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Chapter 34
Quantum Mechanics
1927 Solvay Conference
Wave function
 h/ p
Wavelength of de Broglie wave:
“Displacement” of wave → wave function Ψ
Energy density of EM wave u  E 2
Number density of photon:
N  u  E
Probability of finding photon: p  
Normalization condition:

2
2
2
dV  1
3
Probability wave
|Ψ|2dV at a certain point represents the
probability of finding the particle within volume
dV about the given position and time.
P1  1
2
P2   2
2
P12  1 + 2
2
 P1  P2 !
4
*Coherent & de-coherence
P12  1   2
2
P12  P1  P2
de-coherence
Interference between coherent wave functions
If we can measure which slit the electron passes
Measurement disturbs the state of particle!
Superposition state & Schrödinger cat
5
Uncertainty principle
Measurement disturbs the state of particle
Observe an electron by a photon:
  x ,
  p 
Always an uncertainty in position or momentum
x  
p 
h

(x)(p)  h
h
(x)(p)  
2
The uncertainty principle
6
Heisenberg uncertainty principle
h
(x)(p)  
2
W. K.
Heisenberg
Nobel 1932
The position and the momentum of a particle
can not be precisely determined simultaneously.
1) It is a natural result of duality
2) A criterion of classical and quantum physics
3) The uncertainty is inherent rather than caused
by the limitation of instruments or methods
7
Other forms of uncertainty principle
h
(E )(t )  
2
h
(Lz )( )  
2
Inherent principles in quantum mechanics
Microscopic particles will not stay at rest!
Zero point energy (at 0K) & Dirac sea
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A quantum criterion
Example1: Calculate the uncertainty in position
for (a) a 150-g baseball and (b) an electron both
measured with 50 m/s to a precision of 0.3%.
Solution: (a)
1.055 1034
x 

 4.7 1033 m
p 0.15  50  0.3%
It’s too small, so we can ignore the limitation
(b)
1.055  1034
4
x 


7.7

10
m
31
p 9.11 10  50  0.3%
QM is necessary! 108m/s to precision of 0.3% ?
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Diffraction of electron
Thinking: An electron beam (each with p) falls on
a single slit, show that the central bright fringe
satisfies the uncertainty relation xpx  h
x
Film
Electron
beam
x

y
Single slit
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Probability vs determinism
Probability is inherent in nature, not a limitation
on our abilities to calculate or to measure
Copenhagen interpretation of quantum mechanics
“God does not play dice. ”
—— A. Einstein
“Anyone who is not shocked by quantum theory
has not understood it”
—— N. Bohr
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Schrödinger equation (1)
An equation to determine the wave function Ψ
A free particle (E, p) moves along x axis
can be also treated as a wave:
y ( x, t )  A cos(2 ft 
2 x

Wave function in general
form (complex value) :
h
E
 , f 
p
h
E
)  A cos( t 
  Ae
p
x)
i
 ( Et  px )
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Schrödinger equation (2)
  Ae
i
 ( Et  px )
2
p2
 2 
2
x

i
  E 
t
Nonrelativistic free particle: E  p 2 / 2m
2
2

2 


i
, or 
U   i
2
2
2m x
t
2m x
t
2
in general case, considering the potential energy
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Schrödinger equation (3)
2


U  i
2
2m x
t
2
3D time-dependent S-equation:
2

i
 (
 2  U )  H 
t
2m
H 
2
2m
E. Schrödinger
Nobel 1933
2  U → Hamilton operator
Basic, can’t be derived, check by experiments
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Schrödinger equation (4)
Time-independent case: ( x, y, z, t )   ( x, y, z) f (t )
H ( x, y, z )
1 df (t )

i
E
 ( x, y, z )
f (t ) dt
Time-independent Schrödinger equation:
(
1) E → energy
2
2m
2  U )  E
2) Stationary state / eigenstate
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Solution of S-equation
(
2
2m
2  U )  E
Solve the S-equation to analyze quantum systems
1) Each solution represents a stationary state
2) The system may be in a superposition state
3) The wave function of system should be
continuous, finite and normalized.
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Infinitely deep square well potential
Particle in an infinitely deep square well potential
or “rigid box”
 U ( x)  0, 0  x  L

 U ( x)  , x  0 and x  L
Trapped in the well, can’t escape
Classical case:


U
Free motion until collision
Equal probability at any point
x0
xL
x
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Solution by S-equation (1)


x0
xL
U
 2

 U  E
2
2m x
2
x
Can’t escape, so  ( x)  0, x  0 and x  L
d 2

 E
2
2m dx
2
0 < x < L, U(x) = 0, so:
Or:
d 2
2

k
 0
2
dx
General solution is
where k 
2
2mE
2
 ( x)  A sin kx  B cos kx
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Solution by S-equation (2)


x0
xL
U
 ( x)  A sin kx  B cos kx , k 2 
2mE
2
ψ must be continuous, so
 (0)   ( L)  0
x
 B  0, sin kL  0
2
n
h
k
, n  1, 2, ...  En  n2
2 Quantized!
L
8mL
n
The wave function:  n ( x)  A sin( x)
L
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Quantum properties (1)
2
n
h
 n ( x)  A sin( x), En  n2
L
8mL2
1) The energy of particle is quantized
n: quantum number of state
2) The minimum energy is not zero!


E3  9 E1
Zero point energy
E2  4 E1
Microscopic particles will not
stay at rest!
E1
E0
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Quantum properties (2)

2
dx  1   n ( x) 
2
n
sin(
x)
L
L
3) Figure of wave function
& probability distribution
Like a standing wave
Notice: The probability
is not constant!
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Quantum properties (3)
 n ( x) 
2
n
sin(
x)
L
L
4) de Broglie wavelength?
2
h 2L
nh
h
2
 n 

 pn  2mEn 
En  n
2
pn
n
2L
8mL
In accord with the figure of wave function!
5) Uncertainty principle:
xp  nh 
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Electron in the well
Example2: Calculate the energy of ground state
and first exited state for an electron trapped in an
IDSWP of width L = 0.1nm.
Solution: The ground state (n=1) has energy
h2
18
E1 

6.03

10
J  37.7 eV
2
8mL
First exited state:
E2  4E1  151 eV
λ of photon if jumping from n = 2 to n = 1 ?
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Probability in well
Example3: In the state of  
2
3
sin( x)
L
L
(a) Where does the particle have maximum
probability densities? (b) What is the probability
to find the particle in region 0 < x < L/4 ?
Solution: (a) 
d

dx
2
2
2 2 3
 sin ( x)
L
L
d2
2
 0 and
 0
2
dx
L L 5L
So: x  , ,
6 2 6
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(b) What is the probability to find   2 sin( 3 x)
L
L
the particle in region 0 < x < L/4 ?
L
4
0
P    dx
2
L
4
0
2 2 3
sin ( x)dx
L
L
L
4
0
1
6
[1  cos(
x)]dx
L
L


1 1
 
 0.303
4 6
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Probability for other ψ
Homework: A particle trapped in a special well
has a wave function   Cx( L  x), 0  x  L, and
C is a constant. What is the probability to find
the particle in region 0 < x < L/3 ?
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Finite potential well
If the well potential is not infinitely deep?
Particle can get out even if E < U0 !
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Tunneling through a barrier
Particle can also penetrate a thin barrier ( E < U0 )
This effect is called “quantum tunneling”
Tunneling probability:
T e
where G 
U
U0
2GL
2m(U 0  E )
2
1
3
2
0
L
x
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*Applications of tunneling
1) Radioactive decay
U
238
234
4
Th + He
2) Scanning Tunneling Microscope (STM)
Invented 1981, Nobel 1986
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