# Electromagnetic Wave Packet. The Uncertainty Principle

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19.5Electromagnetic Wave Packet. The Uncertainty Principle

In this Physics tutorial, you will learn:

• What is a wave packet? What are its properties? Why is it used for?
• What does a light wave packet represent?
• What is the spatial extension of wave packet?
• What is the uncertainty relation for photons?
• What does the uncertainty of wave number represent?
• What is the uncertainty principle of Heisenberg?
• What is De Broglie's Wave Packet?
• What are the two mathematical expressions of Heisenberg's uncertainty principle?

## Introduction

As we know, photons have a very high frequency (visible light oscillates in the range of 1014 Hz). In addition, we have explained in other articles that photons are produced when electrons move inside atoms from one layer to another. Do you think it is easy to determine the position of photons at a given instant? Why? What about momentum and energy? Can we be sure of the value of energy and impulse of a particle at a given instant?

What do you understand with "packet"? When do you use packets in daily life?

Besides providing answer to the above questions, this tutorial will unveil the limits of science in certain aspects, especially in measurements. The processes in micro-world do to work in the same way as in macro-world, so this is a point to highlight in this tutorial.

## Recalling the Simple Harmonic Wave (Oscillation)

We know that the position (in two dimensions) of any point of a standing wave is expressed through the sinusoidal equation of simple harmonic motion, that is

y(t) = ymax ∙ sin(ω ∙ t - k ∙ x)

where

y(t) is the y-coordinate of a given point of a wave at any instant t,

x is the x-coordinate of the given point of wave (it is up to us to choose any point we want, so the x-coordinate depends on our choice),

t is the time instant in which we are interested to know the position of the given point of wave,

ω = 2π · f is the angular frequency of wave, and

k = /λ is the wave number - a feature that depends only on the wavelength.

Thus, the above equation can also be written as

y(t) = ymax ∙ sin(2π · f ∙ t - /λ ∙ x)

The shape of the simple harmonic wave in two dimensions is shown below.

For simple harmonic waves, frequency is determined by the features of emitting source, hence it can be considered as constant. The same for wavelength if the wave is propagating in a homogenous medium (where the wave speed is the same throughout the entire medium), given the equation of waves

v = λ ∙ f

Such waves (with constant frequency) are emitted from sources that oscillate unceasingly in a periodic fashion. In these conditions, the wave would propagate in space up to infinity (ideal or standing waves). Obviously, in the real world this cannot occur, as the environment will eventually drain up the wave's energy. However, if we consider short intervals of time up to few minutes, many real waves behave in a similar way as ideal waves. For example, a sound source, a radio emitter etc., can be described using the standing wave pattern.

The approach is different when considering the oscillation of atoms, molecules or other microscopic particles however. Despite they represent EM sources too, their radiation is discrete (with interruptions). A single radiating action produced by atoms is very short; it lasts a few nanoseconds (1 ns = 10-9 s). Therefore, when studying the radiation of atoms, we always refer to "wave packets" instead of individual waves. Likewise, photons are also emitted in wave packets.

## Definition and Features of Wave Packet

A "wave packet" is a fragmented or a discrete wave. It represents a short or a burst wave All sources that operate periodically in short intervals emit wave packets. Therefore, a wave packet has a limited width in space. The spatial extension Δx of a wave packet is given by the equation

∆x ≈ v ∙ ∆τ

where v is the packet's speed and Δτ is the time interval between two consecutive emissions.

### Example 1

Calculate the spatial extension of a radio-wave packet emitted by an AC source in any country in Europe.

### Solution 1

Clues:

(v = c = 300 000 km/s = 3 × 108 m/s)
Δx = ?

We know that the frequency of all AC sources in Europe is f = 50 Hz. Therefore, the source oscillates in time intervals of

T = 1/f = 1/50 Hz = 0.02 s

This value corresponds to the time interval between two consecutive emissions from the source. Thus, we have

∆τ = T = 0.02 s
= 2 × 10-2 s

Hence, since the spatial extension of a wave packet is given by

∆x ≈ v ∙ ∆τ

we obtain after substitutions:

∆x ≈ v ∙ ∆τ
= c ∙ ∆τ
= (3 × 108 m/s) ∙ (2 × 10-2 s)
= 6 × 106 m
= 6000 km

Since the extension of this packet is very large, we can consider this wave as an ideal standing wave.

Now, let's consider another example but this time, the wave source is an atom.

### Example 2

Calculate the spatial extension achieved by a photon emitted by an atom during the time interval of 10-9 s.

### Solution 2

Clues:

Δτ = 10-9 s
(c = 3 × 108 m/s)
Δx = ?

Giving that

∆x ≈ v ∙ ∆τ

we obtain after substitutions:

∆x ≈ v ∙ ∆τ
= c ∙ ∆τ
= (3 × 108 m/s ∙ (10-9 s)
= 0.3 m

From the result, you can see that the EM wave packets emitted by atoms are very concentrated. This allows us visualize the photon as a particle.

Advanced theoretical calculations show that any wave packet does not contain a single a single wavelength but an infinite number of wavelengths instead. These values range from λ - λ/2 to λ + λ/2, where λ represents the mean wavelength of wave packet. The width Δλ of such wave packets relates to its spatial extension through the relation:

∆x ∙ ∆λ ≈ λ2/

This relation is characteristic for any wave group or packet, whose shape is shown in the figure below.

### Example 3

Calculate the wave width of a photon and compare it to the mean wavelength (λ = 600 nm) given that its spatial extension is Δx ≈ 0.3 m.

### Solution 3

Clues:

Δλ = ?
Δx = 0.3 m
λ = 600 nm = 6 × 10-7 m

Applying the equation

∆x ∙ ∆λ ≈ λ2/

we find for the wave width Δλ after substituting the known values:

∆λ ≈ λ2/2π ∙ ∆x
= (6 × 10-7 m)2/2 ∙ 3.14 ∙ (0.3 m)
= 1.91 × 10-13 m

This result means the wave width of photon is much shorter (about 300 000 times shorter) than its average wavelength. This results in an almost continuous type of photons emission, which makes the identification of the discrete nature of light very difficult.

## The Uncertainty Relation for EM Wave Packet (Photon)

Let's assume at a certain instant, a photon is at the distance x from the radiating atom as shown in the figure below.

The exact position of photon is not known; all we know is that the photon lies somewhere between x and x + Δx from the parent atom. Here, the uncertainty Δx corresponds to the spatial extension of photon's wave packet. It is known as the "uncertainty of coordinate".

Likewise, it is clear that the wavelength of photon is uncertain too, as its range of wavelength varies is Δλ. This is another uncertainty, which we call as "uncertainty of wavelength".

The existence of these two uncertainties is a fundamental feature of photon. Given that the wave number k is given by

k = /λ

we have an uncertainty of the wave number, Δk as well. The equation

∆x ∙ ∆λ ≈ λ2/

is usually written in terms of Δx and Δk. But first, let's find the formula of Δk. Thus,

∆k = /λ - /λ + ∆λ
= 2π ∙ (λ + ∆λ) - 2π ∙ λ/λ ∙ (λ + ∆λ)
= 2π ∙ ∆λ/λ ∙ (λ + ∆λ)

Since Δλ << λ (as seen in the example solved earlier), we can approximate the above equation as

∆k ≈ 2π ∙ ∆λ/λ2

Thus,

λ2/ = ∆λ/∆k

Hence, the equation

∆x ∙ ∆λ ≈ λ2/

is transformed into

∆x ∙ ∆λ ≈ ∆λ/∆k

Simplifying both sides by Δλ, we obtain

∆x ≈ 1/∆k

or

∆x ∙ ∆k ≈ 1

The above relation is simpler than the other relation that involves Δλ; hence it is known as the uncertainty relation for photon. In fact, this relation is valid for all wave packets, not only for photons.

### Example 4

Calculate the uncertainty Δk of the wave number of an electromagnetic wave packet if the radiation time is Δτ = 0.01 s.

### Solution 4

Clues:

Δτ = 0.01 s = 10-2 s
(c = 3 × 108 m/s)
Δk = ?

First, we calculate the extension of EM wave packet. We have

∆x = c ∙ ∆τ
= (3 × 108 m/s) ∙ (10-2 s)
= 3 × 106 m

Therefore, we obtain for the uncertainty of wave number Δk

∆x ∙ ∆k ≈ 1
∆k ≈ 1/∆x
= 1/3 × 106 m
= 3.33 × 10-7 m-1

## 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

## Heisenberg's Uncertainty Principle

The state of a particle (for example an electron) is known when we have information at any instant for its position x, momentum p and energy E. For this, we have to take measurements. It is known that every measurement contains an error - systematic or casual it may be (systematic errors depend on the initial conditions of measurement or apparatus while casual errors depend more on the accuracy of the person who takes the measurement). In any case, we cannot pretend to take 100% accurate measurements, so it is obvious that we will have at least two kinds of uncertainties: of position (Δx) and of momentum (Δp), which are inevitable.

In 1927, Werner Heisenberg discovered one of the most important principles in quantum physics - the uncertainty principle which gives the relationship between the two aforementioned individual uncertainties. This principle is based on the rule according which:

"It is impossible to determine with absolute precision both the position and impulse of a particle at the same instant."

Moreover, this principle is not too much related to the accuracy of apparatuses; it would exist even if we had 100% accurate measuring devices. Heisenberg's Uncertainty Principle is a universal law of nature, as the universe has determined certain limits in the accuracy of measurements, beyond which it is impossible to go, despite the advancements in science.

The following relation gives the mathematical form of Heisenberg's Uncertainty Principle:

∆x ∙ ∆p ≈ ℏ

We can transform the above formula to give the relationship between two other uncertainties: that of particle's energy ΔE and the time interval Δτ during which the particle has a given energy. Thus, since any change momentum Δp is equal to the impulse of particle, we can write

∆p = F ∙ ∆τ

where F is the force acting on the particle. When this force is multiplied to the displacement Δx it causes to the particle, we obtain the energy transferred to the object in the form of work done by the force. Hence, we have

∆x ∙ (F ∙ ∆τ) ≈ ℏ
(∆x ∙ F) ∙ ∆τ ≈ ℏ
∆E ∙ ∆τ ≈ ℏ

The meaning of the above relation is as follows:

"It is impossible to determine with absolute precision both the energy of a particle and the instant at which this particle possesses this energy."

Now, let's see a couple of examples in which these two forms of uncertainty principle are applied.

### Example 6

An electron is moving in the X-direction at 3.6 × 105 m/s - a value which we have been able to measure at an accuracy of ± 1%. What is the maximum accuracy at which we can determine the electron's position?

### Solution 6

Clues:

v = 3.6 × 105 m/s
Δv = 1% of v = 0.01 · 3.6 × 105 m/s = 3.6 × 103 m/s
(me = 9.1 × 10-31 kg)
ℏ = 1.055 × 10-34 J · s
Δx = ?

First, let's calculate the uncertainty of electron's momentum. We have

∆p = m ∙ ∆v
= (9.1 × 10-31 kg) ∙ (3.6 × 103 m/s)
= 3.276×10-27 kg ∙ m/s

Now, let's calculate the maximum accuracy of position (which represents the uncertainty of electron's position Δx). We use the Heisenberg principle for this purpose.

∆x ∙ ∆p ≈ ℏ
∆x = /∆p
= (1.055 × 10-34 J ∙ s)/(3.276 × 10-27 kg ∙ m/s)
= 3.22 × 10-8 m
= 32.2 nm

Now, let's see an example where the uncertainty principle in terms of energy and time is applied.

### Example 7

The atom of a chemical element has a 15 eV energy in its excited state for 80 nanoseconds. Calculate the uncertainty of energy for this atom.

### Solution 7

Clues:

E = 15 eV = 15 · 1.6 × 10-19 J = 2.4 × 10-18 J
Δτ = 80 ns = 80 × 10-9 s = 8 × 10-8 s
ΔE = ?

Given that

∆E ∙ ∆τ ≈ ℏ

we have

∆E = /∆τ
= (1.055 × 10-34 J ∙ s)/(8 × 10-8 s)
= 1.312 × 10-27 J

This uncertainty of energy means the energy of atom ranges from E - ΔE to E + ΔE. The quantity ΔE is also known as the natural width of the energetic level E.

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