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Question

Check to see whether the simple linear combination of sine and cosine functions

ψ=Asin(kx)+Bcos(kx)\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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Answered 5 months ago
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Time Independent Schrodinger Eq. (TISE){\textbf{Time Independent Schrodinger Eq. (TISE)}} is defined as follows:

22md2dx2Ψ(x)+V(x)Ψ(x)=EΨ(x)\dfrac{-\hbar^2}{2m}\dfrac{d^2}{dx^2}\Psi(x)+V(x)\Psi(x)=E\Psi(x)

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

22md2dx2Ψ(x)=EΨ(x)\dfrac{-\hbar^2}{2m}\dfrac{d^2}{dx^2}\Psi(x)=E\Psi(x)

Given

Ψ(x)=Asin(kx)+Bcos(kx)\Psi(x)=A\sin (kx)+B\cos(kx)

Since

22md2dx2Ψ(x)=22md2dx2(Asin(kx)+Bcos(kx))=22mddx(Akcos(kx)Bksin(kx))=22m(Ak2sin(kx)Bk2cos(kx))=2(k2)2m(Asin(kx)+Bcos(kx))=h2k22mΨ(x)(1)\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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