Quantum Free Particle

Problem

Consider a quantum free particle characterized by the complex plane wave $\Psi (x,t)=A{e}^{i(kx-\omega t)}$. Show that the free particle satisfies the Schrodinger equation $i\hbar {\frac {\partial \Psi }{\partial t}}={\frac {-{\hbar }^{2}}{2m}}{\frac {{\partial }^{2}\Psi }{\partial {x}^{2}}}+V(x)\Psi (x,t)$ for some potential function $V(x)$.

Solution

The Hamiltonian of a system is $H=T+V$ where $T$ is the kinetic energy and $V$ is the potential energy. Since the quantity $H$ is the total energy, let us rewrite the Hamiltonian as $E={\frac {{p}^{2}}{2m}}+V(x)$.

Now, we take the derivatives:

${\frac {\partial \Psi }{\partial t}}=-i\omega A{e}^{i(kx-\omega t)}=-i\omega \Psi (x,t)$

${\frac {{\partial }^{2}\Psi }{\partial {x}^{2}}}=-{k}^{2}A{e}^{i(kx-\omega t)}=-{k}^{2}\Psi (x,t)$

Since $p={\frac {2\pi \hbar }{\lambda }}$ and $k={\frac {2\pi }{\lambda }}$, where $k$ is the wavenumber and $\lambda$ is the wavelength, we have $k={\frac {p}{\hbar }}$.

Therefore,

${\frac {{\partial }^{2}\Psi }{\partial {x}^{2}}}=-{\frac {{p}^{2}}{{\hbar }^{2}}}\Psi (x,t)$

Next, we multiply $\Psi (x,t)$ to the Hamiltonian:

$E\Psi (x,t)={\frac {{p}^{2}}{2m}}\Psi (x,t)+V(x)\Psi (x,t)$

Notice that the above equation can be expressed as

$E\Psi (x,t)={\frac {-{\hbar }^{2}}{2m}}{\frac {{\partial }^{2}\Psi }{\partial {x}^{2}}}+V(x)\Psi (x,t)$

Since the energy of matter waves is given by $E=\hbar \omega$, we can show that

$E\Psi (x,t)={\frac {\hbar \omega }{-i\omega }}\Psi (x,t)$

Combining all the right parts, we assemble the Schrödinger equation:

$i\hbar {\frac {\partial \Psi }{\partial t}}={\frac {-{\hbar }^{2}}{2m}}{\frac {{\partial }^{2}\Psi }{\partial {x}^{2}}}+V(x)\Psi (x,t)$