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A Little Question On Identical Particles

  1. Jul 18, 2014 #1
    I am learning identical particles recently, but I have some problem interpreting what I am writing down. So if we have two distinguishable particles, absolute value of ψ(x_1, x_2) tells the prob. density of finding the first particle at x_1 and the second at x_2. But for identical particles, it seems to me that it doesn't make sense to--and we can't-- label the particles with numbers. So what is the proper interpretation of the absolute value of ψ(x_1, x_2) in this case? (I guess I understand how the stuff on interchanging the particles works, but just can't wrap my head around the above interpretation)
    Also, for identical particles, how do I calculate the probability density of "finding A particle in x_1 and A particle at x_2"? I mean, classically, it would be the sum of "finding the first at x_1 and the second at x_2" and "finding the first at x_2 and the second at x_1", but this can not be the way it works in QM, right?
    Thank you so much.
     
  2. jcsd
  3. Jul 18, 2014 #2

    tom.stoer

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    In general you could have something like

    [tex]\phi_a(x_1)\,\phi_b(x_2)[/tex]

    That means you have a probability amplitude for the first particle in state a at x1 and the second particle in state b at x2.

    For two identical particles you have to (anti-)symmetrize, i.e.

    [tex]\phi_a(x_1)\,\phi_b(x_2) \pm \phi_b(x_1)\,\phi_a(x_2)[/tex]

    That means you have a probability amplitude for one particle in state a at x1 and one particle in state b at x2. So b/c they cannot be distinguished it does no longer make sense to talk about "the first" and "the second" particle.
     
  4. Jul 18, 2014 #3

    DrDu

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    I rather interprete this as the probability to find one particle at x1 and one particle at x2 without distinguishing between the particles. I.e. you have two detectors, one at x1 and one at x2, and both make click, but you cannot say which particle is responsible for which click.
     
  5. Jul 18, 2014 #4

    DrDu

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    You also noted correctly that labelling the particles with r1 and r2 is doing to much. An alternative is to label the particle by their center of mass R=(r1+r2)/2 and relative position vector r=r2-r1. The center of mass coordinate is unimportant in the following. The particles being indistinguishable means that r can be restricted to lie in a half space, e.g. the one with z>0. Then, due to indistinguishability ## |\psi(x1,x2,0)|=|\psi(-x1,-x2,0)|## on the boundary plane. Hence ## \psi(x1,x2,0)=c\psi(-x1,-x2,0)## where c is of unit modulus. There are arguments that c must be real, whence there are only the two alternatives c=1 and c=-1, corresponding to bosons and fermions, respectively.
    This argument can be extended to the case of more than 2 particles, leading to all representations of the symmetric group of n particles.
     
  6. Jul 18, 2014 #5
    tom.stoer, thanks for your answer. For a moment I thought I got it in terms of the probability amplitude you bring about, but I still got a question.
    So I think that the probability amplitude of "one particle is in state a at x_1, the other in state b at x_2" would be the SUM of the probability amplitudes of the two exclusive events. This helps explain the symmetric one. But how can we explain this interpretation for the anti-symmetric case, where these is a minus sign?
     
  7. Jul 23, 2014 #6

    Jano L.

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    Being identical poses no problem here. "Identical" is a concept used daily in ordinary life to say two things are hard to tell from each other (think of identical xerox copies or a painting and its forgery). In quantum theory it is used with similar meaning - the particles have the same internal characteristics - mass, charge, spin etc. For identical particles, there is no difficulty in interpreting
    $$
    |\psi(x,y)|^2.
    $$
    The Born interpretation of this is probability density that the first particle is at ##x##, the second at ##y##; just a position of the argument in the argument list reveals which particle is connected to it. Identical (two electrons) or not (electron and proton) brings no difficulty here.


    However, being identical is not the same thing as being indistinguishable.


    "Indistinguishable" in quantum theory is used in more than one sense and it is somewhat different from its ordinary use in daily language.

    One often used meaning of "indistinguishable" is "one cannot follow the particles in time and preserve the information which is which". One may initially distinguish two different particles if they are at different, spatially separated sections of the experiment, but if then they are brought close to each other (one is scattered off the other), it is believed that one loses the ability to tell which is which.

    Another meaning of "indistinguishable" is that if the probability density ##\rho(x,y)=|\psi(x,y)|^2## is invariant under interchange of the two arguments, all quantities calculated from it and all predictions derived from it are the same for the particle 1 and for the particle 2. In other words, one cannot distinguish the particles based on the wave function alone.

    It was found long ago that best results of calculations for the atom are obtained with anti-symmetric wave functions, which give probability density of the above kind. The wave function leads to the same predictions (average position, energy etc.) for all electrons in the atom. In this sense, the electrons are indistinguishable particles.

    Some people also use "indistinguishable" in the sense "interchange of two particles is not a real phenomenon" and similar, but I never understood it. Luckily this is not necessary for the theory to work and give useful results.

    If the function describes only two particles, probability that there is particle inside ##dx_1## around ##x_1## and ##dx_2## around ##x_2## is just sum of the two probabilities for mutually exclusive things:
    $$
    Prob. = (|\psi(x_1,x_2)|^2 + |\psi(x_2,x_1)|^2) dx_1 dx_2.
    $$
     
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