1
INFINITE SQUARE WELL
V(x) = 0 0 < x < L V(x) = ∞ x < 0 and x > L
2
FOR REGIONS x < 0 and x > L
)()()()(
2 2
22
xExdx
xd
m
TO KEEP SECOND TERM FINITE
0)( x EIGENFUNCTION AND THUS
0),( tx WAVE FUNCTION OUTSIDE WELL
3
FOR REGION 0 < x < L INSIDE WELL
0)( xV THUS
)()()0()(
2 2
22
xExdx
xd
m
AND
)()(
2 2
22
xEdx
xd
m
4
OR
0)(2)(
22
2
xmE
dx
xd
LET
22 2
mE
k
THEN
0)(
)( 22
2
xkdx
xd
5
IN OPERATOR FORM
0)()( 22 xkD TO FIND SOLUTIONS
0)( 22 kD THUS
ikD 1i AND SOLUTIONS ARE
ikxe
6
REMEMBER LINEAR EQUATION ANY LINEAR COMBINATION IS SOLUTION
i
eex
ixix
2sin
2
cosixix ee
x
THEREFORE ALSO SOLUTIONS
sinkx and coskx
7
GENERAL SOLUTION (EIGENFUNCTION)
kxBkxAx cossin)( WAVE FUNCTION
tE
iekxBkxAtx
)cossin(),( HAVE WAVE FUNCTION IN ALL REGIONS
8
NOW MUST BE WELL BEHAVED SINGLE VALUED ok FINITE ok CONTINUOUS ? DIRIVITIVE CONTINUOUS ?
9
CONTINUOUS
AT x = 0 OUTSIDE WELL
0),0( t INSIDE WELL
10
tE
ieBAt
)0cos0sin(),0( THUS
0)0cos0sin( BA
THUS
0B WAVE FUNCTION
tE
iekxAtx
)sin(),(
11
AT x = L
OUTSIDE WELL
0),( tL INSIDE
tE
iekLAtL
)sin(),(
12
THUS
0sin kLA
0A SO
0sin kL
nkL WHERE
..,.........4,3,2,1,0n
13
REMEMBER
22 2
mE
k
THEREFORE
nLmE
2
2
SOLVE FOR E
14
2222
2 nLmE
2
222
2mLnEn
WHERE
..,.........4,3,2,1,0n
15
16
STILL HAVE WORK TO DO WITH
tE
iekxAtx
)sin(),( WHAT IS VALUE OF A? NEED TO NORMALIZE
1),(),(),( *
x
x
dxtxtxtxP
17
1*
0
*0
*
x
Lx
Lx
x
x
x
dxdxdx
OUTSIDE WELL x < 0 AND x > L Ψ = 0 SO
10
*
Lx
x
dx
18
AND INSIDE THE WELL
tE
iekxAtx
)sin(),( COMPLEX CONJUGATE IS
tE
iekxAtx
)sin(),(*
19
1)sin()sin(0
Lx
x
tE
itE
idxekxAekxA
AND
1 t
Eit
Ei
ee
1)sin(0
22
Lx
x
dxkxA
20
2/
1
sin
1
0
2
2
Lkxdx
A Lx
x
LA
2
SO
tE
iekx
Ltx
)sin
2(),(
21
AND
nkx REPLACE k WITH
L
nk
TO GET
22
tE
i
n eL
xn
LtL
sin
2),(
NORMALIZED WAVE FUNCTION FOR INFINITE SQUARE WELL NOW READY TO FIND EXPECTATION VALUES AND PROBABILITIES
23
EXPECTATION VALUE OF x
dxtxxtxx ),(),(*
0),(),(00
* L
dxtxxtxx
L
tE
itE
idxe
L
xn
Lxe
L
xn
Lx
0
sin2
sin2
24
ONCE AGAIN
1 t
Eit
Ei
ee
L
dxL
xn
Lx
L
xn
Lx
0
sin2
sin2
OR
L
dxL
xnx
Lx
0
2sin2
25
FROM INTEGRAL TABLES
2
22
8
2cos
4
2sin
4sin
a
ax
a
axxaxx
SO FOR n = 1
L
L
Lx
L
Lx
xx
Lx
0
2
2
8
2cos
4
2sin
4
2
24
2 2 LL
Lx
26
EXPECTATION VALUE OF p
PX (momentum) xi
dxx
ip
*
USE n = 1
0sin2
sin2
00
dxeL
x
Lxie
L
x
Lp
tE
iL tE
i
dxL
x
Ldx
di
L
x
Lp
L sin
2sin
20
27
dxL
x
dx
d
L
x
L
ip
L sinsin
20
dxL
x
L
x
LL
ip
L cossin
20
dxL
x
L
x
L
ip
L cossin
202
28
axaxax 2sin2
1cossin
dxL
x
L
ip
L
02
2sin
2
12
dxL
x
L
ip
L
02
2sin
02
cos2 0
2
L
L
xL
L
ip
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