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Almost every solid state physics textbook says crystal momentum is not really physical momentum. For example, phonons always carry crystal momentum but they do not cause a translation of the sample at all.

However, I learned that in indirect-band-gap semiconductors, we need phonons to provide the crystal momentum transfer to allow electron transitions between the top of the valance band and the bottom of the conduction band. Along with absorbing or emitting photons, of course.

Photons do carry physical momentum. For the purpose of momentum conservation, it seems that phonons do carry physical momentum as well.

How can we explain this?

To put it more specifically, I drew a graph to tell the story:

$K$ (capital) is crystal momentum.

For such a transition, the photon provides most of the energy transfer (and a little momentum transfer $hk$; $k$ in lower case), while the phonon provides most of the momentum transfer (and a little energy).

Similar graphs can be found in most solid state physics textbooks. This picture tells me that: either the photon participating in the transition carries crystal momentum, which value is equal to physical momentum $hk$; or the crystal momentum itself is a kind of physical momentum.

However, one can prove that a phonon does not carry physical momentum (here I quote Kittel's "Introduction to Solid State Physics"):

So, how do we explain the momentum transfer in the electron transition aforementioned?

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    $\begingroup$ If you look at it classically, phonons are just waves in the lattice. And they carry energy and momentum in the very same way as do waves in any other medium (e.g. water). But this answer is too obvious so I believe I must have misunderstood your question. $\endgroup$
    – Marek
    Commented Dec 7, 2010 at 10:14
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    $\begingroup$ As for the reason people say that crystal momentum is not a momentum, I guess they are just saying that momentum in the usual sense can take any value (a corollary of the fact that the space is symmetric with respect to arbitrary translations) but in the lattice case it lives in Brillouin zone (because we have only lattice translations here). $\endgroup$
    – Marek
    Commented Dec 7, 2010 at 10:17
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    $\begingroup$ But if crystal momentum is some physical momentum with discrete values, the excitation of phonons should be able to push the entire sample to move. As far as I know we don't see this happening. $\endgroup$
    – skywaddler
    Commented Dec 7, 2010 at 12:15
  • $\begingroup$ Now that is question I would love to see a nice answer for it. Thinking maybe putting a bounty on it. $\endgroup$
    – Bernardo Kyotoku
    Commented Dec 7, 2010 at 12:20
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    $\begingroup$ I don't understand why should the sample move as a whole. When you treat the crystal as a whole then you can safely suppose that its mass is infinite and so that you can safely give it any amount of momentum and it still won't move. You can also get momentum from it without it moving. In this regard, the crystal is like vacuum and phonons are like particles created out of vacuum (in a precise analogy with QFT). $\endgroup$
    – Marek
    Commented Dec 7, 2010 at 13:14

5 Answers 5

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First (as already pointed out in my comments) crystal momentum is in every regard the same as usual momentum except the fact that it takes values only in Brillouin zone (as a consequence of discrete symmetry of lattice; or more precisely its continuum limit). So the answer to your question is: for most purposes crystal momentum is a real momentum.

Now, the term crystal momentum is being used here in two different meanings and that is probably where confusion arises. Your citation uses it as a total momentum of the crystal. This is obviously zero for phonons (which are just harmonic modes of the material) because on average the atoms of the crystal don't move (they just oscillate around stable positions). And that is precisely why nobody uses the term in this way (and I don't understand why your book does).

But locally energy and momentum are still being transfered (jumping from one atom to the next as they interact). So in fact, phonon is a wave that propagates in the material in some direction and carries some energy. Obviously this is a very physical wave with physical energy and physical momentum. It is this latter momentum which is usually referred to as crystal momentum.

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    $\begingroup$ So typically, crystal momentum refers to phonons travelling through the crystal? $\endgroup$
    – Connor
    Commented Mar 8, 2022 at 19:35
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Crystal Momentum can be seen as "less than momentum", that holds a part of the information from what real momentum usually tells, whether or not the other part makes any sense.

Once wave function values is of interest only on lattice sites, for example, as in case of phonons, what matters is how phase changes from one site to another and not elsewhere in space. In such case, 2pi jumps are meaningless, so it is known "up to addition of a reciprocal latice vector"

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I don't see why the crystal should move as a whole. A first argument is the one given by Marek in the comments of the question. But that's also the very principle of a wave: transfer of energy/momentum and no transfer of matter; in this case the ions will vibrate around their equilibrium position in the lattice and the modes of this vibration if a phonon.

Considering this "real momentum"-ness whatever it means this looks strange to me: consider the absorption of a photon (carrying momentum and energy) by the crystal. This process can happen due to the creation of a phonon, or more generally a bunch of them, creating a wave packet. This wave packet will propagate in the crystal (seen as a medium, from the point of view of the phonon this is equivalent to the vacuum from the point of view of the photon) and conserve the momentum.

More generally I think this question is not really interesting from a physical point of view, it is the same as asking if the phonons are real particles or not. I will depend on what you call "real" and following a pragmatic physicist way we call them particles because they have properties that more "naturally real" particles have and they can be treated in the same way...

Edit

I also find that proof a little odd. A real propagating phonon (or plasmon, magnon, whatever) must be seen as a wave packet. In that case the proof given will show that this packet does carry momentum.

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I'm not an expert, but I dont see any contradiction between following statements.

  1. Crystal momentum is not really a physical momentum.
  2. Phonons carry crystal momentum.
  3. Phonons do carry physical momentum as well.
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  • $\begingroup$ Same here. I don't see where the actual complexity in the question should be. And if this simple answer suffices to answer it then it was a pretty bad question in the first place. $\endgroup$
    – Marek
    Commented Dec 7, 2010 at 13:18
  • $\begingroup$ I agree. Either the question is trivial, either it has to be reformulated. $\endgroup$
    – Kostya
    Commented Dec 7, 2010 at 14:09
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Cristal momentum is not a real momentum because this is not an eingevalue of the hamiltonian. Infact you define the cristal momentum according to the periodicity of the bravais lattice. Then the sistem is invariant only under discrete traslation and then k is not a good quantic number.

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    $\begingroup$ This looks like complete nonsense to me. First, momentum is not defined by being an eigenvalue of Hamiltonian. Momentum is a generator of space symmetry of physical laws. Hamiltonian is a generator of time symmetry. There doesn't have to be any relation between the two. Second, why wouldn't crystal momentum be a good quantum number? Quantum numbers come from symmetries and those symmetries don't have to be continuous. $\endgroup$
    – Marek
    Commented Dec 7, 2010 at 17:38
  • $\begingroup$ If the hamiltonian is invariant only under discrete traslation the operator momentum doesn't commute with the hamiltonian $\endgroup$
    – Andy Bale
    Commented Dec 7, 2010 at 17:54
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    $\begingroup$ No, @Andrea. I agree there is a little problem and it is a problem that discrete transformations don't have generators (obviously). But consider 1D lattice with fixed length and lattice spacing going to zero (a continuum limit which is a good approximation to actual 1D crystal) then you'll obtain a good momentum generator that commutes with Hamiltonian (and that is a reason why crystal momentum is a good quantum number too). The only difference to standard momentum is that the crystal momentum will belong to $S^1$ -- a circle. In more dimensions, it will belong to Brillouin zone. $\endgroup$
    – Marek
    Commented Dec 7, 2010 at 18:05
  • $\begingroup$ @marek: yes, I agree with you but if you can measure a lattice length, which is possible for all the crystals known I think, you can define a periodic potential of the lattice and then a periodic hamiltonian. In practice example you consider always the discrete lattice and not the continuum limit because you know the lattice length. Is it wrong? Why? $\endgroup$
    – Andy Bale
    Commented Dec 7, 2010 at 18:50
  • $\begingroup$ depends on what exactly you are doing. If non-vanishing lattice length is important for your calculation and ignoring that would cause problems then sure, work in discrete setting. But when it is not so important, e.g. when dealing with reciprocal lattice, you gain lots of analytic tools by passing to continuum limit (and the limit itself is a very good approximation because lattice length is so small compared to the size of the whole crystal). $\endgroup$
    – Marek
    Commented Dec 7, 2010 at 19:46

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