Partition Function - Simpler computation by changing ensemble I have a question regarding a Derivation presented on the web page http://www.pas.rochester.edu/~stte/phy418S05/hw3.html.
The Problem is to calculate the Partition function for a (discrete) elastic string. The transversal displacement at the nodes $i$ is denoted $y_i$, the Hamiltonian reads
$$H[y_i] = \frac{1}{2} \sum_{i=1}^{L} (y_i - y_{i-1})^2 $$
The string is fixed at then ends, $y_0= 0$ and $y_L= Y$, so the average slope equals $Y/L$.
The partition function reads
$$ Z(Y) = (\prod_{i=1}^L \int_{-\infty}^{\infty} \mathrm{d}y_i) \exp{\frac{-H[y_i]}{kT}} $$
and this could be solved exactly.
But the notes show an easier approach.
A new Ensemble is considered, whereby the average slope can fluctuate too
$$ Z'(\eta) =  \int _{-\infty}^{\infty}
 \mathrm{d}Y Z(Y) \exp{\frac{-\eta Y}{kT}} $$
One is then expected to 
1) compute the free energy
$$ F' (\eta) = -kT ln Z' $$
By then showing that 
2) the average slope equals
$$ <Y> = \frac{\partial{F'(\eta)}}{\partial{ \eta}} $$
one should then be in the position to
3) compute the Legendre transform
$$F(Y) = F'(\eta) - \eta Y $$
Sounds all very good, but I am unable to compute the free energy, step 1).
The author claims this approach to compute $F$ is easier than the direct Partition function Z calculation: yet it seems I still have to integrate over the $Z(Y)$ in the second partition function, must be missing a point here. Maybe some clever integration by parts, or property of the Laplace Transform?
Ok, I could use the fact that in the first ensemble (strings' ends fixed) 
$$ F(Y) = -kT ln Z(Y),$$ but this does not seem that revealing. 
 A: If the string is constrained at $y_L=Y$ then the partition function is
$$Z(Y) = \prod_{i=1}^{L-1} \int_{-\infty}^{\infty}\mathrm{d}y_i \, \exp \left(- \frac{1}{k_{\mathrm{B}}T}\frac{1}{2} \sum_{i=1}^L (y_i - y_{i-1})^2\right) $$ 
This is a multidimensional Gaussian integral that could be calculated directly, but it is not trivial (one would have to compute the eigenvalues of the coupling matrix). However, if one defines the increments $w_i = y_i - y_{i-1}$ then the integral nearly factorizes:
$$Z(Y) = \prod_{i=1}^{L} \int_{-\infty}^{\infty}\mathrm{d}w_i \, \exp \left(- \frac{1}{k_{\mathrm{B}}T}\frac{1}{2} \sum_{i=1}^L w_i^2\right)  \delta \left( \sum_{j=1}^L w_j -Y\right) $$
Here the Dirac delta function is necessary to ensure that all the individual increments add up to $Y$. Here you see the effect of the constraint: it couples all the individual increments together, which makes the evaluation of the partition function difficult (but there is a hint in the problem you linked to).
If the string is not constrained at $y_L$ then the partition function is
$$Z'(\eta) = \prod_{i=1}^{L} \int_{-\infty}^{\infty}\mathrm{d}w_i \, \exp \left(- \frac{1}{k_{\mathrm{B}}T}\frac{1}{2} \sum_{i=1}^L w_i^2\right) 
\exp{\left(-\frac{\eta}{k_{\mathrm{B}}T} \sum_{j=1}^L w_j \right)} $$
In this case the final term is an exponential of the sum of the $w_i$, which factorizes and you can do the integration over each $w_i$ separately. This is easier than computing $Z(Y)$ directly.
