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Question

# Check to see whether the simple linear combination of sine and cosine functions$\psi=A \sin (k x)+B \cos (k x)$satisﬁes the time-independent Schrodinger equation for a free particle (V = 0).

Solution

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${\textbf{Time Independent Schrodinger Eq. (TISE)}}$ is defined as follows:

$\dfrac{-\hbar^2}{2m}\dfrac{d^2}{dx^2}\Psi(x)+V(x)\Psi(x)=E\Psi(x)$

For a free particle where $V=0$, $\textbf{TISE}$ becomes as follows:

$\dfrac{-\hbar^2}{2m}\dfrac{d^2}{dx^2}\Psi(x)=E\Psi(x)$

Given

$\Psi(x)=A\sin (kx)+B\cos(kx)$

Since

\begin{align*} \dfrac{-\hbar^2}{2m}\dfrac{d^2}{dx^2}\Psi(x)&=\dfrac{-\hbar^2}{2m}\dfrac{d^2}{dx^2}\Big(A\sin (kx)+B\cos(kx)\Big)\\\\ &=\dfrac{-\hbar^2}{2m}\dfrac{d}{dx}\Big(Ak\cos (kx)-Bk\sin(kx)\Big)\\\\ &=\dfrac{-\hbar^2}{2m}\Big(-Ak^2\sin(kx)-Bk^2\cos(kx)\Big)\\\\ &=\dfrac{-\hbar ^2\cdot(-k^2)}{2m}\Big(A\sin(kx)+B\cos (kx)\Big)\\\\ &=\dfrac{h^2k^2}{2m}\Psi(x)\dotsc(1) \end{align*}

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