Download Chapter 2

Quantum or classical mechanics?
Quantum mechanics must be used when
• deBroglie wavelength is comparable to or larger
than system size.
λ electron in H atom (~300pm), or 3Å
1Å
The diameter of H atom is
but not H atom (~100pm) in 1L box (~0.1m)
• spacing between energy levels comparable to kT
absorption of light by H atom, but not kinetic
energy of gas atoms in 1L box
1
Boltzmann distribution
ni gi
=
e
nj g j
 ε i −ε j 
−

kT


Let i=2, j=1
See EP2.1
∆ε = ε i , j = ε i − ε j
780
level 1
800
n1
600
400
n2
200
1
2
770
number of particles
Energy
level 2
number of particles
1000
3
4
5
760
750
n1
740
730
0.86
0.88
0.92
2 0.94
What is the ratio of H atoms with the electron in the
2p level to that in the 1s level at 298K? At 50,000K?
kT = 1.382 ⋅ 10−23 ⋅ 298 = 4.1 ⋅ 10−21 J = 0.026eV
∆E1s ,2 p = 43 ⋅ 13.6eV = 10eV
n2 p g 2 p
=
n1s
g1s
 ε 2 p − ε 1s 
−
 kT 

e
3 −397
= ⋅e
1
3 −2.37
= ⋅e
1
= 8 ⋅10
= 0.28
−173
At 298K
At
50,000K
3
Discrete vs quasi-continuous levels
1 kT
≈
• A molecule has 2
in energy per degree
of freedom (translation, rotation, vibration).
• An energy level is broadened by ≈ kT
through collisions with other molecules.
1
1
0.8
0.8
~
>>
0.6
0.6
kT
0.4
0.4
0.2
0.2
5
10
Energy
15
20
25
kT
1
2
Energy
4
In classical physics, waves and particles are distinct
entities
• Waves are characterized by amplitude,
wavelength (λ) and period (T).
• Wave propagation governed by wave
equation.
• Particles characterized by mass and velocity
• Particle motion governed by forces and
Newton’s equations.
5
Waves of a single frequency in one
dimension
Wave moving in +x direction described by
x
t
Ψ ( x, t ) = A sin 2π ( − ) = A sin( kx − ω t )
λ T
2π
2π
wave vector k =
. Frequency ω =
= 2πν
λ
T
The wave amplitude as a function of x and t is
called the wave function.
T is the period (s) and λ is the wavelength
(m).
6
Amplitude depends on both x and t
1
0.5
0
0.5
1.0
1.5
103t (s)
2.0
fixed x
Relative Amplitude
-0.5
-1
T
1
0.5
0
0.5
1
1.5
2
2.5
3
x (m)
fixed t
-0.5
-1
7
Unlike masses, waves can interfere
Two waves of equal A, λ, and T moving
oppositely combine to form standing wave.
0
amplitude
T/8
T/4
3T/8
T/2
5T/8
3T/4
7T/8
T
distance
8
Functional form for standing waves
Combine waves
Ψ( x,t) = A sin(kx −ω t) + sin(kx +ω t) 


Because sin α ± β  = sinα cos β ± cosα sin β




Ψ( x,t) = 2 Asin kx cosω t =ψ ( x )cosω t
The result is the product of a function of x with a
function of t.
Note difference between functional forms for
traveling and standing waves
9
Convenient to represent waves as complex functions
(Euler’s Identity)
Example: because
eiα = exp(iα ) = cos α + i sin α
i ≡ −1
{
i ( kx −ω t +φ )
Ψ ( x, t ) = A cos ( k x − ω t + φ ) = A Re e
Multiply:
e
Differentiate:
if ( x ) ig ( x )
e
=e
}
i  f ( x ) + g ( x )
df ( x ) if ( x )
deif ( x )
=i
e
dx
dx
10
Complex numbers
iθ
Represent as z = a + ib or as re where
iθ
Complex conjugate (z*)
z = a + ib = re
z = ( a + ib ) = ( a − ib ) = re
*
*
− iθ
Magnitude of z= zz * = ( a + ib )( a − ib ) = a 2 + b2
Magnitude 0 ≤ r = a + b phase
2
2
a
θ = cos  
r
a = r cos θ
−1
b = r sin θ
11
Convert 4+4i to reiθ
Magnitude of 4+4i is
Phase is
cosθ =
θ = cos
Can write 4+4i:
 ( 4 + 4i )( 4 − 4i ) 
12
4
4
−1
2
1
2
1
=
=
2
=4
2
or
π
4
iπ
4 + 4i = 4 2 e 4
Think about a circle on X and Y grid but the Y
axis is imaginary. Find 1 + i on that grid.
12
Convert 3e
Because
i 3π
2
to a + ib
3eiα = 3exp(iα ) = 3cos α + i3sin α = a + ib
a = 3cos 3π = 0
2
Therefore 3ei 3π 2
b = 3sin 3π =−3
2
can be written as –3i
Picture a wheel in the complex plane to understand
why this “complex” number is pure imaginary.
13
Look at waves in Maxwell’s Equations
The wave equation for light The electric Field pointing the
y direction is a function of x and time: Ψ ( x, t ) = E y ( x, t )
For light in free space, Maxwell’s Wave Equation is:
∂ Ψ ( x, t )
2 ∂ Ψ ( x, t )
=c
2
2
∂t
∂x
2
2
A comment about partial derivatives. Show that waves
(harmonic, traveling or standing) satisfy this equation.
Practice, Use a wave and show that it is a solution.
Ψ ( x, t ) = Ae
i ( kx −ω t )
= Aeikx e − iω t
What is the intensity (energy)? How do the
frequency and wave vector relate? Compare to
Ψ (t ) = X (t )
H.O. equation of motion of X:
14
Wave (Maxwell’s) equation for light
A sin(kx −ωt +φ )
satisfies the wave equation
2
∂
A sin( kx − ωt + φ )
2
2
c
= − ( ck ) A sin( kx − ωt + φ )
2
∂ x
∂ 2 A sin( kx − ωt + φ )
2
=
−
A sin( kx − ωt + φ )
ω
2
∂t
ω = ± ck
ν=
c
λ
Ψ ( x, t ) = Ae (
i kx −ω t )
= Aeikx e − iω t
Intensity = Ψ ( x, t ) Ψ ( x, t ) = A A = A
*
*
15
2
Energy: Classical Mechanics
The energy in a classical system is a function of the momentum and the
positions of the particles. The function is called the Hamiltonian (after
Hamilton who came up with it). For a simple classical system:
H=
1
2m
p2 + V ( q)
Plugging in the values of the momentum and the positions gives the energy of the
system. The Hamiltonian is the energy written in terms of p and q. The potential
Energy (V) is only a function of the position. Again, apply to the H.O. case and
determine the energy,(E).
The Hamilton form for the energy is very closely related to Newton’s equations:
 dp dV ( x ) 
dp dx dV ( x )
p
+
= v +

dt dt dx
dt
dx


dV ( x )
dp
=F =−
dt
dx
dH
=0=
dt
1
m
16
An Application: The Harmonic Oscillator (HO)
•
•
Let’s apply Hamilton’s principle and Newton’s laws to the equation of
motion of the HO.
Solve for X (and P and E) As a function of time
V = 12 k s x 2
H=
dV ( x )
dp
=F =−
dt
dx
∂2 X
m 2 = −k s X
∂t
X = A cos (ω t + ϕ )
E = 12 A2k s {sin 2 (ω t + ϕ ) + cos2 (ω t + ϕ )}
E = 12 k s A2
X and P are harmonic. The energy is a
constant and depends on the Amplitude
Squared. Can be written in Complex Form
P = − mω A sin (ω t + ϕ )
=A
p 2 + 12 k s x 2
E = 12 A2 {mω 2 sin 2 (ω t + ϕ ) + k s cos2 (ω t + ϕ )}
mω 2 = k s
a (t ) = i
1
2m
1
2m
ks
2
p+
ks
2
x=A
ks − i (ω t +ϕ )
2e
{−i sin (ω t + ϕ ) + cos (ω t + ϕ )}
17
deBrolie’s Hypothesis and A wave equation for it
h
ikx
== h
p = = =k
ψ
x
=
e
( )
2
π
λ
Let’s use the (light) wave equation and make an equation to
get the kinetic energy of a particle (at least in open space)
∂ 2ψ ( x ) ∂ 2 eikx
2
k
ψ ( x)
=
=
−
2
2
∂ x
∂ x
2
∂
ψ ( x)
2
2
2
k
ψ
x
p
ψ ( x)
−=
=
=
=
(
)
(
)
2
∂ x
From classical mechanics:
E=
1
2m
p2 + V ( x )
Then the Quantum Mechanical Analogue Ought to Be:
Eψ ( x ) =
1
2m
p 2ψ ( x ) + V ( x )ψ ( x )
2
=2 d ψ ( x )
−
+ V ( x )ψ ( x ) = Eψ ( x )
2
2m dx
18
Observables and operators
• Observable: quantity that can be measured such as
energy, momentum, position
• Operator: acts on (wave) function to multiply, add,
differentiate, integrate,….
dp ( t )
= F ( x, t )
dt
t2
1
v (t2 ) = v (t1 ) + ∫ F ( x, t ) d t
m t1
Order of operation is important!
19
Eigenfunctions and eigenvalues
Eigenfunctions are a subset of wave functions
• Eigenfunctions: When operated on, the function is
returned, multiplied by a constant.
d ikx
An eigenfunction is
e = ikeikx
specific to a particular
dx
operator!
d
sin kx = k cos kx ≠ k sin kx
dx
• Eigenvalue: the constant multiplying the
eigenfunction. (Eigen is German for
characteristic).
20
Schrödinger Equation is an eigenvalue equation for the total
energy, E
= d ψn ( x)
−
+ V ( x )ψ n ( x ) = Eψ n ( x )
2
2m dx
2
2
Hˆψ n ( x ) = Enψ n ( x )
d
ˆ
P = −i =
dx
1 ˆ2
ˆ
H = 2m P + V ( x )
The Schrödinger
equation is an
eigenvalue problem;
the eigenvalue is the
energy, it is a
constant; there are
many solutions or
Eigenfunctions each
with a different
21
Eigenvalue.
Eigenfunctions of a quantum mechanical
operator are orthogonal
• Orthogonality of vectors defined by
G G GG GG
x⋅ y = x⋅z = y⋅z = 0
• Orthogonality of eigenfunctions defined by
∞
∫ψ
−∞
*
i
( x )ψ j ( x ) dx = 0
unless i = j
22
Show graphically that sin x and cos 3x are
orthogonal functions. Also show graphically
∞
that ∫ ( sin m x )( sin n x ) d x ≠ 0 for n = m = 1
−∞
sin2x
0
sinx cos3x
0
f(x)
cos3x
0
sinx
0
2
4
x
6
8
10
23
Wave functions are normalized if
*( x )ψ ( x )d x =1
ψ
∫
Integrate over interval defined
Normalize a ( a − x ) over interval 0 ≤ x ≤ a
Use a positive constant N to require that ψ ( x ) is
properly normalized.
ψ = Na ( a − x )
a
a
2
N 2 ∫  a ( a − x ) dx = 1
N 2 a 2 ∫  a 2 − 2ax + x 2  dx = 1
0
2 2
0
a
x 
N a  a x − ax +  = 1
3 0

2
5
a
N2 =1
3
2
3
3
N= 5
a
3
ψ = 3 (a − x )
a
24
Normalizing in spherical coordinates
Volume element in spherical coordinates
r = x2 + y2 + z2
r
r
z
y
dr
dφ
r
x
dV
dV = dxdydz = r drd cosθ dϕ
2
25
Normalize e − r over all space
Ψ ( r ) = Ae
−r
0 ≤ r ≤ ∞; 0 ≤ θ ≤ π ; 0 ≤ φ ≤ 2π
Require the function to be normalize:
2π
π
∞
0
0
0
1 = ∫∫∫ Ψ ( r ) d 3r = A2 ∫ dφ ∫ sin θ dθ ∫ r 2 e −2 r dr
∞
4π A2 ∫ r 2 e −2 r dr = 1
Using standard integral
0
2!
1
4π A 3 = 1; A =
π
2
2
Ψ (r) =
1
π
e
−r
is normalized
26
Eigenfunctions of a QM operator form complete set
Like any vector
any wave function
G
G
G
G
v = v x i + v y j +, v z k
∞
ψ ( x ) = ∑ anφn ( x )
n =1
Where φ(x) are the eigenfunctions of a QM operator.
Infinite sum.
Wave functions are normalizable, single valued and
continuous (in short ‘nice’ functions)
27
Limited number of terms often suffices
Fit function using Fourier series
b = 5, and d o = 14
f(x)
f ( x ) = ( 2 x − 1) e
2
f(x)
n=2
x/b
-1
n=3
1
-1
f(x)
1
f(x)
n=4
-1
x/b
x/b
n=6
1
-1
x/b
The fit function dies of outside the window, but the
Fourier Series repeats forever.
1
28
2
− x2