# If the spatial wavefunctions of two identical atoms overlap, but they find themselves in different energy states, are they distinguishable or not?

Say I have two quantum particles, atoms for that matter, that are completely identical in all their physical properties except that they find themselves at different locations and have different energies, i.e., are described by two different wavefunctions. If their spatial wavefunctions overlap, do they need to be treated as distinguishable or indistinguishable particles?

To directly answer your question, two particles are either indistinguishable or they aren't. The fact that they may be well-separated in physical space or occupy different energy states is irrelevant.

If you have a system of $$N$$ indistinguishable particles (in 1D, for the time being), then there is generically no notion of individual wavefunctions for each particle - there is only one system wavefunction which is a function of $$N$$ variables.

For example, let $$N=2$$. A generic wavefunction $$\psi$$ is a function of two variables, which can be interpreted as the position coordinates for the two particles. We then have that $$|\psi(x_1,x_2)|^2dx_1 dx_2$$ is (or is proportional to, depending on normalization) the probability of measuring one particle in the interval $$[x_1,x_1+dx_1]$$ and the other in the interval $$[x_2,x_2+dx_2]$$.

If the two particles are distinguishable, then $$\psi$$ can be pretty much anything which obeys the requisite boundary conditions. If the two particles are indistinguishable, then the system must be in the same state under interchange of the coordinates, which means that $$\psi(x_2,x_1) = e^{i\theta} \psi(x_1,x_2)$$

In nature, we find that there are two kinds of indistinguishable particles - fermions, for which $$\theta=\pi \implies \psi(x_2,x_1)=-\psi(x_1,x_2)$$, and bosons, for which $$\theta = 0 \implies \psi(x_2,x_1)=\psi(x_1,x_2)$$. In principle this need not be so, and in lower dimensional systems anyons can have different values for $$\theta$$, but this is a bit off-topic for your question.

However, given two single-particle wavefunctions $$\alpha$$ and $$\beta$$, one can create a corresponding bosonic wavefunction $$\psi$$ for the two-particle system as follows (fermions are exactly the same, except with a minus sign in between):

$$\psi_B(x_1,x_2) = \frac{1}{\sqrt{2}}\left[\alpha(x_1)\beta(x_2) + \alpha(x_2) \beta(x_1)\right]$$

It doesn't matter if one of the particles is localized to Earth and the other to Jupiter - if you have two indistinguishable bosons, the system wavefunction must look like this.

Here's where the idea of overlapping wavefunctions comes into play. Let's imagine that $$\alpha$$ is localized so that to some domain $$A$$ and $$\beta$$ to some domain $$B$$, by which I mean that if $$x\notin A, \alpha(x)=0$$ (and likewise for $$\beta$$ and $$B$$), and let's say that $$A\cap B=\emptyset$$.

• If $$x_1\in A$$ and $$x_2\in B$$, then $$\psi_B = \frac{1}{\sqrt{2}}\alpha(x_1)\beta(x_2)$$.

• If $$x_2\in A$$ and $$x_1\in B$$, then $$\psi_B = \frac{1}{\sqrt{2}}\alpha(x_2)\beta(x_1)$$.

• Otherwise $$\psi_B = 0$$.

As you can see, in this case the system wavefunction can be treated as a simple product between single particle states, even though in a strict sense it is not. Operators like $$-i\hbar \frac{d}{dx_1}$$ operate on either $$\alpha$$ or $$\beta$$ while leaving the other alone, which ultimately means that we can ignore the effects of the indistinguishability of the particles.

Once $$A\cap B\neq \emptyset$$, then this is no longer true - there will be sets of coordinates $$(x_1,x_2)$$ for which both the first and second terms of $$\psi_B$$ will be nonzero, at which point we cannot ignore the interference between the two.

Our currently accepted theory is the SM, now all elementary particles (let's assume you are talking about elementary particles) in the SM of the same quantum numbers are considered identical. This is the theory.

The answer though in practice usually would be a yes (different energy level elementary particles are distinguishable). We would be able to identify particles with different energy.

This is because when you are asking about distinguishing, then yes, you can distinguish two (identical elementary) particles in an experiment, if they do have different energy levels.

Even if these elementary particles do have all quantum numbers identical, in an experiment, we would be able to distinguish between them if they have different energy levels.

• "Even if these elementary particles do have all quantum numbers identical, in an experiment, we would be able to distinguish between them if they have different energy levels." - My follow-up question would be if this distinguishability relies on their spatial wavefunctions not overlapping? I cannot think of an experiment which would be able to assign with certainty the measured energies to each of two identical particles if their wavefunctions overlap. – Lenus Stueli Jun 12 at 19:56
• @LenusStueli their spatial wavefunctions will not completely overlap (Pauli exclusion principle), maybe partially. In a bubble chamber experiment, the two atoms will leave separate traces, and you will be able to distinguish between them based on their energy levels. – Árpád Szendrei Jun 13 at 6:14

If their individual wavefunctions overlap, then their common wavefunction must obey spin statistics. That will result in an exchange interaction (I.e. force) between them.

The fact that there are two energy states involved doesn’t remove that, because the exchange interaction is about identical particles, not about identical states.