Quantum Mechanics!

The particle in a box is a fairly simple-to-calculate quantum mechanical system that describes a single particle in a box, possessing only translational kinetic energy. That is, a single particle of an ideal gas

We will:

• Solve Schrödinger's equation for a single "particle in a box".
• Find the allowed single-particle energy levels. These depend on the volume of the box)
• Since the particles of an ideal gas do not interact energetically with each other, we'll suppose that a bunch of particles in a box can be described as distributing multipe particles among the energy levels we find.

Schrödinger's equation

In classical mechanics, the energy of a particle moving in one dimension, subject to a potential energy function which depends only on position is given by the "Hamiltonian" of the system: $${\mathcal H}(x) \equiv \frac{p^2}{2m} +V(x) = E.$$ where the first term is the kinetic energy (written in terms of momentum, $p=mv$) and the second term is the potential energy of the particle, and the Energy, $E$, of the particle is a constant.

"Wave mechanics" is one approach to quantum mechanics. In one dimension:

• There is a wave function $\psi(x)$ called the probability amplitude.
• The probability to find a single particle with a position between $x$ and $x+dx$ is $|\psi(x)^2|$.
• Since we're considering a single particle, the total probability to find a particle *somewhere* is $$\int_{-\infty}^{+\infty} |\psi(x)^2|\,dx = 1.$$

Erwin Schrödinger came up with an equation for the time-independent(that is, "equilibrium") probability amplitude, inspired by the classical Hamiltonian, where the momentum is defined in terms of an "operator" $p\rightarrow i\hbar\partial/\partial x$: $$-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x) + V(x)\psi(x) = E\psi(x).$$ And $\hbar\equiv \frac h{2\pi}$ is a constant, where $h$ is Planck's constant, such that the energy of a photon of light of frequency $f$ is $E=hf$.

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

So, inside the box, the potential is 0, so Schrödinger's equation is now: $$-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x) = E\psi(x).$$ $E$, the energy, is a constant in this differential equation. It looks like just the kinetic energy term of the Hamiltonian.

Further constraints on the probability amplitude function:

• The particle can not exist outside the potential well, so 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 $x=L$.

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

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

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

The "normalized" ($\int_0^L |\psi_n(x)|^2\,dx=1$) solutions are: $$\psi_n(x)=\sqrt{\frac 2L}\sin\left( \frac{n\pi}{L}x\right).$$ which look like this:

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

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, then the solutions to the 3-d Schrödinger equation is the product of 1-d solutions: $$\Psi_j(x,y,z)=\psi_{n_x}(x)\psi_{n_y}(y)\psi_{n_z}(z),$$ where the energy depends on the integers $n_x$, $n_y$, and $n_z$ according to: $$\begin{align}E_j &= \frac{\pi^2\hbar^2}{2m}\left(\frac{n_x^2}{L^2}+\frac{n_y^2}{L^2} + \frac{n_z^2}{L^2}\right) = \frac{\pi^2\hbar^2}{2mL^2}n_j^2\\ & = \frac{\pi^2\hbar^2}{2mV^{2/3}}n_j^2.\end{align}$$

The energy is proportional to $n_j^2=n_x^2+n_y^2+n_z^2$. The lowest energy level (ground state) is: $$E_1 = \frac{\pi^2\hbar^2}{2mV^{2/3}}\left(1^2+1^2+1^2\right) = \frac{\pi^2\hbar^2}{2mV^{2/3}}3.$$ There is clearly only one combination of $n_x$, $n_y$, and $n_z$ to get this energy.

The next energy level, $j=2$, occurs when one of the integers is 2 instead of 1, for example, $\{n_x,n_y,n_z\}=\{1,1,2\}\Rightarrow n_2^2 = 1^2+1^2+2^2 = 6$.

There are three combinations such that the sum of squares equals 6. The other two are: $\{1,2,1\}$ and $\{2,1,1\}$ So, the degeneracy of this energy level is $g_2 = 3$.

Because only $\{1,1,1\}$ gives the lowest energy, with $n_1^2=3$, we had $g_1=1$.

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. The energy differences between adjacent energy levels will get smaller. That is, it will become harder and harder to measure the ever-smaller energy differences from transitions between neighboring energy levels, and the energy spectrum will start to look more "continuous": The macroscopic limit of quantum mechanics is a world in which energy is, as far as practical experiments are concerned, a continuous variable.

This also means that, as you increase the size of the container from microscopic to, well, visible to the naked eye, that many, many more energy levels will fit between $U=0$ and $U=k_BT_\text{293 K}$.

Carter estimates for helium, at room temperature, with a one-liter box, a typical atom be might hovering around the $\approx 2\times10^9$'th energy level.