Average speed of a probability density wave and wave packets

[schniefen]

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?)

 

[PeroK]

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.

 

[PeroK]

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.

 

[schniefen]

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?

 

[PeroK]

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.

 

[PeroK]

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]

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?

 

[PeroK]

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 $k$.

 

[schniefen]

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

 

[PeroK]

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]

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

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 $\lim_{||\Delta k|| \to 0}$?

 

[PeroK]

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]

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 $||\Delta k|| \to 0$?

 

[PeroK]

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]

...the delta is just part of the normalization.

Meaning it is equal to $1$, correct?

 

[PeroK]

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.

 

[schniefen]

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

 

[PeroK]

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.