# 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.

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.

• Many thanks. I would love though to see how the first integral could be solved by the couplig matrix method. If you pointed me to a reference I would be grateful. – Smerdjakov May 5 '17 at 9:33
• @user37292 You can rearrange the exponent so that it looks like -(1/2) sum_i,j A_ij y_i y_j + \sum_i J_i y_i + C. Here A is a symmetric (L-1)x(L-1) matrix, J is a vector of length L-1, and C is a constant. The trick is then to do an orthogonal transformation that turns A into a diagonal matrix. This is described in some more detail in this answer. – user8153 May 5 '17 at 21:52