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I would like to determine the number of energy states two free, distinguishable particles in a box of length $L$ have. I would then like to determine the number of states two free, indistinguishable particles, with spin $3/2$ each, have in that box at the elementary level. Finally, determine the number of states in case these two particles with spin $3/2$ each are distinguishable. I am familiar with the following formula for the energy states $$E_n=\frac{(\hbar)^2 \pi^2n^2}{(2mL^2)}$$ but am not sure how to proceed. If the particles are distinguishable, does that entail that one is a fermion whereas the other is a boson? I am not sure. Furthermore, if the two particles have spin $3/2$ each, that means they are both fermions, right? If that is correct, then, due to Pauli's exclusion principle, the two could not be at the same state. I also know that for spin $3/2$ there could be $4$ particles per energy level. But again I am not sure how to coherently process the given data and would appreciate some guidance.

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  • $\begingroup$ As of right now, your question is not very well-defined. Can you show what you have tried and what exactly you are trying to do/got stuck on? $\endgroup$ – Danu Feb 21 '14 at 11:56
  • $\begingroup$ @Danu Is the question itself not well defined or the attempt at solution? I am very new to this topic so I am not sure what else I could possibly add. I have written everything which I deemed relevant to solving this problem, to the best of my understanding. $\endgroup$ – peripatein Feb 21 '14 at 12:08
  • $\begingroup$ In case the spin of each particle is $3/2$, would the number of energy states be $\binom{4}{2}$ for indistinguishable particles and $\frac{4!}{2!}$ for distinguishable particles? $\endgroup$ – peripatein Feb 21 '14 at 12:19
  • $\begingroup$ Distinguishable: all permutations of available states. Indistinguishable: all combinations of available states. $\endgroup$ – Carl Witthoft Feb 21 '14 at 12:58
  • $\begingroup$ @CarlWitthoft Is what I wrote incorrect, namely $\binom{4}{2}$ and $\frac{4!}{2!}$? $\endgroup$ – peripatein Feb 21 '14 at 15:06
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"If the particles are distinguishable, does that entail that one is a fermion whereas the other is a boson?"

No, the particles being indistinguishable from each other does not imply that one is a fermion and the other a boson, they can be any types of particles as long as they aren't the same type.

Eg: One could be an electron and the other a muon (both spin-$\frac{1}{2}$ fermions). One could be a hydrogen atom and the other an alpha particle (both bosons). Or one could be a boson and the other a fermion.

From this question I believe you have a misconception about indistinguishable particles---you seem to imply you think that all fermions are indistinguishable from one another and that all bosons are indistinguishable from one another; this is not the case. The Pauli Exclusion Principle applies to fermions of the same type, not all fermions (and similarly with the exchange symmetry for indistinguishable bosons).

"Furthermore, if the two particles have spin 3/2 each, that means they are both fermions, right?"

Yes, it follows from the spin statistics theorem in quantum field theory that half-integer spin particles are fermions and integer spin particles are bosons. Without going into quantum field theory, however, you just have to take this as given.

http://en.wikipedia.org/wiki/Spin%E2%80%93statistics_theorem

"If that is correct, then, due to Pauli's exclusion principle, the two could not be at the same state."

No, as stated above the particles can both be fermions and yet be distinguishable and, hence, not obey the Pauli Exclusion Principle.

So to answer what I think your main question is:

For a box in which the spectrum of a single particle is $E_n$ where $n$ is a positive, non-zero integer, the energy of two non-interacting, distinguishable particles is $E^{(1)}_n + E^{(2)}_m$ where $n$ and $m$ are both positive, non-zero integers, the superscripts refer to the first and second particle, their spectra may be different because, for example, they may have different masses.

As you've given the spectrum for an infinitely deep well, there are infinitely many single particle bound states and, hence infinitely many bound two particle states. Were you talking about, for instance, a finite potential well there would be only be finitely many single particle bound states. If the number of bound states for the first/second particle is $n^{(1/2)}_{tot}$ the total number of bound states for the two (distinguishable) particle system would be $n^{(1)}_{tot}\times n^{(2)}_{tot}$; these numbers could be different, again, because of the possibility of different masses.

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  • $\begingroup$ First of all, thank you very much! Second, did you read my attempt of the second and third parts of this question, referring to two free, indistinguishable (then distinguishable) particles with spin $3/2$ each (at the elementary level)? $\endgroup$ – peripatein Feb 21 '14 at 13:21
  • $\begingroup$ Is that attempt correct? $\endgroup$ – peripatein Feb 21 '14 at 13:29
  • $\begingroup$ Are you talking about your comments? If so I'm not sure what (4,2) and 4!/2! are meant to be the answer to. Is the question: "for two spin 3/2 free particles, how many different spin configurations are there for each value of energy in both the distinguishable and indistinguishable case?"? $\endgroup$ – JPB Feb 21 '14 at 18:43
  • $\begingroup$ I am not sure whether the formulation actually implies spin configurations. The original question is: how may I determine the number of states two free, indistinguishable particles, with spin 3/2 each, have in that box at the elementary level?; also, how may I determine the number of states in case these two particles, with spin 3/2 each, are distinguishable? Are the answers I wrote in my comments correct? $\endgroup$ – peripatein Feb 22 '14 at 9:47
  • $\begingroup$ Are you quoting the question exactly? If the particles are in the box they are not free; free means no potential term in the Hamiltonian, but the presence of the box is a potential term. Further, if you count all of the possible energy states there are infinitely many in either case. Just talking about a single particle: for a free particle there is one energy state for each value of the momentum (a real number) for particles in the infinite square well, there is one energy state for each value of the principle quantum number, n, a positive, non-zero integer. $\endgroup$ – JPB Feb 22 '14 at 12:32

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