I need help understanding how to do a quantum particle-statistics problem, based on the Example 5.4.1 in Griffiths' Introduction to quantum mechanics.

Let's say that I have 3 identical particles in a 1-D harmonic oscillator potential with total energy $E=\frac{9}{2}\hbar \omega$. I need to find the occupation number configurations, the number of distinct three-particle states there are for each one, and the probability of measuring the energy $\frac{3}{2}\hbar \omega$ when picking a single particle to measure. Let's now say that there are 2 scenarios: one with 3 identical spin 0 bosons and one with 3 identical spin $\frac{1}{2}$ fermions.

Here are the fruits of my labor so far:

  • There are 10 total possible combinations of the $n_i$: \begin{align} (n_1,n_2,n_3) & = (3,0,0),\ (0,3,0),\ (0,0,3)\\ (n_1,n_2,n_3) & = (0,1,2),\ (1,0,2),\ (1,2,0),\ (2,1,0),\ (2,0,1),\ (0,2,1)\\ (n_1,n_2,n_3) & = (1,1,1) \end{align}

  1. For bosons, the occupation number configurations are: \begin{align} (N_0,N_1,N_2,N_3,\ldots) & = (2,0,0,1,0,0,\ldots)\\ (N_0,N_1,N_2,N_3,\ldots) & = (1,1,1,0,0,0,\ldots)\\ (N_0,N_1,N_2,N_3,\ldots) & = (0,3,0,0,0,0,\ldots), \end{align} where $N_n$ represents the number of particle in the $n$th eigenstate of the single-particle hamiltonian. Since these are identical bosons, I think there are 3 distinct three-particle states.

  2. For fermions, the only possible occupation number configuration is: $$(N_0,N_1,N_2,N_3,\ldots) = (1,1,1,0,0,0,\ldots)$$ because we cannot include the first and third row of the total possible combinations (since those rows have particles sharing states, which cannot happen for fermions due to the Pauli Exclusion principle).

    Since these are identical fermions, I think there are 6 distinct three-particle states.

My question: For each of the multi-particle configurations givne above, how do I extract the probability that if I measure the single-particle trap hamiltonian on a given particle, say, the first one, it will read out an energy $\frac32 \hbar\omega$, i.e. the particle will project onto the first excited state of the trap?

I am making the further assumption that the particles are in thermal equilibrium, so that they are in the maximally mixed state within the $E=\frac92\hbar\omega$ eigenspace subject to the restrictions of their given exchange symmetries.


1 Answer 1


Within that formalism, if you have a state with occupations $$(N_0,N_1,N_2,N_3,\ldots)$$ where there is a well-defined number $N_n$ of particles in the $n$th eigenstate, the probability $P_n$ of taking one of the particles and finding it in the $n$th eigenstate really is as simple as $$P_n = \frac{N_n}{N_\mathrm{tot}} = \frac{N_n}{\sum_{k}N_k},$$ i.e. the occupation number of that eigenspace divided by the total number of particles.

In addition to this, within Griffiths' thermal-equilibrium assumption, you need to consider the probability that the particles will be in each of the possible eigenstates that are accessible to their exchange symmetry; here you simply count the number of accessible configurations and take them all with equal probability.

I should note, however, just to throw a wrench into the works, that this only really holds if the spin is completely decoupled from your system, and not really subject to your (baby-steps) thermal equilibrium hypothesis.

If the isn't decoupled, then things become a fair bit more complicated, because the actual states are of the form $\psi_{n,\updownarrow}$, and you can have mixing between the spatial and the spin sector. What this means is that there are states like $$ \psi_{3,\uparrow}\otimes \psi_{0,\uparrow}\otimes \psi_{0,\downarrow}, $$ (i.e. the $(n_1,n_2,n_3)=(3,0,0)$ state in your notation, but with different spins on the different particles) that don't vanish when you antisymmetrize the state, i.e. if you antisymmetrize the state above you get $$ \psi_{3,\uparrow}\otimes (\psi_{0,\uparrow}\otimes \psi_{0,\downarrow}-\psi_{0,\downarrow}\otimes \psi_{0,\uparrow}), $$ plus cyclic combinations of which particle is in the $n=3$ subspace. So things are a fair bit more complicated than what Griffiths lets on.


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