Regularisation of infinite-dimensional determinants

Question

Can a regularisation of the determinant be used to find the eigenvalues of the Hamiltonian in the normal infinite dimensional setting of QM?

Edit: I failed to make myself clear. In finite dimensions, there is a function of $\lambda$ whose roots are the eigenvalues (or inverses of the eigenvalues) of a given operator $U$, namely, the characteristic polynomial $\det (I-\lambda U)$. Is there some way of regularising this determinant to do the same thing in infinite dimensions? In general? Or at least for unitary operators which describe the time evolution of a quantum mechanical system?

link to a related question What does a unitary transformation mean in the context of an evolution equation?

EDIT: Perhaps the question still is not clear. The question was, and still is, ¿if you regularise $\det(I-\lambda U)$ as a complex valued function of $\lambda$, for $U$ a unitary operator, will its zeroes be the values of $\lambda$ such that $I-\lambda U$ fails to be invertible? ¿Has a non-zero kernel? ¿Or does regularising the determinant lose touch with that property of the finite dimensional determinant?

Answer 1

The continuum eigenvalues and eigenvectors of the Schrodinger operator are the limiting low-lying eigenvalues and eigenvectors of the discrete lattice approximations. Given a Schrodinger operator

$$ H= \sum_i A_i \partial_i^2 + V(x_1,....,x_n) $$

Where V is of the appropriate class (smooth is too restrictive--- you can have delta functions too, and random potentials, but I don't know the best possible function class--- it might be any integrable potential, i.e., any potential at all EDIT: of course it can't, as the $-1/r^n$ energy levels run away to be localized on top of the attractive spot. The correct condition on the potential is involved, but you can take it to be continuous for this discussion), you replace the x's by a square lattice of spacing $\epsilon$ and of total size L in each direction with periodic boundaries, replace the $\nabla_i$ by the lattice $\nabla_i$

$$ (H_L \psi) (x) = \sum_i {A_i\over \epsilon^2} (\psi(x_i + \epsilon) - 2\psi(x_i) + \psi(x_{i-1})) + V_L(x) \psi(x) $$

Where $V_L(x)$ is the integral over one lattice volume of the continuum V(x) in an $\epsilon$ box centered at x, and the discrete second derivative is the difference between the forward difference and the backward difference.

Then the approximately smooth low lying eigenvectors of $H_L$ converge to the eigenvalues of H in the continuum limit, and as for the high eigenvectors, who cares, these are lattice artifacts. I am sure that it is possible to prove all this rigorously, although from the physical point of view, if it were not the case, the Schrodinger equation would be physically suspect.

You can see the convergence on a computer, if you simulate a discretized Schrodinger operator. You can prove the convergence of the discrete to continuous propagator relatively easily from the path integral. For the individual eigenvalues and eigenvectors, things will be somewhat more involved. If you want a mathematical proof, I can try to sketch one.

EDIT: Determinant formula

If you look at the eigenvalue equation for the finite dimensional operator $H_L$,

$$det(H_L - \lambda I)$$

you find a finite degree polynomial, whose zeros are the eigenvalues of the equation in the limit $\epsilon\rightarrow 0$, $L\rightarrow\infty$.

Answer 2

As you have already mentioned, there are many ways to understand 'regularization' and it is not very often connected with discrete limit - rather, these are dirty tricks to give a meaning to certain sums/integrals which are clearly divergent. Here, the problem is different - we do not know a priori WHAT should be this divergent object - to have divergence we have to have a limit and we have no limit so far. So, the question is rather about the definitnion than about 'regularization' which might be necessary in later steps.

So, i may suggest a definition - we have an identity for finite dimensional operators (let us assume U unitary) , : $det(I-\lambda U) = exp(Tr(ln(I-\lambda U))$. This is always correct - because $I-\lambda U $ is normal and therefore diagonalizable.

We can expand ln in Taylor series around 1 to obtain (ordinary Taylor series when U is in eigenbasis, no problems with radius of convergence when looked upon U - as modulus of all U's eigenvalues is 1).

$$ det(I-\lambda U) = exp(- \sum_{n=1}^{\infty} \frac{\lambda^n}{n} Tr\, (U^n) ) $$

Now we have an expression explicitly containing a limit and at the same time well defined for U an operator in finite dimensional complex Hilbert space. Note, that the appearance of limit is a side-effect, not intentional. Now, we can ask if this expression makes sense when our space becomes infinite-dimensional. There are theorems that state, that if U is bounded (that is $\exists_{M>0}: \forall_{v \in V}\,\, ||U v||\leq M \, ||v|| $) and trace-class (so that trace always exists and is finite) the above formula is well defined in infinite dimensional case. For unitary operators these requirements boild down to denseness of its range, which will be dense (at least for reasonable hamiltonians generating this unitary trafo). So, the above expression is well defined in infinite-dimensional Hilbert space without nearly any singificant additional hypotheses, nor any 'regularization'. Now, all we have to do is to find zeros of this well-defined 'zeta' function: $$ \zeta(\lambda) = {det(I-\lambda U)} = exp(- \sum_{n=1}^{\infty} \frac{\lambda^n}{n} Tr (U^n) )$$ And, being honest, I haven't got a slightest idea how to do it! However, I am quite sure noone has ever prooven it cannot be done :) . I believe it would be not too difficult to start by proving that all zeros lie on a unit circle (c'mon, we all knew that from the beginning!). Unfortunately i have no time nor ideas to deal with it now. Somebody?

Answer 3

Finding the roots of a polynomial of finite and infinite degree is different. If the roots of a polynomial of a finite degree always exist, then for an infinite degree this is not the case, the example $$exp(\lambda)=\sum_{n=0}^{\infty}\frac{\lambda^n}{n!}=0$$ does not have exact finite roots