Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This is something which I suspect is written up in introductory books on mathematical physics if I knew where to look. Suppose I have some parameters $t_1$, ..., $t_k$ ranging over a neighborhood in $\mathbb{R}^k$. I also have $k$ matrix-valued functions of the $t$'s: $H_1(t_1, \ldots, t_k)$, ... $H_k(t_1, \ldots, t_k)$. These obey both $$[H_i, H_j]=0 \quad (\ast)$$ and $$[\partial_i+H_i, \partial_j+H_j] =0 \quad (\dagger).$$ For those who don't like the language of connections, we can expand $(\dagger)$ as $\partial H_i/\partial t_j - \partial H_j/\partial t_i + [H_i, H_j]=0$ or, in the presence of $(\ast)$, as $$\frac{\partial H_i}{\partial t_j} = \frac{\partial H_j}{\partial t_i}.$$

Equation $(\ast)$ tells us that, assuming the $H_i$ are individually diagonalizable, we can find $u(t)$ a simultaneous eigenvector for all the $H_i$: $$H_i(t) u(t) = \lambda_i(t) u(t). \quad (\ast\ast)$$

Equation $(\dagger)$ tells us that the vector-valued PDE $$\frac{\partial v}{\partial t_i} + H_i v=0 \quad (\dagger \dagger)$$ will have a unique solution $v(t_1, t_2, \ldots, t_n)$ for any initial value.

I'm pretty sure there is supposed to be a relation between the solutions to $(\ast \ast)$ and $(\dagger \dagger)$. What is the right statement, and what is the keyword to read about this situation?

Motivation: I'm trying to work through the papers of Varchenko, Scherbak and others on the KZ equation. I think it would really clear my head to just see this scenario described abstractly without all the details of which operators they are thinking about.

$\def\mg{\mathfrak{gl}_n}$ Edit to spell out the relation. Let $V_1$, $V_2$, ..., $V_n$ be representations of $\mg$. So $U(\mg)^{\otimes n}$ acts on $V_1 \otimes V_2 \otimes \cdots \otimes V_n$. Let $\Omega \in U(\mg) \otimes U(\mg)$ be the Casimir. (Note: The element I learned to call the Casimir was a central element $c$ in $U(g)$. In terms of that element, $\Omega = \Delta(c) - c \otimes 1 - 1 \otimes c$.) Let $\Omega_{ij}$ be $\Omega$ acting in positions $i$ and $j$.

For generic parameters $z_1$, ..., $z_n$, define $H_i = \sum_{j \neq i} \Omega_{ij}/(z_i-z_j)$. Then, as I understand it, the KZ equation is $(\partial_i + H_i) v(z_1, \ldots, z_n)=0$, where $v$ is a function valued in $V_1 \otimes V_2 \otimes \cdots \otimes V_n$. The $H_i$'s obey both $(\ast)$ and $(\dagger)$ (a nice exercise). And people seem to be very interested in solving both $(\ast \ast)$ "diagonalizing the action of the Gaudin subalgebra" and $(\dagger \dagger)$ "solving the KZ equation". So I was hoping to understand how they relate, and why.

share|improve this question
    
Usually one considers the limit of "connections" to "h" (geometric optics or short wave asymtotics). So sections of flat con. are costructed as series where first term made of eigenvectors . Technically you put constant "k" in front of d/dt and k goes to zero so you can forget about d/dt. This is "cortical level" in KZ story. –  Alexander Chervov Apr 25 '12 at 5:06
    
"Critical level" –  Alexander Chervov Apr 25 '12 at 5:07
    
Thanks! But I'm pretty sure that the interesting stuff in, for example, 1102.5368, 0910.4690 or 1004.3253 are all happening without sending $\hbar$ to $0$. –  David Speyer Apr 25 '12 at 5:22
    
You are welcome. I am pretty agree that there is much interesting stuff around the KZ and the Gaudin model, but still when I was working on this I did not see the way how to constuct solutions of KZ from Gaudin hamiltonians in somewhat "nice"/"explicit". Except one very strange case which we discuss at arxiv.org/abs/0711.2236 page 15 section "4.1.1 Application to the Knizhnik-Zamolodchikov equation" –  Alexander Chervov Apr 25 '12 at 6:20
    
I looked at the papers you mention - still I did not see explicit relation with the question you ask and what is discussed there, may be I was looking not very carefully. –  Alexander Chervov Apr 25 '12 at 6:21

4 Answers 4

up vote 3 down vote accepted

Hi David,

I think there is indeed a relation, which I learned precisely from papers of Varchenko among others. All of this is rather classical and can be found e.g. in Etingof-Frenkel-Kirilov book "Lectures on Representation Theory and Knizhnik-Zamolodchikov Equations".

The fact that the $H_i$ satisfies this stronger condition is equivalent to say that for any parameter $\kappa$ the operators $\kappa \partial_i+H_i$ alos satisfies ($\dagger$).

Hence you can take asymptotic expansion of solutions at $\kappa \rightarrow 0$ on some neighbourhood $D$ of some $z_0$, of the form $$e^{S(z)/\kappa} (f_0(z)+O(\kappa))$$

where $S$ is a scalar valued function. Then you can show that, assuming the $H_i(z)$ are simultaneously digonalizable then $f_0$ is a common eigenvector of them, with eigenvalues $\partial_i S$. Conversly given a common eigenvector at some $z_0$ you can construct an asymptotic solution. So the usual trick, widely used in the study of the KZ equation, is to also take some asymptotic limit w.r.t. the variable $z_i$ in such a way that eigenvectors are "easy" to find. The standard example in the KZ case is the asymptotic zone

$$|z_i-z_1| \ll |z_j -z_1|\quad if\quad i < j $$

for which, up to some change of variable, the equation can be written $$\kappa \partial_i f= \left ( \Omega_i/u_i +reg\right)f\quad i=1\dots n-1$$ where $\Omega_i=\sum_{k < i} \Omega_{k,i+1}$ and $reg$ is regular at $u=0$. Then given some common eigenvector $v$ of the $\Omega_i$ with eigenvalues $\mu_i$ there exists a unique solution of the form $$(\prod u_i^{\mu_i/\kappa})(v+r(u))$$ where $r(u)$ is regular at $u=0$ and $r(0)=0$.

I'm not very familiar with D-modules (and by the way I would be happy is someone extends on this), but you can rephrase it as follows: viewing $\kappa$ as a formal variable leads to a filtration on the algebra of differential operators on $V$ (the vector space acted on by the $H_i$) which in turn is nothing but the usual filtration by the degree of differential operators. Taking the associated graded turns the equation

$$(\partial_i+H_i)f=0$$

into the equation

$$(y_i+H_i)f=0$$

Whose solutions are clearly commons eigenvectors of $H_i$. So I'm rather confident that you can say that the spectrum of the $H_i$ for all common eigenvectors is the characteristic variety of the D-module of solutions of the differential equation you started with.

share|improve this answer

I can't comment on the case of several operators $H_i$, but for a single operator, the eigenvector equation $(**)$

$$ H(τ) \psi_τ = λ(τ) \psi_τ $$

and the time-dependent Schrödinger equation $(\dagger\dagger)$

$$ (i\frac{\partial}{\partial t} - H(t)) ψ(t) = 0 $$

are related by the adiabatic theorem.

Not sure if that's what you are looking for, but I would be very surprised if your setting didn't have a similar intuition.


Essentially, the idea of the adiabatic theorem is the following: the eigenvector equation describes, for each parameter $\tau$, an instantaneous eigenvector $\psi_{\tau}$. This gives a solution $\psi_{\tau}(t) = e^{-itλ(τ)} ψ_τ$ to the "instantaneous" Schrödinger equation

$$ (i\frac{\partial}{\partial t} - H(τ)) ψ_τ(t) = 0 $$

where the Hamiltonian $H$ is considered at a fixed time $τ$.

Now, if the Hamiltonian $H(t)$ varies very "slowly" in time, then it is reasonable to expect that the full Schrödinger equation will essentially follow the solutions to the "instantaneous" Schrödinger equation(s). First it evolves like a solution of the instantaneous equation with $H(0)$, then for $H(\Delta t)$ a small time step after, and so on.

This can be made precise by rescaling time to $τ=t/T$ and obtaining an asymptotic expansion

$$ ψ(t) = e^{-i∫λ(τ)dt} ψ_τ + \mathcal O(1/T) $$

in the limit $T\to ∞$ and in the $L^2$ sense. More details can be found wherever you can find details about the adiabatic theorem.

share|improve this answer

Without further assumptions, it does not seem that much can be said. Consider the case k=1. You are asking for a connection between the eigenvalue problem for H(t) and the equation dv/dt+Hv=0. But time-dependent linear ODE systems cannot in general be related to the eigenvalue problem.

On the other hand, the condition $\partial H_i/\partial t_j=\partial H_j/\partial t_i$ implies that $H_i=\partial K/\partial t_i$ for some $K$. Let us now strengthen your assumptions and assume that the $H_i$ commute not only with each other, but also with $K$. Then solutions of ($\dagger\dagger$) can be written as $\exp(-K(t))w$ for fixed $w$, and solutions of ($**$) can be written as $u=\partial v/\partial t_i$, where $v$ is an eigenfunction of $K$.

share|improve this answer

To expand on Greg's answer regarding the adiabatic theorem. You are looking for situations where the adiabatic evolution is exact. This is the case for a Hamiltonian of the form

$H = i\left[\frac{\partial P}{\partial t},P\right]$

where $P$ is a projector onto your chosen instantaneous eigenstate. This comes from T, Kato, J. Phys. Soc. J. Jpn. 5, 435 (1950).

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.