# Solving TISE for particle in the box for Infinite square well

While Solving the TISE for a particle an infinite square well with potential given by: $$U(x) = \left\{ \begin{array}{ll} 0 & \quad -L/2 \leq x \leq L/2 \\ \infty & \quad otherwise \end{array} \right.$$

we get two sets of solutions: $$\psi(x) = \left\{ \begin{array}{ll} A\sin(\frac{n\pi x}{L}) & \quad n = 2,4,6,... \\ B\cos(\frac{n \pi x}{L}) & \quad n = 1,2,3,... \end{array} \right.$$

But when we solve the TISE for potential $$U(x) = 0 \quad for \quad 0 \leq x \leq L$$ we get only one solution i.e. $$\phi(x) = A\sin(\frac{n\pi x}{L}) \space where \space A=\sqrt{\frac{2}{L}}$$.

I want to know what difference does the change of coordinates makes on the system so that for one there are two solutions and for other there is only one?

• if you draw these solutions you will see that both sets are identical. – RogerJBarlow May 10 at 13:16

In the case of a well $$-L/2 < x < L/2$$ the solutions are $$\psi(x) = \begin{cases}A\sin\left(\frac{n\pi x}{L}\right), n = 2, 4, 6,...\\ B\cos\left(\frac{n\pi x}{L}\right), n = 1, 3, 5, ...\end{cases}.$$ Pay attention to the choice of integers $$n$$ - they assure that the boundary conditions are satisfied, i.e. $$\psi(\pm L/2) = 0$$. Indeed, e.g., for $$n=1$$ $$\sin(n\pi x/L)|_{x=\pm L/2} = \sin(\pm \pi/2) = \pm 1$$, i.e., it does not satisfy the boundary condition.

On the other hand, for the well $$0 < x < L$$ we have $$\psi(x) = A\sin\left(\frac{n\pi x}{L}\right), n = 1, 2, 3,...$$ i.e., all the integers are good, and we need not to distinguish the odd an even solutions.

Finally, the energies are in both cases $$E_n = \frac{\hbar^2\pi^2}{2mL^2}n^2, n = 1, 2, 3, ...$$

Conclusion
Choosing a symmetric well allows to distinguish even and odd solutions. Of course, the solutions can be split into even and odd in both cases - more precisely, they are even and odd in respect to the center of the well, which in the latter case is at $$x=L/2$$. The difference is that in the former case one case the potential is chosen to have the same symmetry as the standard trigonometric functions, whereas in the latter case it is not. However, one could solve the equation in terms of displaced trigonometric functions: $$\cos(kx +\pi/2), \sin(kx +\pi/2)$$, obtaining again the separation into the even and odd states.

• But how can one solution be symmetric and even whereas other asymmetric and odd? – user175667 May 10 at 13:16
• What is wrong with it? If you draw the solutions, you will see that they look exactly the same. In the latter case they are still even and odd in respect to the center of the well, which is now at $x = L/2$. The difference is that in one case the potential and the trigonometric functions have the same symmetry, whereas in the othe they do not. However, one could solve the equation in terms of displace trigonometric functions: $\cos(kx +\pi/2), \sin(kx +\pi/2)$. – Vadim May 10 at 13:19

remember that sine and cosine aren't really that different $$\sin\left(x+n\frac \pi 2 \right)=\begin{cases}\pm\sin(x)\quad n \textrm{ even}\\ \pm\cos(x) \quad n \textrm{ odd} \end{cases}$$

so if you rewrite the interval $$0\leq x \leq L$$ as $$-L/2\leq x-L/2\leq L/2$$ and define $$y=x-L/2$$, which just means you translate the x axis, your solution becomes

$$\sin\left(\frac{n\pi x}{L}\right)=\sin\left(\frac{n\pi y+n\pi L/2}{L}\right)=\sin\left(\frac{n\pi y}{L}+n \frac{\pi}{2}\right) =\begin{cases}\pm\sin\left(\frac{n\pi y}{L}\right)\quad n \textrm{ even}\\ \pm\cos\left(\frac{n\pi y}{L}\right) \quad n \textrm{ odd} \end{cases}$$

the signs are accounted for by the normalizing constants.

Sine is just a shifted cosine and vice versa. So substituting $$x\to x-L/2$$ in the case where the well is not symmetric we get: $$\psi(x)=\sin\left(\frac{n\pi x}{L}-\frac{n\pi}{2}\right)\\ = \sin\left(\frac{n\pi x}{L}\right) \cos\left(\frac{n\pi}{2}\right) -\cos\left(\frac{n\pi x}{L}\right) \sin\left(\frac{n\pi}{2}\right)$$

Here we can see that for even $$n$$ only the sine term exists and for odd $$n$$ only the cos term exists.