Question

How to calculate the occupation number for a compound?

Answer 1

Explanation:

hey mate

you just started quantum mechanics basics

so the answer

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If the particles were distinguishable then yes, each state tuple would be a distinct state. Remember what those tuples mean, each entry in the tuple carries a complete set of quantum numbers to distinguish the state for one particle in the system. Since we are working with infinite wells here, the energy level completely specifies the quantum state, so we have one number for each particle (you could imagine a tuple with 3-vector at each entry for a hydrogen atom, for instance).

For clarity, call the first entry in the tuple the "red" particle and the second entry the "blue particle". If these were different particles, then red being in energy level 1 and blue being in energy level 2 would be a different state, but since these particles are indistinguishable, there is no difference between (red in 1, blue in 2) and (red in 2, blue in 1).

To actually construct the wave function, we'd need to know whether we have bosons or fermions. The actual state vector would be (anti)symmetric combination of all the possibilities listed in the rows above for (fermions)bosons. Note: We have Bosons here because Griffiths is including states with more than one particle at the same energy level.

That is to say, for indistinguishable particles each line above is one and only one state, but you must enumerate all the possibilities in order to actually construct the wave function. As an example, the normalized second row wave function for bosons (using the tuple notation):

13–√((13,13,5)+(13,5,13)+(5,13,13))

Using the, perhaps more familiar, notation of wavefunctions you have:

13–√(ψ13(x1)ψ13(x2)ψ5(x3)+ψ13(x1)ψ5(x2)ψ13(x3)+ψ5(x1)ψ13(x2)ψ13(x3))