bulx said:
Apologies
@PeterDonis for my confusing (layman’s!) initial language. Thanks
@Nugatory for setting me straight and
@Hill for the insightful preparation response. I am indeed curious about all this in the context of gates / logic / computing.
This answers my (intended) question, that such an entangled state is mathematically conceivable.
Then, taking:
##\frac {\sqrt 3} 2 \left| HV \right> + \frac 1 2 \left| VH \right>##
Is there any nuance when we use the term “probability” in this context? Is it: if we managed to prepare a trillion of these pairs, randomly selected a particle from each pair and measured its polarization, we can expect the results of those measurements to be more ##H## than ##V##?
In QT it's very important to clearly state which observable you measure. Then the state clearly predicts probabilities for this measurements.
Also note that the notation here is a shortcut, i.e., a product vector like ##|HV \rangle## stands really for ##\hat{a}^{\dagger}(\vec{k}_1,H) \hat{a}^{\dagger}(\vec{k}_2,V)|\Omega \rangle##, where ##|\Omega \rangle## is the vacuum of the (free) electromagnetic field and ##\hat{a}^{\dagger}(\vec{k},H)## is the creation operator for a photon with momentum ##\vec{k}## and polarization state H etc. It describes the situation that you precisely have 1 photon with momentum ##\vec{k}_1## which is H-polarized and one photon with momentum ##\vec{k}_2## that is V-polarized.
So your state more precisely reads
$$|\Psi \rangle=\left ( \frac{\sqrt{3}}{2} \hat{a}^{\dagger}(\vec{k}_1,H) \hat{a}^{\dagger}(\vec{k}_2,V) + \frac{1}{2} \hat{a}^{\dagger}(\vec{k}_1,V) \hat{a}^{\dagger}(\vec{k}_2,H) \right )|\Omega \rangle.$$
Now you can ask precise questions.
The photons are distinguished by their momenta, and you can measure them by puting a detector far enough from the photon source detecting photons coming from the direction ##\vec{k}_1/|\vec{k}_1|## and a detector detecting coming from the direction ##\vec{k}_2/|\vec{k}_2|##, and you can measure whether the photon is in H-polarization or in V-polarization by, e.g., putting a polarizing beam splitter in according orientation before each of the two detectors. In this way you can determining the polarization state of the photon with ##\vec{k}_1## and that of the photon with ##\vec{k}_2##.
Now you can calculate all probabilities concerning the measurement of any polarization state of both photons. It's most simple if you ask for polarizations measuring the states you used to write down the state the photons are prepared in: The probabilities for the four possible outcomes are (now using again the more convenient shorthand notation):
$$\begin{split}
P(H,H) &=|\langle HH|\Psi \rangle=|\langle HH |\sqrt{3}/2 HV \rangle + \langle HH |1/2 VH \rangle|^2=0,\\
P(H,V)&=|\langle HV|\Psi \rangle=|\langle HV |\sqrt{3}/2 HV \rangle + \langle HV |1/2 VH \rangle|^2=3/4,\\
P(V,H)&=|\langle VH|\Psi \rangle=|\langle VH |\sqrt{3}/2 HV \rangle + \langle VH |1/2 VH \rangle|^2=1/4,\\
P(V,V) &= |\langle VV|\Psi \rangle=|\langle VV |\sqrt{3}/2 HV \rangle + \langle VV )|1/2 VH \rangle|^2=0.
\end{split}$$
This tells you that you either find that the photon with momentum ##\vec{k}_1## is H polarized and that with momentum ##\vec{k}_2## is V polarized, which happens with probability ##3/4## or that he photon with momentum ##\vec{k}_1## is V polarized and that with momentum ##\vec{k}_2## is H polarized, which happens with probability ##1/4##. The other two possibilities do not occur, i.e., have probabilities 0.
The photons are entangled concerning the polarization, which implies that if the photon with momentum ##\vec{k}_1## is found to be H polarized, then necessarily the photon with momentum ##\vec{k}## must be V polarized and vice versa.
The single photons are not in a determined polarization state. For the photon with momentum ##\vec{k}_1## the probabilities to find it in either the H or the V polarization state are
$$P_1(H)=P(H,H)+P(H,V)=3/4, \quad P_1(V)=P(V,H)+P(V,V)=1/4,$$
and for the photon with momentum ##\vec{k}_2##
$$P_2(H)=P(H,H)+P(V,H)=1/4, \quad P_3(V)=P(H,V)+P(V,V)=3/4.$$
More precisely the polarization state of the photon with momentum ##\vec{k}_1## is given by a mixed state, which is determined by "tracing" the original two-photon state over the other photon's polarization states, i.e., the socalled reduced state.
The original state is given by the statistical operator
$$\hat{\rho}_{12}=|\Psi \rangle \langle |\Psi \rangle = \frac{3}{4} |HV \rangle \langle HV| + \frac{\sqrt{3}}{4} (|HV \rangle \langle VH|+|VH \rangle \langle HV |) + \frac{1}{4} |VH \rangle \langle VH|.$$
Then the polarization state of photon 1 (i.e., the one with momentum ##\vec{k}_1##) is
$$\hat{\rho}_1=\mathrm{Tr}_2 \hat{\rho}_{12} = \sum_{p_1,p_1', p_2 =H,V} |p_1 \rangle \langle p_1,p_2 |\hat{\rho}_{12}|p_1',p_2 \rangle \langle p_1'|=\frac{1}{4} |H \rangle \langle H| + \frac{3}{4} |V \rangle \langle V|.$$
In the same way you get the polarization state for photon 2:
$$\hat{\rho}_2=\frac{3}{4} |H \rangle \langle H| + \frac{1}{4} |V \rangle \langle V|.$$
You can of course also ask for any other probabilities concerning polarization measurements on the two photons or on each of the single photons.
E.g. you can ask, whether photon 1 is left-circular or right-circular polarized for this you have to put some quarter-wave plate in the right orientation before the polarizing beam splitter. The corresponding polarization states are
$$|L \rangle=\frac{1}{\sqrt{2}} (|H \rangle + \mathrm{i} |V \rangle), \quad |R \rangle=\frac{1}{\sqrt{2}} (|H \rangle -\mathrm{i} |V \rangle).$$
Now you can use directly the reduced statistical operator for photon 1 to get the corresponding probabilities for this measurement
$$P_1(L)=\langle L|\hat{\rho}_1|L \rangle=\frac{1}{2} (\langle H|-\mathrm{i} \langle V|) \hat{\rho}_1 (|H \rangle + \mathrm{i} |V \rangle) = \frac{1}{2} \left (\frac{3}{4}+\frac{1}{4} \right)=\frac{1}{2}.$$
Then of course ##P_1(R)=1/2## too.
) probably not ideal for this, but I don't see why it wouldn't be possible using a coupled qubit system (e.g. two coupled ions or superconducting qubits). It would just be a couple of single qubit gates followed by the abovementioned CNOT, or?
