Quantum Mechanics (Draft 2010 Nov.)

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1 Quantum Mechanics (Draft 00 Nov) For a -dimensional simple harmonic quantum oscillator, V (x) = mω x, it is more convenient to describe the dynamics by dimensionless position parameter ρ = x/a (a = h ) and dimensionless energy ɛ = E/( mω hω) Write down the time-independent Schrodinger equation in terms of ρ derivatives on the eigenfunction u(ρ) Directly show that u 0 (ρ) e ρ satisfies your Schrodinger equation Explain why it is the ground state and give its energy ɛ 0 in the dimensionless unit Directly show that u (ρ) ρe ρ satisfies the Schrodinger equation as the first excited state Also give its energy ɛ The initial wave function u(ρ, 0) is described by u(ρ, 0) = N( ρ ) exp( ρ ), with N as the normalization constant (i) Find the average energy (ii) The momentum operator in the reduced unit is = id/dρ Find its mean value initially (iii) We use a dimensionless time parameter τ = ωt to study how the wave evolves Write down the explicit τ dependence of the wave function (iv) Find as a function of time τ d dρ u(ρ)+ρ u(ρ) =ɛu(ρ) d We can pretend that ω =, m =, h = u d dρ 0(ρ) = ρu(ρ), u dρ 0 (ρ) = (ρ )u(ρ), so u 0 is a solution with ɛ 0 = u 0 has no node, thus it is a ground state Similarly u, with one node, is the first excited solution for ɛ = u(ρ, 0) = N( ρ )e ρ = Nπ (u (ρ) u 0 (ρ)), =N π( + ), N = u(ρ, 0) = 5 (u (ρ) u 0 (ρ)), u(ρ, τ) = 5 (u (ρ)e iτ u 0 (ρ)e iτ ) The average energy is ɛ = = 5 5 u(ρ, τ) = ( ρe iτ e iτ )e ρ 5π,u (ρ, τ) = ( ρe iτ e iτ )e ρ 5π τ = i(d/dρ)u(ρ, τ) = i i 5π 5π ( ( ρ )e iτ +ρe iτ )e ρ ( ρ e +iτ e iτ + ρ e iτ ) e ρ dρ = 5 sin τ The initial τ is zero, but it changes as a sine function 5π In the beginning, a non-relativistic particle of mass m = propagates freely as a wave packet with a mean wave number q = We choose a unit system such that h = Find 7 the group velocity and the phase velocity

2 Let this wave propagates from the remote left toward a repulsive potential V (x) = 9(x ) in the region [, ] The potential vanishes otherwise We can treat the potential as a delta-function potential Give quantitative reasons why Calculate the probabilities of transmission and reflection in the delta potential approximation The momentum is 7, the particle velocity (ie the group velocity) v g = p m = 8 7 K = p = 6 6 ω = = m 79 K h, the phase velocity is v 79 phase = ω = k 7 The wavelength of the particle is π/q = 5π/, which is much greater than the potential width of a size only / The height of the potential is also much greater than K Therefore we can treat the potential as delta function V (x) = Gδ(x) The strength G is given by the integral G = 9(x )dx = 9 We integrate both sides of Schrodinger equation over a small interval around the origin Ae ikx + B ikx, for x<0, ϕ k (x) = Ce ikx + D ikx, for x>0, The B component represents the reflection and the C piece is the transmission Although for an incoming wave from the left, we do not need the D component, we keep it for other future purpose temporarily Continuity of the wave function and its kink in derivatives around the origin gives h m We can solve A and B in terms of C and D, A + B = C + D ik[(a B) (C D)] + G(C + D) = 0 ( A B ) = ( + mig h k mig h k + mig h k mig h k )( C D Now, we require D = 0 for the property of an incoming particle from the left So T (k) C/A = h k/( h k + mig), R(k) B/A = mig/( h k + mig) T 7 = + i = 6 5 = 6% 7 7 Transmission probability is 6%, and the rest 6% is reflected ) For matrices A, B, C, show that [AB, C] =A[B, C] + [A, C]B Let S (or S x ) be the x-component of the spin vector operator, etc Using the algebra of angular momentum, [S x,s y ]=i hs z, and cyclic permutations for a general spin (S =,,, ), simplify [Sx,S z] and [Sy,S z] in terms of S x S y or S y S x Then calculate [S,S z ] The trace of a

3 matrix A is defined as Tr A = m m A m summing over the basis vectors m Show that Tr (AB) = Tr (BA) Based of some earlier steps, show that Tr (S x S y ) = 0 A dark matter (DM) particle of spin S and mass M DM couples to the fixed target nucleus of spin I by a weak spin-dependent contact interaction V = gδ (r)s I What are possible numbers of S I in general? Show all these possibilities for the special case S =, I = 9 Justify the following trace relations, Tr (S i S j )= C S δ ij, similarly Tr (I i I j )=C I δ ij, Determine coefficients C S and C I in terms of general S and I respectively Calculate the transition probability kk, m S,m I V k,m S,m I from an incoming DM plane wave described by e ik r to the outgoing DM wave e ik r The sum adds up all spin states m S and m S of the initial and final spin states of the DM particle, as well as m I and m I of nucleus After average the initial spins of the DM particle and the nucleus, find the total cross section in the Born approximation (Hints: The usual potential scattering in the Born approximation is dσ dω = M π h d rv (r)e i(k f k i ) r The above formula has to be generalized to incorporate the spins of S and I) [AB, C] =ABC CAB = ABC ACB + ACB CAB = A[B, C] + [A, C]B [S y,s z]=s y [S y,s z ] + [S y,s z ]S y = is y S x + is x S y [S x,s z ]=S x [S x,s z ] + [S x,s z ]S y = is x S y is x S y As [S z,s z] = 0, we have [S x + S y + S z,s z] = 0, so [S,S z ] = 0 Tr (AB) = i,j A ij B ji = i,j B ji A ij = Tr (BA), and Tr[A, B] = 0 So Tr(S x S y ) = 0 Let J denote the magnitude number of the sum J = S + I S I = [(S + I) S I ]= h [J(J + ) S(S + ) I(I + )], for discrete choices of J = S + J, S + J,, J S For the special case S = and I = 9, we have two possibilities, (a) J = 5, S I = 9, (b) J =, S I = The overall tracelessness is checked About Tr (S i S j )= C S δ ij, we have verified the case i = x, j = y Other unequal i j cases are also true if we permute indices Symmetry also implies Tr (S x S x ) = Tr (S y S y ) = Tr (S z S z )=C S Tr (S x S x )= Tr S = h (S + )S(S + )

4 We note that Tr (S I) = i,j = Tr [(S i I i )(S j I j )] = h (S+)S(S+)(I+)I(I+) 9 kk, m S,m I V k,m S,m I = g h (S + )S(S + )(I + )I(I + ) 9 Dividing it by section as S+ I+ for the average of initial spins, we obtain the unpolarized cross dσ dω = M DM 6π g S(S + )I(I + ) Variation principle A quantum particle in a two dimensional potential V (r) = V 0 exp( r/a) Let the trial wave function be ϕ(r; β) =Ce βr Determine the normalization C in terms of the attenuation parameter β Find the average position and average momentum x = x, ȳ = y, p x = p x, p y = p y Determine r and p (Hints: d rψ (r) ψ(r) = d r ψ(r) ) What are x and p x? Calculate the product (x x) (p x p x ), and simplify the result in comparison with the Heisenberg uncertainty Find the average kinetic energy and average potential energy Now we set h =, V m 0 = 8, a = Using the variation principle, estimate the ground state energy The optimized β turns out to be a simple number Normalization is given by C (πr)e βr dr =πc! (β) = πc β = C = β π By symmetry x = 0, ȳ = 0, p x =0, p y = 0 r = C (πr )e βr dr =πc! (β) = β p = h C (πr)β e βr dr = h β By symmetry, p x = p y = p = h β Similarly, x = r = Therefore β (x x) (p x p x ) = 8 h, well above the lowest bound h by Heisenberg s uncertainty principle Therefore, the average kinetic energy K = h β, and m V = C (πr)v 0 e (β+/a)r dr = V 0 β /(β + a ) For the given parameters, E = β 8β /(β + ), which is minimized at β = with ( E ) min =, as a good estimate of the ground state energy

5 5 Matrix diagonalization The dynamic of a three-state system of configurations,,, is governed by the Hamiltonian H 0 = i,j i j, which is Work out the Hamiltonian matrix elements i H 0 j, and present the corresponding matrix Find eigen energies as well as the corresponding eigenstates A small perturbation gv = g is applied (g ) Find the st order and nd order corrections of the ground state energy The matrix of Hamiltonian of H 0 is It is easy to see the eigen-column-vectors and eigen-values are N a,e a = ; N b,e b = 0 ; N c 0,E c =0 The normalizations are N a =,N b = 6, N c = As b and c are degenerate in energy, their other superpositions are also acceptable The ground state and the two excited states are a 0 = ( + + ) b 0 = 6 ( + + ) c 0 = (+ ) We find the following matrix elements of the perturbed interaction, 0 a gv a 0 = g 0 b gv a 0 = 8 ( g) 0 c gv a 0 =0g The ground energy is corrected by perturbation as E a = + g 7 g + 5