Particle in a box

Turning away from spin systems, let's look at a fairly realistic quantum model for an ideal gas of non-interacting particles. We will:

  • Solve Schrödinger's equation for a single particle in a box.
  • Find the allowed single-particle energy levels.
  • This will give us the energy levels of a collection of many particles: If the particles interact very little with each other, then the energy of a collection of particles is just the sum of their single-particle energies. We don't need to add any particle-particle interaction terms.
  • Figure out how to count the degeneracy (ways of arranging particles for a given total energy).

With this information in hand, we can make calculations of the equilibrium properties of a fairly realistic gas of particles which obey quantum mechanics instead of classical mechanics.

Schrödinger equation

In classical mechanics, the energy of a particle moving in one dimension is given by the Hamiltonian operator: $${\mathcal H} \equiv \frac{p^2}{2m} +V(x) = E.$$

In quantum mechanics, there is a wave function $\psi(x)$ which is the probabiblity amplitude.

The probability to find a single particle with a position between $x$ and $x+dx$ is $|\psi(x)^2|$.

The time-independent probability amplitude - the probability when the potential energy is not changing with time - satisfies the time-independent Schrödinger equation which strikingly resembles the Hamiltonian operator, with $p\rightarrow i\hbar\partial/\partial x$: $$-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x) + V(x)\psi(x) = E\psi(x).$$

The particle in a box or infinite well problem consists of assuming that the potential is... $$V(x) = \begin{matrix} \infty & x < 0\\ 0 & 0 < x < L \\ \infty & L < x \end{matrix} .$$

So, inside the box, the equation is now: $$-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x) = E\psi(x).$$

  • the wave function must vanish outside of the box.
  • The wave function is continuous across the boundary.

Therefore, the solutions inside the box must vanish at $x=0$ and $x=a$.

The solutions are $\sin$ and $\cos$ functions, but the $cos$ function does not vanish at $x=0$, so

for $0 < x < L$ are $$\psi(x) = A\sin kx,$$

where $$k=n\frac{\pi}{L},\ \ n=1,2,3...$$

Carrying through the second derivative, we find that the allowed energies are (for any positive integer $n$): $$E = \frac{\pi^2\hbar^2}{2mL^2}n^2.$$

3-d box

Now, if the box is a 3-dimensional cubic container, instead of a 1-d container, with extension $L$ in each direction: $$\begin{align}E_j &= \frac{\pi^2\hbar^2}{2m}\(\frac{n_x^2}{L^2}+\frac{n_y^2}{L^2} + \frac{n_z^2}{L^2}\) = \frac{\pi^2\hbar^2}{2mL^2}n_j^2\\ & = \frac{\pi^2\hbar^2}{2mV^{2/3}}n_j^2.\end{align}$$

The lowest energy level (ground state) is: $$E_1 = \frac{\pi^2\hbar^2}{2mV^{2/3}}\(1^2+1^2+1^2\) = \frac{\pi^2\hbar^2}{2mV^{2/3}}3.$$

There is clearly only one combination of $n_x$, $n_y$, and $n_z$ to get this energy.

But when $\{n_x,n_y,n_z\}=\{1,1,2\}\Rightarrow n_2^2 = 1^2+1^2+2^2 = 6$, there are three combinations such that the sum of squares to equal 6. So, the degeneracy of this energy level is $g_2 = 3$.

We had $g_1=1$.

[spot] next levels

What are $n_j^2$ and $g_j$ for $j=3$ and $j=4$?

Macrostate

Each combination of $n_x$, $n_y$, $n_z$ is a different quantum state (microstate). But there are $g_j$ different quantum states that share the same energy.

A description of the macrostate of the system would consist of specifying how many particles $N_j$ are in each energy level $j$ of the system, irregardless of how they're arranged in the quantum states.

As $V$ increases, the energy levels get closer together, quasi-continuously.

Carter estimates for helium, at room temperature, with a one-liter box, a typical $n_j\approx 2\times10^9$

Density of states

For thermodynamic purposes, we need to figure out how to count the number of microstates for each macrostate. One microstate would be an enumeration of how many particles there are in each quantum state. So we start by trying to count the number of quantum states.

If the quantum levels are very close together, we can approximate the discrete function $n_j$ as a continuous function $n(E)$ of a continuously varying $E$.

The density of states $g(E)$ is defined so that the number of quantum states with energies between $E$ and $E+dE$ is $g(E)dE$.

If there is 1 state for every integer $n_x$, $n_y$, $n_z$, then we can imagine these as points spread uniformly in one octant of 3-d space with a density of 1 point per unit volume.

Since $n_j^2=n_x^2+n_y^2+n_z^2$, geometrically $n_j$ is the radial distance $\equiv R$ from the origin to a particular point in "space", $(n_x,n_y,n_z)$.

The number of states $n(E)$ with energy less than E is $\approx$ the volume of that octant of the sphere with radius $R=n_j$, that is $$n(n_j)=\frac 18 \left(\frac 43 \pi n_j^3\right).$$

By solving our energy expression for $n_j$ in terms of $E$ you can show that this is $$n(E)=\frac{\pi}{6}V\(\frac{8m}{h^2}\)^{3/2}E^{3/2}.$$ ...using $\hbar=h/(2\pi)$.

According to this picture, $g(E)dE$ is the number of states (dots) in the shell shown--that is, the volume of the shell shown, which is... $$g(E)dE=n(E+dE)-n(E) \approx\frac{dn(E)}{dE}dE.$$

Taking the derivative of our expression $n(E)$, this becomes: $$g(E)dE = \frac{4\sqrt{2}\pi V}{h^3}m^{3/2}E^{1/2}\,dE.$$

Bosons or fermions?

There are two kinds of real particles: fermions (half-spin) and bosons (integer spin). This $g(E)dE$ is the correct count for spin-0 bosons.

But for spin-1/2 particles (fermions), there are actually 2 quantum states for each one counted above--one with spin up, and one where the particle is oriented with spin down.

So, our final expression for the density of states of a particle in a box will have a spin factor $\gamma_s$ which is 2 for spin-1/2 fermions and 1 for spin-0 bosons: $$g(E)dE = \gamma_s \frac{4\sqrt{2}\pi V}{h^3}m^{3/2}E^{1/2}\,dE.$$