I am struggling to follow a calculation presented in the paper Statistical Mechanics of one-dimensional Ginzburg-Landau fields. An analogous calculation is presented inthe thesis A Study of the Transfer Matrix Method for the Classical Statistical Mechanics of One Dimensional Systems , openly accessible, please refer to equation 19.
A solution to the following integral eigenvalue equation is to be found (I use the notation of the first paper, removing constants I believe are inconsequential):
\begin{gather}
\int \mathrm{d}x_i \exp{ [ -\beta \Delta x f(x_{i+1},x_i) ] } \Psi_n (x_i)
= \exp{[-\beta \Delta x \epsilon_n]} \Psi_n (x_{i+1})
\end{gather}
where
$$ f(x_{i+1}, x_i) = a |x_{i+1}|^2 + b |x_{i+1}|^4 + c \Big| \frac{x_{i+1} – x_i}{\Delta x} \Big| ^2 $$
The left-hand side is re-written by performing a Taylor expansion
\begin{gather}
\int \mathrm{d}x_i \exp{ [ -\beta \Delta x f(x_{i+1},x_i) ] } \big[ \Psi_n (x_{i+1} ) +
(x_{i} -x_{i+1} ) {\Psi} ^{\prime}(x_{i+1}) + \\
\frac{1}{2} (x_{i} -x_{i+1} )^2
{\Psi}^{\prime \prime} (x_{i+1}) +\dots
\big] \\
= \exp{ [-\beta \Delta x \big( a |x_{i+1}|^2 + b |x_{i+1}|^4 \big)
\\
\times (1+\frac{1}{4}\frac{\Delta x}{\beta} \frac{\partial ^2}{\partial x^2_{i=1}} ) \Psi_n (x_{i+1}) ]}
\label{taylor}
\end{gather}
Incidentally, I cannot even reproduce this result as I cannot understand where the factor $\sqrt {\pi}$ from the integral
\begin{equation}
\int_{-\infty}^{\infty} \exp \big(-\frac{x^2}{a}\big) x^2 \mathrm{d}x = \frac{1}{2} \sqrt {\pi} a^{3/2}
\end{equation}
ends up, but that is not my main problem right now.
Going back to the Taylor expansion of the LHS of the integral eigenvalue equation, now a puzzling step is made.
The authors state "formally, the derivative term can be exponentiated", getting to
$$ \exp{[-\beta \Delta x H] } \Psi_n = \exp{[-\beta \Delta x \epsilon_n]} \Psi_n $$
where
$$ H = -\frac{1}{4} \frac{1}{\beta^2} \frac{\partial ^2}{\partial x^2_{i+1}} + a |x_{i+1}|^2 + b |x_{i+1}|^4 $$
I do not understand at all. In the second reference I linked, the step is given for granted.
I have read about exponentiating the derivative operator, as in the definition
\begin{gather}
\exp {[D]} = \sum_{i=0}^{\infty} \frac{D^i}{i!}
\end{gather}
but how that applies to the calculation above, I am not so sure I follow. Ok I see that only even terms contribute to the integral, and $ \exp {[D^2]}$ would exactly pick those. Yet it seems a bit of a liberty to take, I would appreciate it if anybody would clarify why the calculation is rigorously possible.
On a third source The Frenkel Kontorova model I found an equally puzzling calculation, again involving exponentiting operators.
Given the integral eigenvalue equation
$$ \int_{-\infty}^{\infty} K(u,u') \Psi_n (u') \mathrm{d}u' = \lambda_n \Psi_n (u) $$
the kernel $K$ so defined
$$ K(u,u') = \exp{ \Big\{ -\frac{1}{2} \beta [ V(u) + V(u') + g(u-u')^2 ] } \Big\}$$
it is claimed that using the operator identity
$$ \int_{-\infty}^{\infty} \mathrm{d}y \exp [-b(x-y)^2] f(y) = (\frac{\pi}{b})^{\frac{1}{2}} \exp \Big( \frac{1}{4b} \frac{\mathrm{d}^2}{\mathrm{d}x^2} \Big) f(x)
$$
the following holds, (why?)
$$\exp{\big[ -\frac{\beta}{2} V(x) \big]} \exp {(\frac{1}{2 \beta g} \frac{\mathrm{d}^2}{\mathrm{d} x^2})} \exp{\big[ – \frac{\beta}{2} V(x) \big]} \Psi_n (x) = \lambda_n \Psi(x) $$
and "combining three exponentials … into a single one"(??)
$$ \exp{ \Big( \frac{1}{2 \beta g} \frac{\mathrm{d}^2}{\mathrm{d} x^2} -\beta V(x) – \beta W \Big)} \Psi_n (x) = \lambda_n \Psi(x)
$$
where $W$ is to be defined by Taylor expanding the three exponents (also unclear, if it is allowed to use the property of exponentials for an operator too, where would $W$ come from?).
Would be grateful if anybody shared a hint on both these calculations.
EDIT – CHECKING ON A SIMPLER CASE
I thought I would check on a simpler case, maybe it could help clarifying what I am missing.
I will try to solve the integral equation
\begin{gather}
\int \mathrm{d}x_i \exp{ [ -\beta g(x_{i+1},x_i) ] } \Psi_n (x_i)
= \exp{[-\beta \epsilon_n]} \Psi_n (x_{i+1})
\end{gather}
where
$$ g(x_{i+1}, x_i) = ( x_{i+1} – x_i) ^2 $$
Following the method sketched above and detailed in the linked sources, the integral eigenvalue problem is equivalent to the differential equation
$$ \Psi ^{\prime \prime} = \epsilon_n \Psi$$
with solutions, given the boundary conditions considered in the references, $\Psi (0) = \Psi (\pi) = 0$
$$ \Psi (x) = \sin (kx) $$
with $k=0,1,2, \dots, $ and $\epsilon_n = k_n^2$
is this correct?
Actually I could check that
\begin{gather}
\int \mathrm{d}x \exp{ [ -\beta (y-x)^2 ] } \Im [{\exp(ikx)}]
= \sqrt \frac{\pi}{\beta} \Im [{\exp(ikx)}]
\end{gather}
so $\sin(kx)$ seems to be indeed an eigenfunction of the integral operator, as arrived at by the method and the ordinary differential equation. I have no doubt made some mess with constants.
I am not so sure the eigenvalues are correct though.
On top of which, the question would anyhow remain, what is the basis of such "operator exponentiation" calculation.
Best Answer
answering " clarify why the calculation is rigorously possible."
A bit formally, but here's an argument: We want to show that $\int da e^{-\beta a^2}F(a)\sim e^{\nabla^2 \over \beta}F(0) $
lets start by writing the first integral as a Fourier integral:
$\int da e^{-\beta a^2}F(a)=\int da e^{-\beta a^2}\int dk e^{ika}\hat F_k$ (up to factors of $\pi$ etc)
Exchanging limits and completing the square:
$\int dk \int da e^{-\beta (a - i{k\over 2\beta})^2-{k^2\over 4\beta}}\hat F_k$
The $da$ integral gives us a normalization factor that we can ignore ( $ \int da e^{-\beta (a - i{k\over 2\beta})^2 }= \int da e^{-\beta a^2}={Const\over \sqrt\beta}$), and we are left with:
$\int da e^{-\beta a^2}F(a)\sim \int dk e^{-k^2\over 4\beta}\hat F_k$
Now, lets write $e^{\nabla^2 \over \beta}F(x)=e^{\nabla^2 \over \beta}\int dk e^{ikx}\hat F_k=\int dk e^{\nabla^2 \over \beta} e^{ikx}\hat F_k= \int dk e^{-k^2 \over \beta}e^{ikx}\hat F_k$. We pushed the derivative into the integral since we are differenting under the integral sign.
Setting $x=0$ yields
$\int da e^{-\beta a^2}F(a)\sim \int dke^{-k^2 \over \beta} \hat F_k\sim e^{\nabla^2 \over \beta}F(x)|_{x=0}$