Quantum Real Estate

Classical Mechanics Level 5

In classical statistical mechanics, states of a system are distinguished by the values of their extensive variables, i.e. energy, volume, etc. Different values of the extensive variables carry over into different arrangements of the microscopic states of the system. For example, an ideal gas of $N$ particles with energy $E$ is associated with a particular set of states of the gas molecules as represented in the joint $6N$ dimensional space of the positions and momenta of the gas molecules. In general, the entropy of a macrostate is proportional to the natural logarithm of how many microstates can give rise to the given macrostate, i.e. $S\sim\ln \Gamma$ where $\Gamma$ is the "volume" of the microstates in the phase space.

However, without a natural unit with which to measure the "volume", the value of the entropy is arbitrary up to an additive constant. We can fix this by exploring quantum analogues to classical systems.

Consider the particle in a well, a quantum particle that is trapped in a 1d line of length $L$ . The Hamiltonian (total energy) of the particle is given by its kinetic energy, $\hat{H} = \frac{\hat{P}^2}{2m}$ , where $\hat{P}$ is the momentum operator. The allowed energies for the particle are given by $E_n = \frac{n^2\hbar^2\pi^2}{2mL^2}$ where $\hbar$ is Planck's constant divided by $2\pi$ , $m$ is the mass of the quantum particle, and $n$ is a positive integer.

Pretend that $\hat{P}$ is a continuous variable and assume that the position $x$ of the particle in the well varies from $0$ to $L$ . Calculate the volume of the phase space (the space of $x$ and $p$ ) associated with particles of energy level $n$ or lower.

i.e. calculate $\mbox{Vol}\left(E_n\right) = \displaystyle\int_{E\leq E_n}dxdp$

This gives a classical notion of the volume of the phase space occupied by particles of energy level $n$ or below. Next, calculate the number of states of energy $E_n$ or lower.

Divide this classical "volume" of phase space by the number of states of energy less than or equal to $E_n$ to find the "volume" of phase space that is occupied by each eigenstate of the particle. Express your answer in multiples of $\hbar$ .

This gives a value to the smallest division of phase space.

Assumptions

You definitely do not need calculus or any deep knowledge of quantum mechanics for this problem.
Because this system is one dimensional, the phase space "volume" is two dimensional ( $x$ and $p$ ).
Make sure you account for the entire range of values for $p$ that contribute to the volume.

The answer is 6.283185307.

1 solution

Snehal Shekatkar
Feb 3, 2014

This problem, though simple, is somewhat tricky.

Since energy $E_{n}$ is related to momentum $p_{n}$ by,

$\frac{p_{n}^{2}}{2m}=\frac{n^{2}\pi^{2}\hbar^{2}}{2mL^{2}}$

We have,

$p_{n}=\pm \frac{n\pi\hbar}{L}$

It is important to note that two values of momentum (one positive and other negative) correspond to the same value of energy because energy is related to square of the momentum!

Assuming now that $p_{n}$ is continuous variable, we get,

$dp_{n}=\frac{\pi\hbar}{L}dn$

Hence we have,

$Vol(E_{n})=\int_{E\leq E_{n}}dxdp = \frac{\pi\hbar}{L}\int_{0}^{L}\int_{-n}^{n} dxdn = 2n\pi\hbar$

To account for negative values of $p_{n}$ , here limits of integration over $n$ should go from $-n$ to $n$ and not from $0$ to $n$ .

The total number of energy states less than or equal to $n$ is of course $n$ .

Thus dividing $Vol(E_{n})$ by $n$ , in units of $\hbar$ , we get he smalles division to be equal to $\boxed{2\pi}$

Quantum Real Estate

The answer is 6.283185307.

1 solution

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