Question

Free quantum particle. In Quantum Mechanics, is the time-independent Schrodinger's equation for a free particle in one dimension. In this equation. is the wavefnnction of the particle, m is its mass. E is its (kinetic) energy, while is the fundamental Planck constant.

Answer 1

$-\frac{\bar{h}^2}{2m}\frac{d^2\Psi }{dx^2}=E\Psi$

$\Rightarrow \frac{d^2\Psi }{dx^2}=-\frac{2mE}{\bar{h}^2}\Psi$

This equation is of the form,

$\frac{d^2y }{dx^2}=-k^2y$

And its soluton is given by,

$y=Asin(kx)+Bcos(kx)$

In this case,

$k^2=\frac{2mE}{\bar{h}^2}$

$\Rightarrow k=\frac{\sqrt{2mE}}{\bar{h}}$

So, we have our solution,

$\Psi(x) =Asin(kx)+Bcos(kx)$

$\Psi (0)=0$

so we have,

$\Psi(0) =Asin(k*0)+Bcos(k*0)$

$\Rightarrow B=0$

$\Psi(x) =Asin(kx)+Bcos(kx)$

$\Rightarrow {\Psi}'(x) =Akcos(kx)-Bksin(kx)$

${\Psi }'(0)=1$

So, from this we have,

$\Rightarrow {\Psi}'(0) =Akcos(k*0)-Bksin(k*0)$

$\Rightarrow 1=Ak$

$\Rightarrow A=\frac{1}{k}$

So, our wave equation is given by,

$\Psi(x) =\frac{1}{k}sin(kx)\; where\;k=\frac{\sqrt{2mE}}{\bar{h}}$

Or putting the value of k,

$\Psi(x) =\frac{\bar{h}}{\sqrt{2mE}}sin(\frac{\sqrt{2mE}}{\bar{h}}x)$

b)

For spatial period, that is the wavelength, we find the distance between the two alternate points where this function vanishes

Wave Wavelength Wavelength Distance

So,

$\Psi (x)=0$

$\Rightarrow \frac{\bar{h}}{\sqrt{2mE}}sin(\frac{\sqrt{2mE}}{\bar{h}}x)=0$

$\Rightarrow sin(\frac{\sqrt{2mE}}{\bar{h}}x)=0$

$\Rightarrow \frac{\sqrt{2mE}}{\bar{h}}x=\pm n\pi$

Distance between alternate point = 2*pi,

So,

$\frac{\sqrt{2mE}}{\bar{h}}\lambda =2\pi$

$\Rightarrow \lambda =\frac{2\pi\bar{h}}{\sqrt{2mE}}$

c)

Given that,

$E=\frac{p^2}{2m}$

So, we get,

$\Rightarrow \lambda =\frac{2\pi\bar{h}}{\sqrt{2m*\frac{p^2}{2m}}}$

$\Rightarrow \lambda =\frac{2\pi\bar{h}}{p}$