Lecture 19: Quantization of the simple harmonic oscillator · 2009. 10. 21. · Quantization of the...
Transcript of Lecture 19: Quantization of the simple harmonic oscillator · 2009. 10. 21. · Quantization of the...
Lecture 19:Quantization of the simple harmonic
oscillator
Phy851 Fall 2009
Systems near equilibrium
• The harmonic oscillator Hamiltonian is:
• Or alternatively, using
• Why is the SHO so important?– Answer: any system near a stable equilibrium
is equivalent to an SHO
222
2
1
2Xm
m
PH ω+=
22
2
1
2kX
m
PH +=
m
k=ω
A Random Potential
Stable equilibrium points
Definition of stableequilibrium point:
0)( 0 =′ xV
Expand around x0:
K+−′′+
−′+=
200
000
))((2
1
))(()()(
xxxV
xxxVxVxV
0xxy −= 20 2
1)( ykVyV +=
Analysis of energy and length scales
• The parameters available in the SHOHamiltonian are:
• The frequency defines a quantum energy [J]scale via:
• The frequency also defines a quantum lengthscale via:
• This length scale then defines a quantummomentum scale:
oscm ω,,h
222
2
1
2Xm
m
PH oscω+=
oscoscE ωh=
[J _ s], [kg], [s-1]
2
2
oscosc mE
λh
= 2
2
oscosc mλ
ωh
h =osc
osc mωλ
h=
oscosc λ
µh
= oscosc mωµ h=
The SHO introduces asingle new parameter,which must govern all
of the physics
Dimensionless Variables
• To solve the QM SHO it is very useful tointroduce the natural units:– Let
osc
XX
λ=
osc
HH
ωh=
22222
2
2
1
2
1XmP
mH oscosc
oscosc λω
λω +=
hh
PP
P osc
osc h
λµ
==
22
2
1
2
1XPH +=
oscosc
osc
m
mmω
ωλ
hh
hh==
2
2
2
oscosc
oscoscosc mmm ω
ωωλω h
h== 222
22
2
1
2
1XPH oscoscosc ωωω hhh +=
Dimensionless Commutation Relations
• Let’s compute the commutator for the newvariables:
€
X ,P [ ] = X P − P X
λλ
XXXX =→=
PPPPh
h λλ
=→=
€
X ,P [ ] =XλλPh−λPh
Xλ
( )PXXP −=h
1
[ ]PX ,1
h=
€
X ,P [ ] = i
We have stoppedwriting the
subscript ‘osc’
Since the newvariables have no
units, we losethe h
Switch to`Normal’ Variables
• We can make a change of variables:
– It’s more common to use: a, a†
– We use A, A† to stick with our conventionto use capital letters for operators
( )PiXA +=2
1
€
A† =12
X − iP ( )
€
A,A≤[ ] =12
X + iP ( ), 12
X − iP ( )
€
=12
X ,−iP [ ] + iP ,X [ ]( )
€
=12−i X ,P [ ] + i P ,X [ ]( )
1=
€
A,A†[ ] =1
Inverse Transformation
• Inverting the transformation gives:
( )PiXA +=2
1
( )PiXA −=2
1†
€
12
A + A†( ) =12
X + iP + X − iP ( ) = X
€
12
A − A†( ) =12
X + iP − X + iP ( ) = iP
€
X = 12
A + A†( )
€
P = −i2
A − A†( )€
X =λ2A + A†( )
€
P =−ih2λ
A − A†( )
Transforming the Hamiltonian• The Harmonic Oscillator Hamiltonian was:
€
H = 12
P 2 + X 2( )
( )†2
1AAX += ( )†
2AA
iP −
−=
( )( )††2
2
1AAAAX ++=
( )††††2
2
1AAAAAAAAX +++=
( )††††2
2
1AAAAAAAAP +−−−=
€
H = 12
P 2 + X 2( )
€
=12AA† + A†A( )
( )††††
2
1AAAAAAAA −++−=
1†† =− AAAA
1†† +=∴ AAAA
2
1† += AAH
Energy Eigenvalues
• In original units we have:
• Let’s define the energy eigenstates via
Hm
H2
2
λh
=ω
λm
h=
+=2
1†2
AAm
m h
h ω
+=2
1†AAH ωh
εεε =
+2
1†AA
εεε =H
εεδεε ′=′ ,
We expect a discrete spectrum as theclassical motion is bounded
Proof that there is a ground state
• For any energy eigenstate we have:
• The norm of a vector is always a real positivenumber
• Thus we see that:
• So the energy eigenvalues are bounded frombelow by _.
εεεεεε ==H
εεεεεε2
1
2
1 †† +=+ AAAA
2
12+= εA
02≥ψ
2
1≥ε
Setting up for The Big Trick
• Lets look at the commutator:
[ ] [ ]AAAAAAHA †† ,2
1,, =
+=
AAAAAA †† −=
€
= 1+ A†A( )A − A†AAA=
AAHHA =−
( )1−= HAAH
AHAAH −=
††† AHAAH =−
††† AHAAH +=
( )1†† += HAAH
The Big Trick Begins
• Combine these relations with the eigenvalueequation:
• This means that A|ε〉 is proportional to theeigenstate |ε-1〉:
εεε =H
€
H A ε = A H −1( ) ε
€
H A = A H −1( )
€
H A† = A† H +1( )
€
= A ε −1( ) ε
€
H A ε( ) = ε −1( ) A ε( )
( ) 111 −−=− εεεH
1−= εε εcA
( )εε 1†† += HAAH
( )εεε 1†† += AAH
1† += εε εdA
cε is an unknowncoefficient
dε is an unknowncoefficient
Raising and lowering operators
THEOREM:• If state |ε〉 exists then either state |ε-1〉 exists
or cε=0• If state |ε〉 exists then either state |ε+1〉 exists
or dε=0
• Now consider:
• So clearly we must have:
• So cε is only zero for ε=1/2 and dε is only zerofor ε=-1/2
1−= εε εcA 1† += εε εdA
εεε =H
2
1
2
1 2† +=+ εεε cAA
2
12+= εε c
2
1≥ε
εεεc
A=−1 εε
εd
A†1 =+
2
12−= εεc
2
1
2
1 2† −=− εεε dAA
2
12−= εε d
2
12+= εεd
This means that actually de is never zero
Ground State energy
• Let the ground state |ε0〉 have energy:
• Remember our statement:– If state |ε〉 exists then either state |ε-1〉 exists or
cε=0
– Conclusion: either δ = 0, or there is a statelower than the ground state
– For |ε0〉 to be the ground state requires δ = 0
δε +=2
10
2
12
0 0+= εε c
δε =2
0c
2
10 =ε
The second option is obviously a contradiction
Excited states
• If the ground state |ε0〉 exists, then the state|ε0+1〉 exists
• Following this chain of reasoning, we canestablish the existence of states at energies:
εεεd
A†1 =+
2
12+= εεd
K,2
7,2
5,2
3,2
1=ε
This is just a resultwe proved on slide 13
The point is justthat we aren’t
dividing by zero
Are there more states?
• So far we see that a ladder of states mustexist:
• Are there any states in between?– Assume a state exists with
– We have
– So either x=0 or there is a state below theground state! Conclusion: x=0
• If there is a state between 5/2 and 3/2, thena state must exist between 3/2 and 1/2 andthen a state must exist below 1/2, etc…
• So no states between the half integers!
2/12/3 † === εε A
2/1=ε
2/1=ε
2/3=ε
No statesbelow bydefinition
10
2/1
<<
+=
x
xε
xc =−=2
12εε
2/12/1 −=+ xxx
A
2/32
2/5†
=== εεA
M M
This state lies betweenε=1/2 and ε=3/2
x-1/2 < 1/2
Spectrum of the SHO
• We now see that the energy eigenstates canbe labeled by the integers so that:
• We can always go back to our original unitsby putting in the energy scale factor:
K,3,2,1,0;2
1=
+= nnnnH
Kh ,3,2,1,0;2
1=
+= nnnnH ω