Transcript Slide 1

Recap
T.I.S.E
The
behaviour of a particle subjected to a
time-independent potential is governed by the
famous (1-D, time independent, non
relativisitic) Schrodinger equation:
 2  2 ( x)
2m x
2
 E  V  ( x)  0
In
the infinite well, V(x) = 0 for 0 < x < L
The T.I.S.E becomes
 2 ( x)
x 2
2m
  2 E ( x)   B 2 ( x)

1
We have proven that
 ( x)  Asin Bx  C cos Bx
is the solution to the second order homogenous equations
 2 ( x)
x 2
  B 2 ( x)
Where A, C are constants to be determined by ultilising the boundary
conditions pertaining to the infinite well system
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Boundaries conditions

Next, we would like to solve for the constants A,
C in the solution (x), as well as the constraint
that is imposed on the constant B

We know that the wave function forms nodes at
the boundaries. Translate this boundary
conditions into mathematical terms, this simply
means
(x = 0) = (x = L) = 0
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 First,
 Plug
(x = 0) = 0 into  = AsinBx +
CcosBx, we obtain
  x=0) = 0 = Asin 0 + C cos 0 = C
 ie, C = 0
 Hence the solution is reduced to
  x= AsinBx
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Next
(x
we apply the second boundary condition
= L) = 0 = Asin(BL)
Only
either A or sin(BL) must be zero but not
both
cannot be zero else this would mean (x) is
zero everywhere inside the box.
If A = 0, this would be in conflict with the fact that
the integrated probability within the box must be ∫
||2 dx > 0.
The upshot is: A cannot be zero
A
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
This means it must be sinBL = 0, or in other words
 B =Bn = np/ L, n = 1,2,3,…

n is used to characterise the quantum states of
n (x)

B is characterised by the positive integer n, hence
we use Bn instead of B

The relationship Bn = np/L translates into the
familiar quantisation of energy condition:

(Bn = np/L)2  Bn 2  2m En 
2

n 2p 2
2
L
 En  n
2 2
p

2
2m L2
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Hence, up to this stage, the solution is
n(x) = Ansin(npx/L), n = 1, 2, 3,…for 0 < x < L
n(x) = 0 elsewhere (outside the box)

The area
under
the
curves
of |Yn|2
=1 for
every n
The
constant An is yet unknown up to now
We can solve for An by applying another
“boundary condition” – the normalisation
condition that: 
L
2
2

(
x
)
dx


 n
 n ( x)dx  1

0
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Solve for An with normalisation

2
A
n
p
x
2
2
2
2
nL
 n ( x)dx  0  n ( x)dx  An 0 sin ( L )dx  2  1
L
L

thus

We hence arrive at the final solution that
An 
2
L
n(x)
= (2/L)1/2sin(npx/L), n = 1, 2, 3,…for 0 < x < L
n(x) = 0 elsewhere (outside the box)
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Example
An electron is trapped in a
one-dimensional region of
length L = 1.010-10 m.
 (a) How much energy must be
supplied to excite the electron
from the ground state to the
first state?
 (b) In the ground state, what is
the probability of finding the
electron in the region from
x = 0.09010-10 m to 0.110
10-10 m?
 (c) In the first excited state,
what is the probability of
finding the electron between
x = 0 and x = 0.25010-10 m?

0.25A 0.5A 1A
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solutions
(a)
 2p 2
2
2
E

n
E

(
2
)
E0  148eV
E1  E0 

37
eV
2
0
2
2m L
 E | E2  E0 | 111eV
2 2 2 px
2
(b) Pn 1 ( x1  x  x2 )   0 dx 
sin
dx
x

L x1
L
1
x2
x

x2  0.11 A
2px 
x 1
 
sin
 0.0038


L  x1 0.09 A
 L 2p
(c)
For n = 2,
2
2 2px
 2  sin
;
L
L
2 2 2 2px
2
Pn 2 ( x1  x  x2 )   2 dx   sin
dx
L x1
L
x1
x2
x

x2  0.25 A
4px 
x 1
 
sin

L
4
p
L

 x1 0
On average the particle
spend 25% of its time in the
region between x=0 and
x=0.25 A
 0.25
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Quantum tunneling
 In
the infinite quantum well, there are
regions where the particle is “forbidden” to
appear
V infinity
V infinity
I
Forbidden region
where particle
cannot be found
because  = 0
everywhere after
x<0
n=1
II
III
Allowed region
where particle
can be found
Forbidden region
where particle
cannot be found
because  = 0
everywhere after
x>L
x=0)=0
x=L)=0
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Finite quantum well
The fact that  is 0 everywhere x ≤0,
x ≥ L is because of the infiniteness of
the potential, V  ∞
 If V has only finite height, the solution
to the TISE will be modified such that
a non-zero value of  can exist
beyond the boundaries at x = 0 and x
=L
 In this case, the pertaining
boundaries conditions are

V
 I ( x  0)   II ( x  0), II ( x  L)   III ( x  L)
d I
dx

x 0
d II
dx
,
x 0
d II
dx

xL
d III
dx
xL
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


For such finite well, the wave function is not vanished at the boundaries,
and may extent into the region I, III which is not allowed in the infinite
potential limit
Such  that penetrates beyond the classically forbidden regions
diminishes very fast (exponentially) once x extents beyond x = 0 and x = L
The mathematical solution for the wave function in the “classically
forbidden” regions are

 A exp(Cx)  0, x  0
 ( x)  
xL
 A exp(Cx)  0,
The total energy of the particle
E = K inside the well.
V
E2 = K2
E1 = K1
The height of the potential well V is
larger than E for a particle trapped
inside the well
Hence, classically, the particle
inside the well would not have
enough kinetic energy to overcome
the potential barrier and escape
into the forbidden regions I, III
However, in QM, there is a slight chance to find
the particle outside the well due to the quantum
tunelling effect
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 The
quantum tunnelling effect allows a
confined particle within a finite potential
well to penetrate through the classically
impenetrable potential wall
E
Hard
and
high
wall,
V
After many many
times of banging
the wall
E
Quantum tunneling effect
Hard
and
high
wall,
V
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Why tunneling phenomena can
happen



It’s due to the continuity requirement of the wave
function at the boundaries when solving the T.I.S.E
The wave function cannot just “die off” suddenly at the
boundaries
It can only diminishes in an exponential manner which
then allow the wave function to extent slightly beyond the
boundaries
 A exp(Cx)  0, x  0
 ( x)  
xL
 A exp(Cx)  0,


The quantum tunneling effect is a manifestation of the
wave nature of particle, which is in turns governed by the
T.I.S.E.
In classical physics, particles are just particles, hence
never display such tunneling effect
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