1. Let us discuss the harmonic oscillator

2. E = T + V = ½mv 2 + ½kx 2

3. Let us discuss the harmonic oscillator E = T + V = ½mv 2 + ½kx 2 H = p x 2 /2m + ½kx 2

4. Let us discuss the harmonic oscillator E = T + V = ½mv 2 + ½kx 2 H = p x 2 /2m + ½kx 2 By a change of variables

5. Let us discuss the harmonic oscillator E = T + V = ½mv 2 + ½kx 2 H = p x 2 /2m + ½kx 2 By a change of variables x = q(mω) -½ ω = (k/m) ½

6. Let us discuss the harmonic oscillator E = T + V = ½mv 2 + ½kx 2 H = p x 2 /2m + ½kx 2 By a change of variables x = q(mω) -½ ω = (k/m) ½ We get a rather neater equation for the Hamiltonian

7. Let us discuss the harmonic oscillator E = T + V = ½mv 2 + ½kx 2 H = p x 2 /2m + ½kx 2 By a change of variables x = q(mω) -½ ω = (k/m) ½ We get a rather neater equation for the Hamiltonian H = ½ω(p 2 + q 2 )

9. Let’s now consider this set of commutators

10. H = ½ω(p 2 + q 2 ) Let’s now consider this set of commutators NB that I have set ħ = 1 to simplify the relations

11. H = ½ω(p 2 + q 2 ) Let’s now consider this set of commutators NB that I have set ħ = 1 to simplify the relations [q,p] = i [H,q] = ω(-ip) [H,p] = ωiq

12. H = ½ω(p 2 + q 2 ) Let’s now consider this set of commutators NB that I have set ħ = 1 to simplify the relations [q,p] = i [H,q] = ω(-ip) [H,p] = ωiq From this set of quantum mechanical definitions

13. H = ½ω(p 2 + q 2 ) Let’s now consider this set of commutators NB that I have set ħ = 1 to simplify the relations [q,p] = i [H,q] = ω(-ip) [H,p] = ωiq From this set of quantum mechanical definitions We can invent a couple of B-type operators

14. H = ½ω(p 2 + q 2 ) Let’s now consider this set of commutators NB that I have set ħ = 1 to simplify the relations [q,p] = i [H,q] = ω(-ip) [H,p] = ωiq From this set of quantum mechanical definitions We can invent a couple of B-type operators ie operators that follow the form

15. H = ½ω(p 2 + q 2 ) Let’s now consider this set of commutators NB that I have set ħ = 1 to simplify the relations [q,p] = i [H,q] = ω(-ip) [H,p] = ωiq From this set of quantum mechanical definitions We can invent a couple of B-type operators ie operators that follow the form AB – BA = kB

17. [q,p] = i [H,q] = ω(-ip) [H,p] = ωiq

18. AB – BA = kB [q,p] = i [H,q] = ω(-ip) [H,p] = ωiq F + = q + ip

19. AB – BA = kB [q,p] = i [H,q] = ω(-ip) [H,p] = ωiq F + = q + ip [H, F + ] = [H,q] + i[H,p]

20. AB – BA = kB [q,p] = i [H,q] = ω(-ip) [H,p] = ωiq F + = q + ip [H, F + ] = [H,q] + i[H,p] = ω(–ip) + iω(iq)

21. AB – BA = kB [q,p] = i [H,q] = ω(-ip) [H,p] = ωiq F + = q + ip [H, F + ] = [H,q] + i[H,p] = ω(–ip) + iω(iq) = ω(–ip – q) = – ωF +

22. AB – BA = kB [q,p] = i [H,q] = ω(-ip) [H,p] = ωiq F + = q + ip [H, F + ] = [H,q] + i[H,p] = ω(–ip) + iω(iq) = ω(–ip – q) = – ωF + [H, F + ] = –ωF +

23. AB – BA = kB [q,p] = i [H,q] = ω(-ip) [H,p] = ωiq F + = q + ip [H, F + ] = [H,q] + i[H,p] = ω(–ip) + iω(iq) = ω(–ip – q) = – ωF + [H, F + ] = –ωF + and [H, F – ] = +ωF –

24. [H, F + ] = – ωF +

25. HF + – F + H = – ωF +

26. [H, F + ] = – ωF + HF + – F + H = – ωF + Now let’s operate with both sides of this expression on a particular eigenfunction  E n 

27. [H, F + ] = – ωF + HF + – F + H = – ωF + Now let’s operate with both sides of this expression on a particular eigenfunction  E n  ie the n th eigenfunction defined by

28. [H, F + ] = – ωF + HF + – F + H = – ωF + Now let’s operate with both sides of this expression on a particular eigenfunction  E n  ie the n th eigenfunction defined by H  E n  = E n  E n 

29. [H, F + ] = – ωF + HF + – F + H = – ωF + Now let’s operate with both sides of this expression on a particular eigenfunction  E n  ie the n th eigenfunction defined by H  E n  = E n  E n  H F +  E n  – F + H  E n  = – ωF +  E n 

30. [H, F + ] = – ωF + HF + – F + H = – ωF + Now let’s operate with both sides of this expression on a particular eigenfunction  E n  ie the n th eigenfunction defined by H  E n  = E n  E n  H F +  E n  – F + H  E n  = – ωF +  E n  H F +  E n  – E n F +  E n  = – ωF +  E n 

31. [H, F + ] = – ωF + HF + – F + H = – ωF + Now let’s operate with both sides of this expression on a particular eigenfunction  E n  ie the n th eigenfunction defined by H  E n  = E n  E n  H F +  E n  – F + H  E n  = – ωF +  E n  H F +  E n  – E n F +  E n  = – ωF +  E n  H F +  E n  = (E n – ω) F +  E n 

32. [H, F + ] = – ωF + HF + – F + H = – ωF + Now let’s operate with both sides of this expression on a particular eigenfunction  E n  ie the n th eigenfunction defined by H  E n  = E n  E n  H F +  E n  – F + H  E n  = – ωF +  E n  H F +  E n  – E n F +  E n  = – ωF +  E n  H F +  E n  = (E n – ω) F +  E n  So F + has operated on  E n  to produce a new eigenfunction with eigenvalue E n – ω

33. EnEn EnEn

34. EnEn E n – ω EnEn

35. EnEn E n – 2ω EnEn

36. EnEn E n – ω E n – 2ω EnEn Let’s ladder down till we get to the last eigenvalue at which a next application of F + would produce an eigenstate with negative energy which we shall posutlate is not allowed and so F + must annihilate this last eigenstate ie F +  E ↓  = 0

37. EnEn E n – ω E n – 2ω E↓E↓ EnEn Let’s ladder down till we get to the last eigenvalue at which a next application of F + would produce an eigenstate with negative energy which we shall posutlate is not allowed and so F + must annihilate this last eigenstate ie F +  E ↓  = 0 Aaaaghhhhh…… F+F+ E↓E↓

38. F + F – = (q + ip)(q – ip)

39. = q 2 – iqp + ipq +p 2

40. F + F – = (q + ip)(q – ip) = q 2 – iqp + ipq +p 2 = q 2 – i[q,p] + p 2

41. F + F – = (q + ip)(q – ip) = q 2 – iqp + ipq +p 2 = q 2 – i[q,p] + p 2 = q 2 + p 2 + 1

42. F + F – = (q + ip)(q – ip) = q 2 – iqp + ipq +p 2 = q 2 – i[q,p] + p 2 = q 2 + p 2 + 1 F + F – – 1 = q 2 + p 2

43. F + F – = (q + ip)(q – ip) = q 2 – iqp + ipq +p 2 = q 2 – i[q,p] + p 2 = q 2 + p 2 + 1 F + F – – 1 = q 2 + p 2 H = ½ ω(q 2 + p 2 )

44. F + F – = (q + ip)(q – ip) = q 2 – iqp + ipq +p 2 = q 2 – i[q,p] + p 2 = q 2 + p 2 + 1 F + F – – 1 = q 2 + p 2 H = ½ ω(q 2 + p 2 ) H = ½ ω(F + F – – 1)

45. F + F – = (q + ip)(q – ip) = q 2 – iqp + ipq +p 2 = q 2 – i[q,p] + p 2 = q 2 + p 2 + 1 F + F – – 1 = q 2 + p 2 H = ½ ω(q 2 + p 2 ) H = ½ ω(F + F – – 1) H = ½ ω(F – F + + 1)

46. F + F – = (q + ip)(q – ip) = q 2 – iqp + ipq +p 2 = q 2 – i[q,p] + p 2 = q 2 + p 2 + 1 F + F – – 1 = q 2 + p 2 H = ½ ω(q 2 + p 2 ) H = ½ ω(F + F – – 1) H = ½ ω(F – F + + 1) H  E ↓  = ½ ω(F – F + + 1)  E ↓ 

47. F + F – = (q + ip)(q – ip) = q 2 – iqp + ipq +p 2 = q 2 – i[q,p] + p 2 = q 2 + p 2 + 1 F + F – – 1 = q 2 + p 2 H = ½ ω(q 2 + p 2 ) H = ½ ω(F + F – – 1) H = ½ ω(F – F + + 1) H  E ↓  = ½ ω(F – F + + 1)  E ↓  F +  E ↓  = 0

48. F + F – = (q + ip)(q – ip) = q 2 – iqp + ipq +p 2 = q 2 – i[q,p] + p 2 = q 2 + p 2 + 1 F + F – – 1 = q 2 + p 2 H = ½ ω(q 2 + p 2 ) H = ½ ω(F + F – – 1) H = ½ ω(F – F + + 1) H  E ↓  = ½ ω(F – F + + 1)  E ↓  F +  E ↓  = 0 H  E ↓  = ½ ω  E ↓  Zero Point Energy

49. F + F – = (q + ip)(q – ip) = q 2 – iqp + ipq +p 2 = q 2 – i[q,p] + p 2 = q 2 + p 2 + 1 F + F – – 1 = q 2 + p 2 H = ½ ω(q 2 + p 2 ) H = ½ ω(F + F – – 1) H = ½ ω(F – F + + 1) H  E ↓  = ½ ω(F – F + + 1)  E ↓  F +  E ↓  = 0 H  E ↓  = ½ ω  E ↓  Zero Point Energy E (v) = ω (v + ½)

52. They are F ± = q ± ip By inspection we can show that [H, F + ] = – ωF + [H, F – ] = + ωF – These are of the form [A, B] = kB So the Fs are B-type operators

53. Now H can be factorised as H = ½ω(F + F – – 1) H = ½ω(F – F + + 1) H = ½ω(F + F – – 1) H = ½ω(F – F + + 1)

54. Now H can be factorised as H = ½ω(F + F – – 1) H = ½ω(F – F + + 1) H{F +  E n  } = (E n – ω){F +  E n  } H{F –  E n  } = (E n – ω){F –  E n  }

55. Now H can be factorised as H F +  E n  = ½ω(F + F – – 1) F +  E n  H{F +  E n  } = (E n – ω){F +  E n  } H{F –  E n  } = (E n – ω){F –  E n  }

56. A  A’  = A’  A’  AB  A’  – BA  A’  = kB  A’  AB  A’  – BA’  A’  = kB  A’  A{B  A’  } = (A’ + k){B  A’  }

57. Let us discuss the harmonic oscillator E = T + V = ½m 2 + ½kx 2 H = p 2 /2m + ½kx 2 By a change of variables x = q(mω) -½ ω = (k/m) ½ We get a rather neater equation for the Hamiltonian H = ½ω(p 2 + q 2 )

58. Let us discuss the harmonic oscillator E = T + V = ½m 2 + ½kx 2 H = p 2 /2m + ½kx 2 By a change of variables x = q(mω) -½ ω = (k/m) ½ We get a rather neater equation for the Hamiltonian H = ½ω(p 2 + q 2 )

59. Let’s now consider this set of commutators NB that I have set ħ = 1 to simplify the relations [q,p] = i [H,q] = ω(-ip) [H,p] = ωiq From this set of quantum mechanical definitions We can invent a couple of B-type operators ie one that follows the form AB – BA = ±kB

60. [q,p] = i [H,q] = ω(–ip) [H,p] = ω(iq) From this set of quantum mechanical definitions We can invent a couple of B-type operators ie one that follows the form AB – BA = ±kB F + = q + ip [H, F + ] = [H,q] + i[H,p] = ω(–ip) + iω(iq) = ω(–ip – q) = – ωF + [H, F + ] = –ωF + and [H, F – ] = +ωF –

61. [H, F + ] = – ωF + HF + – F + H = – ωF + Operate on a particular eigenfunction  E n  ie the n th eigenfunction defined by H  E n  = E n  E n  with both sides H F +  E n  – F + H  E n  = – ωF +  E n  H F +  E n  – E n F +  E n  = – ωF +  E n  H F +  E n  = (E n – ω) F +  E n 

62. They are F ± = q ± ip By inspection we can show that [H, F + ] = – ωF + [H, F – ] = + ωF – These are of the form [A, B] = kB So the Fs are B-type operators

63. Now H can be factorised as H = ½ω(F + F – – 1) H = ½ω(F – F + + 1) H = ½ω(F + F – – 1) H = ½ω(F – F + + 1)

64. Now H can be factorised as H = ½ω(F + F – – 1) H = ½ω(F – F + + 1) H{F +  E n  } = (E n – ω){F +  E n  } H{F –  E n  } = (E n – ω){F –  E n  }

65. Now H can be factorised as H F +  E n  = ½ω(F + F – – 1) F +  E n  H{F +  E n  } = (E n – ω){F +  E n  } H{F –  E n  } = (E n – ω){F –  E n  }

66. A  A’  = A’  A’  AB  A’  – BA  A’  = kB  A’  AB  A’  – BA’  A’  = kB  A’  A{B  A’  } = (A’ + k){B  A’  }

68. H = ½ω(p 2 + q 2 ) Let’s now consider this set of commutators NB that I have set ħ = 1 to simplify the relations [q,p] = i [H,q] = ω(-ip) [H,p] = ωiq From this set of quantum mechanical definitions We can invent a couple of B-type operators ie one that follows the form AB – BA = ±kB

69. [q,p] = i [H,q] = ω(–ip) [H,p] = ω(iq) From this set of quantum mechanical definitions We can invent a couple of B-type operators ie one that follows the form AB – BA = ±kB F + = q + ip [H, F + ] = [H,q] + i[H,p] = ω(–ip) + iω(iq) = ω(–ip – q) = – ωF + [H, F + ] = –ωF + and [H, F – ] = +ωF –

70. [H, F + ] = – ωF + HF + – F + H = – ωF + Operate on a particular eigenfunction  E n  ie the n th eigenfunction defined by H  E n  = E n  E n  with both sides H F +  E n  – F + H  E n  = – ωF +  E n  H F +  E n  – E n F +  E n  = – ωF +  E n  H F +  E n  = (E n – ω) F +  E n 

71. EnEn E n – ω E n – 2ω E↓E↓ EnEn nn Let’s ladder down till we get to the last eigenvalue at which a next application of F + would produce an eigenstate with negative energy which we shall posutlate is not allowed and that must annihilate this last eigenstate ie F +  E ↓  = 0 Aaaaghhhhh…… F+F+

73. They are F ± = q ± ip By inspection we can show that [H, F + ] = – ωF + [H, F – ] = + ωF – These are of the form [A, B] = kB So the Fs are B-type operators

74. Now H can be factorised as H = ½ω(F + F – – 1) H = ½ω(F – F + + 1) H = ½ω(F + F – – 1) H = ½ω(F – F + + 1)

75. Now H can be factorised as H = ½ω(F + F – – 1) H = ½ω(F – F + + 1) H{F +  E n  } = (E n – ω){F +  E n  } H{F –  E n  } = (E n – ω){F –  E n  }

76. Now H can be factorised as H F +  E n  = ½ω(F + F – – 1) F +  E n  H{F +  E n  } = (E n – ω){F +  E n  } H{F –  E n  } = (E n – ω){F –  E n  }

77. A  A’  = A’  A’  AB  A’  – BA  A’  = kB  A’  AB  A’  – BA’  A’  = kB  A’  A{B  A’  } = (A’ + k){B  A’  }

80. [A, B] = 0 If A and B commute there exist eigenfunctions that are simultaneously eigenfunctions of both operators A and B and one can determine simultaneously the values of the quantities represented by the two operators but if they do not commute one cannot determine the values of the quantities simultaneously

81. [A, B] = 0 [A, B] = k B A I A ’ › = A ’ I A ’ › A { B I A ’ › } = (A ’ + k ) { B I A ’ › }

82. [q, p] = i [H, q] = ω (-ip) [H, p] = ω (-iq)

83. F + = q + i p F − = q − i p [ H, F + ] = − ω F + [ H, F − ] = + ω F − H = − ½ ω ( F + F − − 1 ) H = + ½ ω ( F − F + + 1 )

84. H { F + I E n › } = ( E n − ω ){ F + I E n › } H { F − I E n › } = ( E n − ω ){ F − I E n › }

85. F + I E ↓ › = 0 H I E ↓ › = ½ ω I E ↓ › E (v) = ω (v + ½ )

86. F + = q + i p [H, q] = ω (-ip) F ± = q ± i p F − = q − i p [ H,F ± ] = q ± i p