View Full Version : umklapp processes
ehrenfest
Mar27-08, 10:52 PM
1. The problem statement, all variables and given/known data
Can someone please explain to me how umklapp processes K_1 +K_2 = K_3 +G where G is nonzero conserve momentum? I have read the explanation in Kittel over and over and I just don't understand. I also read http://www.physicsforums.com/showthread.php?t=165385 so don't just give me a link to that.
2. Relevant equations
3. The attempt at a solution
olgranpappy
Mar27-08, 11:06 PM
1. The problem statement, all variables and given/known data
Can someone please explain to me how umklapp processes K_1 +K_2 = K_3 +G where G is nonzero conserve momentum?
your K's are not momentum--they are quasi-momentum. They enter into the theory via bloch's theorem which presumes the existance of an external periodic potential. because of this potential, eigenstates of the hamiltonian (labelled by their quasi-momentum) are not eigenstates of the momentum operator.
Quasi-momentum is not conserved.
Similarly, *in the presence of an external potential* true momentum is not conserved either.
I have read the explanation in Kittel over and over and I just don't understand. I also read http://www.physicsforums.com/showthread.php?t=165385 so don't just give me a link to that.
2. Relevant equations
3. The attempt at a solution
ehrenfest
Mar27-08, 11:37 PM
your K's are not momentum--they are quasi-momentum. They enter into the theory via bloch's theorem which presumes the existance of an external periodic potential. because of this potential, eigenstates of the hamiltonian (labelled by their quasi-momentum) are not eigenstates of the momentum operator.
Quasi-momentum is not conserved.
Similarly, *in the presence of an external potential* true momentum is not conserved either.
OK. So, is there a conservation of momentum equation associated with a given umklapp collision that we can write down or is that not part of the theory?
olgranpappy
Mar28-08, 12:12 AM
the relevant equation is the one you wrote down where quasi-momentum is not-conservered, but is "conserved modulo a reciproal lattice vector". So, for example, if I scatter a particle of energy E and (quasi) momentum \vec p by absorbing a phonon of energy \omega and wave-vector \vec q, then I have for conservation of energy and (non) conservation of quasi-momentum
E_{\rm final}=E(p)+\omega(q)
and
\vec p_{\rm final}=\vec p + \vec q + \vec Q
where Q is a vector of the reciprocal lattice.
genneth
Mar28-08, 05:55 PM
Perhaps another way to look at it is that conservation of momentum is a result of translation invariance -- application of Noether's theorem. So if I have a non-uniform potential through space I should not expect momentum to be conserved. Here, we have the slightly perculiar feature that spatial translation is invariant if you do it by a lattice vector. So we have a variable k which is "conserved up to a reciprocal lattice vector".
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