Download Physics 451 - BYU Physics and Astronomy

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts
no text concepts found
Transcript
Physics 451
Quantum mechanics
Fall 2012
Karine Chesnel
Phys 451
Announcements
Test 1 next week
Mo Sep 24 – Th Sep 27
• Today: Review - Monday: Practice test
Be prepared to present the solution of
your chosen problem during class (~ 5 to 10 min)
Phys 451
EXAM I
• Time limited: 3 hours
• Closed book
• Closed notes
• Useful formulae provided
Review lectures,
Homework
and sample test
Phys 451
EXAM I
1. Wave function, probabilities
and expectation values
2. Time-independent Schrödinger equation
3. Infinite square well
4. Harmonic oscillator
5. Free particle
Phys 451
Review I
What to remember?
Review I
Quantum mechanics
1. Wave function and expectation values
Density of probability
 ( x, t )   ( x, t )

Normalization:


 ( x, t )  1
2
2
Review I
Quantum mechanics
1. Wave function and expectation values

 

p     i
 dx
x 



x 
  xdx
*
*

“Operator” x
“Operator” p

 

Q    Q  x, i
 dx
x 


*
Quiz 9a
What is the correct expression for the operator
T= Kinetic energy?
A.
  



2m  x 
B.
x 2 / 2m
C.
1   

 
2m  x 
2
2
2
2
1

D.
 m 2 x 2
2m x
E.
2

2m x 2
2
Review I
Quantum mechanics
1. Wave function and expectation values
Variance:
x  x  x
2
2
p  p  p
2
2
 x p 
2
2
2
Heisenberg’s
Uncertainty principle
Review I
Quantum mechanics
2. Time-independent Schrödinger equation
2

2
i

V 
2
t
2m x
Here
V ( x)
The potential is independent of time
General solution:
 ( x, t )   ( x) (t )
“Stationary state”
Review I
Quantum mechanics
2. Time-independent Schrödinger equation
2
1 d
1 d 2
i

V
2
 dt
2m  dx
Function of
time only
 (t )  e
Solution:

iE
E
Function of
space only
t
( x, t )   ( x)eiEt /
Stationary state
Quantum mechanics
Review I
2. Time-independent Schrödinger equation
Q
for each
Stationary state
is independent of time
p m v m
d x
0
dt
^
H E
A general solution is

 ( x, t )   cn  n ( x, t )
n 1
where
 n ( x, t )   n ( x)eiEnt /
^
H   cn En
2
n
Review I
Quantum mechanics
3. Infinite square well
d 2
2


k

2
dx
with
0
a
The particle can only exist
in this region
x
k
2mE
Review I
Quantum mechanics
3. Infinite square well
 3 , E3
n 
2
 n
sin 
a
 a

x

Excited states
 2 , E2
Quantization of the energy
Ground state
0
 1 , E1
a
x
n 2 2 2
En 
2
2ma
Review I
Quantum mechanics
3. Infinite square well

 ( x, t )   cn n ( x)e
n 1

 iEn t /

x
n 1
a
  cn sin(n
)e iEnt /

2
n x
cn    n* ( x) ( x, 0)dx 
sin(
)  ( x, 0)dx

a 
a

^
H  n  En n

*
 n   nm
m
^

H   cn En
n 1
2
Quiz 9b
The particle is in this sinusoidal state.
What is the probability of measuring the energy E0 in this state?
A. 0
B. 1
C. 0.5
D. 0.3
0
a
x
E. 1
9
Review I
Quantum mechanics
4. Harmonic oscillator
V(x)
V ( x) 
1 2 1
kx  m 2 x 2
2
2
a 
1
2m 

ip  m x 
• Operator position
• Operator momentum
1

H    a a   
2

x
x
2m
pi
or
m
2
 a  a 
 a  a 
1

H    a a   
2

Review I
Quantum mechanics
4. Harmonic oscillator
Ladder operators:
a 
n 1
Raising operator:
a n  n  1 n1
Lowering operator:
a n  n n1
n 
1
n
 a   0
n!
1

En   n   
2

a
n
 n1
En  
En
En  
Quantum mechanics
Review I
5. Free particle
p2
  E
2m
d 2
2


k

2
dx
 ( x, t )   ( x ) e
1
 ( x, t ) 
2
k
with

 iEt /
 Ae
  ( k )e

i  kx t 
i ( kx t )
dk
2mE
 Be
 i  kx t 
Wave
packet
Quantum mechanics
Review I
Free particle
Method:
1. Identify the initial wave function
 ( x, 0)
2. Calculate the Fourier transform
1
(k ) 
2


 ( x, 0)e ikx dx

3. Estimate the wave function at later times
1
 ( x, t ) 
2

i ( kx t )

(
k
)
e
dk


Quantum mechanics
Review I
5. Free particle
1
 ( x, t ) 
2
Dispersion relation
 (k )

i ( kx t )

(
k
)
e
dk


here
k2

2m
here
Physical interpretation:
• velocity of the each wave at given k:
• velocity of the wave packet:
v phase 
vgroup

k
d

dk
v phase
k

2m
vgroup
k

m
Related documents