Grp Velocity N Uncertainty Principles Slide

PHY-1109 Physics

---

Phase and Group Velocities

A group of waves need not have the same velocity as the waves themselves.

$y=A\cos (\omega t-kx)\text{ (10)}$

The amplitude of the de Broglie waves that correspond to a moving body reflects the probability that it will be found at a particular place at a particular time. It is clear that de Broglie waves cannot be represented simply by a formula resembling Eqn. (10), which describes an indefinite series of waves all with the same amplitude $A$.

Image/Diagram Note: Wave propagation. (Figure showing $y$ vs. $x$ at $t=0$ and $t=t$, illustrating a “Wave group” and “A wave group”).

Instead, we expect the wave representation of a moving body to correspond to a wave packet, or wave group, like that shown here, whose waves have amplitudes upon which the likelihood of detecting the body depends.

A familiar example of how wave groups come into being is the case of beats. When two sound waves of the same amplitude but of slightly different frequencies are produced simultaneously, the sound we hear has a frequency equal to the average of the two original frequencies and its amplitude rises and falls periodically.

The amplitude fluctuations occur as many times per second as the difference between the two original frequencies.

If the original sounds have frequencies of, say, 440 and 442 Hz, we will hear a fluctuating sound of frequency 441 Hz with two loudness peaks, called beats, per second.

Image/Diagram Note: Beats are produced by the superposition of two waves with different frequencies. (Figure showing the superposition of two sine waves resulting in a modulated wave pattern).

A way to mathematically describe a wave group, then, is in terms of a superposition of individual waves of different wavelengths whose interference with one another results in the variation in amplitude that defines the group shape.

If the velocities of the waves are the same, the velocity with which the wave group travels is the common phase velocity.

However, if the phase velocity varies with wavelength, the different individual waves do not proceed together. This situation is called dispersion.

As a result the wave group has a velocity different from the phase velocities of the waves that make it up. This is the case with de Broglie waves.

It is not hard to find the velocity $u$ with which a wave group travels. Let us suppose that the wave group arises from the combination of two waves that have the same amplitude $A$ but differ by an amount $d\omega$ in angular frequency and an amount $dk$ in wave number. We may represent the original waves by the formulas:

$y_1=A\cos (\omega t-kx)$ $y_2=A\cos [(\omega +d\omega )t-(k+dk)x]$

The resultant displacement $y$ at any time $t$ and any position $x$ is the sum of $y_1$ and $y_2$. With the help of the identity:

$\cos \alpha +\cos \beta =2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$

And the relation $\cos (-\theta )=\cos \theta$.

We find that: $y=y_1+y_2=2A\cos \left[\frac{(2\omega +d\omega )t-(2k+dk)x}{2}\right]\cos \left(\frac{d\omega t-dkx}{2}\right)$

Since $d\omega$ and $dk$ are small compared with $\omega$ and $k$ respectively:

$2\omega +d\omega \approx 2\omega$ $2k+dk\approx 2k$

and

$y=2A\cos (\omega t-kx)\cos \left(\frac{d\omega }{2}t-\frac{dk}{2}x\right)\text{ (11)}$

Eqn. (11) represents a wave of angular frequency $\omega$ and wave number $k$ that is superimposed upon it a modulation of angular frequency $\frac{1}{2}d\omega$ and of wave number $\frac{1}{2}dk$.

The effect of the modulation is to produce successive wave groups, as shown in the Figure. The phase velocity $w$ is:

$w=\frac{\omega }{k}\text{ (12)}$

and the velocity $u$ of the wave groups is:

$u=\frac{d\omega }{dk}\text{ (13)}$

Depending on how phase velocity varies with wave number in a particular situation, the group velocity may be less or greater than the phase velocities of its member waves.

If the phase velocity is the same for all wavelengths, as is true for light waves in empty space, the group and phase velocities are the same.

The angular frequency $\omega$ and wave number $k$ of the de Broglie waves associated with a body of mass $m_0$ moving with the velocity $v$ are:

$\omega =2\pi v=\frac{2\pi m_0{c}^2}{h\sqrt{1-v^2/c^2}}\text{ (14)}$

and

$k=\frac{2\pi }{\lambda }=\frac{2\pi m_{0}v}{h\sqrt{1-v^2/c^2}}\text{ (15)}$

Both $\omega$ and $k$ are functions of the body’s velocity. The phase velocity $w$ of de Broglie waves is, as we found earlier:

$w=\frac{\omega }{k}=\frac{c^2}{v}$

This exceeds both the velocity of the body and the velocity of light $c$, since $v<c$.

The group velocity $u$ of the de Broglie waves associated with the body is:

$u=\frac{d\omega }{dk}=\frac{d\omega /dv}{dk/dv}\text{ (16)}$

---

The Uncertainty Principle

We cannot predict the future because of the unpredictability of Present.

Image/Diagram Note: (Figure (d) showing two wave packets, one short and one long, illustrating $\Delta x$ small, $\Delta p$ large, and vice versa).

Image/Diagram Note: How to get de Broglie wave? (Figure showing the Sum of two waves resulting in a Modulated wave).

Uncertainty principle in terms of wave properties of Particle

Let us consider two de Broglie waves:

${\Psi }_1=A\cos (\omega t-kx)$ ${\Psi }_2=A\cos [(\omega +\Delta \omega )t-(k+\Delta k)x]$

Let’s sum it up: $\Psi ={\Psi }_1+{\Psi }_2\approx 2A\cos (\omega t-kx)\cos \left(\frac{1}{2}\Delta \omega t-\frac{1}{2}\Delta kx\right)\text{ (17)}$

The width of each group is half the wavelength ${\lambda }_m$ and it is reasonable to suppose that this width is of the same order of magnitude as the inherent uncertainty $\Delta x$ in the position of the group.

$\Delta x\approx \frac{1}{2}{\lambda }_m\text{ (18)}$

However, ${\lambda }_m=2\pi /k_m$.

The propagation constant of the modulation is: $k_m=\frac{1}{2}\Delta k$

Therefore, ${\lambda }_m=2\pi /\frac{1}{2}\Delta k$.

and $\Delta x\approx 2\pi /\Delta k\text{ (19)}$

The inherent uncertainty of $k$ is $\Delta k$.

$\Delta k\approx 2\pi /\Delta x\text{ (20)}$

The de Broglie wavelength of a particle of momentum $p$ is $\lambda =h/p$.

The propagation constant corresponding to this wavelength is $K=2\pi /\lambda =2\pi p/h$.

Now, $\Delta p=\frac{h\Delta k}{2\pi }\text{ (21)}$

Hence, $\Delta x\Delta p\ge h\text{ (22)}$

Uncertainty principle in terms of particle properties of wave

Image/Diagram Note: (Figure showing an electron being observed by an incident photon, resulting in a change from “Original momentum of electron” to “Final momentum of electron”).

An electron cannot be observed without changing its momentum.

Uncertainty of momentum is $\Delta p=h/\lambda \text{ (23)}$

Uncertainty of position is $\Delta x\approx \lambda \text{ (24)}$

Hence, $\Delta p=\frac{h}{\Delta x}$

Therefore, $\Delta x\Delta p\ge h$

For a single wave group, $\Delta x\approx 1/\Delta k$.

Instead of $\Delta x\approx 2\pi /\Delta k$, However, $\Delta p\approx h\Delta k/2\pi$.

We have the more realistic expression $\Delta x\Delta p\ge h/2\pi$

It is customary to abbreviate $h/2\pi$ by the symbol $\hslash$.

Where, $\hslash =1.054571817\times 10^{-34}\text{ Joule.sec}$.

Summary

The Heisenberg uncertainty principle states that it is impossible to simultaneously measure the x-components of position and of momentum of a particle with an arbitrarily high precision. The product of experimental uncertainties is always larger than or equal to $\frac{1}{2}\hslash$.

The limitations of this principle have nothing to do with the quality of the experimental apparatus but originate in the wave-like nature of matter.

The energy-time uncertainty principle expresses the experimental observation that a quantum state that exists only for a short time cannot have a definite energy.