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I'm solving a quantum mechanics problem for the particle in a potential well, and the equation I have to solve is $\frac{d^2\psi}{dx^2}+k\psi=0$where $k=\frac{2mE}{\hbar^2}$ This seems easy enough to solve. It is a second order linear differential equation with constant coefficient of the form $a\psi''(x)+b\psi'(x)+c\psi(x)=0$, so I thought we were to use the characteristic equation $ar^2+br+c=0$and solve for roots $r_1$ and $r_2$. Doing that, I get $r^2+k=0$and therefore $r=\pm \sqrt{k}$ The general solution is given by $\psi(x)=\exp\left(\sqrt{\frac{2mE}{\hbar^2}}x\right)+\exp\left(-\sqrt{\frac{2mE}{\hbar^2}}x\right)$

However, when I refer to Griffiths' Introduction to Quantum Mechanics, he finds the general solution to be $\psi(x)=A\sin kx+B\cos kx$ Where have I gone wrong? Thanks.

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    Yeah, the things you are exponentiating should be imaginary, otherwise your wavefunction will grow without bound and you will violate your boundary conditions.2012-12-08

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A simple mistake. You need: $r=\pm \sqrt{-k}$ So the general solution is given by: $\psi(x)=a\exp\left(\sqrt{k}ix\right)+b\exp\left(-\sqrt{k}ix\right)$ So by Euler's formula and a change in the constants: $\psi(x)=A\sin \sqrt{k}x+B\cos \sqrt{k}x$

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Solving $r^2+k=0$ yields $r=\pm \sqrt{- k}$, two imaginary values. Let $\sqrt{k} = \omega$.

The solutions take the form $ \psi(x) = ae^{i\omega} + be^{-i\omega}$

and using Euler's formulas for sine and cosine, we can rewrite them as $ \psi(x) = a(\cos\omega + i\sin\omega) + b(\cos\omega - i \sin\omega)$

$ = (a+b)\cos\omega + (a-b)i\sin\omega .$

Because $a$ and $b$ were arbitrary constants, we can use new constants to write the solution set as $ \psi(x) = A\sin(\omega) + B\cos(\omega) $ for complex $A$ and $B$. However, it will be practical to just consider the real solutions of $\psi$.

(Note that the final result you provided is inconsistent with your original notation for the Schroedinger equation. The $k$ in the general solution griffiths gives is the square root of the $k$ as you used it in the Schroedinger equation. I would have set $k^2 = \frac{2mE}{\hbar ^2}$.)

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    $\psi:\mathbb{R}\rightarrow\mathbb{C}$, so all of the constants in this problem are complex. $(a-b)i$ could be real, for example.2012-12-08