Functional determinant


In functional analysis, a branch of mathematics, it is sometimes possible to generalize the notion of the determinant of a square matrix of finite order to the infinite-dimensional case of a linear operator S mapping a function space V to itself. The corresponding quantity det is called the functional determinant of S.
There are several formulas for the functional determinant. They are all based on the fact that the determinant of a finite matrix is equal to the product of the eigenvalues of the matrix. A mathematically rigorous definition is via the zeta function of the operator,
where tr stands for the functional trace: the determinant is then defined by
where the zeta function in the point s = 0 is defined by analytic continuation. Another possible generalization, often used by physicists when using the Feynman path integral formalism in quantum field theory, uses a functional integration:
This path integral is only well defined up to some divergent multiplicative constant. To give it a rigorous meaning it must be divided by another functional determinant, thus effectively cancelling the problematic 'constants'.
These are now, ostensibly, two different definitions for the functional determinant, one coming from quantum field theory and one coming from spectral theory. Each involves some kind of regularization: in the definition popular in physics, two determinants can only be compared with one another; in mathematics, the zeta function was used. have shown that the results obtained by comparing two functional determinants in the QFT formalism agree with the results obtained by the zeta functional determinant.

Defining formulae

Path integral version

For a positive selfadjoint operator S on a finite-dimensional Euclidean space V, the formula
holds.
The problem is to find a way to make sense of the determinant of an operator S on an infinite dimensional function space. One approach, favored in quantum field theory, in which the function space consists of continuous paths on a closed interval, is to formally attempt to calculate the integral
where V is the function space and the L2 inner product, and the Wiener measure. The basic assumption on S is that it should be selfadjoint, and have discrete spectrum λ1, λ2, λ3… with a corresponding set of eigenfunctions f1, f2, f3… which are complete in L2. This roughly means all functions φ can be written as linear combinations of the functions fi:
Hence the inner product in the exponential can be written as
In the basis of the functions fi, the functional integration reduces to an integration over all basisfunctions. Formally, assuming our intuition from the finite dimensional case carries over into the infinite dimensional setting, the measure should then be equal to
This makes the functional integral a product of Gaussian integrals:
The integrals can then be evaluated, giving
where N is an infinite constant that needs to be dealt with by some regularization procedure. The product of all eigenvalues is equal to the determinant for finite-dimensional spaces, and we formally define this to be the case in our infinite-dimensional case also. This results in the formula
If all quantities converge in an appropriate sense, then the functional determinant can be described as a classical limit. Otherwise, it is necessary to perform some kind of regularization. The most popular of which for computing functional determinants is the zeta function regularization. For instance, this allows for the computation of the determinant of the Laplace and Dirac operators on a Riemannian manifold, using the Minakshisundaram–Pleijel zeta function. Otherwise, it is also possible to consider the quotient of two determinants, making the divergent constants cancel.

Zeta function version

Let S be an elliptic differential operator with smooth coefficients which is positive on functions of compact support. That is, there exists a constant c > 0 such that
for all compactly supported smooth functions φ. Then S has a self-adjoint extension to an operator on L2 with lower bound c. The eigenvalues of S can be arranged in a sequence
Then the zeta function of S is defined by the series:
It is known that ζS has a meromorphic extension to the entire plane. Moreover, although one can define the zeta function in more general situations, the zeta function of an elliptic differential operator is regular at.
Formally, differentiating this series term-by-term gives
and so if the functional determinant is well-defined, then it should be given by
Since the analytic continuation of the zeta function is regular at zero, this can be rigorously adopted as a definition of the determinant.
This kind of Zeta-regularized functional determinant also appears when evaluating sums of the form, integration over 'a' gives which it just can be considered as the logarithm of the determinant for a Harmonic oscillator this last value is just equal to, where is the Hurwitz Zeta function.

Practical example

The infinite potential well

We will compute the determinant of the following operator describing the motion of a quantum mechanical particle in an infinite potential well:
where A is the depth of the potential and L is the length of the well. We will compute this determinant by diagonalizing the operator and multiplying the eigenvalues. So as not to have to bother with the uninteresting divergent constant, we will compute the quotient between the determinants of the operator with depth A and the operator with depth A = 0. The eigenvalues of this potential are equal to
This means that
Now we can use Euler's infinite product representation for the sine function:
from which a similar formula for the hyperbolic sine function can be derived:
Applying this, we find that

Another way for computing the functional determinant

For one-dimensional potentials, a short-cut yielding the functional determinant exists. It is based on consideration of the following expression:
where m is a complex constant. This expression is a meromorphic function of m, having zeros when m equals an eigenvalue of the operator with potential V1 and a pole when m is an eigenvalue of the operator with potential V2. We now consider the functions ψm1 and ψm2 with
obeying the boundary conditions
If we construct the function
which is also a meromorphic function of m, we see that it has exactly the same poles and zeroes as the quotient of determinants we are trying to compute: if m is an eigenvalue of the operator number one, then ψm1 will be an eigenfunction thereof, meaning ψm1 = 0; and analogously for the denominator. By Liouville's theorem, two meromorphic functions with the same zeros and poles must be proportional to one another. In our case, the proportionality constant turns out to be one, and we get
for all values of m. For m = 0 we get

The infinite potential well revisited

The problem in the previous section can be solved more easily with this formalism. The functions ψ0i obey
yielding the following solutions:
This gives the final expression