# Degeneracy in $N$-particle Quantum System [closed]

I was recently introduced to the concept of $$N$$ particle systems in Quantum Mechanics, and the concept of indistinguishable and distinguishable particles. While reading the following material online, I came across something I couldn't understand properly.

Question 1 : They have considered non-interacting particles of the same mass. How can, we then consider case 1, where the particles are distinguishable? Shouldn't they have different masses for themselves to be distinguishable? Or have they considered particles of different charges or spins, to make them distinguishable? Also, I see that degeneracy of 1st excited state is 2, because we can distinguish between the two cases that give the same value of energy.

Am I correct in assuming, that the degeneracy here ( in this case), is due to spin, or some other property, since mass is the same in both cases?

Question 2 : The degeneracy of the first excited state for identical bosons is 1, or in other words, the system is non-degenerate. Why is this the case? My assumptions are, for identical particle systems, the states are non-degenerate ( if you discount spin ) because one cannot physically distinguish what particle is in which state, and it is a one-dimensional case.

Question 3 : Am I correct in assuming that N-identical spinless particles, will only have degeneracy if the potential is 2 dimensional or more? I've studied, 1-d potentials are always non-degenerate for single particles. Does the same rule apply to identical particles?

Question 4 : This question is more about the nature and physical intuition behind degeneracy. We say that a particular energy level is degenerate if there exists a number of states that have the same energy. Now, do these states have to be distinguishable?

In the case of multiple particle systems, if the particles are distinguishable, then two states with the same energy can be distinguished between and so, we call that energy degenerate. This can happen even in cases of 1-dimensional system. If the particles are identical and non-distinguishable however, in the case of 1-dimensional system, there is no way for us to distinguish between two different states with the same energy.

Imagine 3 identical non-distinguishable particles in 2 states. We can have the possible arrangements for the first excited state:

$$\psi _1(x_1)\psi _1(x_2)\psi _2(x_3) ;$$

$$\psi _1(x_1)\psi _1(x_3)\psi _2(x_2) ;$$

$$\psi _1(x_2)\psi _1(x_3)\psi _2(x_1)$$

Yet, we don't call this 3 fold degenerate. Is it because these three states are in-distinguishable?

Now imagine the same scenario, for distinguishable particles ( of equal mass )

In this case, the first excited state would be 3-fold degenerate. Is it because, now, the three wave functions can be distinguished from one another?

• Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead. – SolubleFish May 26 at 9:53

Answer 3 There can be some degeneracy even in the $$1D$$ case, with indistinguishable spinless bosonic particles. For example, for the harmonic potential, the energies of the states $$|0\rangle|2\rangle$$ and $$|1\rangle|1\rangle$$ are the same, while the states are distinct.
Answer 4 Particles can be distinguishable or indistinguishable. States are equal (the same) or distinct. An energy level is degenerate if several distinct states have this energy. For example, in a 2-particle system. If the two particles are indistinguishable, then $$\psi_1\psi_2$$ and $$\psi_2\psi_1$$ are the same state, so (if no other states is contributing) the energy is not degenerate. If the particles are distinguishable, then $$\psi_1\psi_2$$ and $$\psi_2\psi_1$$ are not the same state (think "in one case, the blue particle is in the excited state, while in the other it is the red particle which is excited").