Eigenfunctions of the Fourier transform
Mark Reeder February 13, 2012
Fourier transform of e(x/a)
The (complex) Fourier transform F[f ] of a function f (x) is dened by 1 2 We rst derive the Fourier transform
f (x)eix dx.
2 a = e(a/2) , 2
F e(x/a)
(1)
where a > 0 is a constant. Regard this transform as a function of : 1 F () = 2 and dierentiate:
e(x/a) eix dx,
1 1 d F () = d 2 i
xe(x/a) eix dx,
and now integrate by parts, with u = eix , to get 1 a2 d F () = d 2 2 Solving this dierential equation for F () we get F () = F (0) e(a/2) . So it remains to determine F (0). This is essentially the Gaussian integral: 1 F (0) = 2
2
dv = xe(x/a) ,
du = ieix ,
v = (a2 /2)e(x/a) , a2 F (). 2
e(x/a) eix dx =
e(x/a) dx,
which can be computed as follows. The trick is to compute the square of F (0) and use polar coordinates: 1 F (0)2 = 2 so we get
2
e(x/a) dx
1 2
e(x/a) e(y/a) dx dy =
1 2
2 0 0
e(r/a) r dr d =
1 a2 a2 2 = , 2 2 2
a F (0) = , 2
hence
2 a F () = e(a/2) , 2
as claimed in (1). If a = 2 then F () = F ex so the function ex
2 2
/2
= e
/2
/2
is its own Fourier transform.
In general, a smaller a gives a bigger spike in the graph of f (x) and a more spread-out graph of F (). And vice-versa. 2 2 Here are the superimposed graphs of f (x) = e(x/a) (red) and F () = e(a/2) (blue) for a = .5, 1.2, 2, 2.
1.0
1.0
1.0
0.8
0.8
0.8
0.6
0.6
Out[11]=
0.6
0.4
0.4
0.4
0.2
0.2
0.2
-4
-2
-4
-2
-4
-2
1.0
0.8
0.6
Out[15]=
0.4
0.2
-4
-2
The dierential operator viewpoint
2
The calculation of F[ex ] in the rst section is illuminated by a more algebraic point of view. First, let us modify the denition of F so that it takes functions of x to functions of the same variable x: 1 F[f (x)] = 2 Now the derivative formula becomes F[f (x)] = ixf (x). On the other hand if we rst apply F and then dierentiate, we get d 1 F[f ] = dx 2 so F[f ] = iF[xf ] Consider the dierential operator D= Using formulas (2) and (3), we have F[D[f ]] = F[f + xf ] = ixF[f ] + iF[f ] = iD[F[f ]]. d + x, dx D[f ] = f + xf. (3)
f (y)eixy dy.
(2)
(iy)f (y)eixy dy = iF[xf ],
Thus F and D are linear operators such that FD = iDF. (4) If f is a solution of the dierential equation D[f ] = 0, then (4) implies that F[f ] is also a solution. However, every solution of D[f ] = 0 is of the form 2 f (x) = Cex /2 , where C = f (0) is a constant. Therefore there is a constant C such that F[ex To nd C we evaluate at 0: F[ex Thus, the equality F[ex
2 2 2
/2
] = Cex
/2
/2
1 ](0) = 2
ey
/2
1 dy =
2
eu du = 1.
/2
] = ex
/2
(5)
follows from commutation formula (4) and the Gaussian integral.
Eigenvectors of the Fourier Transform
2
In terms of linear algebra, equation (5) asserts that ex /2 is an eigenvector of the operator F, with eigenvalue = 1, 2 on the complex vector space V of functions of the form p(x)ex /2 , where p(x) is a polynomial. The vector space V is innite dimensional, but note that V is a union of nite dimensional subspaces
V =
n=0
Vn ,
where Vn = {p(x)ex
/2
: deg p(x) n} is a subspace of dimension dim Vn = n + 1.
Note that D[p(x)ex so Vn = ker D The function
n+1
/2
] = p (x)ex
n+1
/2
is the subspace of functions in V killed by D
dn x2 /2 e dxn belongs to Vn and Hermite polynomials Hen (x) are dened by Hen (x)ex On the homework, you showed that F[xn ex
2 2
/2
= (1)n
dn x2 /2 e . dxn
2
(6)
/2
] = (i)n Hen (x)ex
2
/2
(7)
so F[Vn ] = Vn for each n. We have found the eigenvector ex for n 0.
/2
V0 ; we now nd a basis of eigenvectors for F in Vn
Dene another operator : V V by [f ](x) = f (x). The Fourier inversion theorem becomes the operator equation F 2 = .
It follows that F 4 = I and the eigenvalues of F on V are powers of i = 1. More precisely, F has order two on the even functions in V and order four on the odd functions in V . So the functions with F-eigenvalues 1 are even and those with F-eigenvalues i are odd. In fact, Hen = xn + terms of lower degree, so equation (7) shows that if F has an eigenvector which lies in Vn but not in Vn1 , then the eigenvalue must be (i)n . Hence for each n 0 there is a unique (up to scalar) function n (x) Vn of the form 2 n (x) = pn (x)ex /2 , where pn (x) is a polynomial of degree n, such that F[n ] = (i)n n . For example, we have 0 = ex
2
/2
1 = 2xex
/2
2 = (4x2 2)ex
/2
3 = (8x3 12x)ex
/2
4 = (16x4 48x2 +12)ex /2 . (8)
To nd n in general we consider a new dierential operator D2 = d2 x2 . dx2
Using equations (2) and (3) we nd that D2 commutes with F: D2 F = FD2 . It follows that F preserves each eigenspace of D2 . That is, if D2 [] = for some C then D2 F[] = F[]. We take = (2n + 1), and consider the equation D2 = (2n + 1), or in other words, + (2n + 1 x2 ) = 0. This has a two dimensional solution space, but only one solution (up to scalar) lies in V . For (x) = p(x)ex solution of (9) in V exactly when p 2xp + 2np = 0. Writing p = ck x , equation (10) is equivalent to the recurrence formula ck+2 = 2(k n) ck . (k + 2)(k + 1)
k
2
(9)
/2
is a (10)
So (10) has a unique polynomial solution p(x) of degree n. Taking cn = 2n (to make the coecients integers) p(x) is given by the Hermite polynomial Hn (x) = (1)n ex
2
dn x2 e = dxn
n/2
(1)m
k=0
n! (2x)n2k . (n 2m)!
This is slightly dierent than the Hermite polynomial Hen dened in (6) above. The two kinds of Hermite polynomials are related by: Hen (x) = 2n/2 Hn (x/ 2). For example, the polynomials H0 , . . . , H4 are the coecients of ex
2
/2
in the list (8).
Now, since F preserves the one-dimensional space of solutions of (9) in V , we have F[Hn (x)ex
2
/2
] = Hn (x)ex
/2
for some constant . But then = (i)n because it is an eigenvalue of F on a function in Vn which is not in Vn1 . Hence n (x) = Hn (x)ex
2
/2
= (1)n ex
/2
dn x2 e dxn
is the desired eigenfunction of F and {0 , 1 , . . . , n } is a basis of Vn consisting of F-eigenvectors. 4
