Does a reflection still transfer momentum to an mirror?

One thing that is mentioned in the previous discussion is the frequency shift experienced by light reflected from a perfect mirror. The previous answers have been dancing around this, and essentially you need to conserve both energy and momentum, resulting in a Doppler shift of the reflected light frequency. Allow me to spell this out a little more, given that this may be a point of confusion.

The key to understanding this is to think of the reflection of a photon as two separate events: absorption and emission. Assume for simplicity that the photon (energy $\hbar\omega_0$) is normally incident to the mirror. The first step is the absorption process, which by conservation of energy and momentum will accelerate the mirror backwards (mirror velocity = $v_0$, momentum = $mv_0$, and kinetic energy = $\frac{1}{2}mv_0^2$) and excites the internal degrees of freedom of the electrons in the mirror to an energy of $\hbar\omega_0$. The photon will then be re-emitted in the opposite direction, reducing the internal energy of the mirror, as well as further accelerating it to velocity $v$. The energy and momentum balance for this emission process is described in [1] and is based on Fermi's original treatment from 1932, which I will reproduce here.

Assume the mirror is moving in the positive direction, while the new photon (energy $\hbar\omega$ and momentum $\hbar\omega/c$) is emitted in the negative direction. Equating the energy and momentum before and after the emission process, we obtain the following equations: \begin{equation} \textrm{Conserve energy:}\qquad \hbar\omega_0+\frac{1}{2}mv_0^2=\hbar\omega+\frac{1}{2}mv^2 \end{equation} \begin{equation} \textrm{Conserve momentum:}\qquad mv_0=mv-\hbar\omega/c \end{equation} Rearrange the equations and divide them to eliminate $m$ and solve for $\omega$: \begin{equation} \frac{v'-v}{2c}=\frac{\omega-\omega_0}{\omega} \end{equation} \begin{equation} \Longrightarrow\qquad \omega_0=\omega(1+\frac{v+v_0}{2c}) \end{equation}

This shows that the frequency of the emitted photon will indeed be red-shifted by a factor of $(1+\frac{v+v_0}{2c})^{-1}$, which is just the Doppler shift ($v\ll c$), with the relevant velocity being the average of the initial and final velocities. In fact, a complete treatment of the reflection problem would include another factor of the Doppler shift from the initial absorption process as well.

Although the mirror has not permanently absorbed the energy of the photon, it has acquired some kinetic energy from the recoil of the reflection process since momentum must be conserved. To conserve energy as well, this small kinetic energy causes a red-shift of the reflected photon by the Doppler factor.

[1] Barnett, S. Journal of Modern Optics, 57, 1445 (2010).

Since the photon reflects, its momentum changes: $p_{ph}'=-p_{ph}$. But total momentum of the system is conserved: $p_m+p_{ph}=p_m'+p_{ph}'$. Thus, the mirror will change its momentum.

But, if the mirror has large mass, then it'll get very small energy from the collision. For zero-mass particle (photon) falling onto the mirror with mass $m_2$, the energy of this particle after reflection is$^\dagger$:

$$\mathscr{E}'_1=\frac{m_2}{1-\cos\theta_1+\frac{m_2}{\mathscr E_1}},$$ where $\theta_1$ is angle of scattering of the particle, $\mathscr E_1$ is its initial energy. We can see that in the limit $m_2\to\infty$, we get $\mathscr E_1'=\mathscr E_1$. What happens with the momentum? Simple: for even tiny mirror velocity its momentum is infinite. So, adding twice the momentum of photon doesn't change the momentum of the mirror.

All this means that such mirror indeed doesn't change its velocity after the collision.

_{$^\dagger$ See Landau, Lifshitz, "The Classical Theory Of Fields", equation (13.9)}