Computation of some functional determinants through Riemann’s ζ-function Sergei Winitzki May 31, 2006
1 Properties of Riemann’s ζ-function Riemann’s zeta function is defined by ∞ X 1 , s n n=1
ζ(s) =
(1)
which converges for Re s > 1, and analytically continued to all s. The resulting function is welldefined in the entire complex plane and has a simple pole at s = 1 (with residue 1). Generally, it is difficult to analyze this function and to establish its properties. It is known that 1 ζ(0) = − , 2
1 ζ 0 (0) = − ln (2π) , 2
ζ(s) =
1 + γ + O(s − 1), s−1
(2)
where γ is Euler’s constant. Other known values are (−1)m+1 (2π)2m B2m , 2 (2m)! B2m ζ(−2m + 1) = − , ζ(−2m) = 0. 2m
(3)
ζ(2m) =
(4)
An integral representation valid for all complex s is Riemann’s formula, 1 Γ(s/2) + ζ(s) = s (s − 1) π s/2
Z
∞
i dt h 1−s s t 2 + t2 t
1
∞ X
n=1
Some useful identities are
e
−n2 πt
!
∞ X
1 −s ζ(s), s = 1−2 (2n − 1) n=1
∞ X (−1)n−1 = 1 − 21−s ζ(s), s n n=1
ζ(1 − s) =
2Γ(s) πs ζ(s). s cos 2 (2π)
.
(5)
(6) (7) (8)
See, for example, the book [1], chapter 1, for a derivation of these properties.
2 Functional determinants In finite dimensions, the determinant of an operator is equal to the product of its eigenvalues (with multiplicities). However, this definition usually does not work in the infinite-dimensional 1
situation. For example, the operator Aˆ ≡ −∂x2 on the interval x ∈ [0, 1] with zero boundary conditions (i.e. in the subspace of twice differentiable functions f (x) such that f (0) = f (1) = 0) has eigenvalues λn = (πn)2 , n = 1, 2, ... (9) each with multiplicity 2. Clearly, the product of these eigenvalues is undefined (diverges). Howˆ ever, we would still like to compute the determinant of the operator A. A standard trick is used to solve this problem: We first define the “generalized” zeta function ζA (s) by the series ∞ X 1 ζA (s) = . (10) λs n=1 n
The series will converge if s is sufficiently large. Then we obtain the analytic continuation of ζA (s) to all (complex) s. The resulting function will be usually analytic at s = 0. Then the determinant of Aˆ is found from the following recipe, d ˆ ζA (s). (11) ln det A = − ds s=0
Note that this recipe would give the correct expression, ln det Aˆ = λ1 λ2 λ3 ..., if the product were well-defined. One considers Eq. (11) as a definition of a “regularized” or “renormalized” determiˆ obtained through the analytic continuation of an auxiliary function ζA (s). nant of an operator A, The justification for this definition of det Aˆ is not completely understood, but it appears that the recipe “works” in the sense that it gives physically sensible answers that could be verified by other calculations. Here we shall concentrate on the computational difficulty inherent in this approach: namely, it is not easy to obtain the analytic continuation of ζA (s) at s = 0. In this note, we show some known examples where the answer can be obtained in closed form. In 0 these examples, we shall reduce ζA (0) to an expression involving specific values of Riemann’s ζ function.
3 First examples Consider again the operator Aˆ ≡ −∂x2 on the interval x ∈ [0, 1] with zero boundary conditions. It is clear that ∞ X 1 1 (12) ζA (s) = 2s = π 2s ζ(2s). n=1 (πn)
Therefore
ln det Aˆ = −
ζ(2s) d = 2ζ(0) ln π − 2ζ 0 (0) = − ln π + ln (2π) = ln 2. ds s=0 π 2s
(13)
A more complicated example is the operator
ˆ ≡ Aˆ + a2 = −∂ 2 + a2 , B x
(14)
again considered on the interval x ∈ [0, 1] with zero boundary conditions. The eigenvalues are λn = a 2 + π 2 n 2 , and the zeta function is ζB (s) =
∞ X
n=1
(a2
n = 1, 2, 3, ...,
(15)
1 s. + π 2 n2 )
(16)
This function can be reduced to Riemann’s zeta funtion ζ(s) using the following trick (motivated by the “heat kernel” formalism; see [2], end of section 2). 2
Recall the definition of Euler’s Gamma function, Z ∞ Γ(s) = ts−1 e−t dt,
(17)
0
and change variable t → λτ to obtain 1 1 = s λ Γ(s)
Z
∞
τ s−1 e−λτ dτ.
(18)
0
Therefore we obtain an alternative representation of the function ζB (s): ∞ Z 1 X ∞ ζB (s) = dτ τ s−1 e−λn τ . Γ(s) n=1 0
(19)
Using this representation, we find
ζB (s) =
∞ Z 2 2 2 1 X ∞ dτ τ s−1 e−(π n +a )τ . Γ(s) n=1 0
(20)
Let us compare this with the analogous representation we would have found for ζA (s): ζA (s) =
∞ Z 2 2 1 X ∞ dτ τ s−1 e−π n τ . Γ(s) n=1 0
(21)
The difference is the extra term with a2 τ in the exponential. Let us expand that term in a series, ! ∞ Z ∞ X 2 2 1 X ∞ a2m τ m s−1 ζB (s) = dτ τ e−π n τ , (22) Γ(s) n=1 0 m! m=0
and interchange the order of summation and integration: ζB (s) = =
Z ∞ ∞ 2 2 1 X X a2m ∞ dτ τ m+s−1 e−π n τ Γ(s) m=0 n=1 m! 0 ∞ ∞ m 1 X X (−1) a2m Γ(m + s) Γ(s) m=0 n=1 m! (πn)2m+2s
∞ m 1 X (−1) a2m Γ(m + s) = ζ(2m + 2s) Γ(s) m=0 m! π 2m+2s
=
∞ 1 1 X (−1)m a2m Γ(m + s) ζ(2s) + ζ(2m + 2s). π 2s Γ(s) m=1 m! π 2m+2s
In this way, we have reduced ζB (s) to an infinite power series in a with coefficients involving ζ(s). In principle, this allows us to compute ζB (s) numerically in an efficient way, since the function ζ(s) is essentially known and the series converges (for small enough a). 0 However, we are interested only in the value ζB (0). To compute that, we recall that sΓ(s) = Γ(s + 1) and hence 1 d 1 s 1 1 = 0, = ⇒ = = 1. (23) Γ(s) Γ(s + 1) Γ(s) s=0 ds s=0 Γ(s) Γ(1)
Then, for bounded f (s) we have
1 d f (s) = f (0), ds s=0 Γ(s) 3
(24)
and so we find 0 ζB (0) = − ln 2 +
Using Eq. (3), we find
∞ m X (−1) a2m ζ(2m) . mπ 2m m=1
0 ˆ = −ζB ln det B (0) = ln 2 +
∞ 2m X (2a) B2m . 2m (2m)! m=1
(25)
(26)
There is a known expansion involving the Bernoulli numbers,
Z
∞ 2m−1 X 1 (2x) 1 = +2 B2m , tanh x x (2m)! m=1
∞ 2m X (2x) dx = ln sinh x = ln x + B2m . tanh x 2m (2m)! m=1
Thus 0 ˆ = −ζB ln det B (0) = ln 2 +
∞ X (2a)2m B2m = ln(e2a − e−2a ). 2m (2m)! m=1
(27) (28)
(29)
4 Further generalizations The success of the calculation in the previous section hinges on the possibility to expand ζB (s) as a series involving ζ(s). Hence, the method can be generalized to operators Zˆ with spectra of the form λn = a0 + a1 n + a2 n2 + ... + ar nr , (30) or even
λn = a1 nα1 + a2 nα2 + ... + ar nαr ,
(31)
where aj and αj are arbitrary real coefficients, and a finite number r of terms is assumed. For instance, consider an operator B with a spectrum λn = a + bnβ . Following the previous calculations, we find ∞ Z 1 X ∞ s−1 −at−bnβ t t e dt ζA (s) = Γ(s) n=1 0 ∞ ∞ Z m 1 X X ∞ m+s−1 (−a) −bnβ t = t e dt Γ(s) n=1 m=0 0 m!
=
∞ ∞ ∞ X 1 X X Γ(m + s) (−a)m Γ(m + s)b−s a m = ζ (βm + βs) . − m+s β(m+s) Γ(s) n=1 m=0 b m! Γ(s)Γ(m + 1) b n m=0
(32)
Note that the resulting series diverges if β < 0 but converges if β > 0 and |a/b| < 1. The derivative of the zeta function is found by splitting of the term m = 0, as before: ζA (s) = b−s ζ(βs) + 0 ζA (0) =
∞ b−s X Γ(m + s) a m ζ (βm + βs) , − Γ(s) m=1 Γ(m + 1) b
∞ X 1 1 a m β ln b + ζ (βm) . − 2 m b m=1
Again, this series converges if |a/b| < 1 and β > 0.
4
(33) (34)
In the case β < 0, a different expansion can be used. As an example, consider the case when an operator A has the spectrum λn = a + bn−α , α > 0. (35) Following the previous calculations, we find ζA (s) =
∞ Z 1 X ∞ s−1 −at−bn−α t t e dt. Γ(s) n=1 0
(36)
Let us now expand the factor exp (−bn−α t) rather than the factor exp (−at): ∞ ∞ Z 1 X X ∞ m+s−1 −at (−b)m ζA (s) = t e dt Γ(s) n=1 m=0 0 m!nαm m ∞ b a−s X Γ(m + s) ζ(αm) − = Γ(s) m=0 m! a m ∞ 1 b a−s X Γ(m + s) ζ(αm). = a−s + − 2 Γ(s) m=1 m! a
This series converges if |b/a| < 1 and α > 0. The derivative of the zeta function is 0 ζA (0)
m ∞ X 1 b 1 = − ln a + − ζ(αm). 2 m a m=1
(37)
Finally, let us consider the curious case of the spectrum of the Hamiltonian of the electron in a Coulomb potential: λn = −αn−2 , α > 0, (38) where the eigenvalue λn has multiplicity 2n2 (including the spin degeneracy). We find ζH (s) =
∞ X
∞ X 2n2 −s −s = 2 (−α) n2s+2 = 2 (−α) ζ(−2s − 2), −2 )s (−αn n=1 n=1
0 ζH (0) = −2ζ(−2) ln (−α) − 4ζ 0 (−2).
(39) (40)
The value ζ 0 (−2) ≈ 0.03 is a transcendental number that can be evaluated numerically, while ζ(−2) = 0. The determinant is then 0 det H = exp (−ζH (0)) = exp (4ζ 0 (−2)) ≈ 1.13.
(41)
The curious result is that the value of the determinant is independent of the Rydberg constant α. More generally, if an operator H has eigenvalues of the form λn = Enα , n = 1, 2, ..., with polynomial degeneracies gn = c1 n2 + ... + ck n2k , containing only even powers of n, then the determinant of H is independent of the scale parameter E. (Thanks are due to Matthew Parry for communicating this statement to the author.)
References [1] H. Bateman and A. Erdélyi, Higher transcendental functions, vol.1, McGraw-Hill, New York, 1953. [2] N. Straumann, “Cosmological phase transitions” (Zuoz lecture notes, 2004), preprint astroph/0409042.
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