Download Lecture 19: Quantization of the simple harmonic oscillator Phy851 Fall 2009

Lecture 19:
Quantization of the simple harmonic
oscillator
Phy851 Fall 2009
Systems near equilibrium
• The harmonic oscillator Hamiltonian is:
P2 1 2
H=
+ kX
2m 2
• Or alternatively, using
k
ω=
m
P2 1
H=
+ mω 2 X 2
2m 2
• Why is the SHO so important?
– Answer: any system near a stable equilibrium
is equivalent to an SHO
Definition of stable
equilibrium point:
A Random Potential
V ′( x0 ) = 0
Expand around x0:
V ( x) = V ( x0 ) + V ′( x0 )( x − x0 )
Stable equilibrium points
1
+ V ′′( x0 )( x − x0 ) 2 + K
2
1
y = x − x0 V ( y ) = V0 + k y 2
2
Analysis of energy and length scales
P2 1
2
H=
+ mωosc
X2
2m 2
• The parameters available in the SHO
The SHO introduces a
Hamiltonian are:
osc single new parameter,
h, m, ω
[J _ s], [kg],
[s-1]
which must govern all
of the physics
• The frequency defines a quantum energy [J]
scale via:
Eosc = hωosc
• The frequency also defines a quantum
scale via:
Eosc
h2
h2
hωosc =
=
2
mλ2osc
mλosc
λosc
length
h
=
mωosc
• This length scale then defines a quantum
momentum scale:
µ osc
h
=
λosc
µ osc = hmωosc
Dimensionless Variables
• To solve the QM SHO it is very useful to
introduce the natural units:
– Let
X
X=
λosc
λosc
P
P=
=
P
µ osc
h
H=
H
hωosc
1 h2
1
2
2
2
2
hωosc H =
P
+
m
ω
λ
X
osc osc
2 mλ2osc
2
h2
h 2 mωosc
=
= hωosc
2
mλosc m h
2
2
mωosc
λ2osc = mωosc
h
= hωosc
mωosc
1
1
2
hωosc H = hωosc P + hωosc X 2
2
2
1 2 1 2
H= P + X
2
2
Dimensionless Commutation Relations
• Let’s compute the commutator for the new
variables:
[ X ,P ] = X P − P X
X =λ X
€
€
P=
h
P
λ
→
X=
→ P=
X
λ
λ
P
h
We have stopped
writing the
subscript ‘osc’
X λP λP X
[ X ,P ] = λ h − h λ
1
= (XP − PX )
h
1
= [X , P ]
h
[ X ,P ] = i
Since the new
variables have no
units, we lose
the h
Switch to`Normal’ Variables
• We can make a change of variables:
A=
1
(X + iP )
2
A† =
1
X − iP )
(
2
– It’s more common to use: a, a†
– We use A, A† to stick with our convention
€ use capital letters for operators
to
 1

1
[ A, A ] =  2 ( X + iP ), 2 ( X − iP )
≤
€
€
€
1
= [ X ,−iP ] + [iP , X ]
2
1
= −i[ X ,P ] + i[ P , X ]
2
=1
(
)
(
)
[ A, A ] = 1
†
Inverse Transformation
1
(X + iP )
2
1
(X − iP )
A† =
2
A=
• Inverting the transformation gives:
1
1
†
A
+
A
=
X + iP + X − iP ) = X
(
(
)
2
2
1
1
†
A − A ) = ( X + iP − X + iP ) = iP
(
2
2
€
€
€
€
1
X=
A + A† )
(
2
−i
P=
A − A† )
(
2
€
€
X=
λ
A + A† )
(
2
−ih
P=
A − A† )
(
2λ
Transforming the Hamiltonian
• The Harmonic Oscillator Hamiltonian was:
1 2
H = (P + X 2 )
2
X=
€
1
A + A†
2
(
X2 =
)
P=
−i
A − A†
2
(
1
A + A† A + A†
2
(
)(
)
)
1
AA + AA† + A† A + A† A†
2
1
P 2 = − AA − AA† − A† A + A† A†
2
1
= − AA + AA† + A† A − A† A†
2
X2 =
(
)
(
)
(
1 2
H = (P + X 2 )
2
1
= ( AA† + A† A)
2
AA† − A† A = 1
∴ AA† = A† A + 1
1
H = A A+
2
€
†
€
)
Energy Eigenvalues
• In original units we have:
h2
H=
H
2
mλ
h 2 mω
=
m h
λ=
h
mω
1
 †
A
A
+


2

1
 †
H = hω  A A + 
2

• Let’s define the energy eigenstates via
H ε =ε ε
1
 †
A
A
+

ε =ε ε
2

ε ε ′ = δ ε ,ε ′
We expect a discrete spectrum as the
classical motion is bounded
Proof that there is a ground state
• For any energy eigenstate we have:
ε H ε =ε ε ε =ε
1
1
†
ε A A+ ε = ε A A ε + ε ε
2
2
2
1
= Aε +
2
†
• The norm of a vector is always a real positive
number
ψ
2
≥0
• Thus we see that:
1
ε≥
2
• So the energy eigenvalues are bounded from
below by _.
Setting up for The Big Trick
• Lets look at the commutator:
1

†
[A, H ]=  A, A A + 2  = A, A† A


= AA† A − A† AA
[
]
= (1+ A† A) A − A† AA
=A
AH − HA = A
€
HA = AH − A
HA = A(H − 1)
H A† − A† H = A†
H A† = A† H + A†
HA† = A† (H + 1)
The Big Trick Begins
• Combine these relations with the eigenvalue
equation:
H ε =ε ε
H A = A( H −1)
H A† = A† ( H + 1)
H A ε = A( H −1) ε
€
€
€
= A(ε −1) ε
H ( A ε ) = (ε −1)( A ε
)
• This means that A|ε〉 is proportional to the
€
eigenstate |ε-1〉: H ε − 1 = ε − 1 ε − 1
(
)
€
A ε = cε ε − 1
cε is an unknown
coefficient
HA† ε = A† (H + 1)ε
HA† ε = A† (ε + 1) ε
A† ε = d ε ε + 1
dε is an unknown
coefficient
Raising and lowering operators
A ε = cε ε − 1
A† ε = d ε ε + 1
A
ε −1 =
ε
cε
A†
ε +1 =
ε
dε
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:
ε H ε =ε
1
1
2
ε A A + ε = cε +
2
2
1
1
2
ε AA − ε = d ε −
2
2
†
†
• So clearly we must have:
2
ε = cε +
1
2
ε = dε
2
1
−
2
• So cε is only zero for ε=1/2 and dε is only zero
for ε=-1/2
This means that actually de is never zero
ε≥
1
2
2
cε = ε −
1
2
2
dε = ε +
1
2
Ground State energy
• Let the ground state |ε0〉 have energy:
1
ε0 = + δ
2
2
1
ε 0 = cε 0 +
2
2
cε 0 = δ
• Remember our statement:
– If state |ε〉 exists then either state |ε-1〉 exists or
cε=0
– Conclusion: either δ = 0, or there is a state
lower than the ground state
The second option is obviously a contradiction
– For |ε0〉 to be the ground state requires δ = 0
1
ε0 =
2
Excited states
• If the ground state |ε0〉 exists, then the state
|ε0+1〉 exists
A†
ε +1 =
ε
dε
2
dε = ε +
1
2
This is just a result
we proved on slide 13
The point is just
that we aren’t
dividing by zero
• Following this chain of reasoning, we can
establish the existence of states at energies:
1 3 5 7
ε = , , , ,K
2 2 2 2
Are there more states?
• So far we see that a ladder of states must
exist:
ε = 3/ 2
ε = 1/ 2
ε = 3 / 2 = A† ε = 1 / 2
†
A
ε = 5/ 2 =
ε = 3/ 2
2
M
M
ε = 1/ 2
No states
below by
definition
• Are there any states in between?
– Assume a state exists with
ε = 1/ 2 + x
0 < x <1
– We have
This state lies between
ε=1/2 and ε=3/2
1
cε = ε − = x
2
A
1/ 2 + x = x −1/ 2
x
2
x-1/2 < 1/2
– So either x=0 or there is a state below the
ground state! Conclusion: x=0
• If there is a state between 5/2 and 3/2, then
a state must exist between 3/2 and 1/2 and
then a state must exist below 1/2, etc…
• So no states between the half integers!
Spectrum of the SHO
• We now see that the energy eigenstates can
be labeled by the integers so that:
1

H n =  n +  n ; n = 0,1,2,3, K
2

• We can always go back to our original units
by putting in the energy scale factor:
1

H n = hω  n +  n ; n = 0,1,2,3, K
2
