Let us discuss the harmonic oscillator
Let us discuss the harmonic oscillator
E = T + V = ½mv2 + ½kx2
Let us discuss the harmonic oscillator
E = T + V = ½mv2 + ½kx2
H = px2/2m + ½kx2
Let us discuss the harmonic oscillator
E = T + V = ½mv2 + ½kx2
H = px2/2m + ½kx2
By a change of variables
Let us discuss the harmonic oscillator
E = T + V = ½mv2 + ½kx2
H = px2/2m + ½kx2
By a change of variables
x = q(mω)-½ ω = (k/m)½
Let us discuss the harmonic oscillator
E = T + V = ½mv2 + ½kx2
H = px2/2m + ½kx2
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 = ½mv2 + ½kx2
H = px2/2m + ½kx2
By a change of variables
x = q(mω)-½ ω = (k/m)½
We get a rather neater equation for the Hamiltonian
H = ½ω(p2 + q2)
H = ½ω(p2 + q2)
H = ½ω(p2 + q2)
Let’s now consider this set of commutators
H = ½ω(p2 + q2)
Let’s now consider this set of commutators
NB that I have set ħ = 1 to simplify the relations
H = ½ω(p2 + q2)
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(p2 + q2)
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(p2 + q2)
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(p2 + q2)
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(p2 + q2)
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
AB – BA = kB
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+
[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 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 – EnF+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 – EnF+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 – EnF+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
En
En – ω
En
En
En – ω
En – 2ω
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
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)
F+F– = (q + ip)(q – ip)
= q2 – iqp + ipq +p2
F+F– = (q + ip)(q – ip)
= q2 – iqp + ipq +p2
= q2 – i[q,p] + p2
F+F– = (q + ip)(q – ip)
= q2 – iqp + ipq +p2
= q2 – i[q,p] + p2
= q2 + p2 + 1
F+F– = (q + ip)(q – ip)
= q2 – iqp + ipq +p2
= q2 – i[q,p] + p2
= q2 + p2 + 1
F+F– – 1 = q2 + p2
F+F– = (q + ip)(q – ip)
= q2 – iqp + ipq +p2
= q2 – i[q,p] + p2
= q2 + p2 + 1
F+F– – 1 = q2 + p2
H = ½ ω(q2 + p2)
F+F– = (q + ip)(q – ip)
= q2 – iqp + ipq +p2
= q2 – i[q,p] + p2
= q2 + p2 + 1
F+F– – 1 = q2 + p2
H = ½ ω(q2 + p2)
H = ½ ω(F+F– – 1)
F+F– = (q + ip)(q – ip)
= q2 – iqp + ipq +p2
= q2 – i[q,p] + p2
= q2 + p2 + 1
F+F– – 1 = q2 + p2
H = ½ ω(q2 + p2)
H = ½ ω(F+F– – 1)
H = ½ ω(F–F+ + 1)
F+F– = (q + ip)(q – ip)
= q2 – iqp + ipq +p2
= q2 – i[q,p] + p2
= q2 + p2 + 1
F+F– – 1 = q2 + p2
H = ½ ω(q2 + p2)
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)
= q2 – iqp + ipq +p2
= q2 – i[q,p] + p2
= q2 + p2 + 1
F+F– – 1 = q2 + p2
H = ½ ω(q2 + p2)
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)
= q2 – iqp + ipq +p2
= q2 – i[q,p] + p2
= q2 + p2 + 1
F+F– – 1 = q2 + p2
H = ½ ω(q2 + p2)
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)
= q2 – iqp + ipq +p2
= q2 – i[q,p] + p2
= q2 + p2 + 1
F+F– – 1 = q2 + p2
H = ½ ω(q2 + p2)
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↓
E(v) = ω (v + ½)
Zero Point Energy
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+ 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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 = ½m2 + ½kx2
H = p2/2m + ½kx2
By a change of variables
x = q(mω)-½ ω = (k/m)½
We get a rather neater equation for the Hamiltonian
H = ½ω(p2 + q2)
Let us discuss the harmonic oscillator
E = T + V = ½m2 + ½kx2
H = p2/2m + ½kx2
By a change of variables
x = q(mω)-½ ω = (k/m)½
We get a rather neater equation for the Hamiltonian
H = ½ω(p2 + q2)
H = ½ω(p2 + q2)
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 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 – EnF+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] = 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+ 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A’ = A’ A’
ABA’ – BAA’ = kBA’
ABA’ – BA’A’ = kBA’
A{BA’} = (A’ + k){BA’}
H = ½ω(p2 + q2)
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 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 – EnF+En = – ωF+En
H F+ En = (En – ω) F+En
n
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 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] = 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+ 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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 IA’
›
›
= A’ IA’
›
A { B IA’ } = (A’ + k ) { B IA’ }
[q , p] = i
[H, q] = ω (-ip)
[H, p] = ω (-iq)
F+ = q +ip
F −= q − i p
[ H , F + ] = − ω F+
[ H , F −] = + ω F −
H = − ½ ω ( F+F − − 1 )
H = + ½ ω ( F −F+ + 1 )
›
›
›
›
H { F + I En } = (En − ω ){ F + I En }
H { F − I En } = (En − ω ){F − I En }
›
F + I E↓ = 0
›
H I E↓
›
= ½ ω I E↓
E (v) = ω (v + ½ )
F+ = q +ip
F −= q − i p
F± = q ±ip
[ H,F ± ] = q ± i
p
[H, q] = ω (-ip)