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Physics Lesson 19.5.4 - De Broglie's Wave Packet. Wave Width and Uncertainty of Momentum

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De Broglie's Wave Packet. Wave Width and Uncertainty of Momentum

The probability wave that is associated to the motion of an electron, neutron etc., is nothing more but a wave packet of spatial extension of Δx and wave's packet width (expressed in terms of wave number) of Δk, which relates to each other by the uncertainty relation

∆x ∙ ∆k ≈ 1

It is believed that at any instant the electron is located somewhere inside the wave packet, so the quantity Δx, in addition to the packet's extension in space also represents the amount of position uncertainty of electron. In other words, Δx shows the space region in which there is a high probability for the electron to be found. It is worth stressing the fact that Δx does not give the dimensions of electron, which are much smaller.

Evidence collected during experiments have shown that any uncertainty in position Δx is results in a corresponding uncertainty in momentum Δp. We will express this uncertainty in momentum in terms of wave's packet width Δk discussed earlier. Thus, since the De Broglie wavelength λ is related to the impulse p of particle through the expression

λ = h/p

while the wave number k is given by

k = /λ

we obtain

k = 2π ∙ p/h

Therefore, we obtain for the width Δk of electron's wave packet:

∆k = 2π ∙ ∆p/h

As explained in the previous tutorial, in quantum physics we often use the reduced Planck's constant instead of the standard constant h, where

ℏ = h/
= 6.626 × 10-34 J ∙ s/2 ∙ 3.14
= 1.055 × 10-34 J ∙ s

In this way, we can write

∆k = ∆p/

The above formula gives the relationship between the width of De Broglie packet Δk and the uncertainty of electron's momentum Δp.

Example 5

The spatial extension of De Broglie's wave packet for electron is 0.002 nm. Calculate the width of electron's wave packet and the uncertainty for its impulse.

Solution 5

Clues:

Δx = 0.002 nm = 2 × 10-12 m
(ℏ = 1.055 × 10-34 J · s)
Δk = ?
Δp = ?

We use the relation

∆x ∙ ∆k ≈ 1

to determine the width Δk of wave packet. Thus,

∆k = 1/∆x
= 1/2 × 10-12 m
= 5 × 1011 m-1

As for the uncertainty Δp of electron's momentum, we have

∆k = ∆p/
∆p = ℏ ∙ ∆k
= (1.055 × 10-34 J ∙ s) ∙ (5 × 1011 m-1)
= 5.275 × 10-24 kg ∙ m/s

You have reached the end of Physics lesson 19.5.4 De Broglie's Wave Packet. Wave Width and Uncertainty of Momentum. There are 5 lessons in this physics tutorial covering Electromagnetic Wave Packet. The Uncertainty Principle, you can access all the lessons from this tutorial below.

More Electromagnetic Wave Packet. The Uncertainty Principle Lessons and Learning Resources

Modern Physics Learning Material
Tutorial IDPhysics Tutorial TitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
19.5Electromagnetic Wave Packet. The Uncertainty Principle
Lesson IDPhysics Lesson TitleLessonVideo
Lesson
19.5.1Recalling the Simple Harmonic Wave (Oscillation)
19.5.2Definition and Features of Wave Packet
19.5.3The Uncertainty Relation for EM Wave Packet (Photon)
19.5.4De Broglie's Wave Packet. Wave Width and Uncertainty of Momentum
19.5.5Heisenberg's Uncertainty Principle

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