2.57 Fall 2004 – Lecture 4
1
2.57 NanotoMacro Transport Processes
Fall 2004
Lecture 4
Quick review of Lecture 3
Photon:
E
h
ν
=
,
/
p
h
λ
=
.
Assuming
Ψ
t
(
r
,t)=
Ψ
(
r
)Y(t), we use separation of variables to solve the Schrödinger
equation
t
i
U
m
t
t
t
∂
∂
Ψ
=
Ψ
+
Ψ
∇
−
=
=
2
2
2
.
The solutions are
[
]
1
1
exp
exp
E
Y
C
i
t
C
i
t
ω
⎡
⎤
=
−
=
−
⎢
⎥
⎣
⎦
=
,
and
(
)
2
2
0
2
U
E
m
−
∇ Ψ +
−
Ψ =
=
,
where the eigen value E represents the total energy of the system.
Heisenberg uncertainty principle states
;
2
2
p
x
t
E
< ∆
>< ∆
>≥
< ∆ >< ∆
>≥
=
=
.
2.3 Example solutions:
Here we determine
( )
u r
G
by the boundary conditions and will not consider the
( , )
u r t
G
case.
2.3.1 Free particles in 1D
In this case, there are no constraints for the particles. The potential energy u=0 so that
2
2
2
0
2
d
E
m dx
Ψ
−
−
Ψ =
=
.
This gives
ikx
ikx
Ae
Be
ψ
−
=
+
,
where
2
/
/
k
mE
p
=
=
=
=
(note
2
/ 2
E
p
m
=
). The final solution is
(
)
(
)
( , )
i
t
kx
i
t
kx
t
x t
Ae
Be
ω
ω
−
+
−
−
Ψ
=
+
,
in which the first term corresponds to negativedirection propagation, the second term is
positivedirection wave. Please also recall problem 2.5 in assignment 2.
2.3.2 Quantum well
Consider the general case of a particle in a onedimensional potential well, which can be,
for example, an electron subject to an electric potential field as shown in the figure. This
is actually the model for thin films. The steadystate Schrödinger equation for the particle
in such a potential profile is
2
2
2
0
2
d
E
m dx
Ψ
−
−
Ψ =
=
(0<x<D);
0
Ψ =
(x<0 or x>D).
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2.57 Fall 2004 – Lecture 4
2
Note: for
u
→ ∞
(x<0 or x>D), only
0
Ψ =
can satisfy the Schrödinger equation.
Same as the free particles, the solution for first equation is still
ikx
ikx
Ae
Be
−
Ψ =
+
,
where
2
2
2
mE
mE
p
k
=
=
=
=
=
=
.
The general boundary conditions are the continuity of the wave functions and their first
derivatives at the boundaries.
The latter derives from the continuity of particle flux at the
boundary. For the current problem, the continuity of the first derivatives is not required
because the wavefunction at the boundaries are already known to be zero. With the
continuity of the wave function at x=0 and x=D, we have
x=0
A+B=0
[
]
[
]
x=D
exp
exp
0
A
ikD
B
ikD
−
+
=
Above equations yield
(
)
0
ikD
ikD
A e
e
−
−
=
.
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 Schrodinger Equation, Uncertainty Principle

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