The momentum of a hole I'm currently working through "A Guide to Feynman Diagrams in the Many-Body Problem" by R.D. Mattuck (self study, not a homework problem) and am stumped by the following problem:
"In a system of free particles, a hole is created in the single particle state $\phi_k=\frac{1}{\sqrt{V}}e^{i\vec{k}\cdot\vec{r}}$. What is the momentum of the hole?"
The answer given is $-\vec{k}$. In the above, $V$ is the normalization volume of the system, $\vec{k}$ is the wavevector and $\vec{r}$ is the position. Natural units ($\hbar=1$) are used throughout this question.
If this were a particle instead of a hole, the momentum would simply be $\vec{p_e}=\vec{k}$. If the answer given in the book is correct, am I then right in thinking we must redefine the momentum operator somehow? If so what motivation do we have for doing this?
In my head these ideas seem to run contrary to what I know from solid state physics: if we have an electron in the valance band and excite it to the conduction band, and if there is no input of momentum during this process, then I always thought we would be left with an electron of wavevector $\vec{k}$ and a hole with wavevector $\vec{k}$ also. My feeling is that this IS correct, but that I am falling victim to the sloppy condensed matter habit of using the terms wave-vector and momentum interchangeably. 
Edit: I have come to the realization that the issue is with the wording of the problem, which is not entirely transparent. It makes more sense if written:
"An electron initially has momentum $\vec{k}_e$ and wavefunction $\phi_{k_{e}}=\frac{1}{\sqrt{V}}e^{i\vec{k}_e\cdot\vec{r}}$. It is removed from the system leaving behind an empty orbital. What is the momentum of the orbital? In the particle-hole picture, what is the momentum of the hole?"
Answer: The orbital left vacant still has wavevector $\vec{k}_e$ which can be seen by applying the momentum operator $\hat{p}=-i\nabla$ to $\phi_{k_{e}}$. The vacant orbital is not, however, the same thing as a hole. If the system originally had zero total momentum, then we have removed some amount $\vec{k}_e$ and the total momentum of the system is now $-\vec{k}_e$. We attribute this momentum to the hole, and say that $\vec{k}_h=-\vec{k}_e$. The hole wavefunction must then be (see Heisenberg, 1931 for proof) $\phi_{k_{h}}=\phi_{-k_{e}}=\frac{1}{\sqrt{V}}e^{-i\vec{k}_e\cdot\vec{r}}$. Applying the momentum operator to this wavefunction gives the correct result.
The confusion arose because, in the question given by Mattuck, it is unclear whether $\vec{k}$ is the wavevector of the empty orbital left behind by the electron or the wavevector of the hole.
 A: 
if we have an electron in the valance band and excite it to the conduction band, and if there is no input of momentum during this process, then I always thought we would be left with an electron of wavevector $\vec k$  and a hole with wavevector $\vec k$  also.

You wrote it yourself : there is no input of momentum during this process. Then why do you want both particles to have the same momentum $\vec k$ ? Since there is no input of momentum the total momentum of the system { hole + electron } is $0$.
Since the electron's wavevector is $\vec k$, the hole's wavevector is obviously $- \vec k$. I don't think that you have to redefine the momentum operator.
You can get a classical picture of your issue. Imagine a bunch of balls (the valence band), they are all going to the same direction (let's say left direction). If you suddenly pick one ball out (you put an electron in the conduction band), then you will see a hole appearing. The hole is not moving, but if you consider its relative motion, the hole is going in the opposite direction, the right direction. Hence the opposite momentum.
A: Momentum does not have a direction associated with it.  A wave vector does.  So the electron and hole can have the same +/- sign for their respective momenta.
