Why can two particles in a box have different quantum numbers and still be indistinguishable? I'm reading from the "Modern Physics" textbook by Randy Harris. 
I'm Chapter 8 on Spin and Atomic Physics, and the book has just introduced how to solve for the wave function of 2 particles in a box. The two particles are identical, and can have different quantum numbers. They do not interact.
From there they go on to explain why we use symmetric and antisymmetric wave functions, because we can't be able to tell which particle is which from the probability density equation, which we ARE able to without using the symmetrical requirement if the two particles have different quantum numbers.
My question is, why can't we distinguish between two particles with different quantum numbers? Doesn't that mean they have different energies, and as such have different wave forms and so appear in different places? The problem used in the book is that the two particles are in a box. But because of their different quantum numbers, one of the two particles can't be found in the middle of the box, while the other can. This should never be the case apparently, so they introduce the symmetrical requirement.
As stated above, I'm confused as to why we're not supposed to be able to tell two particles with different quantum numbers apart.
Thanks for your time.
 A: Consider a wavefunction $\psi(x_1,x_2)$ which describes a two-particle system.  We interpret $|\psi(x_1,x_2)|^2 dx_1 dx_2$ to be the probability of finding particle $1$ in the interval $[x_1,x_1+dx_1]$ and particle $2$ in the interval $[x_2,x_2+dx_2]$ if we perform a simultaneous measurement of their positions.
One might ask if $|\psi(x_1,x_2)|^2 = |\psi(x_2,x_1)|^2$, and in general the answer would be no. For instance, consider two non-interacting, spinless particles in a square well of width $L$, which might exist in the following state:
$$\psi(x_1,x_2) = \frac{2}{L} \sin\left(\frac{\pi x_1}{L}\right) \sin\left(\frac{3\pi x_2}{L}\right)$$
One can plainly see $|\psi(x_1,x_2)|^2$ that the probability of finding particle $1$ in a small neighborhood of $L/2$ and particle $2$ in a small neighborhood of $L/3$ is zero, while the probability of finding particle $1$ in a small neighborhood of $L/3$ and particle $2$ in a small neighborhood of $L/2$ is not.  This must mean that they are in some way distinguishable.  Otherwise, it would be unphysical to ask which particle was where - we could only ask "what is the probability that one of the particles is in a small neighborhood of $L/3$ and the other is in a small neighborhood of $L/2$?"  If swapping the particles changes the probability distribution, then apparently there was something different about them!

If the two particles are indistinguishable, these probability density functions must be invariant under particle interchange - meaning that $|\psi(x_1,x_2)|^2=|\psi(x_2,x_1)|^2$.  As a result, we must have $\psi(x_2,x_1) = e^{i\theta} \psi(x_1,x_2)$ for some real number $\theta$.
Experimentally, it is found that in most cases, quantum mechanical particles fall into two camps - those for which $\theta = 0$, and those for which $\theta = \pi$, called bosons and fermions respectively.  The Spin-Statistics Theorem demonstrates (with some fairly mild assumptions such as relativity and causality) that these are the only two options, and that particles with integer spin have $\theta=0$ while particles with half-odd-integer spin have $\theta = \pi$.   It is worth noting that this theorem holds only for $\geq 3$ spatial dimensions, and much different behavior can be seen in 2D systems.

Lastly, note that I cast the idea of distinguishability in terms of position-space wavefunctions and the associated probability density, but this is not necessary.  The state of the system can be expanded in a basis for any observable, and the indistinguishability of the particles amounts to an invariance of the corresponding probability distribution under particle interchange.  
If, for instance, the probability of particle $1$ having energy $E_A$ and particle $2$ having energy $E_B$ is not symmetric under $1\leftrightarrow 2$, then there must be something distinguishable about them.  In the case of indistinguishable particles, it is only physical to ask for the probability of one having energy $E_A$ and the other having energy $E_B$.  
A: 
why can't we distinguish between two particles with different quantum numbers? Doesn't that mean they have different energies, and as such have different wave forms and so appear in different places?

We could distinguish them, if the particles could be assigned a quantum number or other property that we could later check. This is in principle possible if one particle is electron and the other is proton (mass and electric charge are different) or if there are two electrons, but they are very far from each other so they can't possible exchange the property.
But if these are two electrons too close to each other, there is no way to assign one electron a property and then check later whether it kept that property or passed it over to the other electron. For example, in a multi-electron atom there is no way to check whether the electron that was found near the nucleus at time $t_1$ is the same as the electron that was found at $r=1\text{E-14m}$ away from the nucleus at a later time $t_2$ one second later. Of course, this changes dramatically when $r$ is much greater than a light second - then due to universal speed limit we have a reason to believe the remote electron can't be the first electron.
If two electrons are alone in the same potential hole, then we do not assign them individual psi functions or individual quantum energy numbers. There is only one Schroedinger equation for the whole system, so there is only one psi function describing all electrons inside. (In special cases, we can assign this system pair of quantum numbers, but this is not possible in general.) It turns out the psi function of such system has to be either symmetric or antisymmetric with respect to swap of particle arguments, otherwise different particles will have different probabilities and thus won't be having the same behaviour.
