# Waves, dispersion, group speed

$$f(x-\omega_0 't)=e^0 \int _{-\infty} ^{+\infty} e^{ik(x-\omega_0 't-0)}=\int _{-\infty} ^{+\infty} e^{ik(x-\omega_0 't)}$$
the required term

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$$e^{ik_0\omega_0't}$$

Where did it disapear to in the step between equation 12 and the equation right before it?

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Seriously, is it a mistake in the pdf file?

It could be incoorporated in the function g, I don't know if they introduced it earlier. But my guess is that g(x)=f(x,t=0) so that it is a typo

The second 'x' in the equation should be $\omega _0 '$. Which is an understandable error as k and x often go side by side in such formula.

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I should have added the first part of the "tutorial" too so you see what is g(x) and can actually do the calculations involving equ.2 and 4 to get to equ.12. here it is:

https://www.physicsforums.com/attachment.php?attachmentid=3928

Personally, I see no mistake up to equ. 12 where the exponential term just dispears. It can't have been "incorporated" in g(x), that'd make no sense.

quasar987 said:
I should have added the first part of the "tutorial" too so you see what is g(x) and can actually do the calculations involving equ.2 and 4 to get to equ.12. here it is:

https://www.physicsforums.com/attachment.php?attachmentid=3928
Aha...if that is the definition of g, it exactly accounts for eq 12.

Personally, I see no mistake up to equ. 12 where the exponential term just dispears. It can't have been "incorporated" in g(x), that'd make no sense.
It does, and it is...

$$g(x-\omega '_0 t) e^{i(k_0 x -\omega _0 t)} = f(x-\omega '_0 t,0) e^{ik_0 (\omega '_0 -x)} e^{i(k_0 x -\omega _0 t)}$$

where you see two terms cancel and the result is the given equation.

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OOOOOOOOOOOOOOOOOOOOOOOH. My mistake was that I though

$$f(x, 0) = g(x)e^{ik_0x} \ \ \Rightarrow \ \ f(x-\omega_0't,0) = g(x-\omega_0't)e^{ik_0x}$$

I forgot to substitute the x for $x-\omega_0't$ in the exponential! :yuck:

thx willem.

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I heard that wave packets are important in QM. 2 questions:

1) How important? Do they arise very often?

2) Are the notion of group velocity and dispersion important for quantum mechanical wave packets? I would guess that 'no', since the waves are probabilistic and not physical, so they don't propagate in a medium that we could call "dispersive".

Galileo
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The group velocity is very important in QM (in my opinion at least) since the group velocity of a wave packet allows us to recover the classical expression of a free particle's kinetic energy.

From the Schrodinger equation of a free particle (V=0 everywhere):

$$i\hbar \frac{\partial \Psi}{\partial t}=-\frac{\hbar^2}{2m}\nabla^2 \Psi$$
You can see it admits a plane wave solution of the form:

$$\Psi(\vec r)=A\exp\left(\vec k \cdot \vec r - \omega t\right)$$
if the dispersion relation is:
$$\omega = \frac{\hbar |\vec k|^2}{2m}$$

I'll use the Planck-Einstein relations: $E=\hbar \omega$ and $\vec p = \hbar \vec k$.
Applying them above we immediately see that $E=p^2/2m$ as it should classically.

Now (let's take it in 1D), the phase velocity is:
$$\frac{\omega}{k}=\frac{\hbar k}{2m}=\frac{p}{2m}$$
so it goes at half the speed the it is supposed to go classically. You need the GROUP velocity though:

$$\frac{d\omega}{dk}=\frac{\hbar k}{m}=\frac{p}{m}$$
which corresponds to the classical speed of the particle.

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Where did you pull that dispersion relation from?

And what about the other dispersion relations? For exemple, how was it established that the dispersion relation for waves in deep water is what it is? was it experiemntally?

Galileo
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That plane wave will only be a solution to the Schrodinger equation if that relation between w and k hold. So that's where it comes from. Just plug $\Psi(\vec r)=A\exp\left(\vec k \cdot \vec r - \omega t\right)$ into the Schrodinger equation and see for yourself.

I honestly don't know how the dispersion relation is derived for waves in water. I`m sure sSomeone else may be able to help you there.

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Hans de Vries
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Galileo said:
The group velocity is very important in QM (in my opinion at least) since the group velocity of a wave packet allows us to recover the classical expression of a free particle's kinetic energy.
The relation between the group and phase speed of the de Broglie wave is
very different from that of classical waves, see my write-up here:

Relativistic kinematics of the wave packet:
http://www.chip-architect.com/physics/deBroglie.pdf

Section 1: The the broglie-wave is purely the result of the non-simultaneity of SR
Section 2: The wave packet at rest
Section 3: The moving wave packet
Section 4: The >c phase speed
Section 5: The group speed (recovering the correct time-dilation)
Section 6: The relativistic rotated wave front.

The last part shows that a significant part of both Special Relativity
and Quantum mechanics can be derived from the rule that the wave front
is always at 90 degrees angles with the (physical) group speed, for light
waves as matter waves.

Now illustrated with simulation images.

Regards, Hans