Physics 221B: Solution to HW # 6. 1) Born-Oppenheimer for Coupled Harmonic Oscillators

Transcription

1 Physics B: Solution to HW # 6 ) Born-Oppenheier for Coupled Haronic Oscillators This proble is eant to convince you of the validity of the Born-Oppenheier BO) Approxiation through a toy odel of coupled D haronic oscillators. Since we now how to solve the syste exactly we can copare the exact solution to BO. The D Hailtonian we will investigate is H = p + p + p + x x d) + x x d) ) a) Born-Oppenheier Approxiation In the BO approxiation we are first taing, leaving only x as a dynaical degree of freedo. Then, once we solve the proble for x we use x s energy levels as a potential for x and x. Sending we can ignore the inetic energy ters for x and x, which fixes their values. just to reind ourselves that x and x are fixed, we will call the X and X. Rewriting ), the Hailtonian is now H = p + x X d) + X x d). Cobining into one haronic oscillator we get H = p + x X ) + X + X X d). The second ter is the left-over constant fro copleting the square. We now the solution is E n = ω n + ) + X X d) with ω = ) Now we can treat as a finite ass, freeing X and X to be x and x, and use Eq. ) as the potential for the heavy particles. The Hailtonian is now H n = p + p + ω n + ) + x x d).

2 This has a well nown solution as well. we can go to the center of ass frae, defining to get x = x x µ = H n = P ) + p µ) + + ω n + ) + )x d) where P is the oentu of the center of ass. If we set P = the energy is E n,n = ω n + ) + ω n + ) with ω = and ω = / µ = The BO approxiation is a siple illustration of an iportant concept in physics called an effective theory. When a syste contains two very different frequencies two different energy scales) we can treat the separately i.e. solve for the high frequency odes alone integrate the out of the theory ). Once we solve for the high frequency odes, those degrees of freedo are no longer in our syste, but they have left their ar by changing the potential for the low odes. b) We can solve this proble exactly by diagonalizing the Hailtonian the way we do in classical echanics and then apply our nowledge on the haronic oscillator in quantu echanics. There are several equivalent approaches to doing this I chose to wor with the classical equations of otion. Writing Hailton s equations and cobining the) we get ẍ = x x ) ẍ = x x ) ẍ = x + x x ) This is called integrating out because in the path integral forulation we are perforing the integral over paths with high frequency, leaving an integral over low frequency paths with a different effective) action.

3 Rewriting this in atrix for, ẍ X ẍ = ẍ x x x K X ) Assue a solution of the for x i e iωt. Plugging this in 4) and rearranging gives a hoogeneous equation for X + ω + ω + ω x x x = 4) which coes down to solving for the eigenvalues of the atrix K defined above. The eigenvalue and their eigenvectors vectors coe out to be: ω = with ω = with ω = + with The first ter is just the center of ass otion, and the other two are the two vibrational odes we would have guessed. There is no real need to wor out the inetic ters because having diagonalized the Hailtonian we now we will get the regular p / for the new odes. Finally we get the exact solution for the energy spectru E n,n = ω n + ) + ω n + ) with ω = and ω = +

4 c) Coparing the results of a) and b), we can identify ω of part a) with ω of part b) exactly. We can explain this by noting that in the second ode of our exact solution the light ass does not ove at all. Therefore, the BO approxiation that separates the otion of the heavy and the light asses doesn t iss anything, regarding this ode. When we tae the liit again we see that ω ω. So the BO approxiation is issing corrections of O/). ) Cheistry We choose to ae orthonoral ixed orbitals. For reference, the wavefunctions in atoic units are a) b) s) = π r )e r/ p ) = 4 π re r/ cos θ = 4 π ze r/ p ± ) = 8 π re r/ sin θ e ±iφ p x ) = p ) p + )) = p y ) = i p ) + p + )) = 4 π xe r/ 4 π ye r/ The s) is spherically syetric so won t affect the direction of the pea of the wavefunction. To achieve angles of φ =, π/, 4π/, the non-s) parts of the wavefunctions should go as p x ), py ) p x ), p y ) p x ). Adding in the s) contributions and orthonoralizing, sp ) φ= = s) + p x )) sp ) φ= π sp ) φ= 4π = 6 s) + p y ) p x )) = 6 s) p y ) p x )). The s) pieces appear with the sae weight in all three wavefunctions, as we would hope. Once again, I than Ed Boyda. 4

5 c) Rather than tae the top of the tetrahedron along the z axis, the wily cheists ae a tetrahedron toward the 4 corners of a cube whose faces are centered about the x, y, z axes. Tae the four corners in the xyz) = + + +), +), + ), + ) octants). Adding the s) piece to achieve orthonorality, the sp orbitals are sp ) +++ = s) + p x) + p y ) + p z )) sp ) + = s) p x) p y ) + p z )) sp ) + = s) p x) + p y ) p z )) sp ) + = s) + p x) p y ) p z )). See the atheatica file for plots. Three diensional odels that aided in visualization got extra credit. 5