# Symmetric infinite well potential

Assume we have the following potential :

$$V\left(x\right)=\begin{cases} 0 & -\frac{L}{2}\leq x\leq\frac{L}{2}\\ \infty & else \end{cases}$$

The wave function for a particle in this well potential, is given by :

$$\psi_{even}=\sqrt{\frac{2}{L}}\cos\left(\frac{n\pi x}{L}\right),\thinspace\thinspace\psi_{odd}=\sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)$$

Where $$n$$ is even for the sin function, and $$n$$ is odd for the cos function, as we can see the calculation here

(the last comment)

But I tried to find the odd and even wavefunctions by shifting the wavefunction of a particle in an infinite box (which is not symmetric), and got different result. I'll be glad if someone can explain the difference. Here's my calculation:

We know that for a particle in a potential $$V\left(x\right)=\begin{cases} 0 & 0\leq x\leq L\\ \infty & else \end{cases}$$

The wave function is given by

$$\psi\left(x\right)=\sqrt{\frac{2}{L}}\sin\left(\frac{\pi n}{L}x\right)$$

For a symmetric potential we'll shift the function to the left by $$x\to x+\frac{L}{2}$$ (we can see that for $$x=-L/2$$ and for $$x=L$$ the wave function is $$0$$ so it preserved).

Now $$\psi\left(x\right)=\sqrt{\frac{2}{L}}\sin\left(\frac{\pi n}{L}\left(x+\frac{L}{2}\right)\right)=\sqrt{\frac{2}{L}}\sin\left(\frac{\pi n}{L}x+\frac{\pi n}{2}\right)$$

If we'll write for odd $$n$$ , $$n=2m+1$$ and for even $$n$$, $$n=2m$$, we'll get the solution:

$$\psi\left(x\right)=\sqrt{\frac{2}{L}}\sin\left(\frac{\pi n}{L}\left(x+\frac{L}{2}\right)\right)=\sqrt{\frac{2}{L}}\sin\left(\frac{\pi n}{L}x+\frac{\pi n}{2}\right)=\begin{cases} \psi_{n=2m}=\left(-1\right)^{m}\sqrt{\frac{2}{L}}\sin\left(\frac{\pi n}{L}x\right)\\ \psi_{n=2m+1}=\left(-1\right)^{m}\sqrt{\frac{2}{L}}\cos\left(\frac{\pi n}{L}x\right) \end{cases}$$

Which is different from the previous result (by a factor of $$-1$$ in some of the functions)

What went wrong?

Nothing went wrong, the difference between both results is just a global constant phase, which has no meaning. If $$\psi_n(x)$$ is an eigenfunction of the Hamiltonian, so is $$e^{i \alpha}\psi_n(x)$$. In your case, $$e^{i\alpha}=(-1)^m$$.
• So should we multiply by $(-1)^m$ in order to get the normalize condition? Mar 4 '21 at 10:25
• It is not necessary, as $|(-1)^m\psi_n(x)|^2=|\psi_n(x)|^2$