Download Particle in a Box

particle in a box
Potential problem
 Wave functions
 Significance of wave function
 Normalisation
 The time-independent Schrodinger Equation
 Solutions of the T.I.S.E
Particle in a 1-Dimensional Box
Time Dependent Schrödinger Equation
  2 d 2
 V ( x)  E
2m dx 2
Region I
Region II
Region III
KE
PE
TE
wavefunction is dependent on time & position function:
V(x)=∞
V(x)=0
V(x)=∞
1
 ( x, t )  f (t ) ( x)
Time Independent Schrödinger Equation
L
0
V(x)=0 for L>x>0
V(x)=∞ for x≥L, x≤0
Classical Physics: The particle can
exist anywhere in the box and follow
a path in accordance to Newton’s
Laws.
Quantum Physics: The particle is
expressed by a wave function and
there are certain areas more likely to
contain the particle within the box.
x
  2 d 2 ( x)
 V ( x)  E
2
2m dx
Region I and III:
Applying boundary conditions:
  2 d 2 ( x)
  *  E
2
2m dx
Region II:
  2 d 2 ( x)
 E
2m dx 2
 0
2
Finding the Wave Function
d 2 ( x) 2m

 2 E
dx 2

  2 d 2 ( x)
 E
2m dx 2
This is similar to the general differential equation:
d 2 ( x)

 k 2
2
dx
  A sin kx  B cos kx
So we can start applying boundary conditions:
x=0 ψ=0
0  Asin 0k  B cos 0k
x=L ψ=0
0  Asin kL A  0
Calculating Energy Levels:
2mE
k  2

2
k 2 2
E
2m
h

2
n 2 2 h 2
E 2
L 2m4 2
k 2h2
E
2m4 2
n2h2
E
8mL2
 II  A sin
nx
L
But what is ‘A’?
Normalizing wave function:
L
2
(
A
sin
kx
)
dx  1

0
L
 x sin 2kx 
A  
1
4k  0
2
2
0  0  B *1 B  0
kL  n where n=
Our new wave function:
*
n 

sin
2
L
2L
L
A  
1
n 
2


4
L


Since n=
2 L 
A   1
2
*
A
2
L
Our normalized wave function is:
 II 
2
nx
sin
L
L
Particle in a 1-Dimensional Box
 II 
2
nx
sin
L
L
Applying the
Born Interpretation
 II
2
2
nx 
  sin

L
L 
n=4
n=3
E
x/L
2
n=4
E
n=3
n=2
n=2
n=1
n=1
x/L
de Broglie Hypothesis
In 1924, de Broglie suggested that if waves of wavelength λ
were associated with particles of momentum p=h/λ, then it
should also work the other way round…….
A particle of mass m, moving with velocity v has momentum p
given by:
p  mv 
h

Kinetic Energy of particle
2
2
2
2
p
h
 k
KE 


2
2m 2m
2m
If the de Broglie hypothesis is correct, then a stream of
classical particles should show evidence of wave-like
characteristics……………………………………………
Standing de Broglie waves
Eg electron in a “box” (infinite potential well)
V=
V=0
Electron “rattles” to and fro
V=
V=
V=
V=0
Standing wave formed
wavelengths of confined states
In general, k =nπ/L, n= number of antinodes in
standing wave
2L
3

;k 
3
L
2
  L;k 
L
  2L ; k 

L
energies of confined states
 k
 n
E

2
2m
2mL
2
2
2
En  n 2 E1

E1 
2
2mL
2
2
2
2
Energies of confined states
En  n E1
2

E1 
2
2mL
2
2
particle in a box: wave functions
From Lecture 4, standing wave on a string has form:
y ( x, t )  ( A sin kx) sin( t )
Our particle in a box wave functions represent STATIONARY
(time independent) states, so we write:
 ( x)  A sin kx
A is a constant, to be determined……………
interpretation of the wave function
The wave function of a particle is related to the probability
density for finding the particle in a given region of space:
Probability of finding particle between x and x + dx:
 ( x ) dx
2
Probability of finding particle somewhere = 1, so we have the
NORMALISATION CONDITION for the wave function:

  ( x)

2
dx  1
interpretation of the wave function
Interpretation of the wave function
Normalisation condition allows unknown constants in the wave
function to be determined. For our particle in a box we have
WF:
nx
 ( x)  A sin kx  A sin
L
Since, in this case the particle is confined by INFINITE
potential barriers, we know particle must be located between
x=0 and x=L →Normalisation condition reduces to :
L
  ( x)
0
2
dx  1
normalisation of wave functions
  ( x)
0
 nx 
A  sin 
dx  1
 L 
0
L
L
2
dx  1
2
2
nx
 ( x) 
sin
L
L
2
Solving the SE :an infinite potential well
V ( x)  0
0xL
So, for 0<x<L, the time independent SE reduces to:
 2 d 2 ( x)

 E ( x)
2
2m dx
d 2 ( x) 2mE ( x)

0
2
2
dx

General Solution:
1/ 2
 2mE 
 ( x)  A sin  2 
  
1/ 2
 2mE 
x  B cos 2 
  
x
1/ 2
 2mE 
 ( x)  A sin  2 
  
1/ 2
 2mE 
x  B cos 2 
  
x
Boundary condition: ψ(x) = 0 when x=0:→B=0
1/ 2
 2mE 
 ( x)  A sin  2 
  
x
Boundary condition: ψ(x) = 0 when x=L:
1/ 2
 2mE 
 (0)  A sin  2 
  
n 
E
2
2mL
2
L0
2
2
nx
 ( x)  A sin
L
In agreement with the “fitting waves in boxes” treatment earlier………………..