# Statistical Mechanics: “Pressure” in a one-dimensional quantum system

For statistical mechanics homework I have to solve this problem:

Consider a single quantum-mechanical particle in an infinite one-dimensional well of width L. From elementary quantum mechanics, we know that the spectrum of allowed energies is given by

$$E(n) = \frac{n^2 \hbar^2\pi^2}{2mL^2}$$

Where $$n$$ is an integer greater than 0.

Calculate the partition function, and use it to find the internal energy $$U$$, the heat capacity $$c_L$$, and the equation of state $$f(T,P,L) = 0$$, in both the high temperature and low temperature limits.

I started with the low temperature limit; haven't got to the high temperature part yet. Here's what I have so far:

Letting $$a = \frac{\hbar^2\pi^2}{2mL^2}$$, $$\beta = \frac{1}{T}$$ (with $$k_B = 1$$ for tidiness), the partition function is:

$$Z = \sum_n^\infty \exp(-\beta a n^2)$$

I tried getting a closed-form solution to this but it gave me a headache, so let $$\gamma = \exp(-\beta a)$$:

$$Z = \gamma + \gamma^4 + \gamma^9 + \dots$$ $$Z = \gamma(1 + \gamma^3 + \gamma^8 \dots)$$

We want $$\ln{Z}$$:

$$\ln{Z} = \ln{\gamma} + \ln{(1 + \gamma^3 + \gamma^8 + \dots)}$$ $$\ln{Z} = -\beta a + \ln{(1 + \gamma^3 + \gamma^8 + \dots)}$$

In the low temperature limit, $$\gamma$$ is very small, so expanding $$\ln{(1 + \gamma^3)}$$:

$$\ln{Z} = -\beta a + \gamma^3 = -\beta a + \exp(-3 \beta a)$$

The energy is

$$U = - \frac{\partial \ln{Z}}{\partial \beta} = a[1 + 3 \exp(-3 \beta a)]$$

The heat capacity is

$$c_L = \frac{dU}{dT} = 9a^2\beta^2 \exp(-3 \beta a)$$

(there's a $$k_B$$ somewhere in there but that's not a priority right now)

(Hopefully there aren't any errors in the above)

Anyway, all that is just to give context to the part I'm confused about, which is that I don't have an intuition for what "pressure" means in a system like this. In my notes, I have that the pressure is defined by the relation:

$$P = \frac{1}{\beta} \frac{\partial \ln{z}}{\partial V}$$

I can understand how this dimensionally gives a pressure quantity (Energy/Volume) for a 3-D system, but I don't know whether it applies for this system. The size of this system is characterized by a 1-D quantity $$L$$, and naively using

$$P = \frac{1}{\beta} \frac{\partial \ln{z}}{\partial L}$$

gives a quantity with units of Energy/Distance. Is this the correct analogue for "pressure" in this 1-D system? Like a "linear pressure"? Or do I need to substitute $$V = L^{3}$$ and use the first formula with the $$\frac{\partial}{\partial V}$$ in it?

The intensive parameter that partners with (extensive) length in one-dimensional systems is often called "tension" and given symbols like $\mathcal{T}$ or $\tau$ (the examples of such systems used in traditional texts like Zemansky are generally stretched wires and similar thinks).
As such it has dimensions of $\text{energy}/\text{length} = \text{force}$, and generally has a sign convention such that your particle-in-a-box system will have negative tension.