Gaussian wave packets · Quantum Mechanics

Intuition

A wave packet is a plane wave with an envelope: a phase winding at a steady rate, which gives it a momentum, inside a bump, which gives it a place. With a Gaussian envelope something special happens — the momentum density is a Gaussian too, and the product of the two spreads is exactly $\hbar/2$ , the smallest the uncertainty relation allows. The Gaussian is the least uncertain state there is.

A single ripple travelling across a pond: the crests inside it say how fast it moves, the fading at its edges says where it is. The Gaussian packet is the ripple that is as local and as definite in speed as nature permits at once.

The real part of a Gaussian packet $e^{-x^{2}/4\sigma^{2}}e^{ik_{0}x}$ with $\sigma=1$ and $k_{0}=4$ . The carrier inside sets the average momentum $\hbar k_{0}$ ; the dashed envelope sets the spread $\Delta x=\sigma$ .

The minimum-uncertainty state

A Gaussian wave packet centred at $x_{0}$ with average momentum $p_{0}$ has a Gaussian position density of standard deviation $\sigma$ and a Gaussian momentum density of standard deviation . Its spreads multiply to exactly .

\psi(x)=\frac{1}{(2\pi\sigma^{2})^{1/4}}\,e^{-(x-x_{0})^{2}/4\sigma^{2}}\,e^{ip_{0}x/\hbar}

Its numbers

$\langle x\rangle=x_{0}$ , $\Delta x=\sigma$ ; $\langle p\rangle=p_{0}$ , .

Only Gaussians meet the bound

\Delta x\,\Delta p=\tfrac{\hbar}{2}\ \Longrightarrow\ \psi(x)=C\,e^{-\lambda(x-\langle x\rangle)^{2}/2\hbar}\,e^{i\langle p\rangle x/\hbar}

Equality in the uncertainty relation needs equality in both inequalities of its proof: the Cauchy–Schwarz step forces the momentum deviation to be a multiple of the position deviation, and the step that kept only the imaginary part forces that multiple to be imaginary. What is left is a first-order differential equation, whose solution is a Gaussian times a winding phase.

Proof steps

(\hat{p}-\langle p\rangle)\ket{\psi}=c\,(\hat{x}-\langle x\rangle)\ket{\psi}

Cauchy–Schwarz is an equality only when one vector is a multiple of the other.

c=i\lambda,\qquad \lambda>0

Dropping the real part of the overlap lost nothing only if the overlap is imaginary; normalisability picks the sign.

-i\hbar\psi'-\langle p\rangle\psi=i\lambda\bigl(x-\langle x\rangle\bigr)\psi

Write the operators in the position representation.

\frac{\psi'}{\psi}=\frac{i\langle p\rangle}{\hbar}-\frac{\lambda\bigl(x-\langle x\rangle\bigr)}{\hbar}

Divide by $-i\hbar\psi$ : a separable equation, as in the Differential Equations course.

\psi=C\,e^{i\langle p\rangle x/\hbar}\,e^{-\lambda(x-\langle x\rangle)^{2}/2\hbar}

Integrate: a Gaussian times a plane wave.

Applications

Practice

The Two Spreads

A Gaussian packet of position spread $\sigma$ has momentum spread $\hbar/2\sigma$ . The narrower the packet, the wider its momentum.

\Delta p=\frac{\hbar}{2\sigma}

Try it

A Gaussian packet has $\sigma=2$ nm. What is $\Delta p$ , in units of $\hbar$ per nm?