Conservation of crystal momentum

Question

I am trying to convince myself that crystal momentum is conserved in a periodic lattice modulo a reciprocal lattice vector.

Consider a Hamiltonian $H$ which is periodic under translations of a Bravais lattice vector. The canonical momentum operator $\mathbf{P} = (P_x,P_y,P_z)$ is the generator of translations, so I can write my translation operator as

$$ T(\mathbf{a}) = e^{i \mathbf{a} \cdot \mathbf{P}}, \quad \mathbf{a} \in \mathbb{R}^3.$$

However, for a periodic Hamiltonian, the full symmetry is broken down to translations within the Bravais lattice only. I would express this symmetry as $[ T(\mathbf{a}) , H] =0$ for any Bravais lattice vector $\mathbf{a}$. Now, substituting my translation operator into the commutator, I find

$$ \mathbf{a} \cdot[ \mathbf{P} , H] = 0$$

If my system had the full translation symmetry, I could factor out the $\mathbf{a}$ to conclude that each component of the momentum is conserved: $[P_i, H] = 0$. However, as we are restricted to the Bravais lattice, I can only conclude that $ \mathbf{a} \cdot \mathbf{P}$ is conserved and I would rename $\mathbf{P}$ as the crystal momentum.

I am unsure how I arrive at the fact that the crystal momentum is conserved modulo a reciprocal lattice vector. I imagine it has something to do with assuming I can bring down the exponent in the commutator. I can see why the exponential does not define momentum uniquely, however if I had full translational symmetry, I would be able to say the exponent is conserved. What is different here?

Answer 1

There is no need to expand the exponential at all. Let the lattice have basis $\mathbf{a}_i$. The fact that $$[e^{i \mathbf{a}_i \cdot \mathbf{P}}, H] = 0, \quad [e^{i \mathbf{a}_i \cdot \mathbf{P}}, e^{i \mathbf{a}_j \cdot \mathbf{P}}] = 0$$ indicates that we can simultaneously diagonalize the $e^{i \mathbf{a}_i \cdot \mathbf{P}}$ and $H$. Since $e^{i \mathbf{a}_i \cdot \mathbf{P}}$ is unitary, its eigenvalues are pure phases, so we may define $$e^{i \mathbf{a}_i \cdot \mathbf{P}} |\psi \rangle = e^{i \phi_i} |\psi \rangle.$$ Now, because the $\mathbf{a}_i$ form a basis of $\mathbb{R}^3$, there exist vectors $\mathbf{k}$ so that $$e^{i \phi_i} = e^{i \mathbf{a}_i \cdot \mathbf{k}}.$$ We can then call $\mathbf{k}$ the "crystal momentum". The reason that $\mathbf{k}$ is only defined up to multiples of reciprocal lattice vectors is because we have not specified $\mathbf{k}$ anywhere in this argument, only its exponential. Indeed, if we add a reciprocal lattice vector $\mathbf{b}_j$, then the phases change by $e^{i \mathbf{a}_i \cdot \mathbf{b}_j} = e^{2 \pi i \delta_{ij}} = 1$ by the definition of the reciprocal lattice.

For full translational symmetry, you can take $\mathbf{a}$ infinitesimal and Taylor expand the exponential, giving $[\mathbf{a} \cdot \mathbf{P}, H] = 0$, and then since $\mathbf{a}$ is arbitrary we have $[\mathbf{P}, H] = 0$. But for the lattice translations, expanding the exponential isn't really clean, and it's not necessary either.

Answer 2

I think the answer to your question is that it is more appropriate to say that Crystal Momentum is defined modulo reciprocal lattice translations.

Let us say that the crystal has Bravais lattice vectors $ \{ e_i \} $ , $i=1,...,d$. We can construct the reciprocal lattice vectors $\{f_j\}$ satisfying $e_i . f_j = 2 \pi\delta_{ij}$. A general lattice translation is given by $a = \Sigma n_i$ $e_i$ , $n_i$ $\epsilon$ $\mathbb{Z}$.

These translations are generated by the "crystal momentum", $P = \Sigma P_j \hat{f_j}$.
Here $P_j$ is the component of crystal momentum along the $j$th direction on the reciprocal lattice. The translation operator is $T(a)$ = $T(\{n_i\})$ = $e^{iP.a}$ = $exp ( 2 \pi i\Sigma \frac{n_i P_i}{|f_i|})$ = $exp ( 2 \pi i\Sigma \frac{n_i (P_i + m_i |f_i|) }{|f_i|})$ , for any $m_i$ $\epsilon$ $\mathbb{Z}$ .

The last equality shows that the the same lattice translation is generated if we replace $\{ P_i \}$, the components of crystal momentum in reciprocal space, by $\{ P_i + m_i |f_i| \}$ (Or, $P$ is replaced by $P$ $+$ $\Sigma$ $m_i f_i$).

This means that "crystal momentum" (ie, the generator of lattice translations) is only defined modulo reciprocal lattice vector. Stated another way, the only eigenvalues of crystal momentum that need to be considered are the ones that belong to the first Brillouin zone.

The rest of what you say is correct. From the fact that $[T(\{ n_i \}), H] $ = $0$ $\forall$ possible lattice translations $\{ n_i \}$, we obtain that crystal momentum is conserved.

Also see : Unjustified claim in Kittel about Bloch functions (towards the end) for a 1-d version of the argument presented above, and a derivation of the Bloch theorem.