Do probabilities for entangled polarization states add up to one?

[klw289]

Two photons are moving in opposite directions along the y-axis are in the entangled polarization state
|Ψ> = (1/√2)(|VV>+|HH>)
V is vertical polarization relative to the z axis and measured from an axis n₁ and is defined as θ=θ₁ and H is horizontal polarization to the z axis measured from an axis n₂ and is defined as θ=θ₂.

Would P_VV+P_HH+P_VH+P_HV=1

If P_VV is the probability that photon 1 givees vertical polarization relative to an axis n₁ and photon 2 gives vertical polarization relative to an axis n₂

Am I correct in saying this as there is no way to predict the outcome of a spin measurement and so those probabilities must add to one as they are all the options available and the choice of axis does not matter??

 

[Jilang]

For the state given your third and fourth terms must be zero.

 

[Nugatory]

For the state given your third and fourth terms must be zero.

The detectors are not aligned with the vertical axis and not necessarily at the same angle, in OP's statement of the problem, so those two probabilities need not be zero.

It is correct that the sum of the four probabilities must add to unity - they are mutually exclusive and collectively cover all possible outcomes.

 

[Jilang]

Ah yes I get it now. The V and H have morphed!

 

[vanhees71]

To say it very clearly, the assumed formula is correct. The state is properly normalized to 1, i.e., $\langle \psi|\psi \rangle =\|\psi\|^2=1$. Since $|VV \rangle$, $|HH \rangle$, $|HV \rangle$, and $|VH \rangle$ together build an orthonormal system, i.e.,
$$\langle ab|a'b' \rangle=\delta_{aa'} \delta_{b b'}, \quad \sum_{a,b} |a,b \rangle \langle b,a| =1,$$
you indeed have
$$\langle \psi|\psi \rangle=1=\sum_{ab} \langle \psi |ab \rangle \langle ab|\psi \rangle=\sum_{ab} |\psi_{ab}|^2 = \sum_{ab} P_{ab}.$$
As very often in quantum theory, you can derive a lot from clever insertions of decompositions of the unit operator :-)).