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Solving Schrodinger Equation
• If V(x,t)=v(x) than can separate variables
 2
2m
 2  ( x ,t )
x 2
 V ( x )   i

t
assume  ( x, t )   ( x ) (t )
 2
2m


d 2
dx 2
  2 d 2
2 mdx
2
 V ( x ) (t ) ( x )  i
 V

1

id
dt
G
G is separation constant valid any x or t
Gives 2 ordinary diff. Eqns.
P460 - Sch. wave eqn.
1
d
dt
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  c2 2 ......cn n
each
i   i e iEi t / 
If H Hermitian

i orthogonal

*
i
 j dx   ij

normalized
2
c
 i 1
P460 - Sch. wave eqn.
2
id
dt
 G

 (t )  e  iGt / 
G=E if 2 energy states, interference/oscillation
  2 d 2
 V  E
2
2mdx
 ( x, t )   ( x )e iEt / 
1D time
independent
Scrod. Eqn.
Solve: know U(x) and boundary conditions
want mathematically well-behaved. Do not want:

( x)  

x
 2
x
2
No discontinuities. Usually
 
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 2
d f ( x)
Of ( x) 
dx
• an operator is linear if (to see, substitute in a function)
if O[ f1 ( x)  f 2 ( x)]  Of 1 ( x)  Of 2 ( x)  linear
ex : O 
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 ) 
 iEn t / 
C
u
(
x
)
e
 n n
n

 C ( E )u
E
( x )e
 iEn 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  c2 2 
c1 1e iE1t /   c2 2 e iE2t / 
assume know wave function at t=0
 ( x,0) 
1 
2
5
7
2
7
• at later times the state can oscillate between the two states probability to be at any x has a time dependence
|  ( x, t ) | | c1 1 ( x) |
2
c1c2 (  2 e
*
1
i ( E2  E1 ) t / 
2
 | c2 2 ( x) |
   1e
P460 - Sch. wave eqn.
*
2
2

i ( E1  E2 ) t / 
6
)
Example 3-1
• Boundary conditions (including the functions being mathematically
well behaved) can cause only certain, discrete eigenfunctions
d f ( )
  f ( )
d
with
f ( )  f (  2 )
i
• solve eigenvalue equation
i
1
d f ( )
d f ( )
   eigenvalue or
 i d
f ( )
d
f ( )
int egrate  ln f ( )  i  cons tan t
or
f ( )  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.
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Square Well Potential
• Start with the simplest potential
V ( x )  V0
V ( x)  0
| x |
| x |
a
2
V0  finite or 
a
2
(" in" the well )
For  value
 ( x)  0 for | x |
a
2
 V  is finite
Boundary condition is that  is continuous:give:
 out ( a2 )   in ( a2 )  0 if V0  
V
-a/2
a/2
P460 - Sch. wave eqn.
0
8
Infinite Square Well Potential
• Solve S.E. where V=0
2
2 d 
2 m dx 2
 E
   A sin kx, B cos kx, Ceikx
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 
p2
2m

2k 2
2m

 22n 2
2 ma 2

h 2n 2
8 ma 2
as n  0  E min  E1  0
P460 - Sch. wave eqn.
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Parity
• Parity operator P
x  -x (mirror)
P ( x )   (  x )
• determine eigenvalues
Pu ( x)  u ( x)
P 2u ( x)  Pu ( x)  2u ( x )
but P[ Pu ( x)]  Pu (  x )  u ( x )  2  1    1
even and odd functions are eigenfunctions of P
Odd : Px   x
Even : Px 2  x 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)  12 [ ( x)   (  x)]  12 [ ( x)   (  x )]
 ( x)    ( x)    ( x)


 1


2 (1  P )
P ( x )   ( x )   ( x )
   12 (1  P )
P460 - Sch. wave eqn.
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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

 H ( x, t )
t
• do the same for P when [H,P]=0
i
 ( P )
 H [ P ( x , t )]  P[ H ( x , t )]
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.
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Infinite Square Well Potential
• Need to normalize the wavefunction. Look up in integral tables

2
|

(
x
)
|
dx 


 A
a
2

A2 sin
2
nx
a
dx  1
a
2
2/a
What is the minimum energy of an electron confined to a nucleus? Let a = 10-14m
= 10 F
Emin 
 2 2
2 ma 2

( hc ) 2
8 mc 2 a 2

(1240MeVF ) 2
8.51MeV (10 F ) 2
 4000 MeV  relativist ic
Emin 
m2  p 2 
 k 
hc
2a

redo
m 2  (k ) 2
1240MeV  F
210 F
P460 - Sch. wave eqn.
 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) 
• for the infinite well
n  number of states
dE
8ma 2
n 
E  cE
h2
dn
c
1
c
2ndn  cdE 


dE
2n
2 E
2
• For electron with a = 1mm, what is the number of states within 0.0001
eV about 0.01 eV?
8  511000eV  (. 1cm ) 2
c
 2.7  1012 eV 1
4
2
(1.24  10 eVcm )
dn
1
n 
E 
dE
2
c
1
E 
E
2
P460 - Sch. wave eqn.
2.7  1012
.0001eV  820
.01eV
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 / 2  x  a
A
12 / a
• Find the probability that a measurement of the energy gives the
eigenvalue En
 
Au
n
n
( x)
un 
2
a
sin
nx
a
n
a
An 
 ( x )u
n
a/2
dx  2
0
2

0
24

12 x

a a
2
nx
sin
dx
a
a
1
( 1) n 1
2
n
• With only n=odd only from the symmetry
• The probability to be in state n is then
| An |2 
96
 Pr ob1  .986
 4n 4
Pr ob3 
P460 - Sch. wave eqn.
.986
 .012
34
14
Free particle wavefunction
• If V=0 everywhere then solutions are
  A cos k x , A sin k x , e ikx , e  ikx
E 
p2
2m

 2k 2
2m
• but the exponentials are also eigenfunctions of the momentum
operator

pop  i
x
pop ( eikx )  i  ik eikx  eigenvalue  k  p
pop ( e ikx )  i  ik e ikx     k   p
• can use to describe left and right traveling waves
• book describes different normalization factors
P460 - Sch. wave eqn.
15