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Given a potential well of $V = 0$ on the interval $(0,L)$ and $V = \infty$ outside the well, I am working to solve the Time Independent Schrodinger Equation $$\dfrac{d^2}{dx^2} \psi= \dfrac{-2mE}{\hbar^2}\psi=-k^2\psi.$$ While the sine/cosine solution may be easier, my question is how to do it with exponentials.

With the ansatz $\psi \propto \exp(\alpha x)$ we can write $\alpha^2=-k^2$ so $\alpha = \pm ik$ and the general solution is $$\psi (x)= Ae^{ikx}+Be^{-ikx}.$$

From the boundary condition $\psi (0)=0$, we see that $$\psi (0)= Ae^{0}+Be^{0}=A+B=0 \Rightarrow B = -A$$ so $$\psi (x)= Ae^{ikx}-Ae^{-ikx}.$$ Then, applying the boundary condition at x=L, $$\psi (L)= A(e^{ikL}-e^{-ikL})=0.$$

I am unsure of how to evaluate this last part. $A= 0$ is not valid, so what is the solution to this last equation?

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  • $\begingroup$ Then exp(i k L) = exp(-i k L) ----> exp(2 i k L) = 1 ----> k= n pi/L for n a positive integer. $\endgroup$ – Count Iblis Mar 30 '15 at 20:18
  • $\begingroup$ Note that by the time you're at your last step, you can substitute $A(e^{ikL} - e^{ikL}) = 2A\sinh(ikL) = 2Ai\sin(kL)$ $\endgroup$ – Jerry Schirmer Mar 30 '15 at 20:20
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Solving $$e^{ikL}-e^{-ikL}=0$$ we write: $$e^{ikL}=e^{-ikL}$$ then, dividing both sides of the equation by $e^{-ikL}$ we find that $$\frac{e^{ikL}}{e^{-ikL}}=e^{2ikL}=1$$

From Euler's Formula, $e^{i\theta}=\cos(\theta)+i\sin(\theta)$ the solution is that k is quantized: $k =\frac{n\pi}{L}$ for positive integer n.

Thus $$\psi (x)= A(e^{ikx}-e^{-ikx}) = \sqrt{\frac{2}{a}}(e^{ikx}-e^{-ikx})$$ due to the normalization condition, where k is quantized as above.

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