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The Potential Barrier
- Calculations
Continuity of the wave function and its first derivative
at the boundaries of the barrier.
1)
y
x=- a
2)
y
k 'a
- k 'a
ika
:
Ce
+
De
=
Fe
x=a
3)
dy
dx
4)
dy
dx
: Ae
-ika
+ Be
ika
= Ce
- k 'a
+ De
k 'a
: ikAe-ika - ikBeika = k 'Ce- k 'a - k ' Dek 'a
x=- a
: k 'Ce k 'a - k ' De- k 'a = ikFeika
x=a
The Potential Barrier
- Calculations
I want to calculate the transmission coefficient, so I’m going to calculate A
as a function of F. Thus, I’m going to eliminate the coefficients C and D.
dy
3)
dx
: ikAe-ika - ikBeika = k 'Ce- k 'a - k ' Dek 'a
x=- a
® Ae-ika - Beika = -i kk' Ce- k 'a + i kk' Dek 'a
(5)
Now let’s add together equation 1 and equation 5.
Ae-ika + Beika = Ce- k 'a + Dek 'a
Ae-ika - Beika = -i kk' Ce- k 'a + i kk' Dek 'a
® 2Ae-ika = C (1- i kk' ) e- k 'a + D (1+ i kk' ) ek 'a
(6)
The Potential Barrier
- Calculations
And from equation 4
4)
dy
dx
: k 'Cek 'a - k ' De- k 'a = ikFeika
x=a
® Cek 'a - De- k 'a = i kk' Feika
(7)
Now let’s add and subtract equation7 and equation 2, where equation 2
2)
y
k 'a
-k 'a
ika
:
Ce
+
De
=
Fe
x=a
The Potential Barrier
- Calculations
Adding equation 7 and equation 2:
2Cek 'a = (1+ i kk' ) Feika
ika
e
® C = F2 (1+ i kk' ) k 'a
e
(8)
Subtracting equation 2 from equation 7:
2De- k 'a = (1- i kk' ) Feika
ika
e
® D = F2 (1- i kk' ) - k 'a
e
(9)
The Potential Barrier
- Calculations
Let’s substitute these two equations (equations 8 and 9) into
equation 6 for the coefficients C and D:
ika
ika
é
ù
é
ù k 'a
e
e
-ika
- k 'a
F
k
k'
F
k
k'
2Ae = ê 2 (1+ i k ' ) (1- i k ) k 'a ú e + ê 2 (1- i k ' ) (1+ i k ) k 'a ú e
(10)
e û
e û
ë
ë
And from here we could calculate the transmission coefficient,
T, but let’s clean this up a bit.
So,
(1+ i kk' ) (1- i kk' ) = 1+ i kk' - i kk' +1 = 2 + i kk' - i kk'
(1- i kk' ) (1+ i kk' ) = 1- i kk' + i kk' +1 = 2 - i kk' + i kk'
The Potential Barrier
- Calculations
Now equation 10 becomes:
2Ae-ika
ika
ika
éF
ù
é
ù
e
e
= ê 2 (1+ i kk' ) (1- i kk' ) k 'a ú e- k 'a + ê F2 (1- i kk' ) (1+ i kk' ) k 'a ú e k 'a (10)
e û
e û
ë
ë
2Ae-ika = F2 eika éë( 2 + i kk' - i kk' ) e-2 k 'a + ( 2 + i kk' - i kk' ) e2 k 'a ùû
2Ae-ika = F2 eika éë 2 ( e-2 k 'a + e2 k 'a ) - i ( kk' - kk' ) e-2 k 'a + i ( kk' - kk' ) e2 k 'a ùû
® 2Ae-ika = F2 eika éë 2 ( e-2 k 'a + e2 k 'a ) + i ( kk' - kk' ) ( e2 k 'a - e-2 k 'a ) ùû
What are those functions in parentheses on the right hand side?
The Potential Barrier
- Calculations
From a table of mathematical functions:
(e
-2 k 'a
+ e2 k 'a )
2
2 k 'a
-2 k 'a
e
e
(
)
2
= cosh(2k 'a)
= sinh(2k 'a)
And finally we have something useful:
A = F4 e2ika éë 4 cosh(2k 'a) + 2i ( kk' - kk' ) sinh(2k 'a)ùû
(11)
The Potential Barrier
- Calculations
The hyperbolic functions:
The Potential Barrier
- Calculations
The transmission coefficient:
2
1 A
A* A
=
= *
T F
FF
A e
éë 4 cosh(2k 'a) + 2i ( kk' - kk' ) sinh(2k 'a) ùû
=
F
4
A* e-2ika
k'
k
é
=
4
cosh(2k
'a)
2i
(
k
k ' ) sinh(2k 'a) ù
*
ë
û
F
4
2ika
Now when I multiply these two expressions together the terms
involving the complex exponentials multiply to one.
The Potential Barrier
- Calculations
The transmission coefficient:
2
1 A
A* A
=
= *
T F
FF
A e
éë 4 cosh(2k 'a) + 2i ( kk' - kk' ) sinh(2k 'a) ùû
=
F
4
A* e-2ika
k'
k
é
=
4
cosh(2k
'a)
2i
(
k
k ' ) sinh(2k 'a) ù
*
ë
û
F
4
2ika
Now when I multiply these two expressions together the terms
involving the complex exponentials multiply to one.
The Potential Barrier
- Calculations
The transmission coefficient:
2
2
1 A
A* A
sinh
(2k 'a)
2
k'
k 2
=
= * = cosh (2k 'a) + ( k - k ' )
T F
FF
4
And “lastly” from a table of mathematical formulas:
cosh2 q - sinh2 q = 1
So that
2
1 A
A* A é
sinh
(2k 'a) ù
2
k'
k 2
=
= * = ê1+ sinh (2k 'a) + ( k - k ' )
ú
T F
FF ë
4
û
1
1 k'
k 2ù
é
= 1+ ë1+ 4 ( k - k ' ) û sinh 2 (2k 'a)
T
2
The Potential Barrier
- Calculations
Let’s write the transmission coefficient in terms of the energies:
Start with the expression involving the k’s:
( k ') - k )
(
=
2
1
4
( kk' - )
k 2
k'
4 ( kk ')
2
2
2
4
2
k ') - 2 ( kk ') + k 4
(
=
2
4 ( kk ')
and
1+ 14 ( kk' -
1+
1
4
(
k'
k
-
4
2
k ') - 2 ( kk ') + k 4
(
) = 1+
2
4 ( kk ')
k 2
k'
)
k 2
k'
k ')
(
=
4
+ 2 ( kk ') + k
2
4 ( kk ')
2
4
4 ( kk ') + ( k ') - 2 ( kk ') + k 4
2
=
4
4 ( kk ')
( k ') + k )
(
=
2
4 ( kk ')
2
2
2
2
2
The Potential Barrier
- Calculations
The definitions of k and k’:
So that:
The Potential Barrier
- Calculations
Putting everything together we get the transmission coefficient:
And of course:
R = 1- T