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