# 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: $$g(E)dE = \gamma_s \frac{4\sqrt{2}\pi V}{h^3}m^{3/2}E^{1/2}\,dE.$$