Calculating the correlation function of a vibrating string from its normal modes Given a string of length $L$, its transverse displacement at position $x$ is given by $y(x)$. The string is fixed at both ends, so $y(0) = 0$ and $y(L) = 0$, and $\langle y(x) \rangle = 0$.
The fluctuations of the normal modes are uncorrelated and with energy $1 / 2\beta $, where $\beta = 1/kT$, i.e.
$\langle A_n A_m \rangle = \delta_{nm} \frac{2L}{\beta n^2 \pi^2}$
By expanding $y(x)$ in the normal modes of the string, derive
$\langle y(x_1)y(x_2) \rangle = \frac{x_1(L-x_2)}{\beta L} $ where $x_1 \leq x_2$
So using for the normal modes $y(x) = \sum_n A_n \sin(n\pi x /L)$, then $\langle y(x_1)y(x_2) \rangle$ should go as:
$\langle y(x_1)y(x_2) \rangle = \sum_n \sum_m \int_0^L A_n A_m \sin(n\pi x_1/L) \sin(m\pi x_2/L) P(x) dx$
Where $P(x)$ is the probability distribution of $x$, which I think can be arbitrary.
I'm not clear if there should be integrable terms in the sin functions, but I assumed not since we need to get to $\langle A_n A_m \rangle$. Then the above simplifies to:
$\langle y(x_1)y(x_2) \rangle = \sum_n \sin(n\pi x_1/L) \sin(n\pi x_2/L) \langle A_n^2 \rangle$
But that infinite series just sums to a mess, so I'm not sure.
 A: Note that if we take the second derivative with respect to $x_1$ of the following:
$$S \equiv \sum_{n=1}^\infty \sin\left(\frac{n\pi x_1}{L}\right)\, \sin\left(\frac{n\pi x_2}{L}\right) \, \left(\frac{2L}{\beta n^2\pi^2}\right)$$
we get:
$$\frac{\partial^2 S}{\partial x_1^2} = -\sum_{n=1}^\infty \sin\left(\frac{n\pi x_1}{L}\right)\, \sin\left(\frac{n\pi x_2}{L}\right) \, \left(\frac{2}{\beta L}\right).$$
You should be able to do this sum because it is proportional to a representation of the delta function of $x_1-x_2$ on $(0,L)$:
$$\delta(x-y) = \sum_{n=1}^\infty \frac{2}{ L}\, \sin\left(\frac{n\pi x}{L}\right)\, \sin\left(\frac{n\pi y}{L}\right) $$
Quick proof of that fact:


*

*basis vectors: $\langle x|v_n\rangle = \sqrt{\frac{2}{L}} \sin\left(\frac{n x}{L}\right)$

*identity operator (from completeness of basis): $\mathrm{Id} = \sum_n |v_n\rangle\langle v_n| $

*representation in $x$, $y$: $\langle x | \mathrm{Id} | y \rangle = \sum_{n=1}^\infty \frac{2}{ L}\, \sin\left(\frac{n\pi x}{L}\right)\, \sin\left(\frac{n\pi y}{L}\right) \equiv \delta(x-y)$.


If you extend beyond the interval $(0,L)$ this becomes a Dirac comb.
Notice that this shows that the sum you originally wanted to do is proportional to a Green's function for the second derivative operator. Because $S$ is symmetric under exchange of $x_1\leftrightarrow x_2$, a good choice of basic functions is:
$$S = A + B (x_1 + x_2) + C \frac{|x_1 - x_2|}{2} + Dx_1 x_2.$$
The boundary conditions that fix $A$, $B$, $C$, and $D$ are $S(x_1=0) = 0\ \forall x_2$, $S(x_1 = L) = 0\ \forall x_2$, and $\frac{\partial^2 S}{\partial x_1^2} = -\frac{\pi}{\beta} \delta(x_1 - x_2)$. The last one imposes $C=-\frac{1}{\beta}$. The first one imposes $A = 0$ and $B = \frac{1}{2\beta}$. The second one imposes $D = -\frac{1}{L\beta}$. This means that:
$$S = \frac{1}{2\beta} (x_1 + x_2) - \frac{ |x_1 - x_2|}{2\beta} -\frac{ x_1 x_2}{L\beta}.$$
For $x_1 \le x_2$ this simplifies to:
$$S = \frac{x_1(L - x_2)}{L\beta},$$
as required.
A: Please see:
http://www.claudiug.com/9780444529657/chapter.php?c=3&e=26
This is actually a problem in van Kampen's Stochastic processes in Physics and Chemistry.
