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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?

Thanks in advance.

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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, $e^{i\alpha}=(-1)^m$.

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  • $\begingroup$ So should we multiply by $ (-1)^m $ in order to get the normalize condition? $\endgroup$
    – FreeZe
    Mar 4 at 10:25
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    $\begingroup$ It is not necessary, as $|(-1)^m\psi_n(x)|^2=|\psi_n(x)|^2$ $\endgroup$
    – AFG
    Mar 4 at 10:26

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