Why crystal wave number $k$ is the constant of motion?

Question

I thought if there is an operator $\hat A$ which commutes with Hamiltonian $\hat H$, the eigenvalue of the corresponding observable $A$ should be the constant of motion.

In free space $(V=0)$, $$[\hat H, \hat p]=0,\quad \hat p|p\rangle=\mathbf p|p\rangle$$ so we can find that the eigenvalue $\mathbf p$ is the constant of the motion, which is momentum.

In case of crystal, $\hat p$ does not commute with Hamiltonian, but the translation operator with lattice vector $\mathbf R$ commutes with Hamiltonian. $$[\hat H, \hat T(\mathbf R)]=0$$ Also by Bloch's Theorem, there exists quantum number $\mathbf k$ (which looks so similiar with the wave number $\mathbf k$) that satisfies $$\psi_\mathbf k(\mathbf r+\mathbf R)=e^{i\mathbf k\cdot\mathbf R}\psi_\mathbf k(\mathbf r)$$ So I guessed: $$\hat H\psi_\mathbf k(\mathbf r)=E\psi_\mathbf k(\mathbf r), \quad \hat T(\mathbf R)\psi_\mathbf k(\mathbf r)=\psi_\mathbf k(\mathbf r+\mathbf R)=e^{i\mathbf k\cdot\mathbf R}\psi_\mathbf k(\mathbf r)$$ then $\psi_\mathbf k(\mathbf r)$ would be the common eigenstate for both of them, and the eigenvalues would be $E$ and $e^{i\mathbf k\cdot\mathbf R}$.

But I couldn't come up with the observable concept for the translation operator $\hat T$, so I cannot understand why we could use $\mathbf k$ for the constant of motion even if $\mathbf p$ is not.

Does my statement above shows that the steady state solutions are identified with only $\mathbf k$? If so, I think I'm not familiar with the concept of $\mathbf k$. How can I understand $\mathbf k$ intuitively? And how could it be preserved?

Answer 1

k here does not take on continuous values as momentum does. It is by no means a solution to the quantum mechanical momentum operator. Here it can take on certain descrete values which come from the following equation

Born-Von Karman boundary condition
$$\psi_\mathbf k(\mathbf r+\mathbf L_i)=\psi_\mathbf k(\mathbf r) i= 1,2,3$$ $i= 1,2,3$ corresponds to the directions of primitive traslation vectors, $\mathbf L_i$ are the crystal dimensions. Also we require $$\psi_\mathbf k(\mathbf r+\mathbf L_i)=e^{i\mathbf k\cdot\mathbf L_i}\psi_\mathbf k(\mathbf r)$$ The above two equations require $e^{i\mathbf k\cdot\mathbf L_i}=1$. Here $\mathbf L_i=N_i \mathbf a_i$
$\mathbf a_i$ are the primitive translation vectors and $N_i$ are large numbers. The requirement is satisfied provided
$ \mathbf k= \sum_{i=1}^{3}s_i \mathbf b_i/N_i$
$\mathbf b_i$ are receprocal lattice vectors $s_i$ takes 0 and interger values. So $ \mathbf k$ is nothing more than a quantum number labelled by $s_i$.

This is a note regarding momentum conservation. The transformation that performs the translation is given by T(R)=exp(-iR.P/$\hbar$) That is the hamiltomian H after transformation is H`. Basically the equation goes as H`=exp(-iR.P/$\hbar$) H exp(iR.P/$\hbar$). However there is a relation exp(A)Bexp(-A)=B + [A,B]+[A,[A,B]]/2+.... Applying to the translation H`=H + -iR.[P,H]... Since H is invariant under translation H`=H. Which makes P commute with H and thus a constant of motion. In this case R is restricted and cannot have any value. P is not constant over all space.