Chapter 2
Time-Independent
Schrödinger Equation
2.1
Particle in a time-independent scalar
potential
• Schrödinger equation:
 ( x, t )
 2  2  ( x, t )
i

 V ( x )  ( x, t )
2
t
2m x
• Separating variables:
 ( x, t )   ( x) (t )
d (t )
2
d 2 ( x)
i ( x)

 (t )
 V ( x) ( x) (t )
2
dt
2m
dx
i d (t )
2
d 2 ( x)

 V ( x)  
2
 (t ) dt
2m ( x) dx
2.1
Particle in a time-independent scalar
potential
2
2
i d (t )

d  ( x)

 V ( x)  
2
 (t ) dt
2m ( x) dx
d (t )
i
  (t )
dt
 (t )  eit
 2 d 2 ( x)

 V ( x) ( x)   ( x)
2
2m dx
 ( x, t )   ( x) (t )   ( x)e it
• This is called a stationary solution for a stationary
eigenstate
E  
 ( x, t )   ( x )
2
2
2.1
Particle in a time-independent scalar
potential
2
2
 d  ( x)

 V ( x) ( x)  E ( x)
2
2m dx
• For the Hamiltonian, a linear differential operator:
2
2


Hˆ  
 V ( x)
2
2m x
Hˆ  ( x, t )  E ( x, t )
• This is an eigenvalue equation for H
Hˆ  n ( x)  En n ( x)
n ( x, t )   n ( x)e
 iEn t / 
• Linear superposition of solutions is a solution
 ( x, t )   cn n ( x)e
n
 iEn t / 
 ( x,0)   cn n ( x)
n
2.1
Particle in a time-independent scalar
potential
• Calculating expectation values:
Hˆ 









* ˆ

 Hdx 
2
* ˆ 2
ˆ
H    H dx 

ˆ dx  E 2

Ĥ
E

dx

E

H


*

  Hˆ
2
H
*
*

E

dx
 E   dx  E

2
 Hˆ
*
2

0
• The distribution of energy values in a given
stationary state has zero spread
• Measurements of an energy eigenvalue are certain
to return the given eigenvalue En
2.2
Particle in a 1D infinite square well
• Let us assume:
0
V ( x)  

0 xa
x  0; x  a
• Outside of the well:
 ( x)  0
2
2

d
 ( x)
• Inside of the well:

 E ( x)
2
2m dx
d 2 ( x)
2mE
  2  ( x)
2
dx

• Denoting:
k
2mE

d 2 ( x)
2


k
 ( x)
2
dx
2.2
Particle in a 1D infinite square well
• Solution:
 ( x)  A sin kx  B cos kx
• Continuity condition:
 (0)   (a)  0
Asin 0  B cos 0  B  0
n
 n ( x)  An sin
x
 ( x)  A sin kx
• And:
a
 (a)  A sin ka  0
n
n  1;2;3...
kn 
ka  0; ;2 ;3 ...  n
a
n  1;2;3...
d 2 ( x)
2


k
 ( x)
2
dx
• Therefore:
2.2
Particle in a 1D infinite square well
• Normalizing:
a

0
a
( x) dx   An
2
n
0
• Recall:
x sin 2bx
 sin bxdx  2  4b
2
a

0
2
n
sin
xdx  1
a
2
An
2
2

a
n
 n ( x)  An sin
x
• Therefore:
a
a
2n
2
a
2
2
A
sin
x
n
An a
An x
2
2 n
a

1
An sin
xdx 

n
2
a
2
4
0
a
0
2.2
Particle in a 1D infinite square well
• Solution:
2
n
 n ( x) 
sin
x
a
a
• Recall:
k
2mE

n
kn 
a
• Therefore:
 2 k n2  2 n 2 2
En 

2m
2ma 2
An
2
2

a
n
 n ( x)  An sin
x
a
2.2
Particle in a 1D infinite square well
2

 2 2
sin x
• Ground state:  1 ( x) 
E1 
2
a
• Excited states:
n 1
a
2
n
 n ( x) 
sin
x
a
a
2ma
n 2  2 2
En 
2ma 2
2.2
Particle in a 1D infinite square well
• Properties of eigenstates:
• 1) They are alternately even and odd
• 2) They have n – 1 nodes
• 3) They are orthogonal:

m
* ( x) n ( x)dx   mn
 mn
0 m  n

1 m  n
2.2
Particle in a 1D infinite square well
mn
• Proof of orthogonality:
2
m
n
 m * ( x) n ( x)dx  a  sin a x  sin a xdx
0
a
1  mn 
 m  n 
  cos
x   cos
x  dx
a0  a

 a

a

1
1
mn 
 m  n 

sin 
x  
sin 
x 
 a
 (m  n)
 a
 0
 (m  n)
a
1  sin  m  n  sin  m  n 
 

0

  ( m  n)
( m  n) 
cos( y  z )  cos( y  z )
sin y  sin z 
2
2.2
Particle in a 1D infinite square well
• Properties of eigenstates:
• 1) They are alternately even and odd
• 2) They have n – 1 nodes
• 3) They are orthogonal:
• 4) They are complete:

f ( x)   cn n ( x)
n 1
2 
n

cn sin
x

a n 1
a
2.2
Particle in a 1D infinite square well

• Proof of completeness:
f ( x)   cn n ( x)

n 1
 m * ( x) f ( x)   m * ( x) cn n ( x)



n 1
* ( x) f ( x)dx    cn m * ( x) n ( x)dx
m
n 1

m

* ( x) f ( x)dx   cn  m * ( x) n ( x)dx
n 1

m
* ( x) f ( x)dx   cn mn  cm
n 1
cn   n * ( x) f ( x)dx
2.2
Particle in a 1D infinite square well
• Recall expression for a stationary eigenstate:
n ( x, t )   n ( x)e
 iEn t / 
• Stationary states of the 1D infinite square well:
 n 2  2
i 
2
 2 ma
2
n
n ( x, t ) 
sin
xe
a
a

t


• And the most general solution (spectral
 n 2  2 
decomposition):
t

i 
 ( x, t ) 
2
n
cn sin
xe

a n 1
a
 2 ma 2 


• Using completeness:

 ( x,0)   cn n ( x)
n 1
2
n
cn 
sin
x( x,0)dx

a0
a
a
2.2
Particle in a 1D infinite square well

• Let us recall:

  ( x, t )

2
dx  1
*




iEmt / 
iEnt /  
1     cm m ( x)e
   cn n ( x)e
dx
  n1

 m 1  

iEm t / 
iEn t / 
  cm * e
cn e
 m * ( x) n ( x)dx


m 1 n 1

  cm * cn e
i  Em  En t / 
m 1 n 1
m
2

c
n 1
2
n
1
* ( x) n ( x)dx   mn
 ( x,0)   cn n ( x)
n 1
 mn   cn

n 1



 ( x, t )   cn n ( x)e
n
 iEn t / 
2.2
Particle in a 1D infinite square well

• We can also calculate:




*
Hˆ    * ( x, t ) Hˆ  ( x, t )dx





iEmt / 
iEnt /  
ˆ
    cm m ( x)e
 H   cn n ( x)e
dx
  n1


 m 1
 
iEm t / 
iEn t / 
ˆ  ( x)dx
  cm * e
cn e

*
(
x
)
H
n
 m
m 1 n 1


iEm t / 
iEn t / 
m
n
n
m
 

i  Em  En t / 
m
n n
mn
m 1 n 1
  c * e
m 1 n 1
ce
E
  c * c E e
Hˆ  n  En n


m
* ( x) n ( x)dx

   cn E n
2
n 1
* ( x) n ( x)dx   mn
 ( x,0)   cn n ( x)
n 1


2
ˆ
H   cn E n
n 1
 ( x, t )   cn n ( x)e
n
 iEn t / 
2.2
Particle in a 1D infinite square well

• Synopsizing:
c
m 1
2
n
1

2
ˆ
H   cn E n
n 1
• |cn|2 can be interpreted as the probability of
measuring the energy value En when the system is in
the state described by the wave function Ψ(x,t)
• Expectation value of the energy dos not depend on
time – conservation of energy
 ( x, t )   cn n ( x)e
n
 iEn t / 
2.3
1D harmonic oscillator
• Let us consider 1D system with a potential V(x)
• This potential can be expanded in the vicinity of a
local minimum:
2

1 dV
V ( x)  V ( x0 )   2
2  dx

1
2
 ( x  x0 )  ...  m 2 ( x  x0 ) 2
2
 x  x0
• Then the QM Hamiltonian of the system will be:
2
ˆ
p
1
2 2
ˆ
H
 m xˆ
2m 2
2.3
1D harmonic oscillator
• This Hamiltonian is time-independent, hence we get
a conservative eigenvalue equation:
Hˆ   E
 2 d 2 1
2 2
 m x  ( x)  E ( x)

2
2
 2m dx

2
ˆ
p
1
2 2
ˆ
H
 m xˆ
2m 2
2.3
1D harmonic oscillator
• The Hamiltonian can be parametrized and made
dimensionless by introducing operators
m
ˆ
X
xˆ

• Then:
Pˆ 
1
pˆ
m
2
ˆ
p
1
ˆ
H
 m 2 xˆ 2
2m 2
mPˆ 2 1
 ˆ 2 ˆ 2  Ĥ
2 
2
ˆ

 m
X 
P X
2m
2
m
2

2
2
ˆ
ˆ
P X
ˆ
H
2

2.3
1D harmonic oscillator
• Let’s introduce a definition of a commutator:
• Then:
[ Aˆ , Bˆ ]  Aˆ Bˆ  Bˆ Aˆ
 
xˆf
[ xˆ, pˆ ] f  xˆpˆ  pˆ xˆ  f  xˆpˆ f  pˆ xˆf  xpˆ f 
i x
 f
 f  x
 f  
 if
x
 f
xf   x
x

i x
i x i x
i x i x
[ xˆ , pˆ ]  i
• And:
m
1
 m

1
[ xˆ, pˆ ]
[ Xˆ , Pˆ ]  
xˆ ,
pˆ  

m
m 
 
1 1

i  i
[ Xˆ , Pˆ ]  i
 
2.3
1D harmonic oscillator
• For the new dimensionless Hamiltonian the
eigenvalue problem becomes
Xˆ  iPˆ
Ĥ  
a
2
• Let us introduce the following operators
ˆ
ˆ
X

i
P
†
†
†
a

a
i
(
a

a
)
a

ˆ
• Then Xˆ 
P
2
2
2
• And
 Xˆ  iPˆ Xˆ  iPˆ 
†
†
[
a
,
a
] 1
[ a, a ]  
,

2 
 2
i ˆ ˆ
1 ˆ ˆ

X , X  iPˆ ,iPˆ  Xˆ ,iPˆ  iPˆ , Xˆ  P, X  Xˆ , Pˆ
2
2
i
  i  i   1
2




    
2.3
1D harmonic oscillator
 Xˆ  iPˆ  Xˆ  iPˆ 
• Also a a  





2
2



ˆ  iPˆ
X
1 ˆ 2 ˆ2
a

ˆ
ˆ
ˆ
ˆ
 X  P  iXP  iPX
2
2
ˆ
ˆ
1 ˆ 2 ˆ2
1
X

i
P
2
2
 X  P  i[ Xˆ , Pˆ ]  Xˆ  Pˆ  i  i
a† 
2
2
2
1
ˆ
H
2
1
1
†
ˆ
ˆ
• Therefore
H a a  N 
2
2
1
†
ˆ
• It can be shown that H  aa 
2
†






2.3
Some properties
• Commutators
†
†
†
†
ˆ
[ N , a]  [a a, a]  a aa  aa a  [a , a]a  a
†
†
† †
†
†
†
ˆ
[ N , a ]  [a a, a ]  a aa  a a a  a † [a, a † ]  a †
[ Nˆ , a ]   a
[ Nˆ , a † ]  a †
• Also
[ a, a † ]  1
aa †  a † a  1
aa  Nˆ  1
†
2.3
Some properties
• Since
Ĥ  
• One finds that
1 †
 †
1 †

†
†
†
ˆ
Ha     a a  a     a aa  a  
2
2 


1
1 1
†
†
† ˆ
 a  aa     a   H     
2
2 2


†
†
†
 a†Hˆ   a†  a    a    (  1)a  
Hˆ a†  (  1)a†
2.3
Some properties
• Since
Ĥ  
• One finds that
 † 1
1 

†
ˆ
Ha    aa  a    aa a  a  
2
2 



1
1 1
 †
 a a a     a  Ĥ     
2
2 2


 aĤ  a  a   a   (  1)a 
Hˆ a  (  1)a
2.3
Some properties
• Thereby:
†
†
ˆ
Ha    (  1)a  
Hˆ a  (  1)a



• Because of this, operators a† and a are called ladder
operators; a† – creation (raising) operator and a –
destruction (lowering) operator
2.3
Some properties
• Thereby:
†
†
ˆ
Ha    (  1)a  
Hˆ a  (  1)a



• What happens if we apply the lowering operator
repeatedly?
• Will we reach energy less than zero?
• So, does the ground state exist?
• If it does, what is it’s eigenvalue and eigenvector?
2.3
Ground state
• If the ground state exists then the following
condition must be satisfied:
a 0  0
• Does this equation have solutions?
m
1
ˆ
ˆ
x

i
p
Xˆ  iPˆ

m   0
0 
0
2
2
m
d 

• Explicitly: 
x   0 ( x)  0
dx 
 
• Solution:
m 2

x
m



 0 ( x)  
 e 2
  
1/ 4
2.3
Ground state
m 2

x
m



 0 ( x)  
 e 2
  
1/ 4
2.3
Ground state
• Using the Schrödinger equation:
1
 †
 a a   0   0 0
2

Hˆ  0   0 0
1
a a 0   0   0 0
2
†
1
0 
2
• We found the ground state!
• It is greater than zero!
m 2

x
m



 0 ( x)  
 e 2
  
1/ 4

E0 
2
2.3
Excited states
• We can generate the rest of the spectrum
Hˆ a†  (  1)a†
1  †
†
ˆ
Ha  0    1a  0
2 
3  †
†
ˆ
H a  1    1 a  1
2 
 n  An (a † ) n 0
1

E n    n  
2

1
ˆ
H 0   0
2
3
3
ˆ
H 1   1  A1a † 0
2
2
5
5
ˆ
H 2   2  A2 a † a † 0
2
2
1

Ĥ n   n   n
2

n  0,1,2,...
2.3
Excited states
• We can generate the rest of the spectrum
1
†
ˆ
H a a
2
1
 ˆ 1

 N   n   n   n
2
2


 n  An (a † ) n 0
1

E n    n  
2

Nˆ  n  n n
1

Ĥ n   n   n
2

n  0,1,2,...
2.3
Excited states

• Some properties:


 

d 

 f *  mx   dx  gdx

1
dg 


 mxf * g  f * dx

dx 
2m 

1
df * 


 gmxf *  g
dx

dx 
2m 
1
2m
Xˆ  iPˆ
a 

2
†

f * a † g dx
Xˆ  iPˆ
a
2
Xˆ  iPˆ
†
a 
2
m
1
xˆ  i
pˆ

m 
2
m
 d
x

m dx
2
2.3
Excited states

• Some properties:



 
f * a † g dx


Xˆ  iPˆ
a
2
Xˆ  iPˆ
†
a 
2
d 

 f *  mx   dx  gdx

1
dg 


 mxf * g  f * dx

dx 
2m 

1
df * 


 gmxf *  g
dx

dx 
2m 


1
d 

g  mx  g  f * dx   af *gdx

dx 
2m  


1
2m


 
f * a † g dx 
 af *gdx

2.3
Excited states
• Some properties:


 f * a g dx   af *gdx
†


• Similarly:



f * ag dx 

 a

†

f * gdx
2.3
Excited states
† n


A
(
a
) 0
• Eigenfunctions can be normalized n
n
 1  A1a † 0  c1a † 0

 1 * 1dx  c1


2


 c1
2
†
†
(
a

)
*
a
 0  0 dx
 0 * a(a  0 )dx  c1
†



2
†

*
((
a
a  1) 0 )dx
 0




2
2
†
 c1   0 * (a a 0 )dx   0 *1 0 dx   c1 (0  1)

 

2
c1  1
2.3
Excited states
† n


A
(
a
) 0
• Eigenfunctions can be normalized n
n
 n  cn a † n 1



n
* n dx  cn


2
 (a 
†

n 1

) * a  n 1dx
†
†

c

*
((
a
a  1) n 1 )dx
 cn  n 1 * a(a  n 1 )dx
n
 n1





2
ˆ
 cn   n 1 * ( N n 1 )dx   n 1 *1 n 1dx 

 

2
2
1
2
 cn (n  1)  1  cn  n  1
cn 
n
1
† n
1
 n  An (a )  0 
(a † ) n 0
cn 
n!
n
2
†
2
2.3
Excited states
• Eigenfunctions are orthogonal







* a a n dx  n  m * n dx   (a m ) * a n dx
†
m




  (a † a m ) * n dx  m  m * n dx



m
* n dx   mn
2.3
Eigenfunctions
• The eigenfunctions can be written explicitly
• Recall
 m 
 0 ( x)  
 e
  
1/ 4

m 2
x
2
1
n 
(a † ) n 0
n!
• The creation operator:
a 
†
m
 d
x

m dx
2
2.3
Eigenfunctions
1
† n
 n ( x) 
(a )  0
• Therefore:
n!
n
1 1  m
 d 

  0 ( x)

x

m dx 
n! 2 n  
1     m   m
d 
x  e

 
 
n
2 n!  m      
dx 
n

1/ 4
• The creation operator:
a 
†
m
 d
x

m dx
2
n

m 2
x
2
2.3
Eigenfunctions
1
† n
 n ( x) 
(a )  0
• Therefore:
n!
n
1 1  m
 d 

  0 ( x)

x

m dx 
n! 2 n  
1     m   m
d 
x  e

 
 
n
2 n!  m      
dx 
n

1/ 4
n
3 1/ 4
m 2
 4  m  

x
2

 xe
 1 ( x)   

    


1/ 4
m 2

 m   m 2  2 x
 2 ( x)  
x  1e
 2
 4   


m 2
x
2
2.3
Eigenfunctions
• We have explicit expressions for the eigenfunctions:
 n ( x) 
m
1     m   m
d   2 x 2
x  e

 
 
n
2 n!  m      
dx 
n
1/ 4
n
• They contain the Hermite polynomials
n
 d   x2
H n ( x)   1 e   e
 dx 
n
x2
• Their parity is (–1)n
Charles Hermite
(1822 – 1901)
2.3
Eigenfunctions
• Graphically:
2.4
Free particle
 ( x, t )
 2  2  ( x, t )
i

• For a free particle:
t
2m x 2
2
2
2mE

d
 ( x)
• Eigenproblem:
k


E

(
x
)
2

2
m
dx
2
d  ( x)
2
ikx
 ikx


k

(
x
)

(
x
)

Ae

Be
2
dx
• Stationary solution:
( x, t )  ( Ae  Be
ikx
ikx
)e
iEt / 
 Aeik ( x kt / 2 m )  Be ik ( x  kt / 2 m )

• Normalizing:

  ( A* e

k 

ik  x 
t
2
m



  ( x, t )

 B *e

2
dx 
k 

ik  x 
t
2
m


  ( x, t ) *  ( x, t )dx
 
k 
ik  x 
t
2
m


)( Ae
 Be
k 

ik  x 
t
2
m


  ( A * A  B * B  AB * ei 2 kx  A * Be i 2 kx )dx  

)dx
2.4
Free particle
• The wave function is not normalizable!
• What’s going on?
• A free particle cannot exist in a stationary state
• Free particle with a definite energy is not physical

• Normalizing:

  ( A* e

k 

ik  x 
t
2
m



  ( x, t )

 B *e

2
dx 
k 

ik  x 
t
2
m


  ( x, t ) *  ( x, t )dx
 
k 
ik  x 
t
2
m


)( Ae
 Be
k 

ik  x 
t
2
m


  ( A * A  B * B  AB * ei 2 kx  A * Be i 2 kx )dx  

)dx
2.4
Free particle
• The linear combination of stationary states is a
solution of the Schrödinger equation:
1
( x, t ) 
2

  ( k )e
1
2

  ( k )e
ikx
dk

• And properties of Fourier transform:

1
1
ikx
F (k ) 
f ( x) 
F (k )e dk

2
2 
• One obtains:
 (k ) 
dk

• Using initial condition:
 ( x,0) 
k 

ik  x 
t
 2m 
1
2



f ( x)e ikx dx

ikx

(
x
,
0
)
e
dx


2.4
Wave packet
• Wave packet:
Re ( x)
k 
k 



i
k

x
i
k




 (k0 ) ik0 x 1  0 2  1  0 2  x 
 ( x,0) 
 e
e  e

2
2
2 

 (k0 ) ik0 x 
 k 

e 1  cos
x 
2
 2 

 k 
xd   0
• Interference is destructive when 1  cos
 2

k
 k 
xd   x  2 xd
kx  4
cos
xd   1
2
 2

2.4
Wave packet
• More waves in a packet:
2.4
Wave packet
• More waves in a packet:
kx  1
2.4
Evolution of a free packet
i ( kx t )

(
x
,
t
)

Ae
• For a single wave:
2

k


• Phase velocity: v (k ) 
2m
k
k
v (k ) 
2m
• Three-wave packet:


k  
  
k  
  

i   k 0   x   0 
t
i   k 0   x   0 

 (k0 ) i k0 x 0t  1  2   2   1  2   2 t  
e

( x, t ) 
 e
 e
2
2
2 

 (k0 ) i k0 x 0t  
 
 k

e
1  cos 2 x  2 t 
2



• Maximum occurs when:
 
 k
cos
xM 
t 1
2 
 2
k

xM 
t 0
2
2

xM 
t
k
2.4
Evolution of a free packet
• For multiple waves in a packet :
 ( x, t ) 
• Group velocity:
k 2

2m
1
2

i ( kx t )

(
k
)
e
dk


 d 
vG (k 0 )   
 dk  k  k0
k
 2v (k0 )
vG (k0 ) 
m
δ-function potential
• δ-function:

 f ( x) x  a dx  f (a)

2.5
δ-function potential
2.5
• Schrödinger equation:
 ( x, t )
   ( x, t )
i

  ( x) ( x, t )
2
t
2m x
2
2

d
 ( x)
• Eigenproblem:

  ( x) ( x)  E ( x)
2
2m dx
2
 2mE
d

(
x
)
2
• For x < 0:

   ( x)
2

dx
x
x
x
• Stationary solution:  ( x)  Ae  Be  Be
2
• Similarly, for x > 0:
2
 ( x)  Fex
• From continuity condition
BF
 Be x , ( x  0)
 ( x)   x
Be , ( x  0)
δ-function potential
2.5
• Schrödinger equation:
 ( x, t )
   ( x, t )
i

  ( x) ( x, t )
2
t
2m x
2
2

d
 ( x)
• Eigenproblem:

  ( x) ( x)  E ( x)
2
2m dx
2
2
 Be x , ( x  0)
 ( x)   x
Be , ( x  0)
δ-function potential
2.5
• Schrödinger equation:
 ( x, t )
   ( x, t )
i

  ( x) ( x, t )
2
t
2m x
2
2

d
 ( x)
• Eigenproblem:

  ( x) ( x)  E ( x)
2
2m dx
2
2
• Integrating in the vicinity of 0:


  2  d 2 ( x)


lim 
dx



(
x
)

(
x
)
dx

E

(
x
)
dx

0
2




 0 2m
dx





x
 d ( x)   2m

Be , ( x  0)

lim 

(
0
)

0
 ( x)   x
2

 0 
 dx   
Be , ( x  0)
2.5
δ-function potential
2m
2m
 2mE
 B  ( B )  2 B  0
 2

2

m

2

 2mE



m 2
E
2
2
• Integrating in the vicinity of 0:


  2  d 2 ( x)


lim 
dx



(
x
)

(
x
)
dx

E

(
x
)
dx

0
2




 0 2m
dx





x
 d ( x)   2m

Be , ( x  0)

lim 

(
0
)

0
 ( x)   x
2

 0 
 dx   
Be , ( x  0)
2.5
δ-function potential
2m
2m
 2mE
 B  ( B )  2 B  0
 2

2

m

2

 2mE

• Normalizing:

m 2
E
2
2

m
 ( x) 
e



1    ( x) dx  2 B
2

2
e
 2x

m

2
x
dx 
0
B  
B

m

 Be x , ( x  0)
 ( x)   x
Be , ( x  0)
2
δ-function potential
2.5
• Schrödinger equation:
 ( x, t )
   ( x, t )
i

  ( x) ( x, t )
2
t
2m x
2
2

d
 ( x)
• Eigenproblem:

  ( x) ( x)  E ( x)
2
2m dx
2
2mE
d

(
x
)
2
• For x < 0:
k
 k  ( x)
2

dx
ikx
 ikx
• Stationary solution:  ( x)  Ae  Be
2
• Similarly, for x > 0:
2
 ( x)  Feikx  Geikx
• From continuity condition
A B  F G
δ-function potential
• Integrating in the vicinity of 0:
 d ( x)   2m

lim 
 (0)  0
2

 0 
dx

 

2m
ik ( F  G  A  B)  2 ( A  B)

2
2mE
d

(
x
)
2
• For x < 0:
k
 k  ( x)
2

dx
ikx
 ikx
• Stationary solution:  ( x)  Ae  Be
• Similarly, for x > 0:
 ( x)  Feikx  Geikx
• From continuity condition
A B  F G
2.5
δ-function potential
• Integrating in the vicinity of 0:
 d ( x)   2m

lim 
 (0)  0
2

 0 
dx

 

2m
ik ( F  G  A  B)  2 ( A  B)

F  G  A(1  2i )  B(1  2i )
• Assuming
i
B
A
1 2i
G0
1
F
A
1  2i
A B  F G
2mE
k

m
 2
 k
2.5
δ-function potential
• Reflection coefficient:
2
2
1
R 2 
2 
1 
1  (2 2 E / m 2 )
A
B
• Transmission coefficient:
F
2
1
1
T 2
2 
2
2
1


1  (m / 2 E )
A
R T 1
i
B
A
1 2i
1
F
A
1  2i
A B  F G
2mE
k

m
 2
 k
2.5