Dispersion relations

  • #1
Really struggling with this question guys, any help/advice will be greatly appreciated.
Thanks in advance!

Homework Statement



Sketch dispersion relations for, (i) a proton (ii) a free electron (iii) an electron, in a crystalline solid. Comment on the similarities between (ii) and (iii).



Homework Equations





The Attempt at a Solution



I can't find any information on this topic at all! There isn't even a mention of dispersion relations in my course notes and can't find anything on the net.
I have Electrical Properties Of Materials, Seventh Edition but can't find anything in there either.
 

Answers and Replies

  • #2
674
2
Dispersion relations relate the energy, [itex]E[/itex], to the wavenumber, [itex]k[/itex]. For free particles it is fairly simple. For the crystalline solid there are many methods about getting it, but not sure what you are currently learning.
 
  • #3
Thank you for your response.

I think the class is considered to be a Quantum Physics class. I am an Electrical Engineering student but I have no Physics knowledge so please bear with me. Is k the variable assigned to the wavenumber?

How would we sketch the dispersion relation?

Are we required to use a formula?
 
  • #4
674
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I can't unfortunately give you the answer for the free particles. But since they are free, that means they experience no potential energy. So the total energy is just kinetic energy. So how is kinetic energy written in terms of the wavenumber, [itex]k[/itex].
 
  • #5
I have a formula in my notes;

E = h(bar).(ohmega)

but I don't have anything in terms of k.


I googled 'k' and found the following formula;

k = p / h(bar)

Am i on the right track here?
 
  • #6
674
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Yes. But I am surprised you don't have anything on this in your notes or book.
 
  • #7
I stand corrected. I have just found 2 slides on dispersion relations. Sincerest apologies, must have over-looked them (been studying since I got in from work 6hrs ago)!

Slide 1;

Dispersion Relations

Waves have a wavelength λ and a frequency ν. For simple waves
the relation between λ and ν is simple: their product is a constant.
λ.ν = v, a constant, called the phase velocity

For non-simple waves, the relation ν(λ) is not linear. Two
velocities describe the wave propagation: 1) the phase velocity v
and 2) the group velocity vg (at which information propagates).

We define a frequency variable: ω=2πν (angular frequency)
We define an inverse wavelength: k=2π/λ (wavevector or wavenumber)

Then, v = ω/k and vg= dω/dk

The analytical relation (if it exists) which defines ω in terms of k
is called the dispersion relation, ω(k).


Slide 2;

Dispersion relations for a free electron and a photon

1) For a free electron, the momentum is related to the wavelength
and the energy to the wave frequency of the associated wave:

p = h /λ = h(bar).k and E = hν = h(bar).ω

But the energy and momentum of a massive particle are also
related if the energy is kinetic:
E = 1/2 mv^2 = (p^2)/m

Hence: ohmega = h(bar).k^2 / 2m
This is the dispersion relation of a free electron.

2) For a monochromatic photon,

E = cp

Hence ω = ck is the (simple) dispersion relation of a photon.


I know this is a lot but can you help me make some sense of this please?

Thank you for your help/patience!
 
  • #8
674
2
Make sense of which part?
 
  • #9
Make sense of how to sketch the dispersion relation.

What exactly is the the dispersion relation?
 
  • #10
674
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It is the equation that relates omega to k. So for a free electron you said it was a parabola.
 
  • #11
Sorry but I don't get what you mean. How do we know it is a parabola?
 
  • #12
674
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Hence: ohmega = h(bar).k^2 / 2m
This is the dispersion relation of a free electron.
From your previous post.
 
  • #13
Yeah, but how can we determine this to be a parabola?
 
  • #14
674
2
It is the equation that relates omega to k. So for a free electron you said it was a parabola.
Hence: ohmega = h(bar).k^2 / 2m
This is the dispersion relation of a free electron.
How is omega related to k in the above equation?
 
  • #15
omega = h(bar).k^2 / 2m

But what is k? I'm really struggling with this!
 
  • #16
674
2
'k' is related to the momentum of the electron through:

[tex]p = \hbar k[/tex]

So you can think of a dispersion relation as a relation between energy and momentum.
 

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