# Are phonons eigenstates of the momentum operator?

In the case of electrons in a periodic potential it can be demonstrated that the eigenstates of the Hamiltonian containing the periodic potential are the Bloch functions: $$\Psi_{n \mathbf{k}}(\mathbf{r})=e^{i\mathbf{k}\cdot\mathbf{r}}u_{n\mathbf{k}}(\mathbf{r})$$ where $u_{n\mathbf{k}}(\mathbf{r})$ is lattice periodic: $u_{n\mathbf{k}}(\mathbf{r+R})=u_{n\mathbf{k}}(\mathbf{r})$. The Bloch functions are not eigenstates of the momentum operator: $$-i\hbar\nabla \Psi_{n \mathbf{k}}(\mathbf{r})=\hbar \mathbf{k}\Psi_{n \mathbf{k}}(\mathbf{r})+e^{i\mathbf{k}\cdot\mathbf{r}}(-i\hbar\nabla )u_{n\mathbf{k}}(\mathbf{r}) \tag{0}$$

In the case of Phonons (Eq.(3.17) pag 42 of Ref. [Ulrich Rössler. Solid state theory: an introduction. Springer Science & Business Media, 2009.]) the eigenstates of the dynamical matrix have the form: $$\mathbf{u}(\mathbf{R},b,\alpha)=\mathbf{u}(b,\alpha)e^{+i \cdot \mathbf{q} \cdot \mathbf{R}} \tag{1}$$ where $\mathbf{u}(\mathbf{R},b,\alpha)$ denotes the displacement along the cartesian direction $\alpha$ of the atom $b$ in the unit cell in position $\mathbf{R}$ in the lattice.

The importance of Eq. (1) is that the displacements of the atoms associated to the vibrational mode having wavevectorf $\mathbf{Q}$ in different unit cells differ just by the phase factor $e^{+i \cdot \mathbf{q} \cdot \mathbf{R}}$.

This means that, if we apply the momentum operator (taking an analytic expansion of the discrete variable $\mathbf{R}$ in order to be able to perform the derivative:

$$\hat{\mathbf{p}} \;\mathbf{u}(\mathbf{R},b,\alpha)=-i \hbar \nabla_{\mathbf{R}}\mathbf{u}(b,\alpha)e^{+i \cdot \mathbf{q} \cdot \mathbf{R}} =\hbar \mathbf{q}\;\mathbf{u}(b,\alpha)e^{+i \cdot \mathbf{q} \cdot \mathbf{R}} \tag{2}$$

therefore, due to the independence of the $\mathbf{u}(b,\alpha)$ from $\mathbf{q}$ phonons can be considered eigenstates of the momentum operator, differently from electrons. To my understanding this makes sense since phonons are waves of atomic displacements, hence they have a well defined wave-vector $\mathbf{q}$ and hence, as waves, they carry a momentum $\mathbf{p}=\hbar\mathbf{q}$ (de Broglie particle-wave relation).

I would like to deepen this reasoning, in particular I am not convinced with the "analytic extension" used to take the gradient with respect to the discrete unit cell position $\mathbf{R}$ in Eq. (2).

Does anyone know any good reference where I can deepen this fact or has any useful comment on this reasoning?

## EDIT:

All my reasoning relies on the analytic extension of the discrete variable $\mathbf{R}$. If I define the momentum operator as $-i\hbar\nabla_{\mathbf{R}}$ where $\nabla_{\mathbf{R}}$ acts on the analytically extended discrete variable $\mathbf{R}$ the equation (2) shows that the Bloch functions are eigenstates of the (analytically extended) operator $-i\hbar\nabla_{\mathbf{R}}$.

Then I agree that $\mathbf{q}$ is a crystal momentum defined up to reciprocal lattice vector but, if I restrict myself to the 1st Brillouin zone, why can'I say that in this case the Eq. (1) is an eigenstate of the operator $-i\hbar\nabla_{\mathbf{R}}$? I agree that Eq (2) with the replacement $\mathbf{q}\to\mathbf{q+G}$ still is a solution of the problem but this ambiguity can be removed refolding $\mathbf{q+G}$ to the first Brilluoin zone.

My question comes from the fact that the demonstration that the electron Bloch function are not eigenstates of the momentum operator Eq. (0) does not hold for phonons, Eq. (2).

Why do I need continuous translational symmetry (hence momentum conservation in the strict sense) to speak about eigenstates of the momentum?

• Usually phonons are described as eigenstates of quasimomentum, rather than momentum. – probably_someone Feb 11 '18 at 0:26
• Ok, then this is a matter of names. How do you define the quasi momentum operator? In my reasoning I do not care about the fact that $\mathbf{q}$ is defined up to reciprocal lattice vectors. I have just observed that the momentum operator $-i\hbar \nabla_{\mathbf{R}}$ gives the quasi momentum (or crystal momentum) $\hbar\mathbf{q}$ when applied to the Bloch function for phonons. Moreover I have shown that this is not the case for electrons. – Caos Feb 11 '18 at 12:54

The vector $\hbar \vec k$, where $\vec k$ is the wave vector of phonons in a crystal, is called the crystal momentum in analogy to photons. It is not a real momentum because $\vec k$ is only defined up to integer multiples of the reciprocal lattice vectors. Therefore, also for phonons there is a similar situation as with the quasi-momenta of Bloch-electrons in the crystal. A good but older book to explain this is: A. Anselm, Introduction to Semiconductor Theory, MIR Publishers, Moscow 1981 (you can download it for free).

• Thanks a lot for the answer. I has a look at the book you suggested and it says: "We use the term quasi-momentum for the quantity $\mathbf{q}$ instead of momentum because it displays certain properties not peculiar to the momentum of a free particle; for instance, in collisions the sum of quasi-momenta is conserved to within an arbitrary vector of the reciprocal lattice." However, this does not invalidate the difference between electrons and phonons that I have mentioned in Eq. (0) and Eq. (2). – Caos Feb 11 '18 at 13:09

As stated, the reasoning is just wrong. Momentum conservation comes from continuous translational symmetry; here you only have a discrete translational symmetry. The resulting quantity is crystal momentum, as seen by the fact that it's ambiguous up to multiples of $2\pi/L$. You can also assign the phonon an ordinary momentum by just adding up the momenta of all the lattice ions, but this is generally equal to zero, ignoring relativistic effects.

Edit: to address your second question, this comes down to semantics. Informally, you might say momentum is just a measure of the "oomph" of a particle, the amount it can push something else when it hits it. But in the common usage, spatial momentum is defined to be the conserved quantity associated with continuous spatial translations. (Other momenta are defined the same way, e.g. the field momentum is conserved if the Lagrangian is symmetric under translation of the field.)

Your proposed conserved quantity arises from a different symmetry, so we just give it a different name, usually quasimomentum or crystal momentum. You're free to call it momentum but this can lead to confusion when you talk with other people. For example, the sum of your conserved quantity and the momentum of other objects is not conserved.

The fact that formally you can get the quasimomentum to pop out of a continuous symmetry is deceptive. While it's true that the discrete function $f(n) = n$ can be "continued" to a continuous function $f(x) = x$, this is far from unique (consider $f(x) = x + \sin(\pi x)$). More importantly, it's not physical -- there is nothing there at the in-between points.

• Thanks a lot for the answer. I added an edit to my question. In short: why do we need continuous translational symmetry to speak about eigenstates of the momentum? – Caos Feb 11 '18 at 12:52
• @Caos I edited to address your questions. – knzhou Feb 11 '18 at 15:10