I am trying to understand [this short paper](http://math.caltech.edu/SimonPapers/75.pdf) 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


$$A=-\dfrac{d^2}{dx^2}+V$$


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](http://link.springer.com/book/10.1007%2F978-3-642-66282-9) as justification, but I cannot find anything there that is directly relevant.



Gap-opening perturbations of the periodic Schrödinger operator
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.


