Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
    
Can you please clarify the question? Do you mean, can you discretize the Schrordinger eigenvalue problem to a finite dimensional system, and then find the eigenvalues E of the finite dimensional matrix M using det(M-EI)=0, and recover the correct levels E in the limit that the dimension of the system goes to infinity? –  Ron Maimon Jan 3 '12 at 8:37
    
That would be the best possible answer, but I could settle for less. I'll try to rephrase the question, too. –  joseph f. johnson Jan 3 '12 at 8:50
    
Then your question is not about the determinant at all, but about regulating the Schrodinger operators computationally. This is very easy to do, it is normal in computational studies, but I am not sure how much is rigorously known about the continuum limit. –  Ron Maimon Jan 3 '12 at 12:35
    
I don't know why you say "regularizing". I gave you the "regularized" form below. I think you mean something different, "regularization" means putting the theory on a lattice, or breaking the continuum limit some other way, like doing an infinite sum with additional factors. You are asking whether the determinants converges after appropriate rescaling in the large L small $\epsilon$ limit to a unique function of $\lambda$, and it might, for the appropriate class of potentials. Zeta-function regularization is often used. Convergence isn't necessary for PI. –  Ron Maimon Jan 4 '12 at 18:12
1  
I don't know whether this makes a difference, except in grammar, but I did not say « regularising », I said « regularisation », and that is because that is the normal word for altering the definition of a divergent function (or integral) to make it convergent yet still agree with the original if the original were convergent. E.g. zeta function regularisation of the determinant. Lattice methods are one method of regularisation but not the only one. AFAIK, the zeta function regularisation doesnt work for the characteristic poly, and even if it did, wouldnt have the eigenvalues as roots. –  joseph f. johnson Jan 4 '12 at 22:12

2 Answers 2

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 eigevectors, 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 $\epsilon\rightarrow 0$, $L\rightarrow\infty$.

share|improve this answer
    
You mean $det(H_L - \lambda I)$. This is a finite degree polynomial in $\lambda$ whose eigenvalues are those of $H_L$. –  Ron Maimon Jan 3 '12 at 23:16
    
you seem to be saying that it is the set of zeroes which has a limit? Not the polynomial? –  joseph f. johnson Jan 3 '12 at 23:55
    
@Joseph: yes--- it is not obvious that the polynomial is converging, but for sure the zeros are. But when you get a certain set of zeros, you can write an Euler product formula and make an analytic function which has these zeros, and perhaps this gives a unique correct continuum notion of infinite dimensional determinant, I am not sure. I always think of it regulated. –  Ron Maimon Jan 4 '12 at 7:06
    
It seems, then, that the answer to my question is « no », but you are not sure. –  joseph f. johnson Jan 4 '12 at 8:27
    
@Joseph: The answer seems to be yes, I didn't work out the limit of the polynomial, but an analytic function is usually specified by its infinite set of zeros and some additional constraints, like a polynomial of infinite degree. I am not sure under what conditions the convergence is guaranteed, and convergence of the determinant is not necessary for any of the physics results, but the mathematical topic is Fredholm theory. –  Ron Maimon Jan 4 '12 at 8:39

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?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.