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I am given the following (probably simple) exercise, but I think I misunderstand something:

Let $\psi_{a,b}(r_1,r_2)$ be a two-particle state, calculate the probability density for distinguishable and indistinguishable particles.

I don't quite understand what this exercise wants me to do: If $\psi_{a,b}(r_1,r_2)$ , then this density should be given by $|\ \psi_{a,b}(r_1,r_2)|^2$, if our particles are indistinguishable, then I am not sure whether I am supposed to symmetrize or antisymmetrize ( or do anything else to this state) the wavefunction.

I mean, somehow this exercise wants me to get two different results for the cases, but I don't see which states exactly.

Qmechanic
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Pritam Bemis
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2 Answers2

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The probability density to find one particle at $r_1$ and the other particle at $r_2$ is the absolute square of the wave-function. The question is, so I understand, how does the wave-function look like. If the particles a and b are distinguishable, the wave-function looks as given in the exercise. However, if the particles are indistinguishable fermions, the wave function should be antisymmetrical at the interchange of the particles, i.e. you do antisymmetrization:

$ψ_{a,b}(r_1,r_2) = \frac{\{ψ_{a,b}(r_1,r_2) - ψ_{a,b}(r_2,r_1)\}}{\sqrt{2}}$

On the other hand, if the particles are indistinguishable bosons, you do symmetrization

$ψ_{a,b}(r_1,r_2) = \frac{\{ψ_{a,b}(r_1,r_2) + ψ_{a,b}(r_2,r_1)\}}{\sqrt{2}}$

Abinash Chakraborty
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Sofia
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Think ψa,b(r1,r2) as a variable in a function space. you need to find the subset of these that have the required properties. The set for the distinguishable particles should be different from the set for indistinguishable particles. I suspect notation abuse in the identical/distinguishable theory of particles. its not real bad, but it looks like it nonetheless. f(x,y)=?-f(y,x). Doesn't make much sense to me. How about grouping a set of f(x,y), like ,for example, all f(x,y)=0 for some x and y.

I like to think of an indistinguishable pair of 1 d particles as a single 2d particle. I can then plot the ψ(r1,r2) on the unit square. Fermions have ψ(r1,r2)=0 along r1=r2. I'm tempted to put down other relations but would be drawn to an abuse of notation! Boson wave function would be symmetric with respect to reflection along the r1=r2 line. Fermions antisymmetric. Distinguishable, not symmetric at all withe reflection.

rockandAir8747
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