New answers tagged

8

"Color" is how we sense visible light with different mix of wavelengths. Each wavelength of light in the visible range is a particular color. White light is all the wavelength mixed together at the right balance. What filters do is attenuate selected wavelengths of light. A red filter blocks most blue light, for example. If you start with white light ...


12

What we perceive as colour is basically the frequency, or combination of frequencies, of the light shone on our retinas. The light that comes out of a light bulb (but, note, not of an LED) is a mixture of lights of many different frequencies, actually, of all frequencies in the visible range. This is what we call "white". Filters have the properties of ...


1

I don't know why this hasn't been answered yet, but the simple answer is: no, the wave function doesn't collapse when it goes through the polarizer. The "photon" remains in a superposition of two states.


0

When you discuss polarization of matter due to the propagation of electromagnetic waves you usually consider low-energy photons, meaning photons of long wavelenghts. Given that the inter-lattice spacing $a$ is a few angströms, if you consider for instance visible light of $\lambda \sim$ a few $100$ nm, you have $\lambda >> a$. As the wavevector goes ...


2

But how we can say that photon pass through the second filter if we know that after passing the first one it's polarized vertically? For me there is probability 0 to passing second filter if that filter isn't polarized vertically. Polarizing filters don't just discard photons, they change the polarization of photons. This is simply true, by experiment. If ...


2

Take a look online at the "Dirac three polarizer experiment" for a more comprehensive answer.The key points are well explained by the following illustrations from "http://alienryderflex.com/polarizer/" keep in mind that these illustrations are based upon an initial horizontal polarizer, not a vertical one as stated above, but the concept is the same either ...


0

In reality, you need the quantization field to prepare your atom in a given spin state and have it stay in that state- at zero field, any small perturbation could change it. But in a thought experiment that is not a problem. Still, you have to specify your atom as starting in some initial spin, and if it is starting as usual in an eigenstate of $S_z,$ you ...


2

The PhD. Thesis: Hannah Dunstan Noble, "Mueller Matrix Roots" gives a gentle introduction to the concepts and José J. Gil, "Characteristic Properties of Mueller Matrices", JOSA A, 17, pp328-334 derives necessary and sufficient conditions for a matrix to be a physical Mueller matrix. The situation is not quite as simple as you assume, although the ...


0

The confusion is why there the term $E(\omega_2)E^∗(\omega_2)E(\omega_1)$ has factor of $6$ and $E(\omega_1)E^∗(\omega_1)E(\omega_1)$ has factor of $3$ I understand this confusion as follows. In first term there are two fields and in second term there is only one field. Hence if $E(\omega_1)=E(\omega_2)$ you are actually pumping double power in first case ...


4

In Quantum Field Theory the one particle states are defined as the states of an irreducible unitary representation of the Poincare Group. If this was not true, there would be states of a reducible representation that would not be connected by a Poincare transformation. These states are rather different particles. The Casimirs If we have an irreducible ...


2

According to quantum electrodynamics, the most accurately verified theory in physics, a photon is a single-particle excitation of the free quantum electromagnetic field. More formally, it is a state of the free electromagnetic field which is an eigenstate of the photon number operator with eigenvalue 1. The single-particle Hilbert space of the photon ...


11

Massless particles with spin do not have a "$S_z = 0$" state because they actually do not have spin like massive particles do. They have helicity, which is the value of the projection of the spin operator onto the momentum operator. The reason for this is the representation theory of the group of spacetime symmetry, the Poincaré group. To understand this, ...


12

Photons, as each massless particles, are characterized not by spin (which is defined as total angular momentum at rest, and mathematically corresponds to irreducible representation of the little group of representation), but by helicity $\lambda$ - the projection of total angular momentum on the direction of motion. Actually, the Casimir operator, which ...


16

By definition of spin $S$ it is a positive integer number or zero. Not to confuse with the spin projection possible values $S_z$, which may run from $-S$ to $S$.


5

Spin 1 just means that the spin in any direction can assume values out of {-1,0,1}. The 0 is only possible for massive particles, so the photon can have spin -1 or +1. That's like clockwise and anticlockwise circular polarization


5

A circularly polarized light state can be thought of as a superposition, with equal magnitude weights, of $x$ and $y$ linearly polarized light states, with the $y$ component either leading or lagging the $x$ by a quarter of a period. Therefore, you can extinguish a beam of such light with linear polarizers in several ways, of which two are: Method 1 Pass ...


2

You are correct in asserting that unpolarized light contains a mixture of many polarizations. However, each of these polarizations can be expressed as a combination of horizontally and vertically polarized light. Diagonally polarized light can thus be seen as containing both horizontally as well as vertically polarized light. When a horizontal beam of ...


2

In your final two paragraphs you have it backwards. At Brewster's angle the reflected light is totally polarized, but the total polarization of the transmitted light is usually rather weak. Compare reflection coefficients $r$ and transmission coefficients $t$ from the Fresnel equations: Reflected light is completely polarized at Brewster's angle because ...



Top 50 recent answers are included