Download QM L-7

Particle in a one dimensional Box (infinite square well potential)
• A particle is confined to a one-dimensional
region of space between two impenetrable
walls separated by distance L
– This is a one- dimensional “box”
• The particle is bouncing elastically back
and forth between the walls
– As long as the particle is inside the box, the
potential energy does not depend on its
location. We can choose this energy value to be
zero
• V= 0,
for 0 < x < L,
• but rises to infinity on the outside of the
box, that is V  , x ≤ 0 and x ≥ L
Page 1
Particle in a one dimensional Box (infinite square well potential)
• Since the walls are impenetrable, there is zero
probability of finding the particle outside the box.
Zero probability means that
ψ(x) = 0, for x < 0 and x > L
• The wave function must also be 0 at the walls (x = 0
and x = L), since the wavefunction must be
continuous
– Mathematically, ψ(0) = 0 and ψ(L) = 0
2
 2  x 

 V  x   E  x 
2m x 2
• In the region 0 < x < L, where V = 0, the
Schrödinger equation can be expressed in the form
 2  x 

 E  x 
2m x 2
2
Page 2
Particle in a one dimensional Box (infinite square well potential)
• We can re-write it as
 2  x 
2mE
  2 x
2
x
 2  x 
2


k
x
2
x
k 
2
2mE
2
The most general solution to this differential equation is
ψ(x) = A sin kx + B cos kx
– A and B are constants to be determined by the properties of the
wavefunction as well as boundary and normalization conditions.
Page 3
Particle in a one dimensional Box (infinite square well potential)
1. Sin(x) and Cos(x) are finite and single-valued functions
2.
Continuity condition is: ψ(0) = ψ(L) = 0
• ψ(0) = A sin(k0) + B cos(k0) = 0  B = 0 Not admissible
 ψ(x) = A sin(kx)
• ψ(L) = A sin(kL) = 0  sin(kL) = 0  kL = nπ, n =0, ±1, ±2…
( n = 0 is not admissible because it yields  zero everywhere which means that the
particle is no where).
kL = nπ
kn 
2


L
n
 2 kn
En 
2m
2
 ( n) 2  2 2 
2

 2


h
2
L
n  
n
En 
 
2 
2 
2m
 2mL 
 8mL 
k 
2
2mE
2
Page 4
Particle in a one dimensional Box (infinite square well potential)
The particle can not have arbitrary energy , but
can have certain discrete energy corresponding
to n=1,2,3,…
n2h2
En 
8mL2
Each permitted energy is called ‘eigen value’ of the
particle and constitute the energy level of the
system,
and the integer ‘n’ that specifies an energy level En
is called its principal quantum number.
The wave function  corresponding to each eigen
value is called eigen functions.
Lowest level n = 1, energy not
zero why ?
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n 2 h 2 n 2 2 2
En 

2
8mL
2mL2
Thus the energy of the particle bounded in a box is quantized.
Another conclusion for the motion of particle in a box can also be drawn that the
particle can not have zero energy but has minimum energy and called as zero point
energy. The state corresponding to this energy is called ground state.
If the particle is bound in 1D box of width L. The particle can not have zero
kinetic energy, because the uncertainty in position of the trapped particle in
a box is Δx = L, hence Δp as well as velocity of the particle and their kinetic
energy can not be zero.
According to classical mechanics, when a particle is placed in a box, it can have
zero energy or continuous kinetic energy. Thus the quantization of energy is a
specific result derived from quantum mechanics.
If one assumes v = 0 for a particle in a box, the de-Broglie wave associated with it will
be λ =(h/mv) = , which ia an absurd result because there should be node at the
boundary for the bounded particle.
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Particle in a one dimensional Box (infinite square well potential)
Let us now consider the eigenfunctions of the particle. Substituting B=0 and k=n/L
in the equation of general solution.
Solutions of the Schrödinger equation are
4
E4
E4 = L
L

nx
2
2
 n (x)  Asin
L
To find the value of A, we use the normalization 3

L
condition
2
2
|  n (x) | dx  1
E3


L
A 2  sin 2
0
nx
dx  1
L
A2  L / 2  1
L
2
E2
2
E2 =L
L

2
E1 =2L
A  2/L
2
nx
sin
L
L
E1
2L
n=1,2,3,...
n
Initial 2wavefunctions
 and Energy
2
n 
The normalized eignfunctions of the particle
 n (x) 
E3 =2L
3
k=2π/λ=nπ/L
n 2h 2
En 
for the first four states in
8mL2
a one-dimensional particle in a box
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Particle in a one dimensional Box (infinite square well potential)
Wavefunctions
2
nx
 n (x) 
sin
L
L
n=1,2,3,...
Probability: *n  n
n=1,2,3,...
Note particle most likely
to be found in the middle
for n=1
Energy:
(a) The first two wavefunctions, (b) the corresponding
probability distributions, and (c) a representation of the
probability distribution in terms of the darkness of
shading.
n 2h 2
E=
8mL2
n=1,2,3,...
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No. of antinodes in eigenfunction = n, No. of nodes =No. of antinodes+1
Page 13
Electron in the 10nm Wide Well with Infinite Barriers
n 2h 2
2
En 

n
E1
2
8mL
h2
2 2
where E1 

2
8mL 2mL2
• Calculate E1=?
3.142 (1.05 1034 )2
E1 
2  9.11031  (10 109 )2
E1  6 1022 J  0.00375 eV  3.75 meV
• Assume that a photon is absorbed, and
the electron is transferred from the
ground state (n = 1) to the second
excited state (n = 3)
What was the wavelengths of the photon?
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Electron in the 10nm Wide Well with Infinite Barriers
• E1 = 3.75 meV
Eground  E1  0.00375 eV
Third excited state is E3
E3  3  E1  9  0.00375 eV  0.0338 eV
2
(h )  E3  E1  0.0338  0.00375  0.030 eV
34
c
6.63  10  3  10
h  0.030eV,  

0.030  1.6  1019
1240
λ
 41333 nm  41 μm
0.03
8
Page 15
Particle in the Infinite Potential Well
For the n th state
2
 nx 
n 
sin 

L
L


n 2 2 2
En 
2mL2
En  n 2 E1
Probability to Find particle in the Right Half of the Well
L
L
2
2
|

(x)
|
dx

[
sin
kx]
dx
L/2
L/2 L
2
L
2
2 L1 1
  sin 2 (kx)dx  [
]
L L/2
L 22 2
Page 16
Some trajectories of a particle in a box according to
Newton's laws of classical mechanics (A), and according
to the Schrödinger equation of quantum mechanics (BD). In (B-D), the horizontal axis is position, and the
vertical axis is the real part (blue) or imaginary part (red)
of the wave function. The states (B,C,D) are energy
eigenstates.
Page 17
Page 18
Exercise: Determine < x > for a particle in a box of length ‘L’.
2
nx
 n (x) 
sin
L
L
L
 x    n (x) * x  n (x)dx
0
2
nx
  x sin 2
dx
L0
L
L
2
2L
L


L4
2
Exercise: Determine < p > for a particle in a box of length ‘L’.

2 n
nx
nx
 p    * (x)
 (x)dx 
sin
cos
dx

i x
i L L 0
L
L
0
L
L
L
2
1

k  sin(kx) cos(kx)dx  0 (sin ce  sin ax cos ax dx  sin 2 ax)
Li 0
2a
 2 nx 
2
2
 p  sin

0
(sin
0

sin
n  0, n  1, 2,3,....)

iL 
L 0
L
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