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I have solved $H|\psi\rangle=E_{n}|\psi\rangle$ with $V(x)=0$ from $-a<x<a$ and $\infty$ otherwise. If I propose a solution of the form $\psi(x)=A_{n}e^{ikx}+B_{n}e^{-ikx}$ I arrive to the solution $$ \psi(x) = \frac{1}{\sqrt{L}} \sin \left( \frac{n\pi}{2} \left(\frac{x}{L}-1 \right) \right)$$ for $n$ natural. If I propose a solution of the form $$\psi(x)=A_{n}\sin(kx)+B_{n}\cos(kx) \, .$$ I arrive to the solution $$ \psi(x) =\begin{cases} \frac{1}{\sqrt{a}}\cos \left(\frac{n\pi x}{2a} \right) & \text{n odd} \\ \frac{1}{\sqrt{a}}\sin \left(\frac{n\pi x}{a} \right) & \text{n even } \end{cases} $$

Both solutions with the same energies. However, when plotting both solutions, I see they're not equal. But I cannot find any mistakes on my procedures.

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    $\begingroup$ I think your L =a the width of the well.Try expanding the first solution in terms of sine and cosine function .I think it will be general solution (a combination of two solutions for n odd and even) $\endgroup$ – drvrm May 19 at 20:19
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    $\begingroup$ They are equal up to a phase factor of $-1$, so they are basically equal. Or equivalently $A_n, B_n$ are only fixed by the boundary conditions and normalization up to a phase factor $e^{i\phi}$. $\endgroup$ – lomby May 19 at 20:21
  • $\begingroup$ The solution for n=odd matches my solution, but for n=even, plots are not equal even if consider an extra minus sign. However, the second solution for n=2 matches the first solution for n=4 $\endgroup$ – Juan Pablo Arcila May 19 at 21:05
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Solutions coincide for n=odd and an extra minus sign (wavefunctions are the same up to a phase factor, in this case $e^{i\pi}$). For n=even the solution from the piecewise function is the same than the first one for N=2n. Therefore they describe the same physical solution

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