# 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.
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