Photon in a weighted superposition of states

Consider an experiment that produces photons in an entangled state such as $1/\sqrt{2}(|{H,H}\rangle+|{V,V}\rangle)$. The photons are in a superposition of horizontal and vertical polarization, and the way we analyze this is to say that the photons are in both states at the same time. Though this is odd, I can eventually reconcile it. However, the photons can also be in an entangled state such as $\sqrt{0.2}|H,H\rangle+\sqrt{0.8}|V,V\rangle$. Again, the photons are in both states at the same time, but do we say that they is somehow more in one state than the other? How can we think of this unevenly weighted superposition of states?

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You don't want to use the word "entangled" here as it is generally applied to systems with two or more component each of which may be in a mixed state. "Mixed" and "superposition" are the words generally used for simple systems in non-eigenstates of whatever bass you are using. – dmckee Mar 12 '13 at 4:34
Sorry, @dmckee, I am very confused by all parts of your comment. The OP seem to be asking about the entanglement of 2 photons so what's wrong with the word "entangled"? Also, you seem to be confusing "mixed" and "superpositions". These are different things. "Superposition" is used for pure states (can be expressed via a vector in the Hilbert space), "mixed" is used for mixed states i.e. those that aren't pure (require density matrix). – Luboš Motl Mar 12 '13 at 6:02
@LubošMotl That would be because I paid more attention to the text before the edit than the math. It is clear that Jon does mean "entangled". – dmckee Mar 12 '13 at 6:10
Thanks for the clarification, @dmckee. – Luboš Motl Mar 12 '13 at 6:15

Consider a quantum system with a two-dimensional state space. Let $\Omega$ be an observable on the system with normalized eigenstates $|1\rangle$ and $|2\rangle$ corresponding to eigenvalues $\omega_1$ and $\omega_2$. If a state $|\psi\rangle$ is in some superposition of these states $$|\psi\rangle = a_1|1\rangle + a_2|2\rangle,$$ then the coefficients can be interpreted as probability amplitudes. This means that if a measurement of the observable $\Omega$ is made on the the system, then the probabilities $p_1$ and $p_2$ that the measurement will yield the eigenvalues $\omega_1$ and $\omega_2$ are $$p_1 = |a_1|^2, \qquad p_2 = |a_2|^2$$ This general analysis applies to your example as well. The coefficients should be interpreted as probability amplitudes that if a measurement of the photon polarization is made, then corresponding results (either vertical or horizontal) will be obtained.

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Right. So in my example, since the probability amplitude of $|V,V\rangle$ is higher than that of $|H,H\rangle$, it is more likely that the photons will be measured to be in the vertical state. However, prior to measurement, we must assume the photons are in both states, so if we know that the probability amplitudes are not equal, can we say that the photons are more vertical than horizontal? – Jon Ruffin Mar 12 '13 at 4:47
I'd say that to some extent, the interpretation you're attempting to make is a matter of conceptual taste. However, most physicists would not use the phrase "the photons are in both states" but rather "the photon is in a superposition of states." I can't see anything particularly "wrong" with the intuition that the photons are "more vertical than horizontal," but when it comes down to it, it's most important and meaningful in my opinion to use the precise, operational meaning of probability amplitudes in terms of state preparation and measurement. – joshphysics Mar 12 '13 at 4:57
@Jon: "Can we say that the photons are more vertical than horizontal?" How would you answer any other isomorphic question about uncertain information that we experience in the real world? For example, a woman gets pregnant. You may ask: Can we say that the baby is more female than male? (51+ percent of the newborns are girls.) What's the answer? It's exactly the same thing. We don't know the sex of the baby or polarization of the photons; they're only determined after/via a measurement and probabilities may be predicted in advance. The answers male/fem. and horizontal/vert. are mutually excl. – Luboš Motl Mar 12 '13 at 6:07
Concerning the last comment, one may accurately say that "the expectation value of the vector (well, actually the tensor) giving the polarization axis is more vertical" and "the expectation value of the sex of a newborn is closer to female". But it's misleading to say the same sentence without the "expectation values" because such sentences incorrectly suggest that the horizontal/vertical polarization or male/female sex of the baby (one particular answer) are already known at that moment which they're not. – Luboš Motl Mar 12 '13 at 6:17
Now, the difference is that the sex may be approximately thought of in classical physics because it's a property of macroscopic bodies that quickly decohere etc. In classical physics, one doesn't run into contradictions if he assumes that the answers/properties of objects already objectively exist even before any measurement is made. In quantum mechanics, it is not possible to assume that: all quantities can't be sharply given before measurements (after all, they refuse to commute with each other). But when it comes to "what we can say before the measurement", the situations are identical. – Luboš Motl Mar 12 '13 at 6:18