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Several waves are shown, all with equal amplitude but different. The result of adding these to form a wave packet is also shown. The wave packet is an oscillating wave whose amplitude increases to a maximum then decreases, so that its envelope is a pulse of width Delta x.
Adding together several plane waves of different wavelengths can produce a wave that is relatively localized.

Note that the uncertainty principle has nothing to do with the precision of an experimental apparatus. Even for perfect measuring devices, these uncertainties would remain because they originate in the wave-like nature of matter. The precise value of the product Δ x Δ p depends on the specific form of the wave function. Interestingly, the Gaussian function (or bell-curve distribution) gives the minimum value of the uncertainty product: Δ x Δ p = / 2 .

The uncertainty principle large and small

Determine the minimum uncertainties in the positions of the following objects if their speeds are known with a precision of 1.0 × 10 −3 m/s : (a) an electron and (b) a bowling ball of mass 6.0 kg.

Strategy

Given the uncertainty in speed Δ u = 1.0 × 10 −3 m/s , we have to first determine the uncertainty in momentum Δ p = m Δ u and then invert [link] to find the uncertainty in position Δ x = / ( 2 Δ p ) .

Solution

  1. For the electron:
    Δ p = m Δ u = ( 9.1 × 10 −31 kg ) ( 1.0 × 10 −3 m/s ) = 9.1 × 10 −34 kg · m/s , Δ x = 2 Δ p = 5.8 cm .
  2. For the bowling ball:
    Δ p = m Δ u = ( 6.0 kg ) ( 1.0 × 10 −3 m/s ) = 6.0 × 10 −3 kg · m/s , Δ x = 2 Δ p = 8.8 × 10 −33 m .

Significance

Unlike the position uncertainty for the electron, the position uncertainty for the bowling ball is immeasurably small. Planck’s constant is very small, so the limitations imposed by the uncertainty principle are not noticeable in macroscopic systems such as a bowling ball.

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Uncertainty and the hydrogen atom

Estimate the ground-state energy of a hydrogen atom using Heisenberg’s uncertainty principle. ( Hint : According to early experiments, the size of a hydrogen atom is approximately 0.1 nm.)

Strategy

An electron bound to a hydrogen atom can be modeled by a particle bound to a one-dimensional box of length L = 0.1 nm . The ground-state wave function of this system is a half wave, like that given in [link] . This is the largest wavelength that can “fit” in the box, so the wave function corresponds to the lowest energy state. Note that this function is very similar in shape to a Gaussian (bell curve) function. We can take the average energy of a particle described by this function ( E ) as a good estimate of the ground state energy ( E 0 ) . This average energy of a particle is related to its average of the momentum squared, which is related to its momentum uncertainty.

Solution

To solve this problem, we must be specific about what is meant by “uncertainty of position” and “uncertainty of momentum.” We identify the uncertainty of position ( Δ x ) with the standard deviation of position ( σ x ) , and the uncertainty of momentum ( Δ p ) with the standard deviation of momentum ( σ p ) . For the Gaussian function, the uncertainty product is

σ x σ p = 2 ,

where

σ x 2 = x 2 x 2 and σ p 2 = p 2 p 2 .

The particle is equally likely to be moving left as moving right, so p = 0 . Also, the uncertainty of position is comparable to the size of the box, so σ x = L . The estimated ground state energy is therefore

E 0 = E Gaussian = p 2 m = σ p 2 2 m = 1 2 m ( 2 σ x ) 2 = 1 2 m ( 2 L ) 2 = 2 8 m L 2 .

Multiplying numerator and denominator by c 2 gives

E 0 = ( c ) 2 8 ( m c 2 ) L 2 = ( 197.3 eV · nm ) 2 8 ( 0.511 · 10 6 eV ) ( 0.1 nm ) 2 = 0.952 eV 1 eV .

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Source:  OpenStax, University physics volume 3. OpenStax CNX. Nov 04, 2016 Download for free at http://cnx.org/content/col12067/1.4
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