5.61 Fall 2007
Lectures #12-15
page 1
THE HARMONIC OSCILLATOR
•
Nearly any system near equilibrium can be approximated as a H.O.
•
One of a handful of problems that can be solved exactly in quantum
mechanics
examples
m1
m2
A diatomic molecule
E (electric
field)
µ
(spin
magnetic
moment)
B (magnetic
field)
Classical H.O.
k
m
X0
Hooke’s Law:
X
(
)
f = −k X − X 0 ≡ −kx
(restoring force)
d2x
f = ma = m 2 = −kx
dt
⇒
d2x ⎛ k ⎞
+
x=0
dt 2 ⎜⎝ m⎠⎟
5.61 Fall 2007
Lectures #12-15
Solve diff. eq.:
()
page 2
General solutions are sin and cos functions
( )
( )
x t = Asin ω t + B cos ω t
ω=
or can also write as
()
(
x t = C sin ω t + φ
k
m
)
where A and B or C and φ are determined by the initial conditions.
()
x 0 = x0
e.g.
()
v 0 =0
spring is stretched to position x 0 and released at time t = 0.
Then
()
()
()
x 0 = A sin 0 + B cos 0 = x0
()
v 0 =
So
dx
dt
()
⇒
B = x0
()
= ω cos 0 − ω sin 0 = 0
⇒
A=0
x=0
()
( )
x t = x0 cos ω t
Mass and spring oscillate with frequency: ω =
and maximum displacement
x0
k
m
from equilibrium when cos(ωt)= ±1
Energy of H.O.
Kinetic energy ≡ K
2
2
1
1 ⎛ dx ⎞
1
1
K = mv 2 = m ⎜ ⎟ = m ⎡⎣ −ω x0 sin ω t ⎤⎦ = kx02 sin 2 ω t
2
2 ⎝ dt ⎠
2
2
( )
( )
Potential energy ≡ U
()
f x =−
dU
dx
⇒
()
U = − ∫ f x dx =
1
kx
dx
=
kx
(
)
∫
2
2
=
1 2
kx0 cos 2 ω t
2
( )
5.61 Fall 2007
Lectures #12-15
page 3
Total energy = K + U = E
E=
1 2
kx0 ⎡⎣sin 2 ω t + cos 2 ω t ⎤⎦
2
( )
( )
E=
1 2
kx
2 0
x (t )
x 0(t )
0
t
-x0(t)
U
1 2
kx
2 0
K
E
0
t
Most real systems near equilibrium can be approximated as H.O.
e.g.
Diatomic molecular bond
A
B
X
U
X0
X
A + B separated atoms
equilibrium bond length
5.61 Fall 2007
( )
( )
U X = U X0
Lectures #12-15
dU
+
dX
(X − X )
0
X = X0
Redefine x = X − X 0
(
()
(X − X )
2
0
X = X0
)
(
1 d 3U
+
3! dX 3
(X − X )
3
0
X = X0
)
U X = X0 = U x = 0 = 0
and
dU
U x =
dx
1 d 2U
+
2 dX 2
page 4
1 d 2U
x+
2 dx 2
x=0
1 d 3U
x +
3! dx 3
x=0
x3 + !
2
x=0
U
real potential
H.O. approximation
x
At eq.
dU
dx
=0
x=0
For small deviations from eq.
()
∴ U x ≈
x 3 << x 2
1 d 2U
2 dx 2
x2 ≡
x=0
1 2
kx
2
+!
5.61 Fall 2007
Lectures #12-15
Total energy of molecule in 1D
M = m1 + m2
µ=
m2
X1
XCOM
m1 X 1 + m2 X 2
m1 + m2
xrel = X 2 − X 1 ≡ x
xrel
COM position
relative position
2
2
2
⎞
1 ⎛ dX ⎞
1 ⎛ dX ⎞
1 ⎛ dX
1 ⎛ dx ⎞
K = m1 ⎜ 1 ⎟ + m2 ⎜ 2 ⎟ = M ⎜ COM ⎟ + µ ⎜ ⎟
2 ⎝ dt ⎠
2 ⎝ dt ⎠
2 ⎝ dt ⎠
2 ⎝ dt ⎠
U=
X2
reduced mass
m1 + m2
X COM =
m1
total mass
m1 m2
page 5
2
1 2
kx
2
2
2
⎞
1 ⎛ dX
1 ⎛ dx ⎞
1
E = K + U = M ⎜ COM ⎟ + µ ⎜ ⎟ + kx 2
2 ⎝ dt ⎠
2 ⎝ dt ⎠
2
COM coordinate describes translational motion of the molecule
Etrans
⎞
1 ⎛ dX
= M ⎜ COM ⎟
2 ⎝ dt ⎠
2
QM description would be free particle or PIB with mass M
We’ll concentrate on relative motion (describes vibration)
2
Evib
1 ⎛ dx ⎞
1
= µ ⎜ ⎟ + kx 2
2 ⎝ dt ⎠
2
and solve this problem quantum mechanically.
X
5.61 Fall 2007
Lectures #12-15
page 6
THE QUANTUM MECHANICAL HARMONIC OSCILLATOR
⎡ !2 d 2 1 2 ⎤
Ĥψ x = ⎢ −
+ kx ⎥ ψ x = Eψ x
2
2
⎣ 2m dx
⎦
()
()
K
()
U
Note: replace m with µ (reduced mass) if m1
m2
Goal: Find eigenvalues En and eigenfunctions ψn(x )
Rewrite as:
( ) + 2m ⎡
E −
1
kx
d 2ψ x
⎢
! 2 ⎣
dx 2
2
2
⎤
⎥ ψ
x = 0
⎦
()
This is not a constant, as it was for P-I-B,
so sin and cos functions won’t work.
()
f x = e−α x
TRY:
( ) = −α e
d2 f x
dx
2
or rewriting,
2
2
−α x 2 2
(gaussian function)
+ α 2 x 2 e−α x
2
( ) +α f
d2 f x
dx
()
2
()
= −α f x + α 2 x 2 f x
2
( x) − α
2
()
x 2 f x = 0w
which matches our original diff. eq. if
α=
∴
2mE
!2
and
E=
! k
2 m
α2 =
mk
!2
5.61 Fall 2007
Lectures #12-15
page 7
We have found one eigenvalue and eigenfunction
k
m
ω=
Recall
ν=
or
E=
∴
1
2π
k
m
1
1
!ω = hν
2
2
This turns out to be the lowest energy: the “ground” state
For the wavefunction, we need to normalize:
()
()
ψ x = Nf x = Ne−α x
∫
∞
−∞
()
2
ψ x dx = 1
2
where N is the normalization constant
2
⇒
N
2
∫
∞
−∞
e
−α x2
=1
⇒
⎛α⎞
N =⎜ ⎟
⎝π⎠
14
π α
ψ0
∴
⎛α⎞
x =⎜ ⎟
⎝π⎠
()
E0 =
()
14
e−α x
2
2
()
ψ0 x
1
1
!ω = hν
2
2
E0 =
x
()
1
!ω
2
( )
Note ψ 0 x is symmetric. It is an even function: ψ 0 x = ψ 0 −x
There are no nodes, & the most likely value for the oscillator displacement is 0.
So far we have just one eigenvalue and eigenstate. What about the others?
5.61 Fall 2007
Lectures #12-15
ψ0
⎛α⎞
x =⎜ ⎟
⎝π⎠
()
14
e−α x
1 ⎛α⎞
ψ1 x =
⎜ ⎟
2⎝π⎠
14
1 ⎛α⎞
x =
⎜ ⎟
8⎝π⎠
14
()
ψ2
ψ3
page 8
()
2
2
( 2α x ) e
12
1 ⎛α⎞
x =
⎜ ⎟
48 ⎝ π ⎠
()
( 4α x
14
(8α
2
−α x 2 2
)
− 2 e−α x
32
2
E0 =
1
hν
2
E1 =
3
hν
2
E2 =
5
hν
2
E3 =
7
hν
2
2
)
x 3 − 12α 1 2 x e−α x
2
2
!
!
⎛ km ⎞
α =⎜ 2 ⎟
⎝! ⎠
with
12
These have the general form
()
ψn x =
1
⎛α⎞
1/ 2 ⎜
⎝ π ⎟⎠
14
( 2 n!)
n
(
)
H n α 1 2 x e−α x
2
2
n = 0,1,2,...
Normalization
Gaussian
Hermite polynomial (pronounced “air-MEET”)
( )
H ( y) = 2 y
H ( y) = 4 y − 2
H ( y ) = 8y − 12 y
H ( y ) = 16 y − 48y
H0 y = 1
1
2
2
3
3
4
4
!
( n = 0)
odd ( n = 1)
even ( n = 2 )
odd ( n = 3)
even ( n = 4 )
even
2
+ 12
!
5.61 Fall 2007
Lectures #12-15
page 9
()
()
ψn x
ψ n x
E3 =
()
E2 =
()
E1 =
3!ω
2
()
E0 =
!ω
2
ψ2 x
ψ1 x
ψ0 x
Energies are
7!ω
2
()
ψ3 x
⎛
En = ⎜ n +
⎝
5!ω
2
1⎞
hν
2
⎟⎠
Note E increases linearly with n.
⇒
Energy levels are evenly spaced
⎛
⎛
1⎞
1⎞
En+1 − En = ⎜ n + 1 + ⎟ hν − ⎜ n + ⎟ hν = hν
2⎠
2⎠
⎝
⎝
(
)
There is a “zero-point” energy
E0 =
regardless of n
1
hν
2
E = 0 is not allowed by the Heisenberg Uncertainty Principle.
2
5.61 Fall 2007
Lectures #12-15
page 10
Symmetry properties of ψ’s
ψ 0,2,4,6,....
are even functions
ψ 1,3,5,7,....
are odd functions
Useful properties:
( )=
d odd
dx
( ) ()
ψ ( −x ) = −ψ ( x )
ψ −x = ψ x
(even) ⋅(even) = even
(odd) ⋅(odd) = even
(odd) ⋅(even) = odd
(
d even
(even )
dx
∫ ( odd ) dx = 0
)=
( odd )
∫ (even ) dx = 2 ∫ (even ) dx
∞
−∞
∞
∞
−∞
0
Just from symmetry:
x
n
=
∫
∞
−∞
()
()
ψ n∗ x xψ n x dx = 0
p
n
=
d⎞
∗⎛
ψ
−ih
∫−∞ n ⎜⎝ dx ⎟⎠ ψ n x dx = 0
odd
()
∞
odd
Average displacement & average momentum = 0
IR spectroscopy
⇒
H.O. selection rules
Intensity of vibrational absorption features
n’ = 1
Vibrational transition
hν
n=0
δ+
δ-
5.61 Fall 2007
Lectures #12-15
Intensity
page 11
dµ ∞ ∗
∝
ψ xψ dx
dx ∫−∞ n n '
I nn′
2
1)
Dipole moment of molecule must change as molecule vibrates ⇒
HCl can absorb IR radiation, but N2, O2, H2 cannot.
2)
Only transitions with
(Prove for homework.)
n′ = n ± 1
allowed (selection rule).
QUANTUM MECHANICAL HARMONIC OSCILLATOR &
TUNNELING
Classical turning points
Total energy ET =
Classical H.O.:
E
1 2
kx
2 0
ET
oscillates between K and U.
Maximum displacement x0 occurs when
all the energy is potential.
-x0
x0 =
2ET
k
x0
x
is the “classical turning point”
The classical oscillator with energy ET can never exceed this displacement, since
if it did it would have more potential energy than the total energy.
5.61 Fall 2007
Lectures #12-15
page 12
Quantum Mechanical Harmonic Oscillator.
()
ψ3 x
()
2
ψ 12 x
()212
xψ
()
2
()
2
()
2
ψ2 x
ψ1 x
ψ0 x
1 2
kx
22
12kx
2
At high n, probability
density begins to look
classical, peaking at turning
points.
x
()
ψ3 x
Non-zero probability at x > x 0!
Prob. of (x > x 0, x < -x 0):
2
()
ψ2 x
=
()
()
2
ψ0 x
∞
α
2
2
ψ1 x
2∫
−1 2
2
π1 2
⎛α⎞
ψ 02 x dx = 2 ⎜ ⎟
⎝π⎠
()
∫
∞
1
2
1
2
∫
∞
α
−1 2
()
e− y dy = erfc 1
“Complementary error function”
tabulated or calculated
numerically
Prob. of (x > x 0, x < -x 0) = erfc(1)
= 0.16
Significant probability!
x
2
e−α x dx
5.61 Fall 2007
Lectures #12-15
page 13
The oscillator is “tunneling” into the classically forbidden region. This is a purely
QM phenomenon!
Tunneling is a general feature of QM systems, especially those with very low
mass like e- and H.
E
()
( )
Finite
barrier
()
ψ x ~ e− γ x
ψ x ~ sin kx
V0
()
(
ψ x ~ sin kx + φ
)
x
Even though the energy is less than the barrier height, the wavefunction is
nonzero within the barrier! So a particle on the left may escape or “tunnel” into
the right hand side.
⎡
! 2 d 2
⎤
−
+
V
ψ x = Eψ x
⎢
0⎥
2
2m
dx
⎣
⎦
()
Inside barrier:
or
( ) = ⎡⎢ 2m (V
d 2ψ x
(
γ ∝ V0 − E
)
12
⎢⎣
()
ψ x = Be− γ x
and
)
−E ⎤
⎥ ψ x ≡ γ 2ψ x
2
!
⎥⎦
0
dx 2
Solutions are of the form
Note
()
()
with
()
(
⎡ 2m V0 − E
γ =⎢
!2
⎢⎣
) ⎤⎥
1
2
⎥⎦
γ ∝ m1 2
If barrier is not too much higher then the energy and if the mass is light, then
tunneling is significant.
Important for protons (e.g. H-bond fluctuations, tautomerization)
5.61 Fall 2007
Lectures #12-15
page 14
Important for electrons (e.g. scanning tunneling microscopy)
Nonstationary states of the QM H.O.
System may be in a state other than an eigenstate, e.g.
ψ = c0ψ 0 + c1ψ 1
2
2
c0 + c1 = 1 (normalization), e.g.
with
c0 = c1 =
Full time-dependent eigenstates can be written as
( )
()
Ψ 0 x,t = ψ 0 x e
( )
− iω 0 t
()
Ψ1 x,t = ψ 1 x e
1
2
− iω1t
where
!ω 0 = E0 =
1
1
!ω vib ⇒ ω 0 = ω vib
2
2
!ω 1 = E1 =
3
3
!ω vib ⇒ ω 1 = ω vib
2
2
System is then time-dependent:
1
( )
Ψ x,t =
where
2
e
()
− iω 0 t
c0 t =
()
ψ0 x +
1
2
e
1
− iω 0 t
2
e
− iω1t
()
()
() ( )
() ( )
ψ 1 x = c0 t ψ 0 x + c1 t ψ 1 x
c1 t =
1
2
e
− iω1t
What is probability density?
1⎡ ∗
iω 0 t
iω1t
∗
⎤ ⎡ψ 0 x e− iω 0 t + ψ 1 x e− iω1t ⎤
ψ
x
e
+
ψ
x
e
0
1
⎦⎣
⎦
2⎣
1
1
i ω −ω t
− i ω −ω t
= ⎡ψ 0∗ψ 0 + ψ 1∗ψ 1 + ψ 1∗ψ 0 e ( 1 0 ) + ψ 0∗ψ 1e ( 1 0 ) ⎤ = ⎡⎣ψ 02 + ψ 12 + 2ψ 0ψ 1 cos ω vib t ⎤⎦
⎦ 2
2⎣
( ) ( )
Ψ ∗ x,t Ψ x,t =
()
()
()
()
(
Probability density oscillates at the vibrational frequency!
)
5.61 Fall 2007
Lectures #12-15
(
)
() costψψω
2
Ψ x,t
page 15
ω t = 0, π 4, π 2, 3π 4, π
01
x
()
2
()
2
ψ x
()211 xψ
ψ x
()200 xψ
2ψ 0ψ 1 cos (ω t )
()
012costψψω
x
ω t = 0, π 4, π 2, 3π 4, π
What happens to the expectation value <x>?
5.61 Fall 2007
x =
Lectures #12-15
∫
∞
−∞
page 16
( ) ( )
Ψ ∗ x,t x̂Ψ x,t dx
1 ∞⎡ ∗
ψ 0 x eiω 0 t + ψ 1∗ x eiω1t ⎤⎦ x ⎡⎣ψ 0 x e− iω 0 t + ψ 1 x e− iω1t ⎤⎦ dx
∫
⎣
−∞
2
∞
∞
∞
1 ∞
i ω −ω t
− i ω −ω t
= ⎡ ∫ ψ 0∗ xψ 0 dx + ∫ ψ 1∗ xψ 1 dx + ∫ ψ 1∗ xψ 0 e ( 1 0 ) dx + ∫ ψ 0∗ xψ 1e ( 1 0 ) dx ⎤
−∞
−∞
−∞
⎦⎥
2 ⎣⎢ −∞
()
=
<x >0 = 0
()
()
()
(
= cos ω vib t
<x >1 = 0
)∫
∞
−∞
ψ 0 xψ 1 dx
<x>(t) oscillates at the vibrational frequency, like the classical H.O.!
∫
Vibrational amplitude is
∞
−∞
1
4
⎛α⎞
2
ψ 0 x = ⎜ ⎟ e−α x
⎝π⎠
()
⇒
∴
∫
∞
−∞
ψ 0 xψ 1 dx
1 ⎛α⎞
⎜ ⎟
2⎝π⎠
()
ψ1 x =
2
1
4
⎛α⎞
2
xψ 0 x = ⎜ ⎟ xe−α x 2 = 2α
⎝π⎠
()
( ) ∫
ψ 0 xψ 1 dx = 2α
−1 2
∞
−∞
( )
( )
ψ 02 dx = 2α
−1 2
1
4
( 2α x ) e
12
−α x 2 2
()
ψ1 x
() ( )
−1 2
x t = 2α
−1 2
Relations among Hermite polynomials
Recall H.O. wavefunctions
1
()
ψn x =
1
⎛α⎞4
12
−α x 2
H
α
x
e
n
1/ 2 ⎜
⎝ π ⎟⎠
( 2 n!)
n
(
)
2
Normalization
Gaussian
Hermite polynomial
(
cos ω vib t
n = 0,1,2,...
)
5.61 Fall 2007
Lectures #12-15
( )
H ( y) = 2 y
H ( y) = 4 y − 2
H ( y ) = 8y − 12 y
H ( y ) = 16 y − 48y
page 17
( n = 0)
odd ( n = 1)
even ( n = 2 )
odd ( n = 3)
even ( n = 4 )
H0 y = 1
even
1
2
2
3
3
4
4
2
+ 12
!
!
Generating formula for all the Hn:
Hn
d n − y2
y = −1 e
e
dy n
( ) ( )
n
y2
A useful derivative formula is:
( )=
dH n y
dy
n+1
n y2 d
d n − y2
− y2
−1 2 ye
e
+
−1
e
e
= 2 yH n y − H n+1 y
dy n
dy n+1
( )
n
( )
y2
( )
( )
Another useful relation among the Hn’s is the recursion formula:
( )
( )
( )
H n+1 y − 2 yH n y + 2nH n−1 y = 0
( )
( )
( )
2 yH n y = H n+1 y + 2nH n−1 y above gives
Substituting
( ) = 2nH
dH n y
dy
n−1
( y)
Use these relations to solve for momentum <p>(t)
p =
∫
∞
( ) ( )
Ψ ∗ x,t p̂Ψ x,t dx
−∞
1 ∞⎡ ∗
ψ 0 x eiω 0 t + ψ 1∗ x eiω1t ⎤⎦ pˆ ⎡⎣ψ 0 x e− iω 0 t + ψ 1 x e− iω1t ⎤
∫
⎣
⎦ dx
2 −∞
∞
∞
∞
1 ∞
i ω −ω t
− i ω −ω t
= ⎡ ∫ ψ 0∗ pˆψ 0 dx + ∫ ψ 1∗ pˆψ 1 dx + ∫ ψ 1∗ pˆψ 0 e ( 1 0 ) dx + ∫ ψ 0∗ pˆψ 1e ( 1 0 ) dx ⎤
⎥⎦
−∞
−∞
−∞
2 ⎢⎣ −∞
=
()
<p >0 = 0
()
<p >1 = 0
()
()
5.61 Fall 2007
Lectures #12-15
⎛α⎞
d
ψ0 x = ⎜ ⎟
dx
⎝π⎠
()
∴
∫
∞
ψ 1∗ p̂ψ 0 e (
)
i ω1 − ω 0 t
−∞
To solve integral
∫
∞
−∞
1
4
( −α x ) e
page 18
−α x 2 2
1
2
1
2
⎛α⎞
= −⎜ ⎟ ψ1 x
⎝ 2⎠
()
1
2
⎛ α ⎞ i ω −ω t
⎛ α ⎞ iω t
dx = i! ⎜ ⎟ e ( 1 0 ) ∫ ψ 1∗ψ 1 dx = i! ⎜ ⎟ e vib
−∞
⎝ 2⎠
⎝ 2⎠
∞
− i ω −ω t
ψ 0∗ p̂ψ 1e ( 1 0 ) dx use relations among Hn’s
(
)
2
2
d
d ⎡
d
ψ1 x =
N 1 H1 α 1 2 x e − α x 2 ⎤ = α 1 2 N 1 ⎡ H1 y e − y 2 ⎤
⎦
⎦
dx
dx ⎣
dy ⎣
()
with
y ≡ α1 2 x
( )
d
d
= α1 2
dx
dy
dx = α −1 2 dy
dy = α 1 2 dx
2
2
⎡d
⎤
d
ψ 1 x = α 1 2 N1 ⎢ H1 y e− y 2 − yH1 y e− y 2 ⎥
dx
⎣ dy
⎦
d
H y = 2nH 0 y = 2H 0 y
dy 1
1
1
yH1 y = ⎡⎣ 2nH 0 y + H 2 y ⎤⎦ = H 0 y + H 2 y
2
2
()
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
⎡ 1
⎤
2
2
⎡
⎤
d
1
1
ψ 1 x = α 1 2 N1 ⎢ H 0 y e− y 2 − H 2 y e− y 2 ⎥ = α 1 2 N1 ⎢ ψ 0 x −
ψ2 x ⎥
dx
2
2N 2
⎣
⎦
⎦
⎣ N0
()
( )
∫
∞
−∞
( )
( )∫
− i ω −ω t
− i ω −ω t
ψ 0∗ p̂ψ 1e ( 1 0 ) dx = e ( 1 0 ) −i!
⎡ 1
− i ω −ω t
= e ( 1 0 ) −i! α 1 2 N1 ⎢
⎣ N0
( )
( )
− i ω −ω t
= e ( 1 0 ) −i! α 1 2
Finally
()
∫
∞
−∞
∞
−∞
ψ 0∗
ψ 0∗ψ 0 dx −
1
2
⎛ α ⎞ − iω t
= −i! ⎜ ⎟ e vib
N0
⎝ 2⎠
N1
d
ψ dx
dx 1
1
2N 2
⎤
∗
ψ
ψ
dx
∫−∞ 0 2 ⎥
⎦
∞
()
5.61 Fall 2007
Lectures #12-15
1
⎡
⎛ α ⎞ 2 iω t
1
− iω t
p t = ⎢ i! ⎜ ⎟ e vib − e vib
2⎢ ⎝ 2⎠
⎢⎣
()
(
page 19
)
1
⎤
⎥ = −! ⎛ α ⎞ 2 sin ω t
vib
⎜⎝ 2 ⎟⎠
⎥
⎥⎦
(
Average momentum also oscillates at the vibrational frequency.
)