I'm frustrated that I can't understand something that must be simple and fundamental. I'd appreciate any answer to the question, but also any clarifications of how my presentation of the theorem/proof, or general quantum theory, is incorrect. Apologies for the long question; I could not figure out how to present the question completely without all of these details.

Bloch's Theorem states that, for a system with a periodic Hamiltonian — $H({\bf r} + {\bf R}) = H({\bf r})$ for all ${\bf R} \in A$, where $A$ is the set of all lattice symmetry translations, i.e., all integer linear combinations of the primitive lattice vectors $({\bf a\_1}, {\bf a\_2}, {\bf a\_3})$ — any energy eigenstate satisfying $H | \psi \rangle = E | \psi \rangle$ "can be chosen so that associated with each $\psi$ is a wavevector ${\bf k}$ such that$^1$:"
$$
\psi({\bf r} + {\bf R}) = {\rm e}^{i {\bf k} \cdot {\bf R}} \, \psi( {\bf r})
$$
for each ${\bf R}\in A$. If we write:
$$
\psi({\bf r}) = {\rm e}^{i {\bf k} \cdot {\bf r}} u\_{ {\bf k}}({\bf r})\,,
$$
then we see that
$$
\psi({\bf r} + {\bf R}) = {\rm e}^{i {\bf k} \cdot {\bf r}} {\rm e}^{i {\bf k} \cdot {\bf R}} \, u\_{ {\bf k}}( {\bf r} + {\bf R})\, ,
$$
which implies that $u\_{\bf k}({\bf r} + {\bf R}) = u\_{\bf k}({\bf r})$ for all ${\bf R} \in A$. This is an alternative statement of the theorem: any energy eigenstate can be written as a plane wave times a function, labeled by the wavevector ${\bf k}$, that has the periodicity of the lattice.

Proofs of the theorem, e.g., from Ashcroft and Mermin or the [wikipedia page](https://en.wikipedia.org/wiki/Bloch%27s_theorem#Bloch%27s_theorem), rely on choosing the wavevector to be of the form:
$$
{\bf k} = {\bf x} \cdot {\bf b} = x\_1 {\bf b\_1} + x\_2 {\bf b\_2} + x\_3 {\bf b\_3} \quad ({\bf x} \in \mathbb{R}^3)
$$
where the ${\bf b\_i}$ are the primitive reciprocal lattice vectors satisfying ${\bf a\_i} \cdot {\bf b\_j} = 2 \pi \delta\_{ij}$. The fact that the Hamiltonian commutes with each unitary translation operator $T\_{\bf R}$ implies that energy eigenstates are also eigenstates of these translation operators. Letting
$$
{\bf R} = {\bf n} \cdot {\bf a} = n\_1 {\bf a\_1} + n\_2 {\bf a\_2} + n\_3 {\bf a\_3} \quad ({\bf n} \in \mathbb{Z}^3)
$$
we can write the eigenvalues of the translation operators as
$$
T\_{\bf R} \, \psi({\bf r}) = \exp[i\, 2\pi \, (n\_1 \tilde{x}\_1 + n\_2 \tilde{x}\_2 + n\_3 \tilde{x}\_3)] \, \psi({\bf r})
$$
where the $\tilde{x}\_i$ values specify the eigenvalues of the ${\bf a\_i}$ primitive translations, i.e.,
$$ T\_{\bf a\_i} \, \psi({\bf r}) = \exp[i\, 2 \pi \, \tilde{x}\_i] \, \psi({\bf r})\,.
$$
But, if we choose ${\bf k} = {\bf \tilde{x}} \cdot {\bf b}$, then we can write
$$
T\_{\bf R} \, \psi({\bf r}) = \exp[i\, {\bf k} \cdot {\bf R}] \, \psi({\bf r})
$$
for all ${\bf R}\in A$. Therefore:
$$
 \psi({\bf r} + {\bf R}) = T\_{\bf R} \, \psi({\bf r}) = {\rm e}^{i\,{\bf k} \cdot {\bf R}} \, \psi({\bf r})
$$
for all ${\bf R} \in A$, showing that we can find a ${\bf k}$ such that the theorem is satisfied.

My confusion here is that it seems that the vector ${\bf k}$ is *uniquely* specified given an energy eigenstate $\psi$, i.e., it is given by the values $(\tilde{x}\_1, \tilde{x}\_2, \tilde{x}\_3)$ that yield the eigenvalues of the primitive translations, ${\bf a\_i}$, acting on that eigenstate $\psi$. That interpretation of the proof must be incorrect, however, because it implies a single ${\bf k}$ vector for each energy eigenstate. And, it is my undestanding, from the application of Bloch theory, is that the wavevector ${\bf k}$ should be thought of as a free parameter, an arbitrary real-valued 3-vector label (though its values can be restricted to the "first Brillion zone"). Indeed it is this continous freedom that gives the bands of energies. What am I missing here?

**Possible solution**: If I write the simultaneous energy and translation eigenstates as $| \psi\_{\alpha, {\bf k}}\rangle$ then I must write the action of the translation operator as:
$$
T\_{\bf R} \, |\psi\_{\alpha, {\bf k}} \rangle = \exp[i\, 2\pi \, (n\_1 \tilde{x}\_{\alpha, {\bf k},1} + n\_2 \tilde{x}\_{\alpha, {\bf k},2} + n\_3 \tilde{x}\_{\alpha, {\bf k},3})] \, | \psi\_{\alpha, {\bf k}} \rangle
$$
and then I choose ${\bf k} = {\bf \tilde{x}\_{\alpha, {\bf k}}} \cdot {\bf b}$. This means that the $\tilde{x}\_i$ values are specific to an eigenstate labeled by $(\alpha, {\bf k})$, not just by $\alpha$. Therefore we actually have complete freedom to choose ${\bf k} \in \mathbb{R}^3$. This argument feels a bit circular to me.

1 - From Ashcroft and Mermin, Chapter 8.