How does weak measurement affect quantum state? I'm trying to understand how to describe the quantum state after weak measurement using these two toy examples. Hopefully, these simple examples and your answer will help others who want to learn about weak measurement.


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*Suppose photons are prepared in a superposition of horizontal and vertical polarization: $ \frac{1}{\sqrt{2}}\left(\left|H\right\rangle+\left|V\right\rangle\right) $. The weak measurement is implemented, let's say, by stacking uncoated glass plates oriented at the Brewster angle, so they would transmit the horizontal polarization state with the probability of 100% and reflect the vertical polarization state with the probability of 50%. For the reflected photons, the state collapses to $ \left|V\right\rangle $ because we know that only vertically polarized photons can be reflected (strong measurement). But what is the state of the transmitted photons? Does their state change to $ \frac{\sqrt{3}}{2}\left|H\right\rangle+\frac{1}{2}\left|V\right\rangle $ because 25% of photons have already been reflected, i.e., they have been measured as vertically polarized?

*Suppose photons are prepared in a maximally entangled state $ \frac{1}{\sqrt{2}}\left(\left|H_{s}H_{i}\right\rangle+\left|V_{s}V_{i}\right\rangle\right) $ instead. We conduct the same weak measurement as above on the s photon. If this photon is transmitted, what is the description of the combined system (both the transmitted s and its i pair)? And did the entanglement of the two photons weaken? If it did, how would you express it mathematically?
 A: I'm no expert in weak measurement, so I might be missing something, but I'm pretty sure you can analyze both these examples by considering what the wavefunction should be after the glass optic, applying the projective measurement, and renormalizing. In which case,
1.
If I add a mode label for the three spatial modes involved in the question, (i.e. $|PolarizationState, Spatial Mode Index\rangle$ or so)
$$
\frac{1}{\sqrt{2}}(|H,0\rangle+|V,0\rangle)\rightarrow\frac{1}{\sqrt{2}}\Big(|H,1\rangle+\frac{1}{\sqrt{2}}(|V,1\rangle+|V,2\rangle)\Big)
$$
Then the measurement removes the $|V,2\rangle$:
$$
\propto\frac{1}{\sqrt{2}}\Big(|H,1\rangle+\frac{1}{\sqrt{2}}|V,1\rangle\Big)=\frac{1}{\sqrt{3}}\Big(\sqrt{2}|H,1\rangle+|V,1\rangle\Big)
$$
And in the second example:
$$
\frac{1}{\sqrt{2}}(|H,0\rangle_s|H,0\rangle_i+|V,0\rangle_s|V,0\rangle_i) 
\rightarrow \frac{1}{\sqrt{2}}\Big(|H,1\rangle_s|H,1\rangle_i+\frac{1}{2}(|V,1\rangle_s+|V,2\rangle_s)(|V,1\rangle_i+|V,2\rangle_i)\Big)
$$
Then the projective measurement on the s photon removes the $|V,2\rangle_s$ portion**
$$
\propto\frac{1}{\sqrt{2}}\Big(|H,1\rangle_s|H,1\rangle_i+\frac{1}{2}|V,1\rangle_s(|V,1\rangle_i+|V,2\rangle_i)\Big)
$$
And I don't feel like renormalizing that. 
I hope this helps. Someone should correct me if I'm missing some subtlety that comes from weak measurement.
**I'm assuming that when you label the photons like this, you mean that they are distinguishable in some other degree of freedom, e.g. frequency. If they are indistinguishable photons, you can't measure only the "s" photon, because this label is not meaningful, and my resulting wavefunction above would not be properly symmetrized. 
