Phonon momentum


by Niles
Tags: momentum, phonon
Niles
Niles is offline
#1
Sep19-09, 12:41 PM
P: 1,863
Hi all

My book says:

"The reason that phonons on a lattice do not carry momentum is that a phonon coordinate (except for wavevector K=0) involves relative coordinates of the atoms".

I can't quite figure this statement out. I understand the words, but I cannot see why it is an explanation.

Can you shed some light on this topic?

Best regards,
Niles.
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elduderino
elduderino is offline
#2
Sep19-09, 08:21 PM
P: 57
Hi.

Read into something called the Born approximation for this lattice model where each atom is considered a harmonic oscillator. Basically, the mean position of the vibrating atom (its COM) does not change over time.

Such approximation is needed when making such a model. So you can consider every atom (each with its own fixed co-ordinate system) to be a classical harmonic oscillator, executing small oscillations about its mean position (origin of the relative co-ordinate system for that particular atom).

I guess now it can be understood. Phonons have energy, but no net momentum, as they keep going back and forth about a fixed origin, and in each cycle the momentum cancels itself out.
kanato
kanato is offline
#3
Sep19-09, 11:42 PM
P: 416
I don't really understand that. Phonons definitely have a well-defined pseudomomentum, and Umklapp processes which reduce the total momentum of phonons are responsible for keeping thermal conductivity finite in a perfect crystal. Is there more context to the statement in the book? Are they talking about only k = 0 phonons?

Niles
Niles is offline
#4
Sep20-09, 08:48 AM
P: 1,863

Phonon momentum


k=0 phonons are mentioned in the quote as the only phonons with actual momentum, which is 0.

But you say pseudo-momentum?
kanato
kanato is offline
#5
Sep20-09, 10:48 AM
P: 416
Yes.. in a periodic crystal you don't have continuous translational symmetry, so conservation of momentum doesn't hold. But due to the discrete translational symmetry there is a conservation law of the pseudomomentum vector k, where k is a vector in the first Brillouin zone.
Niles
Niles is offline
#6
Sep21-09, 02:57 AM
P: 1,863
Quote Quote by kanato View Post
Phonons definitely have a well-defined pseudomomentum, and Umklapp processes which reduce the total momentum of phonons are responsible for keeping thermal conductivity finite in a...
Ok, so Umklapp processes change the total phonon pseudomomentum. But then how can we talk about conservation of phonon-momentum?
kanato
kanato is offline
#7
Sep21-09, 03:02 PM
P: 416
Conservation of total momentum you can't talk about, it doesn't exist in a system with an external potential. Pseudomomentum is conserved, even in Umklapp processes, but pseudomomentum is only well defined up to the edge of the first Brillouin zone, or more precisely, any function in the lattice of f(k) = f(k + K) where K is any integer combination of reciprocal lattice vectors. Only the value of k (the pseudomomentum in the first Brillouin zone) is conserved, if some process adds multiple pseudomomenta and gets a value outside the first BZ it will be translated back in by a vector K.


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