_This example is mathematically equivalent to the general case for [particle beams with spins two possible spins](Two%20spin%20ensemble.md) We merely specify the [circular polarization basis.](photon%20polarization.md#circular%20polarization) Thus, we may shorten the explanation._  Consider two light sources, $1$ and $2$, each emit [circularly polarized photons](photon%20polarization.md#circular%20polarization), that are each either _left_ or _right_ polarized. The photons are then combined to form a [mixed state](mixed%20state.md) consisting of photons from light source $1$ and light source $2$ expressed as $\{|\psi_1\rangle,|\psi_2\rangle\}.$ There's a probability to encounter a photon from source $1,$ $p_1$ and the probability to encounter a photon from source $2,$ $p_2$. This is equivalently a [[Two spin ensemble]] and thus the full state must be described by the following [density matrix](mixed%20state.md#Density%20Matrix),  ^6cca11 # Measuring polarization In order to tell which state the photon is in one needs to perform a [measurement on the mixed state](mixed%20state.md#Measurement%20on%20a%20mixed%20state). This measurement is represented by the observable $\hat{\varepsilon}$ in the [right-left basis](polarization%20operator.md#Circular%20Polarization). This is the photon helicity operator. ^a5c583 ## Probabilities Here there's a probability associated with each [polarization state](photon%20polarization.md#circular%20polarization), and a probability associated with sources $1$ and $2$, such that each photon is in state $|\psi_1\rangle$ or $|\psi_2\rangle.$ Thus we define the probability that we detect a photon with a spin state $|\phi\rangle$ as  ^91f482 Thus the probability that a photon from source is right-polarized is as follows, $P_R = p_1|\langle R |\psi_1\rangle|^2+p_2|\langle R|\psi_2\rangle|^2$ ^dd2fd3 and similarly $P_L = p_1|\langle L |\psi_1\rangle|^2+p_2|\langle L|\psi_2\rangle|^2$ ^12ca0f is the probability that a photon is left-polarized. ## Expectation value We use the above [probabilities](ensemble%20of%20circularly%20polarized%20photons.md#Probabilities) to derive the following [[expectation value]]: $\langle\hat{\varepsilon}\rangle_{\psi_1,\psi_2} = +\hbar P_R+(-\hbar)P_L$$+\hbar[p_1(|\langle R|\psi_1\rangle|^2)+ p_2(|\langle R|\psi_2\rangle|^2]-\hbar[p_1(|\langle L|\psi_1\rangle|^2)+p_2(|\langle L|\psi_2\rangle|^2)]$$=\hbar[p_1\langle \psi_1 |R\rangle\langle R|\psi_1\rangle+p_2\langle \psi_2 |R\rangle\langle R|\psi_2\rangle]-\hbar[p_1\langle \psi_1 |L\rangle\langle L|\psi_1\rangle+p_2\langle \psi_2 |L\rangle\langle L|\psi_2\rangle]$$=p_1\langle \psi_1 |\hbar(|R\rangle\langle R|-|L\rangle\langle L|)|\psi_1\rangle+p_2\langle \psi_2 |\hbar(|R\rangle\langle R|-|L\rangle\langle L|)|\psi_2\rangle$ Recall that $\hat{\varepsilon}$ in the [right-left basis](polarization%20operator.md#Circular%20Polarization) may be given as $\hat{\varepsilon}=\hbar(|R\rangle\langle R|-|L\rangle\langle L|).$ Thus $\langle\hat{\varepsilon}\rangle_{\psi_1,\psi_2}=p_1\langle \psi_1 |\hat{\varepsilon}|\psi_1\rangle+p_2\langle \psi_2 |\hat{\varepsilon}|\psi_2\rangle$ --- # Related Examples * We may also consider a version of this problem where the combined beam of light contains polarized photons in [superpositions](ensemble%20of%20circularly%20polarized%20photons%20with%20superpositions.md) * This example describes a type of [spin ensemble.](Two%20spin%20ensemble.md) * This example can also be used to model [unpolarized light.](quantum%20mechanical%20description%20of%20monochromatic%20unpolarized%20light.md) #QuantumMechanics/QuantumOptics #QuantumMechanics/TwoLevelSystems #PhysicalExamples/PhysicalExamplesInQuantumMechanics/PhysicalExamplesInQuantumOptics