Summary Notes of Free Particle and Momentum Operator

Free Particle

Function of a standing wave is:

\[\Psi(x) = A * e^{i k x} + B * e^{-i k x}\]

Solve for the energy of the wave by applying the hamiltonian to Ψ

Apply the operator on the wavefunction

\[H\Psi = -\frac{\hbar^2}{2m} \nabla^2(A e^{i k x} + B e^{-i k x})\] \[= -\frac{\hbar^2}{2m}(A(i k)^2 e^{i k x} + B(-i k)^2 e^{-i k x})\] \[= -\frac{\hbar^2}{2m}(-A k^2 e^{i k x} - B k^2 e^{-i k x})\] \[E\Psi = -\frac{\hbar^2}{2m}(-k^2)(A e^{i k x} + B e^{-i k x})\]

where $E$ is the Energy (eigenvalue) and the wavefunction is returned (eigenfunction)

\[E = \frac{\hbar^2 k^2}{2m}(A e^{i k x} + B e^{-i k x})\]

The free particle has no boundaries and is not confined in space, the spatial frequency, $k$, and $E$ values are not restricted and can be any number. Therefore, since they are not restricted, the system is not quantized.

Momentum Operator

Define momentum ($\rho$) via the de Broglie relation where $\lambda = \frac{2\pi}{k}$ relating the wavelength ($\lambda$) to the spatial frequency ($k$).

\[\rho = \frac{h}{\lambda}\] \[\rho = \frac{h}{\lambda} = h*k\] \[\frac{2\pi}{\lambda} = \hbar * k\] \[\rho = \hbar * k\]

Momentum operator ($\rho$):

The momentum operator makes changes to the wavefunction. Ask for momentum and get back the eigenfunction ($\Psi$) and eigenvalue ($\rho$) from operation.

Confirm this with an example where $\rho = \frac{\hbar}{i} \frac{d}{dx}$.

$\hat{\rho} \Psi =$\rho(A e^{i k x}) = \frac{\hbar}{i} \frac{d}{dx}(A e^{i k x}) = \frac{\hbar}{i}A(i k)e^{i k x} = \hbar k A e^{i k x} = \rho A e^{i k x}$$

This returns the scalar momentum ($\rho$) times the eigenfunction ($A e^{i k x}$).

Now confirm for $\rho^2$ where the Energy for classical physics is

\[E = \frac{1}{2}mv^2 = \frac{\rho^2}{2m}\] \[\hat{\rho^2} = \left(\frac{\hbar}{i}\frac{d}{dx}\right)^2 = \frac{\hbar^2}{i^2}\frac{d^2}{dx^2} = -\hbar^2 \frac{d^2}{dx^2}\] \[\frac{d^2}{dx^2} = \nabla^2\]

confirms that this is a reasonable match to classical physics

\[\frac{\hat{\rho^2}}{2m} = H = -\frac{\hbar^2}{2m} \nabla^2\]