Quantum Mechanics · Lesson 08

A particle in a potential

For a particle on a line the Hamiltonian is kinetic energy plus potential energy, and the Schrödinger equation becomes a partial differential equation for the wavefunction. For a free particle its solutions are plane waves whose frequency grows as the square of the wavenumber, and a packet built from them drifts at the classical velocity while it spreads out — slowly for a grain of sand, astonishingly fast for an electron.

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i ℏ \frac{d}{d t} ∣ ψ ⟩ = H^∣ ψ ⟩

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Intuition

For a particle on a line the Hamiltonian is kinetic energy plus potential energy, and the Schrödinger equation becomes a partial differential equation for the wavefunction. For a free particle its solutions are plane waves whose frequency grows as the square of the wavenumber, and a packet built from them drifts at the classical velocity while it spreads out — slowly for a grain of sand, astonishingly fast for an electron.

Runners who start together at slightly different speeds drift apart as the race goes on. A wave packet is a crowd of plane waves with slightly different momenta, and it spreads for the same reason.

The probability density of a free Gaussian packet at $t=0$ and later. The centre has moved at the velocity and the packet has spread, keeping its area: a free particle's position becomes less certain as it goes, while its momentum distribution does not change.

H^∣ E ⟩ = E ∣ E ⟩

U^(t) = e^{- \frac{i H ^ t}{ℏ}}

ψ (t) = e^{- \frac{i E t}{ℏ}} ψ (0)

0

p_{0}/m

The Schrödinger equation on a line

With $\hat{H}=\hat{p}^{2}/2m+V(\hat{x})$ and the momentum operator of the last chapter, the Schrödinger equation becomes a partial differential equation for $\psi(x,t)$ . For a free particle, $V=0$ , the plane waves are solutions with frequency $\omega=\hbar k^{2}/2m$ .

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

The free particle

Plane waves $e^{i(kx-\omega t)}$ solve the free equation with $\hbar\omega=\hbar^{2}k^{2}/2m$ : energy is the kinetic energy of the momentum $\hbar k$ .
The crests move at the phase velocity $\omega/k=\hbar k/2m$ ; a packet as a whole moves at the group velocity $d\omega/dk=\hbar k/m=p/m$ , the classical velocity.
A Gaussian packet of initial width $\sigma_{0}$ spreads as $\sigma(t)=\sigma_{0}\sqrt{1+\bigl(\hbar t/2m\sigma_{0}^{2}\bigr)^{2}}$ .
Its momentum distribution never changes, since momentum is conserved for a free particle.
The Differential Equations course met partial differential equations only by name; this is the one the rest of the course solves.

Plane waves solve the free equation

\psi=e^{i(kx-\omega t)}\ \text{solves}\ i\hbar\partial_{t}\psi=-\tfrac{\hbar^{2}}{2m}\partial_{x}^{2}\psi \iff\ \hbar\omega=\frac{\hbar^{2}k^{2}}{2m}

Differentiate the plane wave once in time and twice in space. Each derivative brings down a factor, and the equation becomes a relation between the frequency and the wavenumber: the dispersion relation, which is the energy of a free particle written through $E=\hbar\omega$ and $p=\hbar k$ .

Proof steps

i\hbar\,\partial_{t}\psi=i\hbar(-i\omega)\psi=\hbar\omega\,\psi

One time derivative brings down $-i\omega$ .

-\tfrac{\hbar^{2}}{2m}\partial_{x}^{2}\psi=-\tfrac{\hbar^{2}}{2m}(ik)^{2}\psi=\tfrac{\hbar^{2}k^{2}}{2m}\psi

Two space derivatives bring down $(ik)^{2}=-k^{2}$ .

\hbar\omega=\frac{\hbar^{2}k^{2}}{2m}

The two sides agree exactly when this holds.

E=\frac{p^{2}}{2m}\quad\text{with}\quad E=\hbar\omega,\ p=\hbar k

It is the classical kinetic energy, in the language of waves.

Applications

Practice

Kinetic Plus Potential

For a particle on a line the Hamiltonian is $\hat{p}^{2}/2m+V$ , and $\hat{p}$ is $-i\hbar$ times the derivative. The Schrödinger equation becomes a partial differential equation for the wavefunction.

\hat{H}=-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}+V(x)

Try it

Which is the Schrödinger equation for a particle of mass $m$ in a potential $V(x)$ ?