# Average speed of a probability density wave and wave packets

• schniefen
In summary: You would need to normalize the integrand to have a unit amplitude. For example, you could divide by ##\hbar## to get a normalized value.
schniefen
Poster has been reminded to always show effort in schoolwork thread starts
Homework Statement
See attached images.
Relevant Equations
##\omega=2\pi f, k=\frac{2 \pi}{\lambda}, E=\hbar \omega, p=\hbar k##
Show that ##v_{av}=\frac{\hbar k_2 + \hbar k_1}{2m}## is equal to ##v_{av}=\frac{\omega_2 - \omega_1}{k_2-k_1}##. Which of the identities listed above (if any) would make the sign change between ##k_2## and ##k_1##?

One can attain a "wave packet" by superposing two or more sinusoidal waves with different wave numbers and amplitudes. How come such a "wave packet" is given by an integral? What area does this integral represent? How come it is not given by, for example, ##\sum_i^n A(k_i) e^{i(k_ix-\omega t)} ## (or by the formula for the Fourier series?)

Last edited:
schniefen said:
View attachment 252915 One can attain a "wave packet" by superposing two or more sinusoidal waves with different wave numbers and amplitudes. How come such a "wave packet" is given by an integral? What area does this integral represent? How come it is not given by, for example, ##\sum_i^n A(k_i) e^{i(k_ix-\omega t)} ## (or by the formula for the Fourier series?)

You could use a finite sum. The integral over ##k## represents a limit of such finite sums.

schniefen said:
Homework Statement: See attached images.
Homework Equations: ##\omega=2\pi f, k=\frac{2 \pi}{\lambda}, E=\hbar \omega, p=\hbar k##

View attachment 252914 Show that ##v_{av}=\frac{\hbar k_2 + \hbar k_1}{2m}## is equal to ##v_{av}=\frac{\omega_2 - \omega_1}{k_2-k_1}##. Which of the identities listed above (if any) would make the sign change between ##k_2## and ##k_1##?

I'm afraid I don't understand this question.

Regarding the first question, one needs to show that ##\frac{\hbar k_2 + \hbar k_1}{2m}=\frac{\omega_2 - \omega_1}{k_2-k_1}##. This is part of a problem where the wave function is given by ##Ae^{i(k_1x-\omega_1 t)}+ e^{i(k_2x-\omega_2 t)}## and the probability density function by ##2|A|^2(1+\cos{((k_2-k_1)x-(\omega_2 - \omega_1)t)})##. Since ##v_{av}=\frac{\omega}{k}=\frac{\omega_2 - \omega_1}{k_2-k_1}## and ##p_{av}=\frac{\hbar k_2 + \hbar k_1}{2}##, another expression for ##v_{av}## is simply ##\frac{p_{av}}{m}=\frac{\hbar k_2 + \hbar k_1}{2m}##.

Regarding the second question, in the attached image they integrate from negative to positive infinity. This wouldn't be a finite sum, would it?

schniefen said:
Regarding the second question, in the attached image they integrate from negative to positive infinity. This wouldn't be a finite sum, would it?

Ultimately, yes, that integral is the limit of a sequence of finite sums. Each sum has a greater number of values of ##k## and extends further in terms of the maximum value of ##|k|##.

That's simply what integral calculus represents.

schniefen said:
Regarding the first question, one needs to show that ##\frac{\hbar k_2 + \hbar k_1}{2m}=\frac{\omega_2 - \omega_1}{k_2-k_1}##.

What have you tried?

You could start with the right-hand side and work on that. Using the relationship between ##k## and ##\omega##.

It should come out quite easily.

schniefen
PeroK said:
Ultimately, yes, that integral is the limit of a sequence of finite sums. Each sum has a greater number of values of ##k## and extends further in terms of the maximum value of ##|k|##.

That's simply what integral calculus represents.

What does the integrand ##A(k) e^{i(kx-\omega t)}dk## represent? ##dk## represents a small interval between ##k_1## and ##k_2##, but how would the integral then represent waves of different wave numbers? The ##k## in the exponent of the integrand is what one wants to change, isn't it?

schniefen said:
What does the integrand ##A(k) e^{i(kx-\omega t)}dk## represent? ##dk## represents a small interval between ##k_1## and ##k_2##, but how would the integral then represent waves of different wave numbers? The ##k## in the exponent of the integrand is what one wants to change, isn't it?

##A(k)## represents the "density" of each value of ##k##. This is the continuous, integral limit of the "weight" of each value of ##k## in a finite sum.

The coefficients ##a_k## in a finite sum become a continuous density function ##A(k)## in the integral.

For example, you could imagine ##A(k)## as a Gaussian or Normal distribution about some mean value of ##k##. You could approximate this as closely as you like with a finite sum of a weighted values of ##k##.

How would one write the integral out as a sum, that is ##\int_{-\infty}^{\infty} A(k) e^{i(kx-\omega t)}dk##?

schniefen said:
How would one write the integral out as a sum, that is ##\int_{-\infty}^{\infty} A(k) e^{i(kx-\omega t)}dk##?
##\sum_{j=1}^{n} A(k_j) e^{i(k_jx-\omega_j t)}##

Where the ##k_j## are a suitable set of values of ##k##. For example ##-100, -99, \dots 99, 100##.

That might give a good approximation. But, of course, you can make the ##k_j## as close to each other as you want. And you could take an infinite sum if you wanted to.

Note that technically you'd need a normalization factor as well.

schniefen
PeroK said:
##\sum_{j=1}^{n} A(k_j) e^{i(k_jx-\omega_j t)}##

Where the ##k_j## are a suitable set of values of ##k##. For example ##-100, -99, \dots 99, 100##.

That might give a good approximation.

Note that technically you'd need a normalization factor as well.

That clarifies it. How do the infinite limits in the integral "translate" to the sum? Would this correspond to a Riemann integral? If so, where is the ##\Delta k_j## term and ##\lim_{||\Delta k|| \to 0}##?

Last edited:
schniefen said:
That clarifies it. How do the infinite limits in the integral "translate" to the sum? Would this correspond to a Riemann integral? If so, where is the ##\Delta k_j## term and ##\lim_{||\Delta k|| \to 0}##?
Technically you take the proper integral over some large interval ##[-K, +K]## and then take the limit again as ##+K \rightarrow +\infty## etc. That gives the infinite or "improper" integral.

schniefen
PeroK said:
Technically you take the proper integral over some large interval ##[-K, +K]## and then take the limit again as ##+K \rightarrow +\infty## etc. That gives the infinite or "improper" integral.
Makes sense. It is still unclear where the ##\Delta k_j## appears in the sum, i.e. that turns into ##dk## in the limit ##||\Delta k|| \to 0##?

Last edited:
schniefen said:
Makes sense. It is still unclear where the ##\Delta k_j## appears in the sum, i.e. that turns into ##dk## in the limit ##\lim_{||\Delta k|| \to 0}##?
Assuming the ##k_j## are evenly spread the delta is just part of the normalization. You can put it in explicitly:

##\sum_{j=1}^{n} A(k_j) \Delta k e^{i(k_jx-\omega_j t)}##

schniefen
PeroK said:
...the delta is just part of the normalization.

Meaning it is equal to ##1##, correct?

schniefen said:
Meaning it is equal to ##1##, correct?

I was thinking ahead to when you want to normalize the wave packet. But, actually, putting in the ##\Delta k## is sufficient at this stage.

By the normalization, do you mean ##|\psi|^2##?

schniefen said:
By the normalization, do you mean ##|\psi|^2##?
Yes. And, in fact, the only way you can create a normalized wave function from these plain waves is by a continuous, integrated linear combination of them. Any finite sum is not normalizable.

## 1. What is the difference between a probability density wave and a wave packet?

A probability density wave describes the probability of finding a particle in a certain location, while a wave packet describes the spatial distribution of a particle's wavefunction. In other words, a probability density wave is a measure of the likelihood of finding a particle at a specific point, while a wave packet shows how the particle's wavefunction is spread out in space.

## 2. How is the average speed of a probability density wave calculated?

The average speed of a probability density wave is calculated by taking the derivative of the wave's phase with respect to time. This is known as the group velocity and represents the speed at which the wave packet as a whole moves through space.

## 3. Can the average speed of a probability density wave exceed the speed of light?

No, according to the theory of relativity, no object or information can travel faster than the speed of light. This applies to the average speed of a probability density wave as well.

## 4. How does the average speed of a probability density wave change over time?

The average speed of a probability density wave remains constant if the wave is in a vacuum. However, if the wave encounters a medium with varying refractive index, such as in optics, the average speed may change. In quantum mechanics, the average speed of a wave packet may also change due to interactions with other particles.

## 5. What is the significance of the average speed of a probability density wave?

The average speed of a probability density wave plays a crucial role in understanding the behavior of waves in various systems, such as in quantum mechanics and optics. It also has practical applications, such as in the design of electronic devices and communication systems. Additionally, the average speed of a probability density wave is a fundamental concept in the study of wave-particle duality and the behavior of matter on a quantum level.

• Introductory Physics Homework Help
Replies
4
Views
1K
• Electromagnetism
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
3K
Replies
1
Views
2K
Replies
3
Views
1K
Replies
6
Views
1K
Replies
1
Views
4K
Replies
3
Views
3K
• Quantum Physics
Replies
14
Views
5K