The momentum operator · Quantum Mechanics

Intuition

De Broglie said a particle of momentum $p$ is a wave of wavelength $h/p$ . A wave $e^{ikx}$ has wavelength $2\pi/k$ , so its momentum should be $\hbar k$ — and the operator that returns exactly that when it acts on the wave is $-i\hbar$ times the derivative. Momentum measures how fast the phase of the wavefunction turns as you move along the line.

On a spiral staircase, how steep the climb feels depends on how tightly it winds. The momentum of a wavefunction is how tightly its phase winds along the line.

The real part of a plane wave $e^{ikx}$ , one wavelength marked. Its wavelength is $\lambda=2\pi/k$ , and by de Broglie its momentum is $h/\lambda=\hbar k$ — exactly the eigenvalue of on it.

Momentum is a derivative

In the position representation the momentum operator is $-i\hbar$ times the derivative. Plane waves are its eigenfunctions, with eigenvalues $\hbar k$ , which is de Broglie's relation read backwards. It is Hermitian on normalisable wavefunctions, by integration by parts.

(\hat{p}\psi)(x)=-i\hbar\,\frac{d\psi}{dx},\qquad \hat{p}\,e^{ikx}=\hbar k\,e^{ikx}

What follows

$\langle p\rangle=\int\psi^{*}\bigl(-i\hbar\psi'\bigr)dx$ . A real wavefunction, or a real one times a constant phase, has .

The momentum operator is Hermitian

\int\phi^{*}\bigl(-i\hbar\,\psi'\bigr)dx=\int\bigl(-i\hbar\,\phi'\bigr)^{*}\psi\,dx

Integrate by parts to move the derivative from $\psi$ to $\phi^{*}$ . The boundary term vanishes because normalisable wavefunctions die away at infinity; the sign from integrating by parts and the sign from conjugating $-i$ cancel, so the operator lands on the bra side unchanged.

Proof steps

\int\phi^{*}\psi'\,dx=\bigl[\phi^{*}\psi\bigr]_{-\infty}^{\infty}-\int(\phi^{*})'\psi\,dx

Integration by parts.

\bigl[\phi^{*}\psi\bigr]_{-\infty}^{\infty}=0

Normalisable wavefunctions vanish at infinity.

\int\phi^{*}(-i\hbar\psi')\,dx=i\hbar\int(\phi')^{*}\psi\,dx

Multiply by $-i\hbar$ ; the minus signs combine.

i\hbar(\phi')^{*}=\bigl(-i\hbar\,\phi'\bigr)^{*}

Conjugating $-i\hbar$ gives $i\hbar$ .

\int\phi^{*}(\hat{p}\psi)\,dx=\int(\hat{p}\phi)^{*}\psi\,dx

So $\hat{p}$ acts the same way on either side: it is Hermitian.

Applications

Practice

Plane Waves Have Definite Momentum

The derivative of $e^{ikx}$ is $ik$ times itself, so $-i\hbar$ times the derivative returns $\hbar k$ times the wave.

-i\hbar\frac{d}{dx}e^{ikx}=\hbar k\,e^{ikx}

Try it

What is $\hat{p}\,e^{3ix}$ ?