# Using Fredholm theory to evaluate a ratio of operator determinants

This is part of a proof from Sidney Coleman's "Aspects of Symmetry," page 340.

$$(\partial_{t}^2 - W(t))\psi = \lambda \psi$$

where W is a bounded function of time, and the operator acts on the space of functions vanishing at $\pm T/2$. We define the determinant of the operator as the product of its eigenvalues:

$$\det(\partial_{t}^2 - W) = \prod \lambda_n$$.

Now, the ratio

$$\frac{ \det\left({\partial_{t}^2 - W^{(1)} - \lambda}\right)}{ \det\left({\partial_{t}^2 - W^{(2)} - \lambda}\right)}$$

is a meromorphic function of $\lambda$ with a simple zero at each $\lambda_n^{(1)}$ and a simple pole at each $\lambda_n^{(2)}$. The proof then claims that by elementary Fredholm theory, the ratio goes to one as $\lambda \to \infty$. What does Coleman mean here by Fredholm theory? Most of what I have found on Fredholm is about finding solutions to certain integral equations, which doesn't seem relevant here.

I am open to either proofs of this statement using the relevant theory, or reference suggestions for where to learn about the appropriate theory. Thanks!

• This cannot be right without putting some constraint on how $\lambda$ goes to infinity, since the $\lambda_n^{(1)}$ and $\lambda_n^{(2)}$ presumably go to infinity as well (the operator is unbounded!). – Michael Renardy Jul 27 '17 at 22:42
• The claim is that $\prod (\lambda + \lambda^{(1)}_j)/(\lambda + \lambda^{(2)}_j)\to 1$, and for this one needs to know the asymptotics of $\lambda_j$. I assume you consider operators on a bounded interval, and then this topic has been well studied. It has nothing to do with Fredholm theory. – Christian Remling Jul 27 '17 at 22:43
• Also, of course your "definition" of $\det (d^2/dt^2-W)$ is a very badly divergent product; only the ratio can possibly make sense. – Christian Remling Jul 27 '17 at 22:47
• @MichaelRenardy: In fact, $\lambda_j\to-\infty$. The OP is certainly suppressing assumptions, a general Schrodinger operator need not have any eigenvalues at all. – Christian Remling Jul 27 '17 at 22:49
• @MichaelRenardy yes, sorry, I forgot to mention that the operator is indeed bounded. – Shayne Jul 27 '17 at 22:51

As I already mentioned in a comment, the product "defining" $\det(D^2-W)$ is badly divergent, but we can consider the ratio, interpreted as $\prod\frac{\lambda-\lambda_n(W_1)}{\lambda-\lambda_n(W_2)}$.
Then this is quite easy for bounded potentials $W_j$: the general asymptotics of the eigenvalues are $|\lambda_n(W) + n^2\pi^2/T^2|\le \|W\|_{\infty}$. This follows by just comparing them with those of $W=0$.
Thus $$\log \prod \frac{\lambda-\lambda_n(W_1)}{\lambda-\lambda_n(W_2)} = \sum \log \frac{\lambda-\lambda_n(W_1)}{\lambda-\lambda_n(W_2)} = \sum O(1/(\lambda+n^2))$$ indeed goes to $0$, by dominated convergence.
For more general $W$, this would require more precise asymptotic formulae.