Let us discuss the harmonic oscillator Let us
![[H , F+] = – ωF+ HF+ – F+H = – ωF+ Now let’s](https://slidetodoc.com/presentation_image/3863ac9d5172e3e5026bb40d481df4b6/image-26.jpg "[H , F+] = – ωF+ HF+ – F+H = – ωF+ Now let’s")
![[H , F+] = – ωF+ HF+ – F+H = – ωF+ Now let’s](https://slidetodoc.com/presentation_image/3863ac9d5172e3e5026bb40d481df4b6/image-27.jpg "[H , F+] = – ωF+ HF+ – F+H = – ωF+ Now let’s")
![[H , F+] = – ωF+ HF+ – F+H = – ωF+ Now let’s](https://slidetodoc.com/presentation_image/3863ac9d5172e3e5026bb40d481df4b6/image-28.jpg "[H , F+] = – ωF+ HF+ – F+H = – ωF+ Now let’s")
![[H , F+] = – ωF+ HF+ – F+H = – ωF+ Now let’s](https://slidetodoc.com/presentation_image/3863ac9d5172e3e5026bb40d481df4b6/image-29.jpg "[H , F+] = – ωF+ HF+ – F+H = – ωF+ Now let’s")
![[H , F+] = – ωF+ HF+ – F+H = – ωF+ Now let’s](https://slidetodoc.com/presentation_image/3863ac9d5172e3e5026bb40d481df4b6/image-30.jpg "[H , F+] = – ωF+ HF+ – F+H = – ωF+ Now let’s")
![[H , F+] = – ωF+ HF+ – F+H = – ωF+ Now let’s](https://slidetodoc.com/presentation_image/3863ac9d5172e3e5026bb40d481df4b6/image-31.jpg "[H , F+] = – ωF+ HF+ – F+H = – ωF+ Now let’s")
![[H , F+] = – ωF+ HF+ – F+H = – ωF+ Now let’s](https://slidetodoc.com/presentation_image/3863ac9d5172e3e5026bb40d481df4b6/image-32.jpg "[H , F+] = – ωF+ HF+ – F+H = – ωF+ Now let’s")
![[q, p] = i [H, q] = ω(–ip) [H, p] = ω(iq) From this](https://slidetodoc.com/presentation_image/3863ac9d5172e3e5026bb40d481df4b6/image-60.jpg "[q, p] = i [H, q] = ω(–ip) [H, p] = ω(iq) From this")
![[H , F+] = – ωF+ HF+ – F+H = – ωF+ Operate on](https://slidetodoc.com/presentation_image/3863ac9d5172e3e5026bb40d481df4b6/image-61.jpg "[H , F+] = – ωF+ HF+ – F+H = – ωF+ Operate on")
![[q, p] = i [H, q] = ω(–ip) [H, p] = ω(iq) From this](https://slidetodoc.com/presentation_image/3863ac9d5172e3e5026bb40d481df4b6/image-69.jpg "[q, p] = i [H, q] = ω(–ip) [H, p] = ω(iq) From this")
![[H , F+] = – ωF+ HF+ – F+H = – ωF+ Operate on](https://slidetodoc.com/presentation_image/3863ac9d5172e3e5026bb40d481df4b6/image-70.jpg "[H , F+] = – ωF+ HF+ – F+H = – ωF+ Operate on")
![[A , B] = 0 If A and B commute there exist eigenfunctions that](https://slidetodoc.com/presentation_image/3863ac9d5172e3e5026bb40d481df4b6/image-80.jpg "[A , B] = 0 If A and B commute there exist eigenfunctions that")
- Slides: 86
Let us discuss the harmonic oscillator
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 = px 2/2 m + ½kx 2
Let us discuss the harmonic oscillator E = T + V = ½mv 2 + ½kx 2 H = px 2/2 m + ½kx 2 By a change of variables
Let us discuss the harmonic oscillator E = T + V = ½mv 2 + ½kx 2 H = px 2/2 m + 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 = px 2/2 m + 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 = px 2/2 m + 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)
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 = k. B
AB – BA = k. B
[H , F+] = – ωF+ HF+ – F+H = – ωF+ Now let’s operate with both sides of this expression on a particular eigenfunction En
[H , F+] = – ωF+ HF+ – F+H = – ωF+ Now let’s operate with both sides of this expression on a particular eigenfunction En ie the nth eigenfunction defined by
[H , F+] = – ωF+ HF+ – F+H = – ωF+ Now let’s operate with both sides of this expression on a particular eigenfunction En ie the nth eigenfunction defined by H En = En En
[H , F+] = – ωF+ HF+ – F+H = – ωF+ Now let’s operate with both sides of this expression on a particular eigenfunction En ie the nth eigenfunction defined by H En = En En H F+ En – F+H En = – ωF+ En
[H , F+] = – ωF+ HF+ – F+H = – ωF+ Now let’s operate with both sides of this expression on a particular eigenfunction En ie the nth eigenfunction defined by H En = En En H F+ En – F+H En = – ωF+ En H F+ En – En. F+ En = – ωF+ En
[H , F+] = – ωF+ HF+ – F+H = – ωF+ Now let’s operate with both sides of this expression on a particular eigenfunction En ie the nth eigenfunction defined by H En = En En H F+ En – F+H En = – ωF+ En H F+ En – En. F+ En = – ωF+ En H F+ En = (En – ω) F+ En
[H , F+] = – ωF+ HF+ – F+H = – ωF+ Now let’s operate with both sides of this expression on a particular eigenfunction En ie the nth eigenfunction defined by H En = En En H F+ En – F+H En = – ωF+ En H F+ En – En. F+ En = – ωF+ En H F+ En = (En – ω) F+ En So F+ has operated on En to produce a new eigenfunction with eigenvalue En – ω
En
En En – ω En – 2ω 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
En En – ω En – 2ω F+ E↓ Aaaaghhhhh…… E↓ 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
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
They are F± = q ± ip By inspection we can show that [H , F+] = – ωF+ [H , F–] = + ωF– These are of the form [A , B] = k. B 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+ En } = (En – ω){F+ En } H{F– En } = (En– ω){F– En }
Now H can be factorised as H F+ En = ½ω(F+F– – 1) F+ En H{F+ En } = (En – ω){F+ En } H{F– En } = (En– ω){F– En }
Let us discuss the harmonic oscillator E = T + V = ½m 2 + ½kx 2 H = p 2/2 m + 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 2/2 m + 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)
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 = ±k. B
[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 = ±k. B 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 En ie the nth eigenfunction defined by H En = En En with both sides H F+ En – F+H En = – ωF+ En H F+ En – En. F+ En = – ωF+ En H F+ En = (En – ω) F+ En
They are F± = q ± ip By inspection we can show that [H , F+] = – ωF+ [H , F–] = + ωF– These are of the form [A , B] = k. B 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+ En } = (En – ω){F+ En } H{F– En } = (En– ω){F– En }
Now H can be factorised as H F+ En = ½ω(F+F– – 1) F+ En H{F+ En } = (En – ω){F+ En } H{F– En } = (En– ω){F– En }
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 = ±k. B
[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 = ±k. B 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 En ie the nth eigenfunction defined by H En = En En with both sides H F+ En – F+H En = – ωF+ En H F+ En – En. F+ En = – ωF+ En H F+ En = (En – ω) F+ En
n En En – ω En – 2ω 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 F+ Aaaaghhhhh…… E↓
They are F± = q ± ip By inspection we can show that [H , F+] = – ωF+ [H , F–] = + ωF– These are of the form [A , B] = k. B 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+ En } = (En – ω){F+ En } H{F– En } = (En– ω){F– En }
Now H can be factorised as H F+ En = ½ω(F+F– – 1) F+ En H{F+ En } = (En – ω){F+ En } H{F– En } = (En– ω){F– En }
[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