A proof of Gel'fand-Yaglom theorem I am trying to understand the Gel'fand-Yaglom theorem.
The Gel'fand-Yaglom theorem is the following.
Let us consider the eigenvalue problem of the operator $-\partial^2+W(t)$ with the eigenfunctions of Dirichlet boundary conditions
$$\psi(-T)=0,\ \psi(T)=0.$$
Then let $\psi_\lambda^1(t)$, $\psi_\lambda^2(t)$ be the solutions to
$$[-\partial^2+W_1(t)]\psi_\lambda^1(t)=\lambda\psi_\lambda^1(t);$$
$$[-\partial^2+W_2(t)]\psi_\lambda^2(t)=\lambda\psi_\lambda^2(t);$$
with the conditions 
$$\psi_\lambda^1(-T)=0,\ {\psi_\lambda^1}'(-T)=1,$$
and similar for $\psi_\lambda^2(t)$.
Now we have the statement:
$$\frac{\det(-\partial^2+W_1(t)-\lambda)}{\det(-\partial^2+W_2(t)-\lambda)}=\frac{\psi_1(T)}{\psi_2(T)}.\tag{1},$$
where the determinant of a differential operator is defined as
$$\det(-\partial^2+W(t))=\prod_n\lambda_n$$
with $\lambda_n$ being the eigenvalues of $-\partial^2+W(t)$.
Coleman gave a very elegant proof.

$\psi_\lambda^1$, $\psi_\lambda^2$ are the eigenfunctions when and only when $\psi_\lambda^1(T)=0$, $\psi_\lambda^2(T)=0$. The left-hand side of Eq.(1) is a meromorphic function of $\lambda$, with a simple zero at each eigenvalues $\lambda_n^1$ and a simple pole at each $\lambda_n^2$. By elementary Fredholm theory, it goes to one as $\lambda$ goes to infinity in any direction except along the positive real axis. The right-hand side is a meromorphic function with exactly the same zeros and poles. By elementary differential-equation theory, it also goes to one in the same limit. Thus the ratio of two sides is analytic function of $\lambda$ that goes to one as $\lambda$ goes to infinity in any direction except along the positive real axis. That is, the ratio of two sides is one.

What I am confused is about the role of the condition ${\psi_\lambda^1}'(-T)=1$ and ${\psi_\lambda^2}'(-T)=1$. At which step does this condition play its role in the above proof. What I guess is the step that "the right-hand side of Eq.(1) goes to one as $\lambda$ goes to infinity". But I think the initial condition 
$$\lim_{t\rightarrow -T}\frac{\psi_\lambda^1(t)}{\psi_\lambda^2(t)}=1$$
suffices. 
That is, we actually need ${\psi_\lambda^1}'(-T)={\psi_\lambda^2}'(-T)$ and both derivatives does not need to be 1. Actually, this condition seems to be better because $\psi_\lambda$ is the eigenfunctions when and only when $\psi_\lambda(-T)=\psi_\lambda(T)=0$ and there is no requirement for its derivatives at $-T$. Is this opinion right? 
 A: The condition
$$
\psi'(-T)=1
$$
can be replaced by any non-zero constant and will not affect the theorem.
In general, $\psi_\lambda(T)$ is not $0$ except $\lambda=\lambda_n$. This can be proved by considering
$$
\left(\partial^2+W(t)\right)\psi_n(t)=\lambda_n\psi_n(t)
$$
with Dirichlet boundary conditions and
$$
\left(\partial^2+W(t)\right)\psi^1_\lambda(t)=\lambda\psi_\lambda^1(t).
$$
with Jacobi field conditions. Cancel the potential, we have
$$
\partial(\psi_n\partial\psi^1_\lambda-\psi^1_\lambda\partial\psi_n)=\psi_n\partial^2\psi^1_\lambda-\psi^1_\lambda\partial^2\psi_n=(\lambda-\lambda_n)\psi_n\psi^1_\lambda.
$$
Integrate it:
$$
\psi^1_\lambda(T)\partial\psi_n(T)=(\lambda-\lambda_n)\int_{-T}^{T}\mathrm{d}t\psi_n\psi^1_\lambda.
$$
This locates the zeros of $\psi_\lambda^1(T)$.
The limit $\lambda\rightarrow\infty$ makes the potential $W_1$ and $W_2$ unimportant, thus $\psi^1$ and $\psi^2$ satisfy the same equation. And
$$
\lim_{\lambda\rightarrow\infty}\frac{\psi^1_\lambda(T)}{\psi^2_\lambda(T)}=1.
$$
