Let us discuss the harmonic oscillator

E = T + V = ½mv 2 + ½kx 2

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

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

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

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

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 )

Let’s now consider this set of commutators

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

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

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

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

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

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

[q,p] = i [H,q] = ω(-ip) [H,p] = ωiq

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

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

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)

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 +

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 +

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 –

[H, F + ] = – ωF +

HF + – F + H = – ωF +

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

[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, 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 + ] = – 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 + ] = – 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 + ] = – 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 

[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

EnEn EnEn

EnEn E n – ω EnEn

EnEn E n – 2ω EnEn

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

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↓

F + F – = (q + ip)(q – ip)

= q 2 – iqp + ipq +p 2

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

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

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

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

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

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

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

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

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

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)

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  }

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  }

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

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 )

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 )

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

[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 –

[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 

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

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)

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  }

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  }

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

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

[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 –

[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 

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+

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

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)

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  }

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  }

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

[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

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

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

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 )

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

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

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