Summary Notes of Free Particle and Momentum Operator

Free Particle

Function of a standing wave. Where k tells about the p and direction of the wave.

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

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

apply the operator on the wavefunction

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

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, therefore \(k\) and \(E\) values can be anything. No restrictions/boundaries means not quantized.

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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 \(p = \frac{\hbar}{i} \frac{d}{dx}\).

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

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

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

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

confirms that this is a reasonable match to classical physics

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

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Page Author: Juliette