Given the equation ##\frac{d^2 \psi (x)}{{dt}^2}+\frac{2m}{{\hbar}^2}(E-V(x))=0## the general solution is:(adsbygoogle = window.adsbygoogle || []).push({});

$$\psi (x)=A_1 e^{ix \sqrt{\frac{2m}{{\hbar}^2}(E-V(x))}} +A_2 e^{-ix \sqrt{\frac{2m}{{\hbar}^2}(E-V(x))}}$$

If we have an infinite potential well: ## V(x)=\begin{cases} \infty \quad x\ge b\\ 0 \quad a < x < b \\ \infty \quad x \le a \end{cases}## then would that mean I take the limit of x to a and b and both should equal zero? So:

$$\lim_{x\rightarrow a}\left(A_1 e^{ix \sqrt{\frac{2m}{{\hbar}^2}(E-V(x))}} +A_2 e^{-ix \sqrt{\frac{2m}{{\hbar}^2}(E-V(x))}}\right)=0+\lim_{x\rightarrow a}\left( A_2 e^{x \sqrt{\frac{2m}{{\hbar}^2}(V(x)-E)}}\right)$$ where ## \lim_{x\rightarrow a}\left( A_2 e^{x \sqrt{\frac{2m}{{\hbar}^2}(V(x)-E)}}\right)## diverges. But how could that be zero? Would this mean that ##A_2## must be Zero?

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# I Schrödinger's Equation Infinite Potential Well

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