Show that R = l x − 1 − x 2 D and L = l x + 1 − x 2 D , where D = d / d x , are raising and lowering operators for Legendre polynomials [compare Hermite functions, (22.1) to (22.11) and Bessel functions, Problems 22.29 and 22.30]. More precisely, showthat R P l − 1 ( x ) = l P l ( x ) and L P l ( x ) = l P l − 1 ( x ) . Hint : Use equations (5.8d) and (5.8f). Note that, unlike the raising and lowering operators for Hermite functions, here R and L depend on l as well as x , so you must be careful about indices. The L operator operates on P 1 , but the R operator as given operates on P l − 1 to produce l P 1 . [If you prefer, you could replace l by l + 1 to rewrite R as ( l + 1 ) x − 1 − x 2 D ; then it operates on P l to produce ( l + 1 ) P l + 1 . Assuming that all P l ( 1 ) = 1 , solve L P 0 ( x ) = 0 to find P 0 ( x ) = 1 , and then use raising operators to find P 1 ( x ) P 2 ( x ) .
Show that R = l x − 1 − x 2 D and L = l x + 1 − x 2 D , where D = d / d x , are raising and lowering operators for Legendre polynomials [compare Hermite functions, (22.1) to (22.11) and Bessel functions, Problems 22.29 and 22.30]. More precisely, showthat R P l − 1 ( x ) = l P l ( x ) and L P l ( x ) = l P l − 1 ( x ) . Hint : Use equations (5.8d) and (5.8f). Note that, unlike the raising and lowering operators for Hermite functions, here R and L depend on l as well as x , so you must be careful about indices. The L operator operates on P 1 , but the R operator as given operates on P l − 1 to produce l P 1 . [If you prefer, you could replace l by l + 1 to rewrite R as ( l + 1 ) x − 1 − x 2 D ; then it operates on P l to produce ( l + 1 ) P l + 1 . Assuming that all P l ( 1 ) = 1 , solve L P 0 ( x ) = 0 to find P 0 ( x ) = 1 , and then use raising operators to find P 1 ( x ) P 2 ( x ) .
Show that
R
=
l
x
−
1
−
x
2
D
and
L
=
l
x
+
1
−
x
2
D
,
where
D
=
d
/
d
x
,
are raising and lowering operators for Legendre polynomials [compare Hermite functions, (22.1) to (22.11) and Bessel functions, Problems 22.29 and 22.30]. More precisely, showthat
R
P
l
−
1
(
x
)
=
l
P
l
(
x
)
and
L
P
l
(
x
)
=
l
P
l
−
1
(
x
)
.
Hint: Use equations (5.8d) and (5.8f). Note that, unlike the raising and lowering operators for Hermite functions, here
R
and
L
depend on
l
as well as x, so you must be careful about indices. The
L
operator operates on
P
1
, but the
R
operator as given operates on
P
l
−
1
to produce
l
P
1
.
[If you prefer, you could replace
l
by
l
+
1
to rewrite
R
as
(
l
+
1
)
x
−
1
−
x
2
D
; then it operates on
P
l
to produce
(
l
+
1
)
P
l
+
1
.
Assuming that all
P
l
(
1
)
=
1
,
solve
L
P
0
(
x
)
=
0
to find
P
0
(
x
)
=
1
,
and then use raising operators to find
P
1
(
x
)
P
2
(
x
)
.
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