step potential
DESCRIPTION
asdasdTRANSCRIPT
1
STEP POTENTIAL
V(x) = 0 FOR x < 0 V(x) = V0 FOR x > 0 TOTAL ENERGY OF PARTICLE = E
2
PARTICLES IN REGION I (x < 0) MOVING IN +x PARTICLES HAVE MASS m FIND REFLECTION COEFICIENT REMEMBER PROBABILITY FLUX
x
txtx
dx
txtx
m
itxS
),(),(
),(),(
2),(
**
3
DEFINE REFLECTION COEFICIENT
),(
),(
txS
txSR
xI
xI
THEREFORE WE NEED THE WAVE FUNCTION WE WILL TAKE A SHORTCUT THE POTENTIAL V IS A FUNCTION OF X ONLY
4
REGION I (x < 0) TIME INDEPENDENT SWE V = 0
II E
dx
d
m
0
2 2
22
or
02
22
2
II mE
dx
d
5
LET
21
2
mE
k
0212
2
II k
dx
d
SOLUTION
xikxikI BeAe 11
6
REGION II (x > 0) V=V0 TISE
IIIIII EV
dx
d
m
02
22
2
REARRANGE
7
02
022
2
IIII EV
m
dx
d
LET
)(2
022 EVm
k
0222
2
IIII k
dx
d
SOLUTION
xkxkII DeCe 22
8
WAVE FUNCTIONS FOR EACH
tE
ixikxikI eBeAe
)( 11
tE
ixkxkII eDeCe
)( 22
9
FORCE TO BE WELL BEHAVED FINITE? ALL TERMS MUST BE FINITE AS
x xkCe 2 THEREFORE C MUST BE ZERO SO
10
tE
ixkII eDe
)( 2 WAVE FUNCTION CONTINUOUS AT x = 0
),0(),0( tt III
tE
iktE
iikik eDeeBeAe )()( 000 211
THUS
11
)()( DBA (1.) DIRIVATIVE MUST BE CONTIUNOUS
),0(),0( tx
tx III
12
)( 11 xikxiktE
iI BeAe
dx
de
x
)( 2xktE
iII De
dx
de
x
OR
13
)( 1111
xikxiktE
iI BeikAeike
x
)( 22
xktE
iII Deke
x
AT x = 0
)()( 02
01
01
11 IIktE
iikiktE
iDekeBeikAeike
14
DkBikAik 211 )(
Dik
kBA
1
2)(
Dk
kiBA
1
2)( 2.
ADD EQUATIONS 1 AND 2
15
)1(21
2
k
kiDA
)1(2 1
2
k
ki
DA
SUBTRACT EQUATIONS 2 FROM EQUATION 1
Dk
kiDBABA
1
2)()()(
16
Dk
kiB )1(2
1
2
)1(2 1
2
k
ki
DB
THEREFORE EIGENFUNCTIONS
17
xikxikI e
k
ki
De
k
ki
D11 )1(
2)1(
2 1
2
1
2
AND
)( 2xkII De
18
WAVEFUNCTIONS
tE
ie
11
tE
ie
22
ONLY ONE UNKNOWN D
19
ANALYIZE
tE
ixikxikI eBeAe
)( 11
OR
)( 11t
Eixikt
Eixik
I eBeeAe
20
OR
)()()( 11 t
Exkit
Exki
I BeAe
OR
)()( 11 txkitxkiI BeAe
SINCE
21
E
LET
)( 1 txkiI Ae
EQUATION FOR WAVE MOVING IN PLUS (+) x DIRECTION
I WAVE FN FOR +x DIRECTION
22
AND
)( 1 txkiI Be
EQUATION FOR WAVE MOVING IN MINUS (-) x DIRECTION
I WAVE FN FOR -x DIRECTION
23
REFLECTION COEFICIENT
I
I
S
SR
WHERE
x
txtx
dx
txtx
m
itxS
),(),(
),(),(
2),(
**
24
FOR +x DIRECTION
xdxm
iS I
II
II
**
2
)( 1 txkiI Ae
)(1
1 txkiI Aeikx
25
)(** 1 txkiI eA
)(*1
*1 txkiI eAik
x
PUT INTO PROBABILITY FLUX EQN
26
)(1
)(* 11
2txkitxki
I AeikeAm
is
)(*
1)( 11 )(
2txkitxki eAikAe
m
i
1)()( 11 txkitxki ee
etc.
27
THEREFORE
)()(2 1
*1
* ikAAikAAm
isI
*1
* )(22
AAm
kikAA
m
isI
MOMEMTUM
mvkp
28
*
1 AAvSI
SAME PROCEEDURE WITH
)( 1 txkiI Be
WAVE MOVING IN – x DIRECTION
29
*1BBvSI
AND THEREFORE
*1
*1
AAv
BBv
S
SR
I
I
30
)1(2 1
2
k
ki
DA
)1(2 1
2
k
ki
DB
)1(2 1
2*
*
k
ki
DA
)1(2 1
2*
*
k
ki
DB
31
)1(2
)1(2
)1(2
)1(2
1
2*
1
2
1
2*
1
2
k
ki
D
k
ki
D
k
ki
D
k
ki
D
R
1*
*
DD
DDR
32
PLOT WAVE FUNCTION
tE
ie
11
AND
tE
ie
22
33
WHERE IS PARTICLE BORN
*),( txP REGION x > 0
xkxktE
ixktE
ieDDeDeeeDtxP 222 2**),(
THIS IS NOT EQUAL TO ZERO POSSIBLE FOR PARTICLE TO BE IN REGION x > 0
34
NEW PROBLEM WITH ENERGY OF PARTICLES GREATER THAN V0