Schroeder's Thermal Physics: Multiplicity of a Single Ideal Gas Molecule

Question

I'm reading Daniel Schroeder's An Introduction to Thermal Physics. Please see the below pictures for the text from which I have questions:

1. On pp. 69, I don't get the underlined text. What does he mean by there being a finite number of independent wave functions if both position and momentum space are limited. For, say, a particle in a box, all the wave functions of the TISE are sinusoidal of the form - $\sin \Bigg ( \displaystyle \frac{2 \pi n x}{\displaystyle L} \Bigg )$. They're all orthogonal but the set of all such functions is countably infinite, no?
2. More Importantly I am not able to understand his description of the number of distinct position states and how he uses it to calculate the multiplicity of a single monoatomic ideal gas in a container at fixed energy and volume. I don't see the rationale/reasoning behind. This derivation proceeds to the first few lines of pp. 70 as well.

Please ignore the highlighted text toward the end of pp. 70.

Thanks.

Answer 1

We need to count the number of linearly independent wavefunctions available to a particle. With the constraints that the region is finite and its energy is limited, this number of states is always a finite number.

Despite this, there are many different sets of linearly independent wavefunctions we can operate with. So choose the ones with definite energy values.

Kinetic energy depends on the square of the momentum. The wavefunctions must satisfy the boundary conditions, that is, in a 1 dimensional box, they must go to zero at both ends. So only certain wavelengths are permitted, with discrete values 2L, 2L/2, 2L/3....

$E_n$ = $h^2$$n^2$/8m$L^2$ where n is any positive integer.

Any other wavefunction can be written as a linear combination of definite energy wavefunctions. These definite energy wavefunctions are linearly independent.

So counting the number of definite energy wavefunctions is a way to count "all" the states in the box.

Inside a 3 D box, we multiply 3 1 dimensional definite energy wavefunctions to create a 3 Dimensional definite energy wavefunction.

$\psi$(x, y, z) = $\psi_x (x)$$\psi_y (y)$$\psi_z$(z)

These products aren't all the definite energy wavefunctions, but the others can be written as linear combinations of these.

If the box is a cube we have

E = $h^2$/8m$L^2$$[n^2$$_x$ + $n^2$$_y$ + $n^2$$_z$]

Most of the energy levels are degenerate, which corresponds to multiple linearly independent states that must be counted separately. The number of linearly independent states that have a given energy is known as the degeneracy of the level. That is, (n-fold) degeneracy.

You have a certain number of distinct position states, each of which can have a certain number of distinct momentum states associated with it, so the total number of distinct states is the product of the two.

So a molecule could be at position X = 5, with momentum P = 7, and so on. At least that's how I read it until I look up a few books. He then extends this idea to 3D.

Answer 2

Concerning (1): The states you wrote have energy $\frac{\hbar^2}{2m}(\frac{2\pi n}{L})^2$ and in this case there is indeed infinitely many linearly independent wavefunctions.

In the book, the author begins by considering a single gas atom of fixed energy $U$ inside a box. Then he wants to count the number of linearly independent wavefunctions at this particular energy. This corresponds to a sphere in momentum space: $p_x^2+p_y^2+p_z^2=p^2$, and the number of such linearly independent wavefunctions is finite.

I think he stresses that you need independence to get a finite number, because given two energy eigenstates $|\phi\rangle$,$|\psi\rangle$, one can always take "silly" linear combinations:

$\alpha |\phi\rangle+(1-\alpha)|\psi\rangle$.

Hope this helps.

Edit: about (2):

This is traditionally done for $N$ particles but let me give an informal sketch:

Using count_to_10's notation: $E_=(n_x^2+n_y^2+n_z^2) h^2/8mL^2$. We now want to count how many states have energy smaller than $\bar{E}$.

First of all some notation: let $\Phi(\bar{E})$ denote the number of states with energy $\leq \bar{E}$ and $\omega(\bar{E})$ denotes the number of states with energy in the range $[\bar{E},\bar{E}+\delta E]$.

Looking at the formula for the energy: $(n_x^2+n_y^2+n_z^2)=8mL^2 E / h^2$. This is precisely stating that the vector $(n_x,n_y,n_z)$ has fixed length (squared norm). Hence states with energy smaller than $\bar{E}$ are given by points in $n$ space inside the sphere of radius $(8mL^2 E / h^2)^{1/2}$. The total number of such points may be approximated by the volume of the sphere:

$\Phi(\bar{E})=\frac{1}{8}\frac{4}{3}(\frac{8mL^2\bar{E}}{h^2})^{\frac{3}{2}}$.

The $1/8$ is due to the fact that we only consider positive $n$'s, hence only one octant.

Now that we know the number of state with energy $\leq \bar{E}$, we may get the number of states with energy in $[\bar{E},\bar{E}+\delta E]$ by taking the derivative wrt. $\bar{E}$:

$\omega(\bar{E})=\frac{1}{4} L^3 (\frac{8m}{h^2})^{3/2} E^{\frac{1}{2}} \delta E$.

Here you see the "position volume" given by $L^3$ and the rest may be thought as coming from the volume spanned by allowed momentas. Anyway in think the author wrote this formula as a handwaving argument and will prove the exact result later in the book.

P.s.: the above trick may be understood as follows:

$\omega(E)=\Phi(E+\delta E)- \phi(E)= \frac{\Phi(E+\delta E)-\Phi(E)}{\delta E} \delta E \approx \frac{d\Phi}{dE}\delta E$.