Second Derivative of Eigenvalue Consider the potential $V_t=tx$ and the corresponding Schrödinger operator
$H_t=- \frac{\partial ^2}{\partial x^2}+V_t$ on $L^2([0,R])$ with Neumann or Dirichlet boundary conditions. Let $\lambda_j(t)$ and $u_j(t,x)$ be the eigenvalues and corresponding eigenfunctions. 
I'm reading a paper which uses the formula for the second derivative of the first eigenvalue with respect to t, namely:
$\frac{\partial^2 }{\partial t^2} \lambda_1(t)=-\sum_{j\neq1}(\lambda_j(t)-\lambda_1(t))^{-1}|<u_1(t,x),txu_j(t,x)>|^2$.
They refer to M. Reed and B. Simon, Methods of modern mathematical physics. IV, but I can't find any proof of this formula in there. Plus, I've no idea how to prove this myself. Does anyone know a proof or has an idea how to get there?
Thanks in advance
 A: This is a regular Sturm-Liouville problem. Whether you use Dirichlet or Neumann conditions, there is an infinite set of eigenvalues
$$
               \lambda_{t,1} < \lambda_{t,2} <\cdots < \lambda_{t,n} < \cdots
$$
with corresponding real eigenfunctions $\phi_{j}(x)$ such that
$$
                     \phi_{t,j}''+ tx\phi_{t,j}=\lambda_{t,j}\phi_{t,j},\;\;\;
                                  \int_{0}^{R}\phi_{t,j}^{2}\,dx = 1.
$$
The eigenspaces are one-dimensional for each $\lambda_{t,j}$; so $\phi_{t,j}$ is unique up to a multiplicative factor of $\pm 1$ if you choose $\phi_{t,j}$ to be real. These eigenfunctions $\{ \phi_{t,n}\}_{n=1}^{\infty}$ form a complete orthonormal basis of $L^{2}[0,R]$, which guarantees that $f = \sum_{n=1}^{\infty}(f,\phi_{t,n})\phi_{t,n}$ converges in $L^{2}[0,R]$ for every $f \in L^{2}[0,R]$. The resolvent $R_{t}(\lambda)=(\lambda I-H_{t})^{-1}$ is a common way to study perturbations:
$$
               R_{t}(\lambda)f =\sum_{n=1}^{\infty}\frac{(f,\phi_{t,n})}{\lambda-\lambda_{n}}\phi_{t,n}.
$$
It's easy to verify that $(\lambda I-H_{t})R_{t}(\lambda)f=f$. The resolvent $R_{t}(\lambda)f$ has simple poles (or removable singularities) at every eigenvalue of $H_{t}$. If $C_{t,n}$ is a circular contour centered at $\lambda_{t,n}$ and not enclosing any other eigenvalue $\lambda_{t,k}$, then
$$
   \frac{1}{2\pi i}\oint_{C_{t,n}}\lambda R_{t}(\lambda)f\,d\lambda = \lambda_{t,n}(f,\phi_{t,n})\phi_{t,n},\;\;\;
   \frac{1}{2\pi i}\oint_{C_{t,n}}R_{t}(\lambda)f\,d\lambda = (f,\phi_{t,n})\phi_{t,n}.
$$
If $(f,\phi_{t,n})\ne 0$, this leads to an expression for $\lambda_{t,n}$ which is independent of $f$:
$$
 \lambda_{t,n} 
    =\frac{\frac{1}{2\pi i}\oint_{C_{t,n}}\lambda(R_{t,n}(\lambda)f,f)\,d\lambda }
          {\frac{1}{2\pi i}\oint_{C_{t,n}}(R_{t,n}(\lambda)f,f)\,d\lambda }.
$$
Let $d_{t}(\lambda)$ be the distance from $\lambda$ to the set of eigenvalues $\{ \lambda_{t,1},\lambda_{t,2},\lambda_{t,3},\cdots\}$. Then
$$
  \|R_{t,n}(\lambda)f\|^{2}
    =  \sum_{n=1}^{\infty}\frac{|(f,\phi_{t,n})|^{2}}{|\lambda-\lambda_{t,n}|^{2}}
   \le \frac{1}{d_{t}(\lambda)^{2}}\sum_{n=1}^{\infty}|(f,\phi_{t,n})|^{2} = \frac{1}{d_{t}(\lambda)^{2}}\|f\|^{2}.
$$
Therefore, $\|R_{t}(\lambda)\| \le 1/d_{t}(\lambda)$. It is this estimate which allows a straightforward perturbation expansion. If $A$ is a bounded linear operator on a Hilbert space, and $\|A\| < 1$, then $I-A$ can be shown to be invertible by using a geometric expansion
$$
       (I-A)^{-1} = I + A + A^{2} + A^{3} + \cdots\;.
$$
To see how to apply this, let $C_{t,n}$ be the circle of radius $r_{t,n}/2$, where $r_{t,1}$ is $|\lambda_{t,1}-\lambda_{t,2}|$ and is $\min\{|\lambda_{t,n}-\lambda_{t,n-1}|,|\lambda_{t,n}-\lambda_{t,n+1}|\}$ otherwise.  For $\lambda$ on $C_{t,n}$, we have $\|R_{t}(\lambda)\| \le 2/r_{t,n}$. We want to perturb $H_{t}$ to obtain $H_{t+\epsilon}$, and choose $\delta > 0$ small enough that $\lambda I-H_{t+\epsilon}$ remains invertible for $\lambda$ on $C_{t,n}$ for all $\epsilon \in [-\delta,\delta]$. This will guarantee that there is exactly one eigenvalue inside $C_{t,n}$ for $H_{t+\epsilon}$, and we'll be able to find a single quotient of contour integrals on $C_{t,n}$ which yields $\lambda_{t+\epsilon,n}$ for all $|\epsilon| \le \delta$. Let $M_{x}$ be the operator of multiplication by $x$ on $L^{2}[0,R]$. Then $\|M_{x}\|\le R$ follows trivially. Furthermore,
$$
     (\lambda I-H_{t+\epsilon})R_{t}(\lambda)=\{(\lambda I-H_{t})-\epsilon M_{x}\}R_{t}(\lambda)=I-\epsilon M_{x}R_{t}
$$
is invertible for $\lambda$ on $C_{t,n}$ if $\|\epsilon M_{x}R_{t}(\lambda)\| \le 2|\epsilon|R/r_{t,n} < 1$. And, in that case,
$$
     R_{t+\epsilon}(\lambda)  = R_{t}(\lambda)\{I+\epsilon (M_{x}R_{t})+\epsilon^{2}(M_{x}R_{t})^{2}+\epsilon^{3}(M_{x}R_{t})^{3}+\cdots\}
$$
So $R_{t+\epsilon}(\lambda)$ exists for all $\lambda$ on $C_{t,n}$ and for all $|\epsilon| \le \delta$, provided $\delta$ is chosen so that
$$
                              0 <\delta < r_{t,n}/2R.
$$
If we further restrict $\delta$ so that $\int_{C_{t,n}}(R_{t+\epsilon}(\lambda)\phi_{t,n},\phi_{t,n})\,d\lambda \ne 0$ for $|\epsilon| \le \delta$, then we obtain
$$
 \lambda_{t+\epsilon,n} 
    =\frac{\frac{1}{2\pi i}\oint_{C_{t,n}}\lambda(R_{t+\epsilon,n}(\lambda)\phi_{t,n},\phi_{t,n})\,d\lambda }
          {\frac{1}{2\pi i}\oint_{C_{t,n}}(R_{t+\epsilon,n}(\lambda)\phi_{t,n},\phi_{t,n})\,d\lambda },\;\;\; |\epsilon| \le \delta.
$$
The choice of $f=\phi_{t,n}$ is a convenient one because it (a) guarantees that the denominator is non-zero for $\epsilon=0$ and (b) $R_{t}(\lambda)\phi_{t,n}=\frac{1}{\lambda-\lambda_{t,n}}\phi_{t,n}$ simplifies the perturbation expressions. For example, the second order term that you have expressed concern about involves the inner-product
$$
     \epsilon^{2}(R_{t}(\lambda)M_{x}R_{t}(\lambda)M_{x}R_{t}(\lambda)\phi_{t,n},\phi_{t,n}) = \epsilon^{2}(R_{t}M_{x}R_{t}\phi_{t,n},M_{x}R_{t}^{\star}\phi_{t,n})
$$
$$
\begin{align}
    & = \epsilon^{2}(R_{t}M_{x}\frac{1}{\lambda-\lambda_{t,n}}\phi_{t,n},M_{x}\frac{1}{\overline{\lambda}-\lambda_{t,n}}\phi_{t,n}) \\
    & = \frac{\epsilon^{2}}{(\lambda-\lambda_{t,n})^{2}}(R_{t}(\lambda)M_{x}\phi_{t,n},M_{x}\phi_{t,n}) \\
    & = \frac{\epsilon^{2}}{(\lambda-\lambda_{t,n})^{2}}
         \left(\sum_{k=1}^{\infty}\frac{(M_{x}\phi_{t,n},\phi_{t,k})}{\lambda-\lambda_{t,k}}\phi_{t,k},M_{x}\phi_{t,n}\right) \\
    & = \frac{\epsilon^{2}}{(\lambda-\lambda_{t,n})^{2}}\sum_{k=1}^{\infty}\frac{|(M_{x}\phi_{t,n},\phi_{t,k})|^{2}}{\lambda-\lambda_{t,k}}.
\end{align}
$$
When you integrate this term over $C_{t,n}$, all terms $1/(\lambda-\lambda_{t,k})$ are holmorphic in the curve except for $k=n$; the integral evaluation is the derivative of such terms evaluated at $\lambda_{k}$, which leads to the expression
$$
            -\epsilon^{2}\sum_{k=1,k \ne n}^{\infty}\frac{\int_{0}^{R}x\phi_{t,n}\phi_{t,k}\,dx}{(\lambda_{t,n}-\lambda_{t,k})^{2}}.
$$
So, for example, the denominator has the expansion
$$
    \frac{1}{2\pi i}\oint_{C_{t,n}}(R_{t+\epsilon}(\lambda)\phi_{t,n},\phi_{t,n})\,d\lambda = 1-\epsilon^{2}\sum_{k=1,k \ne n}^{\infty}\frac{\int_{0}^{R}x\phi_{t,n}\phi_{t,k}\,dx}{(\lambda_{t,n}-\lambda_{t,k})^{2}}+O(\epsilon^{3}).
$$
(This first order terms vanish in this case.) For the numerator expression you have an extra multiplicative factor of $\lambda$ in the integral, and that changes things.
