# Homework Help: Solid State - Phonons at Brillouin Zone Boundary

1. Apr 1, 2013

### Lengalicious

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

2. Relevant equations
{3.9b}
$$A[2\mu -m\omega ^2 ]=2\mu Bcos(\frac{ka}{2})$$
$$B[2\mu -M\omega ^2 ]=2\mu Acos(\frac{ka}{2})$$

3. The attempt at a solution

All I can think of is setting $$k =\frac{\pi}{a}$$ so that

$$B[2\mu -M\omega ^2 ]=A[2\mu -m\omega ^2 ]$$

solve for omega squared and take the negative root as the accoustic phonon frequency which I imagine to be =0 to show that the phonons are stationary in the accoustic branch. What I'm not entirely sure of is how to initially manipulate the amplitude Eq's, I guess I somehow have to sub part of the equation representing the small mass particles into the equation representing the large mass particles (as the optic phonons are dependent of the small mass particles and visa versa) and manipulate, some guidance would be great.

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2. Apr 5, 2013

### TSny

Since you are only interested in the zone boundary, go ahead and substitute $k = \pi/a$ into the above equations. So:
$$A[2\mu -m\omega ^2 ]=0$$
$$B[2\mu -M\omega ^2 ]=0$$
To get a non-trivial solution for $A$ and $B$, you need to assume at least one of $A$ or $B$ is nonzero. What can you conclude if you let $A \neq 0$?

3. Apr 5, 2013

### Lengalicious

Well if A does not equal zero then once you equate the equations B cannot equal zero? :S

4. Apr 5, 2013

### TSny

Start with $A[2\mu -m\omega ^2 ]=0$ and the assumption that $A$ is not zero. What does this equation tell you about the frequency $\omega$?

5. Apr 5, 2013

### Lengalicious

I guess frequency squared is inversely proportional to the mass so the smaller the particle mass the larger the frequency? And visa versa, the optical branch frequency > accoustic branch so it would imply the optical branch should affect light particles? Not sure

6. Apr 5, 2013

### TSny

Note that if $A \neq 0$ then the equation $A[2\mu -m\omega ^2 ]=0$ can only be satisfied if $\omega$ has a particular value. With the value of $\omega$ now fixed, move on to the equation $B[2\mu -M\omega ^2 ]=0$ and see what that tells you about $B$.

7. Apr 5, 2013

### Lengalicious

Ok, to figure that out I'll need to understand the first bit, how come A being non-trivial can only be satisfied if omega is fixed? Not really understanding that bit :S

8. Apr 5, 2013

### TSny

Suppose you have two two numbers x and y and you know two things: x≠0 and xy=0.

What can you conclude about y?

Apply that to $A[2\mu -m\omega ^2 ]=0$

9. Apr 5, 2013

### Lengalicious

ok yeh, so [2*mu - m*w^2] must equal 0, but then m AND omega should be a specific value such that m*w^2 = 2*mu right, not just omega? Cause i'm assuming the mass is a variable

10. Apr 5, 2013

### Lengalicious

Also I understand now, since omega is fixed and m does not equal M. . . [2*mu - M*w^2] does not equal 0 in the 2nd equation like it does in the first?

11. Apr 5, 2013

### TSny

No, the masses m and M are considered fixed masses for a given lattice. They are not variables.

12. Apr 5, 2013

### Lengalicious

Yeh my bad :), so basically all this implies that B=0 and A does not

13. Apr 5, 2013

### Lengalicious

So, B = [2*mu - m*w^2]

((B-2*mu)/m)^1/2 = w, am I on the right lines?

14. Apr 5, 2013

### TSny

No, I'm afraid that isn't right at all.

Earlier, you made the correct deduction that when A ≠ 0 then 2*mu - m*w^2 = 0 and B = 0 . The equation 2*mu - m*w^2 = 0 can be used to find w for this mode.

Repeat similar arguments for the other mode where B ≠ 0.

15. Apr 5, 2013

### Lengalicious

Ok, well thanks for all the help, much appreciated :)

16. Apr 6, 2013

### Lengalicious

Ok sorry, so I get w = (2*mu/m)^1/2, but I don't see how this is indicative of the small mass particles being stationary in the accoustic branch?

17. Apr 6, 2013

### TSny

What do the numbers A and B represent physically?

18. Apr 6, 2013

### Lengalicious

Amplitudes, so B = 0 means the amplitude for the particles in the accoustic branch is zero?

19. Apr 6, 2013

### TSny

B is the amplitude of oscillation for which particles? Particles with mass m or particles with mass M?

You have found that if A ≠ 0 then B = 0.

You also have found that if A ≠ 0, then ω = (2*mu/m)^1/2. Does this frequency correspond to the acoustic branch or the optical branch?

In order to answer this, you might want to work out the other case where you start with the assumption that B≠ 0.

20. Apr 6, 2013

### Lengalicious

well I imagine that in the sake of answering the question, B=0 would suggest the amplitude of 'm' mass particles is zero considering those are the ones that are supposed to be stationary, but yes I'll attempt setting B to not equal 0

21. Apr 6, 2013

### Lengalicious

so, ω = (2*mu/m)^1/2 when B=0 & ω = (2*mu/M)^1/2 when A=0.

I'm still not really seeing it =/. So the frequency is independent of the large mass at B = 0 and visa versa. I guess the one independent of the larger mass is for the optical branch and the one independent of the smaller mass is for the acoustic branch? so, ω = (2*mu/m)^1/2 when B=0 suggests that the amplitude of the particles in the optical branch is zero?

22. Apr 6, 2013

### TSny

The original equations of motion are
$$A[2\mu -m\omega ^2 ]=2\mu Bcos(\frac{ka}{2})$$
$$B[2\mu -M\omega ^2 ]=2\mu Acos(\frac{ka}{2})$$
These show that A is the amplitude for motion of the particles with the smaller mass m and B is the amplitude for the mass M particles.

At the zone boundary, you have found that if the small mass particles are moving (i.e., A ≠ 0) then B must be zero. So, the large mass particles are not moving. You also found that in this case the frequency is ω = (2*mu/m)^1/2.

The other possibility is for the large mass particles to be moving (i.e., B ≠ 0). Then you find that A = 0 and ω = (2*mu/M)^1/2.

Now, one of these cases corresponds to the acoustic branch while the other case corresponds to the optical branch. Can you see which case is the acoustic branch? (You should have learned which branch corresponds to the lower frequency and which branch corresponds to the higher frequency.)

Last edited: Apr 6, 2013
23. Apr 6, 2013

### Lengalicious

Ok, that makes a lot more sense now, and in my notes it is said that the frequency in the optical branch is proportional to 1/(m)^1/2 so that would be the larger frequency
so basically, ω = (2*mu/m)^1/2 corresponds to the optical branch and the deduction has shown that the particles with mass M are stationary & visa versa.

Yeah that makes so much more sense to me now, thank you very much indeed :)