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The wave function for a particle in a infinite potential well located at $0$ and $L$ is $$\psi_n=\sqrt{\frac2L}\sin\frac{n\pi x}L$$

I want to find our wave function for the same situation where the walls are located at $-L$ and $+L$

I can write the Schrödinger equation, but I am unable to apply the boundary conditions, so I am unable to find the wave function.

How can I find the wave function in case of walls at $-L$ and $+L$?

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I will answer this question by copying a passage out of the very good QM book bei Shankar (see here) which I can recommend wholeheartedly to anyone trying to learn quantum mechanics. I considered writing up a solution myself but since the passage answers your question exactly (ok, not quite exactly since the calculation is performed for the infinite wells being located at $-\frac{L}{2}$ and $\frac{L}{2}$ but that doesn't change anything important) it would have been a waste of time and effort to do so. If you have any questions about the derivation I'm happy to help you.

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It is very simple, there is no need to solve it again since you already know the solution of infinite square well. The idea is very simple, the trap looks the same, but wider now $2L$. So first you should extend it by a factor of two and then move it back to the center, it is given by the rescaling and translation: $$L \mapsto 2L \tag{1}$$ $$x \mapsto x-L \tag{2}$$

So the solution is: $$\psi_{n}=\sqrt{\frac{1}{L}}\sin\left(\frac{n\pi x}{2L}-\frac{n\pi}{2}\right) \tag{3}$$ $$E_{n}=\frac{n^2\pi^2 \hbar^2}{8mL^2}$$ As expected. Note that $\sin(\pi/2-x)=\cos(x)$ so that you can split it into odd and even series, but the Eq (3) given above is more succinct.

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  • $\begingroup$ I think it useful to note that there is a lowest possible energy level. The uncertainty principle, knowing the distance between the walls, allows us to compute the minimum momentum the particle can have. Since $E = p^2/2m$, you have the lowest energy. The rest is as hwlau has it. $\endgroup$ – Paul J. Gans Aug 17 '13 at 1:12

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