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# Disclaimer

Peeter’s lecture notes from class. May not be entirely coherent.

# Review. Classical phase space calculation

From this we calculated , and

Fudging with a requirement that , we corrected this as

Now let’s do the quantum calculation.

**Quantum calculation**

Recall that for the solutions of the Quantum free particle in a box, as in (Fig 1), our solutions are

Fig 1: 1D Quantum free particle in a box

where , and

.

In three dimensions, with we have

and

so that

How do the multiplicities scale by energy? We’ll have expect something like (Fig 2).

Fig 2: Multiplicities for free quantum particle in a 3D box

Provided the energies are large enough, we can approximate the sum with

So

**Harmonic oscillator in 1D.**

Our phase space is of the form (Fig 3).

Fig 3: 1D classical SHO phase space

Where the number of states in this classical picture are found with:

Rescale

so that we have

How about the quantum calculation?

We have for the energy

graphing the counts (Fig 4), we again have stepping as a function of energy, but no multiplicities this time

Fig 4: 1D quantum SHO states per energy level

we make the continuous approximation for the summation again, and throwing away the zero point energy, we have

**Why ?**

We have a problem with out counting here. Consider some particles in a box as in (Fig 5).

Fig 5: Three particles in a box

- particle at
- particle at
- particle at

or

- particle at
- particle at
- particle at

This is fine in the classical picture, but in the quantum picture with an assumption of indistinguishability, no two particles (say electrons) cannot be labelled in this fashion.

\paragraphAndIndex{Gibbs paradox}

Suppose we double the volume as in (Fig 6), then our total entropy for the bigger system would be

Fig 6: Gibbs volume doubling argument. Two identical systems allowed to mix

We have

This is telling us that each particle could be in either the left or the right side, but we know that this uncertainty shouldn’t be in the final answer. We must drop this term.

So, if we assume that these particles are identical, and divide by , then we find