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Let f*(x) a" ((i¾ f^(¾ ))Ø or (f*)^(¾ ) = i¾ f^(¾ ). Then f* " L2(R), and
" " " "
ëø öø = ëø öø
dxöø ëø dxöø ëø
ìø
+" x2lf(x)l2 ÷ø ìø +" ¾ 2lf^(¾ )l2d¾ ÷ø ìø +" x2lf(x)l2 ÷ø ìø +" l(i¾ )f^(¾ )l2d¾ ÷ø
íø øø íø øø íø øø íø øø
" " " "
" "
öø
= ëø dxöø ëø
ìø
+" x2lf(x)l2 ÷ø ìø +" l(f*)^(¾ )l2d¾ ÷ø
íø øø íø øø
" "
" "
ëø
= 2À dxöø ëø dxöø
ìø
+" x2lf(x)l2 ÷ø ìø +" lf*(x)l2 ÷ø (by Parseval s identity).
íø øø íø øø
" "
Since
" "
2 2
îø
1
* *
ïø x (x) f(x) + f*(x) f(x) úø = ïø x Re (x) f(x) úø
(f )dxùø îø (f )dxùø
+" 2 +"
ðø ûø ðø ûø
" "
" "
2 2
îø ùø ôø ôø
= ïø Re x f(x) f*(x) dx úø d" ôø x f(x) f*(x) dx ôø
+" +"
ðø ûø ôø ôø
" "
" "
d" ëø dxöø ëø dxöø
ìø
+" x2lf(x)l2 ÷ø ìø +" lf*(x)l2 ÷ø
íø øø íø øø
" "
(by the Cauchy-Schwartz inequality).
We will show that
" "
x (x) f(x) + f*(x) f(x) dx
(f )dx =
+" * +" lf(x)l2
" "
from which the result follows, for then
" " " "
ëø öø = 2À ëø
dxöø ëø dxöø ëø dxöø
ìø
+" x2lf(x)l2 ÷ø ìø +" ¾ 2lf^(¾ )l2d¾ ÷ø ìø +" x2lf(x)l2 ÷ø ìø +" lf*(x)l2 ÷ø
íø øø íø øø íø øø íø øø
" " " "
"
2
îø
1
*
e" 2À ïø x (x) f(x) + f*(x) f(x) úø
(f )dxùø
+" 2
ðø ûø
"
18.ÊHeisenbergÕsÊInequality.
61
" "
îø ùø2 À ëø "
ëø 1 öø
= À ïø dxúø dxöø
ìø
2 +" lf(x)l2 ûø = 2 +" lf(x)l2 ÷ø ìø +" lf^(¾ )l2d¾ ÷ø
2À
ðø íø øø íø øø
" " "
" "
1 ëø ëø öø .
= dxöø
ìø
4 +" lf(x)l2 ÷ø ìø +" lf^(¾ )l2d¾ ÷ø
íø øø íø øø
" "
To complete the proof, we assume that f is continuous and piecewise smooth, This
assumption can be removed since functions in L1(R) are the uniform limit of such
functions.
Then from the property of Fourier transforms,
f*(x) = f'(x)
wherever the derivative exists. Then for any interval [a, b] ,
b
d
b lf(b)l2 a lf(a)l2 =
+" dx (x lf(x)l2)dx
a
b
=
(x f '(x) f(x) + x f(x) f '(x) + lf(x)l2 )dx
+"
a
b b
= x (x) f(x) + f*(x) f(x) dx
(f )dx +
+" * +" lf(x)l2
a a
The assumption f " L2(R) implies that b lf(b)l2 ’! 0 as b ’! " and a lf(a)l2 ’! 0 as a ’! "
since otherwise lf(x)l > c lxl 1/2 as lxl ’! " , which is not integrable. Taking the limit as
b ’! " and a ’! " ,
" "
0 = x (x) f(x) + f*(x) f(x) dx
(f )dx +
+" * +" lf(x)l2
" "
as required.
As for the case of equality in Heisenberg s inequality, this holds if and only if f(x) f*(x) is
real and f*(x) = K x f(x) for some complex constant K. That is,
f(x) f*(x) = f(x) K x f(x) = x lf(x)l2 K is real.
Therefore K is real.
The differential equation
f'(x) = f (x) = K x f(x)
has solutions of the form
Kx2
2
f(x) = c e , c any real constant,
18.ÊHeisenbergÕsÊInequality.
62
Kx2
2
and f(x) = c e " L2(R) if and only if K > 0. Therefore equality holds in Heisenberg s
2
inequality only if f(x) = c e kx for constants c " R and k > 0.
Kx2
2
Conversely, let f(x) = e for constant K > 0. Then
" " " "
ëø öø = 2À ëø
dxöø ëø dxöø ëø dxöø
ìø
+" x2lf(x)l2 ÷ø ìø +" ¾ 2lf^(¾ )l2d¾ ÷ø ìø +" x2lf(x)l2 ÷ø ìø +" lf*(x)l2 ÷ø
íø øø íø øø íø øø íø øø
" " " "
" "
ëø
= 2À dxöø ëø dxöø
ìø
+" x2lf(x)l2 ÷ø ìø +" lK x f(x)l2 ÷ø
íø øø íø øø
" "
"
2
= 2À K2 ëø dxöø
ìø
+" x2lf(x)l2 ÷ø
íø øø
"
"
Kx2 2
= 2À K2 ëø
ìø ÷ø
+" x2e dxöø
íø øø
"
"
2
îø ùø
x
Kx2
( 2Kxe
)
= 2À K2 ïø dxúø
+" ëø 2Köø
íø øø
ðø ûø
"
"
2
îø ùø
x
Kx2
( 2Kxe
)
= 2À K2 ïø dxúø
+" ëø 2Köø
íø øø
ðø ûø
"
"
2
îø
x d Kx2
( )ùø
= 2À K2 ïø dxúø
+" ëø 2Köø dx e
íø øø
ðø ûø
"
"
2 2
îø ùø
1
= 2À K2 ïø dxúø
+" ëø 2Köø e Kx ûø (integration by parts)
íø øø
ðø
"
"
À îø Kx2 2
ùø
= 2 ïø
+" e dxúø
ðø ûø
"
"
À îø t2 2 2
ùø À
= ïø
2K +" e dtúø = 2K .
ðø ûø
"
Whereas,
" " " "
1 ëø öø = 1 ëø ëø
dxöø ëø dxöø 2À dxöø
ìø
4 +" lf(x)l2 ÷ø ìø +" lf^(¾ )l2d¾ ÷ø 4 +" lf(x)l2 ÷ø ìø +" lf(x)l2 ÷ø
íø øø íø øø íøìø " øø íø øø
" " "
"
2
À
= ëø dxöø
ìø
2 +" lf(x)l2 ÷ø
íø øø
"
18.ÊHeisenbergÕsÊInequality.
63
"
À îø Kx2 2
ùø
= 2 ïø
+" e dxúø
ðø ûø
"
and equality holds.
The case of a `" 0, ± `" 0, follows by observing that F(x) = e i± x f(x + a) satisfies the same
hypotheses as f(x) and " f = " F and " f^= " F^ for any a `" 0, ± `" 0.
a 0 ± 0
As a consequence of the inequality f f^ 1
(" )(" ± )e" , we see that it is impossible for
a 4
both " f and " f^ to be simultaneously small. That is, if one of " f or " f^ is very small
a ± a ±
then the other must be large.
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