# Quantum Real Estate

**Classical Mechanics**Level 5

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.

**Your answer seems reasonable.**Find out if you're right!

**That seems reasonable.**Find out if you're right!

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