Topics Covered: Motion in a central potential, spherical harmonic oscillator, hydrogen atom, orbital electric and magnetic dipole moments

Transcription

1 PHYS85 Quantum Mechanics I, Fall 9 HOMEWORK ASSIGNMENT Topics Covered: Motion in a central potential, spherical harmonic oscillator, hydrogen atom, orbital electric and magnetic dipole moments. [ pts] A particle of mass M and charge q is constrained to move in a circle of radius r in the x y plane. (a) If no forces other than the forces of constraint act on the particle, what are the energy levels and corresponding wavefunctions? If the particle is forced to remain in the x-y plane, then it can only have angular momentum along the z-axis, so that L = L z e z and L = L z. The kinetic energy can be found two ways: Method : Using our knowledge of angular momentum. We start by choosing φ as our coordinate H = L I = L z Mr () so that the eigenstates are eigenstates of L z i partial φ, from which we see know that the energy levels are then E m = m, where m = Mr, ±, ±, ± , and the wavefunctions are φ m = π e imφ. Method : Solution from first principles. We start by choosing s as our coordinate, where s is the distance measured along the circle. The classical Lagrangian is then L = Mṡ the canonical momentum is p s = s L = Mṡ. The Hamiltonian is then H = pṡ L = p s M () (3) promoting s and p s to operators, we must have [S, P s ] = i, so that in coordinate representation, we can take S s, and P s i s, which gives the energy eigenvalue equation is then H = M s (4) or equivalently M s ψ(s) = Eψ(s) (5) s ψ(s) = ME ψ(s) (6)

2 This has solutions of the form: single-valuedness requires which means where m is any integer. This gives so that ψ(s) e ±i ME s (7) ψ(s + πr ) = ψ(s) (8) ME πr = πm (9) E = m Mr () ψ m (s) = eims/r πr () Both methods agree because s = r φ. (b) A uniform, weak magnetic field of amplitude B is applied along the z-axis. What are the new energy eigenvalues and corresponding wavefunctions? Using the angular momentum method, we now need to add the term qb M L z to the Hamiltonian to account for the orbital magnetic dipole moment, which gives H = L z Mr qb M L z () so that the eigenstates are still L z eigenstates, ψ m (φ) = eimφ π, where m =, ±, ±,..., but the degeneracy is lifted so that E m = m Mr qb M m (3) (c) Instead of a weak magnetic field along the z-axis, a uniform electric field of magnitude E is applied along the x-axis. Find an approximation for the low-lying energy levels that is valid in the limit qr E /Mr. Hint: try expanding around the potential about a stable equilibrium point. Here we need to add the electric monopole energy. The electrostatic potential of a uniform E-field along e x is φ( r) = E x, so that the potential energy is U = qe x. The full Hamiltonian of the particle is then given by L z H = Mr qe r cos(φ) (4) The stable equilibrium point is at φ =. Expanding to second-order about the equilibrium then gives H = Mr φ qe r + qe r φ (5) This is just a harmonic oscillator Hamiltonian, with M eff = Mr, and ω = qe Mr, so that the energy levels are ( qe E n = qe r + n + ) (6) Mr

3 where n =,,,.... This approximation must be valid only when the level spacing is small compared to the depth of the cos potential, so that qe qe r (7) Mr which is equivalent to Mr qe r (8) 3

4 . [ pts] Write out the fully-normalized hydrogen wavefunctions for all of the 3p orbitals. Expand out any special functions in terms of elementary functions. You can look these up in a book or on-line, but keep in mind that you will be penalized if your expression is not properly normalized. We have ψ n,l,m (r, θ, φ) = 8(n l )! n(a n) 3 (n + l)! e r/a n ( ) r l ( ) L (l+) r a n n l Yl m (θ, φ) (9) a n Using Mathematica, I then get for n = 3 and l =, ψ 3,, (r, θ, φ) = ψ 3,, (r, θ, φ) = ψ 3,, (r, θ, φ) = Normalization checks out: 8a 7/ e r/3a (6a r)r sin θe iφ () π 8a 7/ e r/3a (6a r)r cos θ () π 8a 7/ π e r/3a (6a r)r sin θe iφ () Untitled- In[83]:= Clear@y, n, l, m, r, q, f, ad In[863]:= 8 Hn - l - L! y@n_, l_, m_, r_, q_, f_d := $%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ %%%%%%%%%% n Ha nl 3 Hn + ll! ExpA -r ÅÅÅÅÅÅÅÅ a n E i k j ÅÅÅÅÅÅÅÅ r l y z LaguerreLAn - l -, l +, ÅÅÅÅÅÅÅÅ r E SphericalHarmonicY@l, m, q, fd a n { a n In[864]:= y3 = FullSimplify@y@3,,, r, q, fdd Out[864]= "####### ÅÅÅÅÅ r a 3 - ÅÅÅÅÅÅ 3 a +Â f r H-6 a + rl Sin@qD ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 8 a è!!! p In[865]:= y3 = FullSimplify@y@3,,, r, q, fdd Out[865]= "####### ÅÅÅÅÅ r a 3 - ÅÅÅÅÅÅ 3 a "##### ÅÅÅ H6 a - rl r Cos@qD p ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 8 a In[866]:= y3m = FullSimplify@y@3,, -, r, q, fdd Out[866]= "####### ÅÅÅÅÅ r a 3 - ÅÅÅÅÅÅ 3 a -Â f H6 a - rl r Sin@qD ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 8 a è!!! p In[87]:= Integrate@Conjugate@y3D y3 r Sin@qD, 8f,, p<, 8q,, p<, 8r,, <, Assumptions Ø a > D Out[87]= In[87]:= Integrate@Conjugate@y3D y3 r Sin@qD, 8f,, p<, 8q,, p<, 8r,, <, Assumptions Ø a > D Out[87]= 4

5 3. [ pts] Numerically compute the matrix elements of the z-component of the orbital electric and magnetic dipole moments for the,, and transitions in hydrogen. Be sure to show your work. For the electric dipole moments, we need to compute e i Z f = e i R cos Θ f. The selection rules are m f = m i and L f = L i ±. Of these three transitions, only satisfies these selection rules. Using the wavefunction from., and mathematica, and taking a = 5. m and e =.6 9 C, we find ez = (3) π π ez = dr r dθ cos θ dφ ψ,,(r, θ, φ)r cos θψ,, (r, θ, φ) = Cm (4) ez = (5) For the magnetic dipole moments, we need µ = e m e L z, so the selection rule is m i = m f. The dipole moment is then µ = e m e m l. This gives zero for all transitions. Note that when spin is included, there will can be non-zero magnetic dipole transitions between these levels. 5

6 4. [5 pts] Based on the classical relation E = T +V, where E is the total energy, T is the kinetic energy, and V is the potential energy, what is the probability that the velocity of the relative coordinate exceeds the speed of light for a hydrogen atom in the s state? What about the s state? Based on these answers, which of the two energy levels would you expect to have a larger relativistic correction? Using H = T + V and T = mv, we find v = m (E V ) so for the hydrogen system with principle quantum number n this gives v (r) = ] [ m ma n + e 4πɛ r Setting this equal to c and solving for r c gives r c (n) = ma n e πɛ (m a c n + ) with the parameters (from Google) m = 9. 3 kg, a = 5.9 m, e =.6 9 C, ɛ = 8.85 C N m, c = 3. 8 ms, and =.5 34 Js, we find: For n = : r c () = m For n = : r c () = m So we see that dependence on n is very weak. The probability to be within this radius, however, depends strongly on n. For n =, we have for n = we have P (r < r c ()) = P (r < r c ()) = rc() rc() rc()/a dr R(r) = 4 dx e x x = 8. 3 rc()/a dr R(r) = dx e x x ( x ) = 4. 3 Therefore we would expect the ground-state to have the larger relativistic correction. 6

7 5. [ pts] Consider the Earth-Moon system as a gravitational analog to the hydrogen atom. What is the effective Bohr radius (give both the formula and the numerical value). Based on the classical energy and angular momentum, estimate the n and m quantum numbers for the relative motion (take the z-axis as perpendicular to the orbital plane). The Bohr radius for Hydrogen is given by a = 4πɛ me From wikipedia I found M M = 7.35 kg, M E = kg, r M = m, and v M =. 3 ms To compute the Bohr radius for the moon, we just need to make the substitutions This gives The classical energy is Solving for n gives m µ = M MM E = M M + M E = 7.6 kg e 4πɛ GM M M E = (6.67 )( )(7.35 ) = a M = GMM M = m E E = µv E GM MM E r M n = E = µa M n = J µa M E = To calculate m, we take L z = µv M r M and us m = L z = µv Mr M = Just for fun: For a transition from n to n, the energy released is [ ] E = µa M n (n ) = µa M (n ) n n (n ) = µa M n n (n ) µa M This gives a numerical result of E =.76 4 J. With λ = π c/ E we find λ = 7. 4 m. Using lyr = m we find that λ =.75 light years. The lunar month is 7. days, or.74 years. Coincidence? n 3 7