Density operator in momenta representation using Fourier transform I'm determining the density operator in momentum space and by working out the Fourier transform from the coordinate representation I get to:
$$\hat n_q=\sum_{kk'ss'}\left<k,s|e^{-iq\hat r}|k's'\right>a^\dagger_{ks}a_{k's'}$$
How does the exponential defined above shifts the momenta according to:
$$\hat n_q=\sum_{kk'ss'}\delta_{ss'}\left<k-q|k'\right>a^\dagger_{ks}a_{k's'}$$
How did the exponential act on the bra and shifted the momenta?
 A: Momentum generates translations in position, which means that $e^{i \hat{q} r}$ shifts position by $r$. And this is completely symmetric: position generates translations in momentum, which means that $e^{i q \hat{r}}$ shifts momentum by $-q$. 
To see this explicitly, note that
$$\hat{r} = i \frac{\partial}{\partial p}$$
and so, expanding in a Taylor series,
$$e^{i q \hat{r}} \tilde{\psi}(k)  = \sum_{n=0}^\infty \frac{(iq \hat{r})^n}{n!} \tilde{\psi}(k) = \sum_{n=0}^\infty \frac{(-q)^n}{n!} \tilde{\psi}^{(n)}(k) = \tilde{\psi}(k-q)$$
where the last step follows from the definition of Taylor series. So for momentum eigenstates,
$$e^{i q \hat{r}} |k \rangle = |k-q \rangle.$$
A: One way to approach it is by direct calculation (in the position representation):
$$\langle k |e^{-iqx}|k'\rangle = 
\int_{-\infty}^{+\infty}dx \frac{1}{\sqrt{2\pi}}e^{-ikx} e^{-iqx} \frac{1}{\sqrt{2\pi}}e^{ik'x} = \delta (k+q-k') = 
\int_{-\infty}^{+\infty}dx \frac{1}{\sqrt{2\pi}}e^{-i(k+q)x} \frac{1}{\sqrt{2\pi}}e^{ik'x} = \langle k+q |k'\rangle.$$
(Note that I get a different sign - this may have to do with how the momentum eigenstates are defined in your book.)
A more elegant and convincing approach is to think of it in the momentum representation, where the position operator corresponds to differentiating in respect to momentum:
$$\hat{x} = i\hbar\frac{d}{dp} = i\frac{d}{dk},$$ ($p = \hbar k$) that is 
$$e^{-iq\hat{x}} = e^{q\frac{d}{dk}},$$
which is a shift operator for $k$-dependent functions:
$$ e^{q\frac{d}{dk}}f(k) = \sum_{n=0}^{n=+\infty}\frac{1}{n!}\frac{d^n}{dk^n}f(k) = f(k+q).$$
