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Answer to (b): the Poynting vector is defined as $$\mathbf{S}=\mathbf{E}\times\mathbf{H},$$ which has magnitude proportional to $|\mathbf{E}||\mathbf{H}|$. For a plane wave $|\mathbf{E}|\propto \exp[i(\mathbf{k}\cdot\mathbf{x}-\omega t)]$, $\mathbf{E}$ and $\mathbf{H}$ are perpendicular and both orthogonal to the direction of propagation. This means that $\... 1 I think what they are doing is representing the (real part of) the complex vector space in which a (pure) qubit is embedded. Remember that the state of a$d$-dimensional system, which we usually denote via a complex unit vector$|\psi\rangle\in\mathbb C^n$, is more precisely defined as an equivalence class of such vectors, where different unit vectors are ... 0 From the solid state point of view one can think of many quasiparticles analogous to photons, and described formally in a similar way, which would have something else in place of electric field. The most obvious example is phonons, which are essentially quantized elastic vibrations of the lattice, and which have three polarizations: two transverse and one ... 1 Electromagnetic waves, hence photons, are described by a vector field$^*$. A vector field has 3 (2S+1) components and corresponds to spin S=1. In the case of electromagnetism only the 2 spin components parallel and antiparallel to propagation are allowed because of charge conservation. The direction of this vector field can rotate about the direction of ... 7 1. Short answer Yes, there exists the analog of electric force for the other gauge bosons (the electric field is defined by having the electric force, so we need the classical analogs of other fundamental forces to define the corresponding classical field). In fact, for each gauge boson (including weak force carriers$W^{\pm}$and$Z^0$, gluons in strong ... 1 In addition to @ProfRob’s answer, I’ll just point out that refractive index is in general a complex number. And, the Fresnel Equations apply to complex refractive index too. So the answer is: You’re right, the reflection phase is not always$\pi$. In principle, it can be anything! 2 You have to adopt a sign convention when defining the Fresnel coefficients. Usually you say that a positive reflection coefficient means the wave component tangential to the interface keeps the same direction. For s-polarised light that is the E-field, but for p-polarised light it is the H-field. Note that you cannot just say "the direction of the E-... 1 If you apply an electric field E' on a material. Then polarisation happens inside that material. This polarisation will create a new electric field that will be equal to the field caused by$\rho_b \, and \, \sigma_b$(That's why we are able to find those two factors from polarisation only.). Now if you try to measure the electric field practically what you ... 4 The formula$2s+1$refers to massive particles, which mediate short-range forces, unlike electromagnetism. Massless particles have only two polarization states, for any spin. So, the analog to the two polarization states of the electromagnetic field, for higher spin fields, is still two polarizations. It is the exact same thing. For massive bosons like$Z,W^\...

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That is because the cross-sectional area of the diagonal slice is larger than that of the perpendicular slice by a factor of $\sec{\theta}$, which exactly cancels out the $\cos{\theta}$ in the dot-product. The area of the oblique surface, which is an ellipse, is $\pi \, a\, b$, where $a$ and $b$ are the two axes of the ellipse. If the diameter of the tube is ...

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Filter 1 allows the component of the light vibrating in the same direction of the grating though. If the angle of the first polarizer is 90 degrees (parallel to the y-axis), then the percent of the photon’s energy passing through will equal the sine of the photon’s angle of vibration (polarization mode). With an equal distribution of polarization modes, ...

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In general, if you want to know the polarization of light experimentally, you need to make a series of measurements in different bases/directions using a different filter each time. Each measurement indeed will destroy a single photon, so in order to fully know the polarization of light you need to measure an ensemble/collection of photons. I think that this ...

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