zonde said:
Of course. It's just sometimes people focus on entangled photons so much that they start to perceive them as the only "natural state" for twin photons. So I wanted to bring up product state twin photons just to be sure that we are on the same page.
I tried to think over your claim in post #54. I'm not sure I understand your (and probably not only your) view about symmetrization. I have seen the explanations about swapping the particles and getting the same situation. Well if we include position into particle description then we are not swapping anything physical, we just swap meaningless labels in representation. But swapping one representation with other should not have any physical consequences.
And then it seems that in QFT we get very elegant treatment of this situation because in QFT particles don't have arbitrary labels. There particle state is either occupied or not so there is nothing to swap.
On the other hand there is absolutely real experimentally observable phenomena called Hong-Ou-Mandel interference which shows that there is something interesting about identical (indistinguishable) photons.
So I view symmetrization as property of dynamical process where two indistinguishable photons at initial states 1 and 2 end up in final states 3 and 4. And the physical property of symmetrization is that this process can not be described as (photon from state 1 transitions to state 3 and photon from state 2 transitions to state 4) OR (photon from state 1 transitions to state 4 and photon from state 2 transitions to state 3).
If I remember correctly I got this explanation form Feynman's "QED: The Strange Theory of Light and Matter".
And this explanation seems to be consistent with HOM interference.
My point is much simpler. Photons are indistinguishable bosons. A convenient (generalized) basis for free photons is the momentum-helicity or momentum-linear-polarization-state basis (you cannot easily talk about position representations for photons since photons do not have position observables to begin with). I denote them with ##|\vec{p},\lambda \rangle##, where ##\lambda \in \{-1,1 \}## is the helicity of the photon (the projection of the total angular-momentum on the direction of ##\vec{p}##). This refers to plane-wave left- and right-circular polarized em. waves.
Now photons are bosons, and thus the correct Hilbert space is a bosonic Fock space, i.e., for each ##(\vec{p},\lambda)## there's an annihilation operator ##\hat{a}(\vec{p},\lambda)## which obeys the bosonic commutation relations ##[\hat{a}^{\dagger}(\vec{p},\lambda),\hat{a}^{\dagger}(\vec{p}',\lambda')]=\delta^{(3)}(\vec{p}-\vec{p}') \delta_{\lambda \lambda'})##. A complete set of (generalized) orthonormal basis vectors are common eigenstates of the occupation-number observables ##\hat{N}(\vec{p},\lambda)=\hat{a}^{\dagger}(\vec{p},\lambda) \hat{a}(\vec{p},\lambda)##,
$$|\{n(\vec{p},\lambda ) \}_{\vec{p} \in \mathbb{R}^3,\lambda \in \{-1,1\}} \rangle = \prod_{\vec{p},\lambda} \frac{1}{\sqrt{n(\vec{p},\lambda)!}} [\hat{a}^{\dagger}(\vec{p},\lambda)^{n(\vec{p},\lambda)}|\Omega \rangle,$$
where ##|\Omega \rangle## is the vacuum (ground) state, for which all ##n(\vec{p},\lambda)=0##.
An ##n##-photon Fock state is thus from the totally symmetrized part of ##n##-fold Kronecker product of the single-photon Hilbert space.
E.g., if for a photon pair you want the polarization-singlet state ##|1,-1 \rangle-|-1,1 \rangle##, the spatial (momentum) part must also be antisymmetric. I.e., written in the product basis you have
$$|\psi^{-} \rangle=\frac{1}{2} (|\vec{p}_1 \rangle \otimes |\vec{p}_2 \rangle-|\vec{p}_2 \rangle \vec{p}_1)(|1 \rangle \otimes |-1 \rangle)-|-1 \rangle \otimes |1 \rangle.$$
Multiply this out and you get the completely symmetrized state in the tensor-product notation. This is very cumbersome. With the creation operators it's much more convenient
$$|\psi^- \rangle = \frac{1}{2} (\hat{a}^{\dagger}(\vec{p}_1,1) \hat{a}^{\dagger}(\vec{p}_2,-1) - \hat{a}^{\dagger}(\vec{p}_1,-1) \hat{a}^{\dagger}(\vec{p}_2,1 ) \Omega \rangle.$$
The HOM (Hong-Ou-Mandel) effect in its most simple form can be described as follows. Two photons of the same frequency enter a (symmetric) beam splitter. Let's look at plane-wave modes. Let the incoming two photons be created by ##\hat{a}_{\lambda_1}^{\dagger}## and ##\hat{b}_{\lambda_2}^{\dagger}## respectively. Here we use the abbreviation
$$\hat{a}_{\lambda_1}=\hat{a}(\vec{p}_a,\lambda_1), \quad \hat{b}_{\lambda_2} = \hat{a}(\vec{p}_b,\lambda_2).$$
The incoming two-photon state is
$$|\psi_0 \rangle=\hat{a}_{\lambda_1}^{\dagger} \hat{b}_{\lambda_2}^{\dagger}.$$
The "scattering matrix" describing the balanced symmetric beam splitter is given by the unitary matrix
$$U_{\text{BS}}=\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & \mathrm{i} \\ \mathrm{i} & 1 \end{pmatrix}.$$
This transforms the creation operators of the incoming to those of the outgoing photons
$$\begin{pmatrix} \hat{a}_{\lambda}^{\prime \dagger} \\ \hat{b}_{\lambda}^{\prime \dagger} \end{pmatrix}= U_{\text{BS}} \begin{pmatrix} \hat{a}_{\lambda}^{\dagger} \\ \hat{b}_{\lambda}^{\dagger} \end{pmatrix}=\frac{1}{\sqrt{2}} \begin{pmatrix} \hat{a}_{\lambda}^{\dagger} + \mathrm{i} \hat{b}_{\lambda}^{\dagger} \\ \mathrm{i} \hat{a}_{\lambda}^{\dagger}+\hat{b}_{\lambda}^{\dagger} \end{pmatrix}.$$
Thus after the beam splitter the state is
$$|\psi_2 \rangle=\hat{a}_{\lambda_1}^{\prime \dagger} \hat{b}_{\lambda_2}^{\prime \dagger}|\Omega \rangle = \frac{1}{2} (\hat{a}_{\lambda_1}^{\dagger}+\mathrm{i} \hat{b}_{\lambda_1}^{\dagger})(\mathrm{i} \hat{a}_{\lambda_2}^{\dagger} + \hat{b}_{\lambda_2}^{\dagger})|\Omega \rangle=\frac{\mathrm{i}}{2} (\mathrm{i} \hat{a}_{\lambda_1}^{\dagger} \hat{a}_{\lambda_2}^{\dagger}+\mathrm{i} \hat{b}_{\lambda_1}^{\dagger} \hat{b}_{\lambda_2}^{\dagger} - \hat{a}_{\lambda_2}^{\dagger} \hat{b}_{\lambda_1}^{\dagger}+\hat{a}_{\lambda_1}^{\dagger} \hat{b}_{\lambda_2}^{\dagger})|\Omega \rangle .$$
Now if the photons have different parity, i.e., are distinguishable, you get with probability 1/2 both photons in the same mode (either a or b) and with probability 1/2 in different modes (one in a and one in b). This is as expected from the classical case.
If, however ##\lambda_1=\lambda_2=\lambda## the output state is
$$|\psi_2 \rangle=\frac{1}{\sqrt{2}}(|2_{a,\lambda} \rangle + |2_{b,\lambda}),$$
i.e., both photons always end up in the same mode (with probability 1/2 in a and with probability 1/2 in b).
This is a specific quantum effect of indistinguishable photons and cannot explained in the classical wave picture.