I am trying to understand this short paper and I am getting stuck right at the end.

Let $V(x)$ be $C^\infty$ and 1-periodic (that is, $V(x)=V(x+1)$). We are considering the operator


on $L^2([0,1],dx)$ where $A$ has periodic boundary condition $f'(1)=f'(0)$, $f(1)=f(0)$.

Suppose that for some $n$ even, the $n+1$st eigenvalue of $A$ coincides with the $n$th eigenvalue of $A$. We call that eigenvalue $\hat E$. In other words, there is a multiplicity 2 eigenvalue of $A$ at $\hat E$.

It is known that all solutions $u\in L^2 (\mathbb R, dx)$ to the Schrödinger equation $$-u''(x)+V(x)u(x)=\hat E u(x)$$ are periodic. Let $u_1(x)$ be the solution with initial conditions $u(0)=0, u'(0)=1$ and let $u_2(x)$ be the solution with initial conditions $u(0)=1, u'(0)=0$. It is easy to find a 1-periodic function $W(x)$ in $C^\infty$ that satisfies

$$\int_0^1 W(x) u_1(x)^2 dx\neq \int_0^1 W(x) u_2(x)^2 dx.$$

Let $\lambda$ be a small constant. Supposedly, when we perturb $V$ by $\lambda W$ in the definition of $A$ (that is, replace $V$ with $V+\lambda W$) we can guarantee that for sufficiently small $\lambda$ the $n+1$st eigenvalue of $A$ no longer coincides with the $n$th eigenvalue. Why is that true?

The paper cites Kato's Perturbation Theory for Linear Operators as justification, but I cannot find anything there that is directly relevant.


Here's a slightly different (direct) method that gives similar conclusions: Since we have a double periodic eigenvalue at $\widehat{E}$ (let me in fact assume that $\widehat{E}=0$, for convenience), the transfer matrix $$ T_0(x) = \begin{pmatrix} u_1(x) & u_2(x) \\ u'_1(x) & u'_2(x) \end{pmatrix} $$ satisfies $T_0(1)=1$. Also, $$ T'_0(x) = \begin{pmatrix} 0 & 1 \\ V(x) & 0 \end{pmatrix} T_0(x) , $$ and the perturbed transfer matrix $T$ solves a similar equation, with $V$ replaced by $V+\lambda W$. Let me write $T=T_0Y$; then $Y$ solves $$ Y'(x) = \lambda M(x)Y(x), \quad Y(0)=1, \quad\quad\quad\quad (1) $$ with $$ M(x) = W(x)T_0^{-1}(x)\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}T_0(x) = W\begin{pmatrix} -u_1u_2 & -u_2^2 \\ u_1^2 & u_1u_2 \end{pmatrix} . $$ I now want to solve (1) approximately (for small $\lambda$) by iteration. This gives $$ Y(1) = 1 +\lambda\int_0^1 M(t)\, dt + \lambda^2 \int_0^1 dt\, M(t)\int_0^t ds\, M(s) + O(\lambda^3) . $$ I'm interested in $\textrm{tr}\, T(1)=\textrm{tr}\, Y(1)$. The linear term has trace zero, and then a calculation shows that $$ \textrm{tr}\, Y(1) = 2 + \lambda^2\left[ \left( \int_0^1 Wu_1u_2 \right)^2 - \int_0^1 Wu_1^2 \int_0^1 Wu_2^2 \right] + O(\lambda^3) . $$ I can now choose a $W$ that makes $\int Wu_1^2 = 0$, but $\int Wu_1u_2\not=0$ (this is different from the assumptions you wanted to make). Then $\textrm{tr}\, Y(1)>2$ for all small $\lambda\not=0$, so we're inside an open gap.

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  • $\begingroup$ This works well enough for my purposes. Thanks Christian! $\endgroup$ – Darren Ong Jul 20 '16 at 2:59

The eigenvalues and eigenvectors of $A+\lambda W$ are holomorphic functions of $\lambda$ (Kato p.387 "Continuation of the eigenvalues", notice footnote 1).

Suppose the eigenvalues of $A+\lambda W$ are equal in a neighbourhood of $\lambda = 0$. This is the case of no splitting, so every eigenvector of $A$ has a continuation with the same eigenvalue $E(\lambda)$. In particular, $u_1$ has a continuation $u_1(\lambda)$ so

$$(u_1,(A+\lambda W)u_1(\lambda)) = E(\lambda)(u_1,u_1(\lambda)).$$

By substituting the Taylor series for $u_1(\lambda)$ and $E(\lambda)$ and equating coefficients we find that the the first order coefficient in $E(\lambda)$ is $\frac{(u_1, Wu_1)}{\Vert u_1\Vert^2}$. Similarly we find this is also equal to $\frac{(u_2, Wu_2)}{\Vert u_2\Vert^2}$. Simon argues that $W$ can be chosen such that these are unequal, which contradicts the no splitting assumption. In other words, we can find a $W$ that the splits the eigenvalues.

(This is a bit confusing, since in the case of splitting only a particular pair of linear combinations of $u_1$ and $u_2$ have continuations.)

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  • $\begingroup$ Here we are perturbing a degenerate eigenvalue, so your formula doesn't apply in this form. $\endgroup$ – Christian Remling Oct 22 '16 at 0:45

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