Physics 2D Lecture Slides
Week of Feb 22, 2010
Sunil Sinha
UCSD Physics
Potential Barrier
U
?
E<U
Transmitted
x
Description of Potential
U=0 x< 0
(Region I )
U=U 0<x<L
(Region II)
U=0 x>L
(Region III)
Consider George as a “free Particle/Wave” with Energy E incident from Left
Free particle are under no Force; have wavefunctions like
Ψ= A ei(kx-wt) or B ei(-kx-wt)
Tunneling Through A Potential Barrier
Region I
U
E<U
II
Region III
Prob
?
•Classical & Quantum Pictures compared: When E>U & when E<U
•Classically , an particle or a beam of particles incident from left
encounters barrier:
•when E > U  Particle just goes over the barrier (gets transmitted )
•When E<U  particle is stuck in region I, gets entirely reflected, no
transmission (T)
•What happens in a Quantum Mechanical barrier ? No region is
inaccessible for particle since the potential is (sometimes small) but
finite
Beam Of Particles With E < U Incident On Barrier From Left
Region I
Region III
II
U
A
B
Incident Beam
Reflected Beam
F
Transmitted Beam
0
L
x
Description Of WaveFunctions in Various regions: Simple Ones first
In Region I : # I ( x, t ) = Aei ( kx "! t ) + Bei ( " kx "! t ) = incident + reflected Waves
! 2k 2
with E = !! =
2m
|B|2
define Reflection Coefficient : R =
= frac of incident wave intensity reflected back
|A|2
In Region III: # III ( x, t ) = Fei ( kx "! t ) + Gei ( " kx "! t ) = transmitted
Note : Gei ( " kx "! t ) corresponds to wave incident from right !
This piece does not exist in the scattering picture we are thinking of now (G=0)
So # III ( x, t ) = Fe
i ( kx "! t )
|F|2
represents transmitted beam. Define T =
|A|2
Unitarity Condition $ R + T= 1 (particle is either reflected or transmitted)
Wave Function Across The Potential Barrier
In Region II of Potential U
! 2 d 2# ( x )
TISE: + U# ( x ) = E# ( x )
2
2m dx
d 2# ( x ) 2m
%
= 2 (U $ E )# ( x )
2
dx
!
=! 2# ( x )
2m(U-E)
with ! =
; U>E % ! 2 > 0
!
Solutions are of form # ( x ) ' e ±! x
2
& II ( x, t ) = Ce +! x $i" t + De $! x $i" t 0<x<L
To determine C & D % apply matching cond.
& II ( x, t ) = continuous across barrier (x=0,L)
d & II ( x , t )
= continuous across barrier (x=0,L)
dx
Continuity Conditions Across Barrier
At x = 0 , continuity of " (x) $
A+B=C+D (1)
d" (x)
At x = 0 , continuity of
$
dx
ikA # ikB = ! C # ! D (2)
Similarly at x=L continuity of " (x) $
Ce #! L + De +! L = FeikL
(3)
d" (x)
at x=L, continuity of
$
dx
-(! C)e #! L + (! D)e! L = ikFeikL (4)
Four equations & four unknowns
Cant determine A,B,C,D but if you
Divide thruout by A in all 4 equations :
$ ratio of amplitudes % relations for R & T
That's what we need any way
Potential Barrier when E < U
Expression for Transmission Coeff T=T(E) :
Depends on barrier Height U, barrier Width L and particle Energy E
"1
# 1% U2
$
&
2m(U " E )
2
T(E) = '1+ )
* sinh (! L) ( ; ! =
4
E
(
U
"
E
)
!
+
,
.
and
R(E)=1- T(E).........what's not transmitted is reflected
General Solutions for R & T:
Above equation holds only for E < U
For E>U, α=imaginary#
Sinh(αL) becomes oscillatory
This leads to an Oscillatory T(E) and
Transmission resonances occur where
For some specific energy ONLY, T(E) =1
At other values of E, some particles are
reflected back ..even though E>U !!
That’s the Wave nature of the
Quantum particle
Tunneling across a barrier
Oxide layer
Wire #1
Wire #2
Q: 2 Cu wires are seperated by insulating Oxide layer. Modeling the Oxide
layer as a square barrier of height U=10.0eV, estimate the transmission coeff
for an incident beam of electrons of E=7.0 eV when the layer thickness is
(a) 5.0 nm (b) 1.0nm
Q: If a 1.0 mA current in one of the intwined wires is incident on Oxide layer, how much of
this current passes thru the Oxide layer on to the adjacent wire if the layer thickness is 1.0nm?
What becomes of the remaining current?
# 1% U
$
&
2
T(E) = '1+ )
* sinh (! L) (
- 4 + E (U " E ) ,
.
2
"1
2m(U-E)
2mE
!=
,k =
!
!
"1
Oxide layer
Wire #1
Wire #2
# 1% U2
$
&
2
T(E) = '1+ )
* sinh (! L) (
4
E
(
U
"
E
)
+
.
/
! , m = 511 keV/c 2
Use " =1.973 keV.A/c
e
2me (U " E )
2 , 511kev / c 2 (3.0 ,10"3 keV )
! -1
0! =
=
=
0.8875A
!
"
1.973 keV.A/c
Substitute in expression for T=T(E)
"1
# 1 % 102 &
2
! -1 )(50A
! ) $( = 0.963 ,10"38 ( small )!!
T= '1+ )
si
nh
(0.8875
A
*
4
7
(10
"
7)
+
.
/
! T=0.657 ,10-7
However, for L=10A;
Reducing barrier width by , 5 leads to Trans. Coeff enhancement by 31
orders of magnitude !!!
Q=Nq e
! N =6.25 "1015 electrons
t
N T =# of electrons that escape to the adjacent wire (past oxide layer)
1 mA current =I=
N T = N .T = (6.25 "1015 electrons ) " T ;
! T=0.657 "10-7 ! N = 4.11"10 ! I = 65.7 pA !!
For L=10 A,
T
T
Current Measured on the first wire is sum of incident+reflected currents
and current measured on "adjacent" wire is the IT
Oxide thickness makes all the difference !
That’s why from time-to-time one needs to
Scrape off the green stuff off the naked wires
QM in 3 Dimensions
• Learn to extend S. Eq and its
solutions from “toy” examples
in 1-Dimension (x) → three
orthogonal dimensions (r ≡
x,y,z)
! ˆ ˆ
ˆ
r = ix + jy + kz
• Then transform the systems
– Particle in 1D rigid box  3D
rigid box
– 1D Harmonic Oscillator  3D
Harmonic Oscillator
z
• Keep an eye on the number
of different integers needed
to specify system 1 3
(corresponding to 3 available
degrees of freedom x,y,z)
y
x
Quantum Mechanics In 3D: Particle in 3D Box
z
z=L
Extension of a Particle In a Box with rigid walls
1D → 3D
⇒ Box with Rigid Walls (U=∞) in
X,Y,Z dimensions
U(r)=0 for (0<x,y,z,<L)
Ask same questions:
• Location of particle in 3d Box
• Momentum
• Kinetic Energy, Total Energy
• Expectation values in 3D
y
y=0
y=L
x
To find the Wavefunction and various
expectation values, we must first set up
the appropriate TDSE & TISE
The Schrodinger Equation in 3 Dimensions: Cartesian Coordinates
Time Dependent Schrodinger Eqn:
!2 2
!# ( x, y, z, t )
$ " # ( x, y , z , t ) + U ( x, y , z ) # ( x, t ) = i !
.....In 3D
2m
!t
2
2
2
!
!
!
"2 = 2 + 2 + 2
!x !y !z
z
y
x
!2 2 % !2 ! 2 & % !2 ! 2 & % !2 ! 2 &
So $ " = ' $
+ $
+ $
= [K ]
2 ( '
2 ( '
2 (
2m
) 2m !x * ) 2m !y * ) 2m !z *
=
[K x ]
+
[K x ]
+
[K x ]
so [ H ]# ( x, t ) = [ E ]# ( x, t ) is still the Energy Conservation Eq
Stationary states are those for which all probabilities are constant in time
and are given by the solution of the TDSE in seperable form:
" -i+ t
"
# ( x, y, z, t ) = # (r , t ) =, (r)e
This statement is simply an extension of what we derived in case of 1D
time-independent potential
Particle in 3D Rigid Box : Separation of Orthogonal Spatial (x,y,z) Variables
!2 2
TISE in 3D: " ! ( x, y, z ) + U ( x, y, z )! ( x, y, z ) = E! ( x, y, z )
2m
x,y,z independent of each other , write ! ( x, y, z ) = ! 1 ( x)! 2 ( y )! 3 ( z )
and substitute in the master TISE, after dividing thruout by! =! 1 ( x)! 2 ( y )! 3 ( z )
and noting that U(r)=0 for (0<x,y,z,<L) )
# ! 2 1 % 2! 1 ( x) $ # ! 2 1 % 2! 2 ( y ) $ # ! 2 1 % 2! 3 ( z ) $
( = E = Const
'&
(+'&
(+'&
2
2
2
* 2m ! 1 ( x) %x + * 2m ! 2 ( y ) %y
+ * 2m ! 3 ( z ) %z
+
This can only be true if each term is constant for all x,y,z )
! 2 % 2! 1 ( x)
! 2 % 2! 2 ( y )
! 2 % 2! 3 ( z )
&
= E1! 1 ( x) ; &
= E2! 2 ( y ) ; &
= E3! 3 ( z )
2
2
2
2m %x
2m %y
2m %z
With E1 + E 2 + E 3 = E=Constant
(Total Energy of 3D system)
Each term looks like particle in 1D box (just a different dimension)
So wavefunctions must be like ! 1 ( x) , sin k1x , ! 2 ( y ) , sin k2 y , ! 3 ( z ) , sin k3 z
Particle in 3D Rigid Box : Separation of Orthogonal Variables
Wavefunctions are like ! 1 (x) " sin k1x , ! 2 ( y) " sin k2 y , ! 3 (z) " sin k3 z
Continuity Conditions for ! i # ni$ = ki L
! !
Leads to usual Quantization of Linear Momentum p="k .....in 3D
% $"(
% $"(
% $"(
px = ' * n1 ; p y = ' * n2 ; pz = ' * n3 (n1 , n 2 , n 3 = 1,2,3,..+)
& L)
& L)
& L)
Note: by usual Uncertainty Principle argument neither of n1 , n 2 , n 3 = 0! (why ?)
1
$ 2"2 2
2
2
2
2
2
Particle Energy E = K+U = K +0 =
( px + p y + pz ) =
(n
+
n
+
n
)
1
2
3
2
2m
2mL
Energy is again quantized and brought to you by integers n1 , n 2 , n 3 (independent)
!
and ! ( r)=A sin k1x sin k2 ysin k3 z (A = Overall Normalization Constant)
E
E
-i t
!
! -i " t
,( r,t)=! ( r) e = A [sin k1x sin k2 ysin k3 z] e "
Particle in 3D Box :Wave function Normalization Condition
E
E
-i t
!
! -i " t
!( r,t)=" ( r) e = A [sin k1x sin k2 ysin k3 z] e "
E
E
i t
i t
* !
* !
"
! ( r,t)=" ( r) e = A [sin k1x sin k2 ysin k3 z] e "
!
* !
! ( r,t)!( r,t)=A2 [sin 2 k1x sin 2 k2 ysin 2 k3 z]
Normalization Condition : 1 =
$$$ P(r)dx dydz
#
x,y,z
L
1= A
2
$ sin
x=0
3
2
L
2
k1x dx
$ sin
L
2
k2 y dy
y=0
+2.
!
# A = - 0 and !( r,t)=
, L/
$
z=0
3
2
% L( % L( % L(
sin k3z dz = A ' * ' * ' *
& 2)& 2)& 2)
2
2
E
-i t
+2.
"
[sin
k
x
sin
k
ysin
k
z]
e
- 0
1
2
3
L
, /
Particle in 3D Box : Energy Spectrum & Degeneracy
E n1 ,n 2 ,n3
! 2!2 2
2
2
=
(
n
+
n
+
n
1
2
3 ); n i = 1, 2, 3...#, ni $ 0
2
2mL
Ground State Energy E111
3! 2 ! 2
=
2mL2
6! 2 ! 2
Next level % 3 Excited states E 211 = E121 = E112 =
2mL2
Different configurations of " (r)=" (x,y,z) have same energy % degeneracy
z
z=L
y=L y
x=L
x
Degenerate States
E 211 = E121 = E112
Ground State
6! 2 ! 2
=
2mL2
E111
ψ
ψ
z
z
y
y
x
x
E211
E121
E112