383
E. Dirac δ-Function
(ii)
δ(x) = δ(−x)
i.e. an even function.
(E.34)
Proof
∫
f (0) = δ(x)f (x) dx.
(E.35)
If f (x) is an odd function, f (0) = 0. Thus δ(x) must be an even function.
∑
1
(iii)
δ(f (x)) =
δ(x − a i )
(E.36)
|df /dx| x=ai
i
where a i are the roots of f (x) = 0.
Proof
The δ-function is only non-zero when its argument vanishes. Thus we are
concerned with the roots of f (x) = 0. In the vicinity of a root
f (a i ) = 0
(E.37)
we can make a Taylor expansion
( )
df
f (x) =
+ (x − a i )
+ · · · .
(E.38)
dx x=ai
Thus the δ-function has non-zero contributions from each of the roots a i of
the form
[
( )
]
∑
df
δ(f (x)) =
δ (x − a i )
.
(E.39)
dx x=ai
i
Hence (using property (i)) we have
∑
1
δ(f (x)) =
δ(x − a i ).
(E.40)
|df /dx| x=ai
i
Consider the example
δ(x
2
− a
2 ).
(E.41)
Thus
f (x) = x
2
− a
2 = (x − a)(x + a)
(E.42)
with two roots x = ±a (a > 0), and df /dx = 2x. Hence
δ(x
2
− a
2 ) =
1 [δ(x − a) + δ(x + a)].
(E.43)
2a
(iv)
xδ(x) = 0.
(E.44)
This is to be understood as always occurring under an integral. It is obvious
from the definition or from property (ii).