# Dispersion relations

1. Aug 10, 2010

### stagger_lee

Really struggling with this question guys, any help/advice will be greatly appreciated.

1. The problem statement, all variables and given/known data

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).

2. Relevant equations

3. 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.

2. Aug 10, 2010

### nickjer

Dispersion relations relate the energy, $E$, to the wavenumber, $k$. 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. Aug 10, 2010

### stagger_lee

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. Aug 10, 2010

### nickjer

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, $k$.

5. Aug 10, 2010

### stagger_lee

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. Aug 10, 2010

### nickjer

Yes. But I am surprised you don't have anything on this in your notes or book.

7. Aug 10, 2010

### stagger_lee

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. Aug 10, 2010

### nickjer

Make sense of which part?

9. Aug 11, 2010

### stagger_lee

Make sense of how to sketch the dispersion relation.

What exactly is the the dispersion relation?

10. Aug 11, 2010

### nickjer

It is the equation that relates omega to k. So for a free electron you said it was a parabola.

11. Aug 11, 2010

### stagger_lee

Sorry but I don't get what you mean. How do we know it is a parabola?

12. Aug 11, 2010

### nickjer

From your previous post.

13. Aug 11, 2010

### stagger_lee

Yeah, but how can we determine this to be a parabola?

14. Aug 11, 2010

### nickjer

How is omega related to k in the above equation?

15. Aug 11, 2010

### stagger_lee

omega = h(bar).k^2 / 2m

But what is k? I'm really struggling with this!

16. Aug 11, 2010

### nickjer

'k' is related to the momentum of the electron through:

$$p = \hbar k$$

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