1 / 45

Let’s consider the consequences of this commutator further [A,B] = 0

Let’s consider the consequences of this commutator further [A,B] = 0. Let’s consider the consequences of this commutator further [A,B] = 0 Here is the S equation for the H atom. Let’s consider the consequences of this commutator further [A,B] = 0 Here is the S equation for the H atom.

patrickjones
Download Presentation

Let’s consider the consequences of this commutator further [A,B] = 0

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Let’s consider the consequences of this commutator further [A,B] = 0

  2. Let’s consider the consequences of this commutator further [A,B] = 0 Here is the S equation for the H atom

  3. Let’s consider the consequences of this commutator further [A,B] = 0 Here is the S equation for the H atom

  4. Let’s consider the consequences of this commutator further [A,B] = 0 Here is the S equation for the H atom 1 2 B A

  5. Let’s consider the consequences of this commutator further [A,B] = 0 Here is the S equation for the H atom 1 2 B A R

  6. Let’s consider the consequences of this commutator further [A,B] = 0 Here is the S equation for the H atom 1 r12 2 B A R

  7. Let’s consider the consequences of this commutator further [A,B] = 0 Here is the S equation for the H atom 1 r12 2 r1A B A R

  8. Let’s consider the consequences of this commutator further [A,B] = 0 Here is the S equation for the H atom 1 r12 2 r1A r2B B A R

  9. Let’s consider the consequences of this commutator further [A,B] = 0 Here is the S equation for the H atom 1 r12 2 r1A r2B r1B B A R

  10. Let’s consider the consequences of this commutator further [A,B] = 0 Here is the S equation for the H atom 1 r12 2 r1A r2A r2B r1B B A R

  11. Let’s consider the consequences of this commutator further [A,B] = 0 Here is the S equation for the H atom Now let us consider something that may seem a bit odd the permutation operators P12 or PAB and their effect on H

  12. Let’s consider the consequences of this commutator further [A,B] = 0 Here is the S equation for the H atom Now let us consider something that may seem a bit odd the permutation operators P12 or PAB and their effect on H P12 permutes the coordinates of particles 1 and 2

  13. Let’s consider the consequences of this commutator further [A,B] = 0 Here is the S equation for the H atom Now let us consider something that may seem a bit odd the permutation operators P12 or PAB and their effect on H P12 permutes the coordinates of particles 1 and 2 the electrons

  14. Let’s consider the consequences of this commutator further [A,B] = 0 Here is the S equation for the H atom Now let us consider something that may seem a bit odd the permutation operators P12 or PAB and their effect on H P12 permutes the coordinates of particles 1 and 2 the electrons PAB permutes the coordinates of particle A and B

  15. Let’s consider the consequences of this commutator further [A,B] = 0 Here is the S equation for the H atom Now let us consider something that may seem a bit odd the permutation operators P12 or PAB and their effect on H P12 permutes the coordinates of particles 1 and 2 the electrons PAB permutes the coordinates of particle A and B the protons

  16. I think it is obvious that the Hamiltonian is unaffected and so Pij and the Hamiltonian commute ie [H, Pij] = 0

  17. I think it is obvious that the Hamiltonian is unaffected and so Pij and the Hamiltonian commute ie [H, Pij] = 0 Thus P12 Ψ(1,2) = pΨ(1,2)

  18. I think it is obvious that the Hamiltonian is unaffected and so Pij and the Hamiltonian commute ie [H, Pij] = 0 Thus P12 Ψ(1,2) = pΨ(1,2) but also P12 Ψ(1,2) = Ψ(2,1)

  19. I think it is obvious that the Hamiltonian is unaffected and so Pij and the Hamiltonian commute ie [H, Pij] = 0 Thus P12 Ψ(1,2) = pΨ(1,2) but also P12 Ψ(1,2) = Ψ(2,1) and P12P12 Ψ(1,2) = P12 Ψ(2,1) = Ψ(1,2)

  20. I think it is obvious that the Hamiltonian is unaffected and so Pij and the Hamiltonian commute ie [H, Pij] = 0 Thus P12 Ψ(1,2) = pΨ(1,2) but also P12 Ψ(1,2) = Ψ(2,1) and P12P12 Ψ(1,2) = P12 Ψ(2,1) = Ψ(1,2) So as P12P12 Ψ(1,2) = p2Ψ(1,2)

  21. I think it is obvious that the Hamiltonian is unaffected and so Pij and the Hamiltonian commute ie [H, Pij] = 0 Thus P12 Ψ(1,2) = pΨ(1,2) but also P12 Ψ(1,2) = Ψ(2,1) and P12P12 Ψ(1,2) = P12 Ψ(2,1) = Ψ(1,2) So as P12P12 Ψ(1,2) = p2Ψ(1,2) p2 = 1 and p = ± 1

  22. I think it is obvious that the Hamiltonian is unaffected and so Pij and the Hamiltonian commute ie [H, Pij] = 0 Thus P12 Ψ(1,2) = pΨ(1,2) but also P12 Ψ(1,2) = Ψ(2,1) and P12P12 Ψ(1,2) = P12 Ψ(2,1) = Ψ(1,2) So as P12P12 Ψ(1,2) = p2Ψ(1,2) p2 = 1 and p = ± 1 So two different types of quantum particles exist … those for which the total wave function on interchange stays the same i.e. p = +1 or changes sign p = – 1

  23. It is found empirically that p = +1 for integral spin particles

  24. It is found empirically that p = +1 for integral spin particles D 1 N 1 Photons 1

  25. It is found empirically that p = +1 for integral spin particles D 1 N 1 Photons 1 and follow Bose-Einstein statistics and are called Bosons

  26. It is found empirically that p = +1 for integral spin particles D 1 N 1 Photons 1 and follow Bose-Einstein statistics and are called Bosons p = –1 for half-integral spin particles

  27. It is found empirically that p = +1 for integral spin particles D 1 N 1 Photons 1 and follow Bose-Einstein statistics and are called Bosons and p = –1 for half-integral spin particles electrons ½ protons ½ chlorine nuclei 3/2

  28. It is found empirically that p = +1 for integral spin particles D 1 N 1 Photons 1 and follow Bose-Einstein statistics and are called Bosons and p = –1 for half-integral spin particles electrons ½ protons ½ chlorine nuclei 3/2 and follow Fermi-Dirac statistics and are called Fermions

  29. ↑ ↓ H2 Protons ↑ ↑ ↑ ↑ ↓ ↓↑ ↓↓ ↓  ↑ ↑and ↓↓  are already symmetric Opposing off-diagonals can form a symmetric and an antisymmetric combination  ↑ ↓± ↓↑

  30. αβ H2 Protons α α αβ α β β α β β α α αβ + β α β β αβ – β α Three symmetric and one antisymmetric wavefunctions

  31. ↑ → ↓ N2 I = 1 particles ↑ ↑ ↑ → ↑ ↓ ↑ → →↑ →→ → ↓ ↓↑ ↓→ ↓↓ ↓  ↑ ↑→ →↓↓  are already symmetric Opposing off-diagonals can form symmetric and antisymmetric combinations eg  ↑ →± →↑

  32. N2 I = 1 particles +1 0 -1 +1+1 +1 0 +1 -1 +1 0 -1 0 +1 00 0 -1 -1 +1 -1 0 -1 -1 +1+10 0−1+1 are already symmetric Opposing off diagonals can form symmetric and antisymmetric combinations eg +10±0+1

  33. N2 I = 1 particles +1 0 -1 +1+1+10 +1-1 +1 0 -1 0+100 0-1 -1+1 -10 -1-1 +1+10 0−1+1 are already symmetric Opposing off-diagonals can form symmetric and antisymmetric combinations +10±0+1 in pairs

  34. Spin I I I -1 1 - I - I 2I+1 I I -1 1 - I - I 2I+1

  35. There are (2I+1) x (2I+1) functions all-together of which 2I+1 are diagonal and thus already symmetric. There are (2I+1)2 - (2I+1) off-diagonal functions which can form ½[(2I+1)2 - (2I+1)] symmetric combinations and ½[(2I+1)2 - (2I+1)] antisymmetic combinations

  36. 2I+1 = n Symmetric ½(n2 – n) + n Antisymmetric ½(n2 –n) S ½(n2 – n) + n = A ½(n2 –n) S ½(n – 1) + 1 = A ½(n – 1) S I + 1 = A I

  37. I I -1 1 - I - I 2I+1 I I -1 1 - I - I 2I+1

  38. It is found empirically that p = +1 for integral spin particles D 1 N 1 Photons 1 and follow Bose-Einstein statistics and are called Bosons p = –1 for half-integral spin particles electrons ½ protons ½ chlorine nuclei 3/2

  39. Let’s consider the consequences of this commutator further [A,B] = 0 Here is the S equation for the H atom

  40. Men’n” Men’n”

  41. http://www.umich.edu/~chem461/QMChap10.pdf

  42. ↑ ↓ ↑ ↑ ↑ ↑ ↓ ↓↑ ↓↓ ↓  ↑ ↑and ↓↓  are already symmetric Opposing off-diagonals can form a symmetric and an antisymmetric combination  ↑ ↓± ↓↑

More Related