Introduction: From Classical to Quantum Momentum
Classically, momentum is \(p = mv\). In quantum mechanics, momentum becomes a differential operator. Its form is dictated by the de Broglie relation and the structure of Hilbert space.
Definition
In position representation, the momentum operator is:
\[\hat{p} = -i\hbar\frac{d}{dx}\]It differentiates the wavefunction and multiplies by \(-i\hbar\).
Why This Form?
The de Broglie relation says \(p = \hbar k\) for wavelength \(\lambda = 2\pi/k\). A plane wave \(e^{ikx}\) should have definite momentum \(p = \hbar k\):
\[\hat{p}e^{ikx} = -i\hbar\frac{d}{dx}e^{ikx} = -i\hbar(ik)e^{ikx} = \hbar k \cdot e^{ikx}\]The plane wave is indeed an eigenstate with eigenvalue \(\hbar k\). ✓
Worked Examples
Example 1: Momentum Eigenstates
Solve \(\hat{p}|p_0\rangle = p_0|p_0\rangle\) in position representation:
\[-i\hbar\frac{d\psi_{p_0}}{dx} = p_0\psi_{p_0}\] \[\frac{d\psi}{dx} = \frac{ip_0}{\hbar}\psi \Rightarrow \psi_{p_0}(x) = Ae^{ip_0 x/\hbar}\]Plane waves are momentum eigenstates!
Example 2: Expectation Value of Momentum
\[\langle\hat{p}\rangle = \int_{-\infty}^{\infty} \psi^*(x)\left(-i\hbar\frac{d\psi}{dx}\right) dx\]For a real, symmetric \(\psi(x)\) centered at origin: \(\langle\hat{p}\rangle = 0\).
Example 3: Momentum of a Gaussian
For \(\psi(x) = Ne^{-(x-x_0)^2/2a^2}e^{ip_0 x/\hbar}\):
\[\langle\hat{p}\rangle = p_0\]The phase factor \(e^{ip_0 x/\hbar}\) gives the wavepacket average momentum \(p_0\).
The Quantum Connection
The momentum operator generates spatial translations:
\[e^{-ia\hat{p}/\hbar}\psi(x) = \psi(x - a)\]This deep connection between momentum and translation comes from Noether's theorem: translation symmetry implies momentum conservation. In quantum mechanics, the operator that generates a symmetry is the conserved quantity.