Why Must Δpx Be Much Less Than px for a Non-Spreading Wave Packet?

[ognik]

Homework Statement

Consider a wave packet satisfying the relation $\Delta x \Delta p_x \approx \hbar$
Show that if the packet is not to spread appreciably while it passes through a fixed position, the condition $\Delta p_x << p_x$ must hold.

Homework Equations

$p_x = m \ddot{x}$

The Attempt at a Solution

I feel I am missing some information, I wote down $(x - x_0)(p - p_0) = (x - x_0)m(x - x_0) \approx \hbar$ ... can't see how to apply the inequality usefully?

 

[blue_leaf77]

What is $p_x$ in this case, the average momentum? Does the wavepacket propagate in free space?

 

[DrClaude]

Homework Equations

$p_x = m \ddot{x}$

Really?

 

[ognik]

Really?

Sorry, Latex typo, should have been $\dot{x}$

 

[ognik]

What is $p_x$ in this case, the average momentum? Does the wavepacket propagate in free space?

I think it is the momentum at a fixed point x - of the 'envelope' of the wave packet?

So what I was thinking is that the spread of the packet, $= 2|x - x_0|$, must be approx. the same at position x as it was at $x_0$, but then I would have expected $p \approx p_0$?

Note that in the preceding text, $\omega (\vec{k})$ was Taylor-expanded; the quadratic & higher terms were ignored so that the wave packet does not disperse. I am not sure if the text assumes the same for this exercise?

 

[blue_leaf77]

I think it is the momentum at a fixed point x - of the 'envelope' of the wave packet?

You cannot apply such a notion because for a fixed time $t$, because the uncertainty of a measurement of momentum everywhere along the $x$ axis is the same (remember that to calculate an uncertainty, one has to integrate over all $x$).

Note that in the preceding text, ω(⃗k) \omega (\vec{k}) was Taylor-expanded; the quadratic & higher terms were ignored so that the wave packet does not disperse.

I think the problem asks you to verify that statement in the text in a mathematical manner. You need to show that if the center momentum of the wavepacket is appreciably bigger than the width (in the momentum space), then you can indeed omit the higher powers in the expansion of $\omega(k_x)$.

 

[ognik]

You cannot apply such a notion because for a fixed time $t$, because the uncertainty of a measurement of momentum everywhere along the $x$ axis is the same (remember that to calculate an uncertainty, one has to integrate over all $x$).

I have been thinking about this (always dangerous) but I think I've just made myself confused. Please give me a simple explanation of what $\Delta p_x$ (the spread of momentum ?) and $p_x$ (centre momentum, but what does it represent?) are, and how they inter-relate?

 

[blue_leaf77]

Please give me a simple explanation of what ΔpxΔpx \Delta p_x (the spread of momentum ?)

It's the width of $|\phi(k_x)|^2$ (up to the definition used for the width) and it's obviously related to the width of $\phi(k_x)$.

pxpx p_x (centre momentum, but what does it represent?)

$p_x$ in this problem is defined to be a representative of the center momentum of $|\phi(k_x)|^2$ (or of $\phi(k_x)$). Good candidates for this parameter could be the average momentum $\langle \phi |p_x|\phi \rangle$ or the group velocity $\frac{d\omega}{dk_x}\bigg|_{k_x=\kappa}$ with $\kappa$ being the point aound which $\omega(k_x)$ is Taylor-expanded.

how they inter-relate?

There are no relationships between them because they are free parameters, i.e. you can change one of them without changing the other one.

 

[ognik]

Have been going round in circles with this.
$\bar{k_x}$ is the value of $k_x$ where $| \phi(k_x) |^2$ is a maximum, and the average momentum $p_x = \hbar \bar{k_x}$

But $\Delta p_x = \hbar \Delta k_x$ is from $2 * (\bar{k_x} - k_x)$ where we find $k_x$ at $\frac{1}{2} | \phi(k_x) |^2$

How to take this further eludes me ...