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The solution is:

$A{e}^{+\sqrt{{\mathrm{c}}_{\mathrm{x}}\mathrm{x}}}+B{e}^{-\sqrt{{\mathrm{c}}_{\mathrm{x}}\mathrm{x}}}$

The infinity well problem is one of the important problems in quantum mechanics that help us to understand other phenomena.

The wave function outside the well is zero.

The given solution is:

$A{e}^{+\sqrt{{\mathrm{c}}_{\mathrm{x}}\mathrm{x}}}+B{e}^{-\sqrt{{\mathrm{c}}_{\mathrm{x}}\mathrm{x}}}$

It can also be written as:

$Aexp\left(+\sqrt{c_{x}x}\right)+Bexp\left(-\sqrt{c_{x}x}\right)$

These equations at two points $x_1,x_2$can be written as:

role="math" localid="1659763359765" $$\begin{gathered} Aexp\left(+\sqrt{c_x{x}_1}\right)+Bexp\left(-\sqrt{c_x{x}_1}\right) \\ Aexp\left(+\sqrt{c_x{x}_2}\right)+Bexp\left(-\sqrt{c_x{x}_2}\right) \end{gathered}$$

Solve each equation for B/A as:

role="math" localid="1659763407252" $$\begin{gathered} \frac{\mathrm{B}}{\mathrm{A}}=\exp \left(2\sqrt{{\mathrm{c}}_{\mathrm{x}}}{\mathrm{x}}_1\right) \\ \frac{\mathrm{B}}{\mathrm{A}}=\exp \left(2\sqrt{{\mathrm{c}}_{\mathrm{x}}}{\mathrm{x}}_2\right) \end{gathered}$$

From the above two expressions, we can conclude that,

$\exp \left(2\sqrt{{\mathrm{c}}_{\mathrm{x}}}{\mathrm{x}}_2\right)=\exp \left(2\sqrt{{\mathrm{c}}_{\mathrm{x}}}{\mathrm{x}}_1\right)$

This is possible only if$x_1=x_2$ . (i.e., only one point) but this assumption does not happen in general cases. Thus, this is quite a contradiction to the assumption that the function is zero at two different points$x_1$ and $x_2$.