Developing Wave Equation

Download Report

Transcript Developing Wave Equation

Solving Schrodinger Equation
• If V(x,t)=v(x) than can separate variables

2

2m
2
 ( x ,t )
2
x

2


2
d
d
2 mdx

2
dx

2
2

t
 ( x , t )   ( x ) ( t )
assume

2m
 V ( x )   i
2
d
 V ( x ) ( t ) ( x )   i  
 V

1

i d 
 dt
dt
 G
G is separation constant valid any x or t
Gives 2 ordinary diff. Eqns.
P460 - Sch. wave eqn.
1
Solutions to Schrod Eqn
• Gives energy eigenvalues and eigenfunctions (wave functions).
These are quantum states.
• Linear combinations of eigenfunctions are also solutions. For
discrete solutions
 ( x , t )  c1  1  c 2  2 ...... c n  n
each
i   ie
 iE i t / 
If H Hermitian


 i orthogonal
  j dx  
*
i
ij

normalized

c
2
i
P460 - Sch. wave eqn.
 1
2
i d 
 G
dt

 ( t )  e  iGt
/
G=E if 2 energy states, interference/oscillation
  d 
2
2
2 mdx
2
1D time
 V  E
independent
 ( x, t )   ( x )e
 iEt / 
Scrod. Eqn.
Solve: know U(x) and boundary conditions
want mathematically well-behaved. Do not want:

( x)  

 
x


2
x
No discontinuities. Usually
2
except if V=0 or  =0
 
in certain regions
P460 - Sch. wave eqn.
3
Linear Operators
• Operator converts one function into another
Of ( x ) 
f ( x)  x
Of ( x ) 
d f ( x)
2
dx
• an operator is linear if (to see, substitute in a function)
if O [ f 1 ( x )  f 2 ( x )]  Of
ex : O 
1
( x )  Of
2
( x )  linear
d
dx
• linear suppositions of eigenfunctions also solution if operator is
linear……use “Linear algebra” concepts. Often use linear algebra to
solve non-linear functions….
P460 - Sch. wave eqn.
4
Solutions to Schrod Eqn
• Depending on conditions, can have either discrete or continuous
solutions or a combination
 ( x, t ) 

C nu n ( x )e
 iE n t / 
n

 C ( E )u
E
( x )e
 iE n t / 
dE
• where Cn and C(E) are determined by taking the dot product of
an arbitrary function  with the eigenfunctions u. Any function
in the space can be made from linear combinations
P460 - Sch. wave eqn.
5
Solutions to Schrod Eqn
• Linear combinations of eigenfunctions are also solutions. Assume two
energies
 ( x , t )  c1  1  c 2  2 
c1 1 e
 iE 1 t / 
 c 2
e
2
 iE 2 t / 
assume know wave function at t=0
 ( x ,0 ) 
2
7
1 
5
7

2
• at later times the state can oscillate between the two states probability to be at any x has a time dependence
|  ( x , t ) |  | c1
2
c1 c 2 ( 1 
*
e
2
1
( x) |
 i ( E 2  E1 ) t / 
2
 | c 2
2
  2 1 e
P460 - Sch. wave eqn.
*
( x) |
2

 i ( E1  E 2 ) t / 
6
)
Example 3-1
• Boundary conditions (including the functions being mathematically
well behaved) can cause only certain, discrete eigenfunctions
i
d f ( )
d
  f ( )
f ( ) 
with
• solve eigenvalue equation
i
1
d f ( )
f ( )
d
int egrate
   eigenvalue
 ln
f ( ) 
or
f (  2 )
or
d f ( )
f ( )
  i d 
f (  )   i   cons tan t
f (0)e
 i 
• impose the periodic condition to find the allowed eigenvalues
e
 i ( 2  )
 1    0 ,  1,  2 , etc
P460 - Sch. wave eqn.
7
Square Well Potential
• Start with the simplest potential
V ( x )  V0
V ( x)  0
| x |
| x |
V0 
a
2
a
2
finite
or 
(" in " the well )
For  value
 ( x )  0 for | x | 
a
2
 V   is
finite
Boundary condition is that  is continuous:give:

out
(
a
2
) 
(
in
a
2
)  0 if V 0  
V
-a/2
a/2
P460 - Sch. wave eqn.
0
8
Infinite Square Well Potential
• Solve S.E. where V=0
2

2m
d 
2
dx
2
 E
   A sin kx , B cos kx , Ce
Boundary condition quanitizes k/E, 2 classes
Odd
Even
=Bcos(knx)
=Asin(knx)
kn=n/a
kn=n/a
n=1,3,5...
n=2,4,6...
(x)=(-x)
(x)=-(-x)
En 
2
p
2m

2
2
 k
2m

as n  0  E min

2
2
n
2 ma
2

2

h
2
n
8 ma
2
2
 E1  0
P460 - Sch. wave eqn.
9
 ikx
Parity
x  -x (mirror)
• Parity operator P
P ( x)   ( x)
• determine eigenvalues
Pu ( x )   u ( x )
P u ( x)  Pu ( x)   u ( x)
2
2
but P [ Pu ( x )]  Pu (  x )  u ( x )  
 1    1
2
even and odd functions are eigenfunctions of P
Odd
:
Even :
Px   x
Px
 x
2
2
P sin x   sin x
P
P cos x  cos x
P
 

x

2
x
2


x

2
x
2
• any function can be split into even and odd
 ( x) 
1
2
 ( x)  
[ ( x )   (  x )] 

( x)  
P ( x)  


( x)
(x) 

( x)
P460 - Sch. wave eqn.
1
2
[ ( x )   (  x )]



1
2
(1  P )



1
2
(1  P )
10
Parity
• If V(x) is an even function then H is also even then H and P commute
[ H , P ]  HP  PH
 0
• and parity is a constant. If the initial state is even it stays even, odd
stays odd. Semi-prove:
• time development of a wavefunction is given by
i

t
 H  ( x, t )
• do the same for P when [H,P]=0
i
(P )
t
 H [ P  ( x , t )]  P [ H  ( x , t )]
• and so a state of definite parity (+,-) doesn’t change parity over time;
parity is conserved (strong and EM forces conserve, weak force does
not)
P460 - Sch. wave eqn.
11
Infinite Square Well Potential
• Need to normalize the wavefunction. Look up in integral tables
a
2


| ( x) |
2
A
2
sin
nx
a
2
dx  1
a
2



dx 
A 
2 / a
What is the minimum energy of an electron confined to a nucleus? Let a = 10-14m
= 10 F
E min

 4000
E min


2

2
2 ma
2

MeV
m
 k 
2
hc
2a
( hc )
8 mc
2
2
a
2

( 1240 MeVF

2

2
8  . 51 MeV  ( 10 F )
 relativist
 p
)
m
1240 MeV  F
2 10 F
P460 - Sch. wave eqn.
2
ic
2
redo
 ( k )
2
 60 MeV
12
Infinite Square Well Density of States
• The density of states is an important item in determining the
probability that an interaction or decay will occur
• it is defined as
dn
 (E ) 
n  number
of
states
dE
• for the infinite well
n
2

8 ma
h
2 ndn  cdE

dn
2
E  cE
2
c

dE
2n

1
c
2
E
• For electron with a = 1mm, what is the number of states within 0.0001
eV about 0.01 eV?
c 
8  511000 eV  (. 1 cm )
n 
(1 . 24  10
dn
dE
E 
4
eVcm )
1
c
2
E
2
 2 . 7  10
2
E 
P460 - Sch. wave eqn.
1
2
12
2 . 7  10
eV
1
12
. 0001 eV
 820
. 01 eV
13
Example 3-5
• Particle in box with width a and a wavefunction of
 ( x )  A( x / a )
0  x  a / 2
 ( x )  A (1  x / a )
A 
12 / a
a / 2  x  a
• Find the probability that a measurement of the energy gives the
eigenvalue En
 

un 
An u n ( x )
2
a
sin
nx
a
n
a
An 

a /2
( x ) u n dx  2
0
 2

12
x
a
a
0
24

1
n
2
(  1)

2
sin
a
nx
dx
a
n 1
• With only n=odd only from the symmetry
• The probability to be in state n is then
| An | 
2
96
 4n 4
 Pr ob 1  . 986
Pr ob 3 
P460 - Sch. wave eqn.
. 986
3
4
 . 012
14
Free particle wavefunction
• If V=0 everywhere then solutions are
  A cos kx , A sin kx , e ikx , e  ikx
E 
p
2
2m

2
 k
2m
2
• but the exponentials are also eigenfunctions of the momentum
operator

p op   i 
p op ( e
p op ( e
ikx
x
)   i   ik e
 ikx
ikx
)   i    ik e
 eigenvalue
 ikx
 k  p
    k   p
• can use to describe left and right traveling waves
• book describes different normalization factors
P460 - Sch. wave eqn.
15