Heisenberg Uncertainty Principle

Consider a long string which contains a wave that moves with a fairly well-defined wavelength across the whole length of the string. The question, "Where is the wave?", does not seem to make much sense, since it is spread throughout the length of string. A quick snap of the wrist and the string produces a small bump-like wave which has a well defined position. Now the question, "What is the wavelength?", does not make sense, since there is no well defined period. This example illisturates the limitation on measuring the wavelength and the position simultaneously. Relating the wavelength to momentum yields the de Broglie equation, which is applicable to any wave phenomenon, including the wave equation:

$p=\frac{h}{\lambda}=\frac{2\pi \hbar}{\lambda}$

The above discussion suggests that there is a relation between the uncertainty in position Δx of a particle and that of the canonically conjugate momentum Δp, both defined as the standard deviations of the respective measurements. This relation is known as the Heisenberg Uncertainty Principle:

$\Delta x\,\Delta p \geq \frac{\hbar}{2}$

Similar uncertainty relations exist for other non-commuting observables; we will derive a general uncertainty relation, of which the above is a special case, in a later section.

Two variables are said to be canonically conjugate if they are related by a Fourier transform. More specifically, they are variables such that, when one takes the Fourier transform of the wave function in terms of one of the pair, the result is the wave function in terms of the other. For example, as implied above, position and momentum are canonically conjugate variables:

$\Phi(p,t) = \frac{1}{\sqrt{2\pi\hbar}}\int^{\infty}_{-\infty}e^{-ipx/\hbar}\Psi(x,t)\,dx;$ $\Psi(x,t) = \frac{1}{\sqrt{2\pi\hbar}}\int^{\infty}_{-\infty}e^{ipx/\hbar}\Phi(p,t)\,dp.$

A similar pair of transforms holds for the time-independent case:

$\phi(p) = \frac{1}{\sqrt{2\pi\hbar}}\int^{\infty}_{-\infty}e^{-ipx/\hbar}\psi(x)\,dx;$ $\psi(x) = \frac{1}{\sqrt{2\pi\hbar}}\int^{\infty}_{-\infty}e^{ipx/\hbar}\phi(p)\,dp$

It is precisely this relationship that leads to the uncertainty principle. We may illustrate this by considering a Gaussian wave packet in position space:

$\psi(x)=\frac{1}{\sqrt{\Delta x\sqrt{2\pi}}}e^{-x^2/4(\Delta x)^2}$

If we go to momentum space, we will find that

$\phi(p)=\frac{1}{\sqrt{\Delta p\sqrt{2\pi}}}e^{-p^2/4(\Delta p)^2},$

where

$\Delta x\,\Delta p=\frac{\hbar}{2}.$

This is just the Heisenberg Uncertainty Principle quoted above, with the inequality turned into an equality. While the relation obtained for Gaussian wave packets will not necessarily hold true in general, we will see later that the uncertainties will always be given by the standard deviations of the observables.

We thus have a true physical constraint on a wave packet - if we compress it in one variable, it expands in the other! If it is compressed in position (i.e., localized) then it must must be spread out in wavelength. Similarly, if it is compressed in momentum, it is spread out in space.

The natural question to ask now is what the physical meaning of the Heisenberg Uncertainty Principle. Suppose that we prepare N systems in the same state. For half of them, we measure their positions x; for the other half, we measure their momenta px. No matter how we prepare the state of these systems, the distributions obtained obey these inequalities. These are intrinsic properties of the quantum description of the state of a particle; the uncertainty principle has nothing to do with the accuracy of the measurements. We will still see this relation hold even if each measurement is done with infinite precision. They also have nothing to do with the perturbation that a measurement causes to a system, inasmuch as each particle is measured only once.

In other words, the position and momentum of a particle are defined numerically only within limits that obey these inequalities. There exists some “fuzziness” in the numerical definition of these two physical quantities. If we prepare particles all at the same point, they will have very different momenta. If we prepare particles with a well-defined momentum, on the other hand, then they will be spread out in a large region of space.

Since we can no longer describe a particle as having both a definite position and a definite momentum at the same time, we see that classical mechanics, strictly speaking, no longer holds, since it relies on being able to specify a definite position and momentum for a given particle.

The Heisenberg Uncertainty Principle gives us a way to formulate the classical limit. Note that the product of the uncertainties in the position and momentum is on the order of $10^{-34}\,\text{J}\cdot\text{s}$, and typical uncertainties in position are of the order of a nanometer. In a macroscopic system, the length and momentum scales are much larger than the uncertainties, and thus we would find it very difficult, if not impossible, to detect these uncertainties in practice. We therefore expect classical mechanics to work very well for macroscopic systems.

Problems

(1) (Eugen Merzbacher, Quantum Mechanics 3rd edition, Exercise 2.7)

Make an estimate of the lower bound for the distance Δx, within which an object of mass m can be localized for as long as the universe has existed ($\approx 10^{10}\,\text{yr}$). Compute and compare the values of this bound for an electron, a proton, a one-gram object, and the entire universe.

A picture of the box considered in Problem 2. From W. Greiner, Quantum Mechanics.

(2) (W. Greiner, Quantum Mechanics)

Consider a box with a particle (a nucleon) in it. The width of the box is l (see illustration). Determine the magnitude of the kinetic energy of the particle.

(3) Verify the Heisenberg Uncertainty Principle for a Gaussian wave packet; i.e.,

$\psi(x)=\left (\frac{\pi }{a}\right )^{-1/4}e^{-ax^{2}/2}.$

Show, in fact, that the inequality appearing in said relation becomes an equality for this case.