d)  Sketch a graph of the energy levels as a function of the external magnetic field strength, B, including the effects of both H' and H'B.  Identify the curves with the corresponding states identified in part (b).

Solution:

 Concepts: Addition of angular momentum, common eigenbasis for J12, J22, J2, Jz. Reasoning: We find the eigenstate and eigenvalues of H' and then the corrections to these energy eigenvalues due to H'B. Details of the calculation: (a)  The operators  Sp2, Sn2, S2, and Sz all commute, and a common eigenbasis {|sp,sn;s,ms>} exists.  We have the singlet state |00> = ½½(|+-> - |-+>) and the triplet states  |11> = |++>,  |10> = ½½(|+-> + |-+>),  |1-1> = |-->. Sp∙Sn = ½(S2 – Sp2 – Sn2),  Spz + Snz = Sz,  the vectors {|s1,s2; s,ms>} are also eigenvectors of Sp∙Sn and  Spz + Snz. So Sp2, Sn2, S2, and Sz commute with H', Spz and Snz do not commute with H'.(b)  = (k/2)ħ2[s(s+1) – sp(sp+1) – sn(sn+1)].For the singlet state <0,0|H|0,0> = -(3k/4)ħ2.For the triplet states <1,ms|H|1,ms> = (k/4)ħ2.The original ground state split in into two energy levels.  The level (3k/4)ħ2 is non-degenerate and the level -(k/4)ħ2 is three-fold degenerate.(c)  Spz + Snz = Sz,  <1,ms|H'B|1,ms> = bBzmsħ,  ms = -1, 0, 1. The energy levels now are -(3k/4)ħ2,  (k/4)ħ 2 - bBħ,  (k/4)ħ 2,  and  (k/4)ħ 2 + bBħ.  (d)

Problem 5:

A harmonic oscillator in two dimensions has the unperturbed Hamiltonian
H = ½m(px2 + py2) + ½mω2(x2 + y2).
It is subjected to the perturbation H1 = Δxy, where Δ is the strength of the perturbation.
(a)  Write the eigenstates of the unperturbed oscillator in terms of the eigenstates of the 1-dimensional harmonic oscillator.  What are the eigenvalues of H?
(b)  Evaluate the first order corrections to the energies of the three states lowest in energy when Δ > 0.

Solution:

 Concepts:First order perturbation theory for non-degenerate and degenerate states Reasoning:Separation of variables is possible. The 2D harmonic oscillator eigenfunctions are products of the of the 1D harmonic oscillator eigenfunctions.  The ground state is non-degenerate, and the first excited state is two-fold degenerate.For the ground state ΔE = <0,0|H1|0,0> is the first order energy correction, using non-degenerate perturbation theory.  The first order corrections to the degenerate first excited state are found by diagonalizing H1 in the subspace of degenerate states. Details of the calculation:(a)  Let |n> and |m> denote the eigenstate of the 1D harmonic oscillator in x and y, respectively.The eigenstates of the 2D oscillator are {|n,m>}.H = Hx + Hy.  The operators px , x  and py, y, act in different subspaces.  The energy of the state |n,m> is Enm = (½ + n)ħω + (½ + m)ħω = (n + m + 1)ħω.For the ground state E00 = ħω.  For the first excited state E10 = E01 = 2ħω.   (b)  For a 1D Hamiltonian Hx = px2/(2m) + mω2x2/2 define a = αx + iβpx,  a† = αx - iβpx,  with α =√(mω/(2ħ)),  β =1/√(2mωħ). Then [a, a†] = 1 and Hx = ħω (a†a + ½).We have x = (a + a†)/(2α),  px = -i(a - a†)/(2β).For eigenstates |n> of Hx we have a†|n> = (n + 1)½|n + 1>, a|n> = n½|n - 1>. Similarly for the y-direction we have y = (b + b†)/(2α),  py = -i(b - b†)/(2β).For eigenstates |m> of Hy we have b†|m> = (m + 1)½|m + 1>, b|m> = m½|m - 1>.For the ground state ΔE = Δ<0,0|xy|0,0> = Δ<0x|x|0x><0y|y|0y>  = 0.  The first order energy correction for the ground state is zero. In the subspace spanned by {|1,0>, |0,1>} the matrix elements of xy are<1,0|xy|1,0> = <0,1|xy|0,1> = 0,  <1,0|xy|0,1> = <1x|x|0x><0y|y|1y> = (1/(2α)2) = <0,1|xy|1,0>. det(H' - ΔE) = ΔE2 -Δ2/(2α)4 = 0. The eigenvalues are ΔE = ±Δ/(2α)2. The first order energy corrections for the first excited states are ΔE = ±Δ/(2α)2,The degeneracy is removed.