Problem Set 6 215B-, Winter 2019

Due: Monday March 11, 2019 by 5pm Put homework in mailbox labelled 215B on 1st floor of Broida (by elevators).

1.) VARIATIONAL METHOD FOR 1d OSCILLATOR

To get a feel for how well the variational approximation works, it is useful to try it out on a simple soluble model - here the 1d Harmonic oscillator with Hamiltonian,

Pˆ2 1 H = + mω2Xˆ 2. (1) 2m 2 Using the following variational wave functions: −α|x| a.) ψα(x) = e , −αx2 b.) ψα(x) = e ,

c.) ψα(x) = α − |x| for |x| ≤ α and ψ = 0 otherwise, estimate the energy of the harmonic oscillator, and discuss how your estimate compares with the exact result, E0 =hω/ ¯ 2. A variational bound on the energy of the first , E1, can be obtained by using an odd variational (why?). Estimate E1 using −α|x| d.) ψα(x) = xe , −αx2 e.) ψα(x) = xe , and compare to the exact result. The quality of your variational estimate for the ground state wavefunction can be determined by evaluating the overlap between the (normalized) variational state |ψvari and the exact (normalized) ground state |0i as

p 2 O = |hψvar|0i| . (2)

Bonus: Evaluate O for the three variational states (a), (b) and (c) and discuss.

2.) FOR A NON-LINEAR OSCILLATOR

Consider the Hamiltonian Hˆ = Hˆ0 + Hˆ1, where Hˆ0 is the (beloved!) Harmonic oscillator Hamiltonian, 1 1 Hˆ = pˆ2 + mω2xˆ2, (3) 0 2m 2 and Hˆ1 is a perturbation produced by additional cubic and quartic terms, which can be expressed in the dimensionally transparent form as, 1 Hˆ = mω2xˆ2[g (ˆx/x ) + g (ˆx/x )2], (4) 1 2 3 0 4 0

1/2 where x0 = (¯h/2mω) is a characteristic length and g3 and g4 are dimensionless coupling constants. (If Hˆ0 is regarded as an approximate Hamiltonian for a particle near the minimum of a general potential well, V (x), then Hˆ can be regarded as a more accurate representation, obtained by taking terms throughx ˆ4 in the expansion of V (x) about it’s minimum at x = 0.

A.) Calculate the change in the ground state energy and wave function to linear order in g3 and g4.

B.) Calculate the change in the ground state energy to second order in g3 and g4. † Hint: It will be useful to expressx ˆ in terms of oscillator raising and lowering operators asx/x ˆ 0 =a ˆ +a ˆ . 2

3.) TWO-LEVEL SYSTEM Physical systems that can be modeled in terms of a simple 2-level quantum mechanical system are common, a classic example being two nearby tunneling states for an impurity atom in a solid. More generally, the two states can often represent two adjacent energy levels in the spectrum of a more complicated system. ∗ The Hamiltonian for a two-state system can be represented as a 2 × 2 Hermitian matrix, Hij = Hji with i, j = 1, 2. In the absence of time reversal symmetry breaking terms the Hamiltonian is real, and hence symmetric, and in general is given by,

(0) (0) H11 = E1 ,H22 = E2 ,H12 = H21 = ∆, (5)

(0) (0) where E1 and E2 are the “unperturbed” energies, and ∆ is a (real) tunneling matrix element. For ∆ = 0, the (0) T (0) T energy are given by, φ1 = (1, 0) and φ2 = (0, 1) .

A.) Solve this problem exactly to find the energy eigenfunctions ψ1 and ψ2, and associated eigenvalues, E1 and E2.

(0) (0) B.) Assuming that ∆ << |E1 − E2 |, use time independent perturbation theory to solve for the eigenfunctions ψ1, ψ2 up to first order in ∆, and the eigenvalues up to second order. Compare with the exact expressions.

C.) Show that, up to an additive constant, Hij is equivalent to the Hamiltonian of a -1/2 particle (with gyromagnetic ration γ) in a magnetic field,

H = −γS~ · B,~ (6)

~ (0) with B = Bxxˆ + Bzzˆ, and express Bx and Bz in terms of Ei and ∆. (0) repulsion: When Ei correspond to the adjacent energy levels in a very complicated quantum system (eg. a high Z-nucleus, or electron states in a disordered solid) with mixing ∆, it is often useful to presume that these parameters are essentially random. In this case, the (exact) energy level spacing E = |E1 − E2| will also be random. To be concrete, assume that the (“fictitious”) magnetic field B~ is uniformly distributed in the Bx − Bz plane, with |B~ | ≤ B0. D.) Under this assumption, compute the energy level spacing distribution P (E) (where P (E)dE = probability that energy spacing is in the interval E to E + dE), and show that it vanishes as E → 0, varying as P (E) ∼ En. Determine the exponent n. The vanishing probability for degenerate energy levels is referred to as “level repulsion” - an extremely useful concept in complicated quantum systems.

4.) VAN DER WAALS INTERACTION The van der Waals interaction is a weak interaction between two neutral atoms, which are separated by a distance much larger than their size. It arises from the interaction between fluctuating electric on the two atoms. Here, consider the van der Waals interaction between two atoms, with their protons fixed in space, at a distance r along the z− axis from one another. In terms of ~r1 (a vector from the first proton to its electron) and ~r2 (from the 2nd proton to its electron) the Hamiltonian can be written as Hˆ = Hˆ0 + Hˆ1 with

2 2 ˆ 1 ~ 2 1 ~ 2 e e H0 = P1 + P2 − − , (7) 2m 2m r1 r2

e2 e2 e2 e2 Hˆ1 = + − − , (8) r |~r + ~r2 − ~r1| |~r + ~r2| |~r − ~r1| with ~r = rzˆ. In the ground state of H0, the two electrons are confined within a of their protons, so that ri ∼ a0.

A.) For large separation, r >> a0, show that H1 can be approximated as, e2 H = (~r · ~r − 3z z ) + O(1/r4), (9) 1 r3 1 2 1 2 which corresponds to the interaction between two electric dipoles, e~ri, separated by ~r. 3

B.) Treating this approximate form of H1 as a perturbation, show that the first order correction to the ground state energy of H0 vanishes. C.) Show that the 2nd order correction is non-vanishing and corresponds to an attractive van der Waal interaction between the two Hydrogen atoms, which varies as 1/r6.