What is this $\xi$ character in the context of the infinite well? this might be a question that does not have an answer. I am working out the formula for energies for the anti-symmetric wave functions for the infinite square well. I need to determine the value for $\xi_0$ which yields a bound state. I am struggling to understand what $\xi_0$ is. 
From notes and asking my teacher, I am told that $\xi = a\sqrt{\frac{2mE}{\hbar^2}}$; where $m$ is mass, $E$ is the total energy of the particle, $
\hbar$ is planks constant, and $a$ is the length or width of the square well.
Question: What does $\xi$ represent? 
Context to $\xi$:
It seems like an arbitrary value. Although, from the context of the square well problem I can find something similar. Take the time independent scrodinger equation and set $U(x)=0$, because it has no potential energy while inside the walls of the well. 
$$-\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2} + U(x)\psi(x)= E\frac{d\psi(x)}{dt}\\
[U(x) = 0]\\
-\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2} = E\frac{d\psi(x)}{dt}$$
Then divide both sides by $-\frac{\hbar^2}{2m}$.
$$\frac{d^2\psi(x)}{2m}=-\frac{2mE}{\hbar^2}\frac{d\psi(x)}{dt}$$
For convenience, make the definition $k^2\equiv \frac{2mE}{\hbar^2}$. Then $ak = \xi$.
The anti-summetric solution looks like the following, inside the well
$$\psi_n(x)=\sqrt{\frac{2}{L}}sin(\frac{n\pi x}{a});\qquad0<x<a$$
$n$ is an integer. The quantized energy states that form because at $\psi(0)=\psi(a)=0$ and $\psi$ has to be normalized produces
$$E_n = \frac{n^2\pi^2\hbar^2}{2ma^2}$$
 A: The de Broglie wavelength of a free quantum particle is
$$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}.$$
Therefore, up to a constant factor, we just have
$$\xi = \frac{a}{\lambda}.$$
That is, $\xi$ is a dimensionless measure of how wide the well is, compared to the de Broglie wavelength of the particle. 
It's useful for work with $\xi$ instead of $a$ because it lets us "factor out" a bunch of constants we don't want to carry around, and the final answer will be something nice like "$\xi = 1$" instead of something messy like "$a = 4.7239 \times 10^{-14} \text{ meters}$". Phrasing the answer solely in terms of $\xi$ will also automatically tell you how the value of $a$ should change as other parameters, like $m$ and $E$, change as well. 
A: Your given sine wave eigenfunctions are anti-symmetric in the well only for $n$ is even. The parameter $\xi = a\sqrt{\frac{2mE}{\hbar^2}}=ka$ is equal to the product of wave vector k and the width of the well a which is the dimensionless argument of the sine wave function.The boundary conditions of the well restrict the values of  $\xi$ to $\xi_n =ka=n\pi x$ with the symmetric solutions for $n=1,3,5,..$ and the anti-symmetric solutions for $n=2,4,6,...$. All these sine eigenfunctions are bound states.
