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A question re. 'single-particle' states

  1. Dec 27, 2013 #1

    Seasons greetings to everyone :-)

    I've been revising statistical mechanics and have stumbled across an area that I've always been a little 'hazy' on.
    By the term 'single-particle' state, is it meant that this is a particular quantum state that one (or more) particle(s) can occupy, a term which holds true regardless of whether the wave-function describing the whole system is separable?

    My understanding, quite possibly incorrect, is that single-particle states are the quantum states that each individual particle occupy (of course, it may be that more than one individual particle occupies the same quantum state), regardless of whether the wave-function describing the system can be expressed as a product of the wave-functions describing each single-particle state, or not. In cases where it can be, it is then possible to express the energy eigenvalues of the microstates of the system as a sum of the energy eigenstates corresponding to each single-particle state.

    Please enlighten me if this is incorrect.

    Many thanks.
  2. jcsd
  3. Dec 27, 2013 #2

    Simon Bridge

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    A "single particle state" is, strictly, a state of a system with only one particle in it.
    However, in many particle systems, it is possible for many particles to occupy a particular single-particle state ... i.e. Bose-Einstein statistics.

    Note: iirc the term is not jargon, so watch the context.

    I have found Sze Tan quite clear: http://home.comcast.net/~szemengtan/ [Broken]
    Last edited by a moderator: May 6, 2017
  4. Dec 27, 2013 #3
    Thanks for your speedy response.

    So would it be correct to consider a multi-particle system in a given microstate as a collection of sub-systems, each containing a single-particle in a particular state, where it is possible that there is more than one 'copy' of the same sub-system?
    (Or is this type of description only valid if one considers cases in which the particles are weakly interacting, such they can be treated as independent?)
    Last edited: Dec 27, 2013
  5. Dec 27, 2013 #4


    Staff: Mentor

    No - because of the peculiar quantum aspect of entanglement.

    Basically this is if you have two particles that can be in state |a> and |b> then they can be entangled in a way that doesn't allow them to be considered as two separate systems ie it is in a superposition of particle 1 in state |a> with particle 2 in state |b> and particle 1 in state |b> with particle 2 in state |a> ie c1*|a>|b> + c2*|b>|a>.

    Just as an aside there is a view that this is the rock bottom essence of QM:

    Basically it would seem that the two most reasonable generalized probability models for modelling physical systems is ordinary probability theory and QM - but QM is the only one that allows entanglement.

  6. Dec 27, 2013 #5


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    It is only correct for non-interacting particles. For a finite number of interacting particles, the single-particle states are just used to make basis vectors for the Hilbert space, and the state is then some superposition of basis vectors.

    http://www.itp.phys.ethz.ch/education/fs12/cqp/chapter04.pdf [Broken]
    Last edited by a moderator: May 6, 2017
  7. Dec 27, 2013 #6

    Simon Bridge

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    Since you are reviewing the basics - you are best advised to avoid extending the concepts you are reviewing beyond the material actually in the course notes for that level. You are reviewing a very simple model that will be built on later on.

    Baby steps.
  8. Dec 28, 2013 #7
    Ok, I'll keep reading.
    Thank you for your help guys, much appreciated!
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