Question on the creation of an entangled photon pair in a specific realization of the EPR-Bohm Experiment

In a student experiment the following assembly is used to create an entangled photon pair.

From the left, a laser beam polarized along $frac{hat{x}+hat{y}}{sqrt{2}}$ is incident on a pair of BBO crystals. The optical axis of the first crystal is directed along $hat{y}$ whereas in the second crystal it is directed along $hat{x}$. Due to the birefringence of the crystals, the incoming beam is split into its ordinary component (polarized orthogonal to the optical axis) and its extraordinary component (polarized along the optical axis). In each of the crystals, some of the incident photons with extraordinary polarization will be converted to photon pairs with ordinary polarization. This effect is known as spontaneous parametric down-conversion of type I:
$$
textit{e} longrightarrow textit{o} + textit{o} qquad text{(SPDC type I)}
$$
The wave vectors of the emitted photons are on opposite sides of a cone, which is shown in the figure. Note that any photon pair produced in the first crystal will have its polarization aligned along the x-axis while pairs produced in the second crystal will be polarized along the y-axis.

In the photon picture, this experiment is described the following way: The incident photons are in the state
$$
|Psi_prangle = cos{(frac{pi}{4})} |V_prangle + e^{i phi} sin{(frac{pi}{4})} |H_prangle
$$
where $|V_prangle$, $|H_prangle$ denote the basis states with polarization along the y-axis and the x-axis respectively. $Phi$ represents a possible phase shift between the linearly polarized components of the incident beam. The incident photons are converted according to
begin{align*}
|V_prangle &longrightarrow |H_srangle |H_irangle
|H_prangle &longrightarrow e^{i Phi’} |V_srangle |V_irangle
end{align*}
where $|V_srangle$, $|H_srangle$ and $|V_irangle$, $|H_irangle$ are the basis states of polarization of the emitted photons.
Now, it is claimed that we have produced entangled pairs of photons. In order to see this, we apply the conversion rules given above to the state of the incident photons, which gives us
$$
|Psi_{DC}rangle = cos{(frac{pi}{4})} |H_srangle |H_irangle + e^{i (Phi+Phi’)} sin{(frac{pi}{4})} |V_srangle |V_irangle
$$
and by choosing $Phi$ appropriately, we can even obtain a Bell state.

However, I don’t see why the resulting state of the two-photon system should be a pure state. My argument is as follows: When an incident photon enters the first BBO crystal, a measurement of its polarization state is performed. The measurement operators are the projectors in the x- and y-direction respectively. If we observe an extraordinary polarization, i.e. polarization along the y-axis, SPDC produces a pair of photons in the state $|H_srangle|H_irangle$. Otherwise, nothing happens. The second crystal repeats this procedure with swapped axes, i.e. producing a pair of photons in the state $|V_srangle|V_irangle$ if the incident polarization is found along the x-axis. The whole apparatus effectively measures the direction of incident polarization and, depending on the result, spits out a photon pair in one of the two states mentioned. However, we failed to record the reult of the measurement, thereby losing information on the state of the photon pair. Therefore, the outgoing photons are in the ensemble state described by
$$
rho = p(V) , |H_prangle |H_irangle langle H_p|langle H_i| + p(H) , |V_prangle|V_irangle langle V_p|langle V_i|
$$
with
begin{align*}
p(V) &= langle Psi_p| P_hat{y} |Psi_prangle = text{probability of measuring vertical polarization of the incident photon}
p(H) &= langle Psi_p| P_hat{x} |Psi_prangle = text{probability of measuring horizontal polarization of the incident photon}
end{align*}

Now, this is clearly not what we were looking for. The emitted photon pair is in the state $|H_prangle |H_irangle$ some of the time and in the state $|V_prangle |V_irangle$ some of the time, but not in an entangled state. Where have I gone wrong?

Any help is appreciated.