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I understand mathematically how $\sqrt{2/L} \ \sin\left({k_n x}\right)$ satisfies the boundary conditions for the infinite square well in terms of the fact that $\psi(0) = \psi(a) = 0$, and excuse the naiveté of such a question, but if I pull up the image of $\sqrt{2/L} \ \sin\left({k_n x}\right)$ on an online calculator, it is this:

enter image description here

So, is $\sqrt{2/L} \ \sin\left({k_n x}\right)$ a solution, or:

$\psi_n(x) = \begin{cases} \sqrt{2/L} \ \sin\left({k_n x}\right), & \text{$0 \le x \le L$} \\ 0 & \text{otherwise} \end{cases}$

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    $\begingroup$ The second one is the correct answer as you have guessed ;) $\endgroup$ Nov 3 '18 at 15:25
  • $\begingroup$ @SahandTabatabaei Why do we then typically state the solution is merely $\sqrt{2/L} \ \sin\left({k_n x}\right)$? Is the piecewise function implicit? Is this an abuse of notation? $\endgroup$
    – sangstar
    Nov 3 '18 at 15:26
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    $\begingroup$ It's usually implicitly assumed that the only range of the $x$ axis that we're interested in is the $(0,L)$ interval, since outside this range the wavefunction is trivially zero. $\endgroup$ Nov 3 '18 at 15:29
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The wavefunction

$\psi_n(x) = \begin{cases} \sqrt{2/L} \ \sin\left({k_n x}\right), & \text{$0 \le x \le L$} \\ 0 & \text{otherwise} \end{cases}$

is the solution. What you've plotted is $\sqrt{2/L} \ \sin\left({k_n x}\right)$ for $-\infty \le x \le \infty$.

The actual wavefunction will look like

enter image description here

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