Correct. Metals are not molecular. And there definately is a difference between reflection off a metal and reflection off a dielectric; now that you mention it. Reflection off a metal is referred to as Reflectance and it is a property that is unique to a surface that is conducting - like what you said. But before I delve into this some housekeeping is in order. k is the direction vector (also called the propagation vector). It points in the direction of a plane wave travelling through free space. We will need this vector. u is the unit vector normal to the interface pointing in the direction from the incident to the transmitting medium. E the electric field vector and B the magnetic field vector. Notice that u is normal to the interface regardless of the direction of the electric field. Okay, now lets get into it. Lets assume for starters a basic example: An electric field E perpendicular to the plane-of-incidence and that B is parallel to it. So from electromagnetism (remember E = vB) so.. k x E = vB k . E = 0 Now I wont go into the maths but what we try to do here is work out the formula for the reflection coefficient will depend on... 1. the angle of incidence and transmission 2. the refractive indicies of the two media This makes sense. But the formula is a little complicated. Ultimately you end up with E_r / E_t (subscripts are for reflection and transmission). This the the quotient of the electric field vector of the reflecting wave and the electric field vector of the transmitted wave. Now you have to express each vector as a sum of its angular components (because we know the angle changes through a medium). Reflection Coefficient = [n_i cosθ_i - n_t cosθ_t]/[n_i cosθ_i + n_t cosθ_t] You may recognise these expressions for E as the Fresnel Equations. Perhaps you might look this up! However, in this thread we are not really interested in the mathematics but rather the physical implications. In particular we are interested in the reflection (and transmission; they come hand in hand). Now we get into the meaty bits. Consider a beam of light incident on a surface. This illuminates a spot of area A. The power per unit area crossing the surface (in a vacuum of course) whose normal is parallel to the Poynting vector (S) is given by... S = c²ε_0 E x B This is from second year electromagnetism (first year advanced physics perhaps). From this relationship we work out the radiant flux density (a very important concept used throughout advanced electromagnetism at uni). I = <S> = 1/2 cε_0 E² This is the average energy per unit time crossing the area on the surface normal to S. We let I_i , I_r, and I_t be the incident, reflected and transmitted flux densities respectively. Can you see that the cross section of a circular wavefront impinging on a surface will trace out a circular illumination? The cross-sectional area of the incident beam will be Acosθ_i. And for the relflected and transmitted cross-sections they will be Acosθ_r and Acosθ_t. Similarly the incident power is I_i Acosθ_i and so on for reflection and transmission. This is the energy per unit time flowing in the beams, and it is therefore the power arriving in our cross-sections. Now we are in a position to define reflectance (R) to be the ratio of the reflected power (flux) to the incident power. R = flux_r / flux_i = I_r Acosθ_r / I_i Acosθ_i Now we know what Relfectance is and how it is related to the Fresnel equations of electric fields. Reflectance can be applied to any media. But now we want to apply it to metals. METALS You know that the main characteristic of a metal is its conducting property; that is, the presence of a great number of free electric charges. For metals these charges are electrons and their motion is electric current. The current per unit area is the conductivity of the medium ρ. Housekeeping done. Now imagine an electric field plane wave impinging on a metal surface. Lets calculate the reflectance of metal... R = I_r / I_i. Take n_i = 1 (air) and n_t = n (a complex index). Therefore R = [(n-1)/(n+1)][(n-1)/(n+1)]* Since n = n_r - in_i R = [(n_r - 1)² + n_i²]/[(n_r + 1)² + n_i²] Notice how the conductivity of the metal goes to zero. Whereas in the case of a dielectric (with real indices). But why are the indices complex for metals? Envision the metal as a medium full of driven, damped oscillators (electrons affected by an electric field). Some of these oscillators correspond to free electrons and will have zero restoring force whereas others may be bound to the atom. The conduction electrons are however predominant. Now I did an assignment on electron oscillators being driven by an time-varying electric field. What you have is a second order non-homogeneous differential equation and when you solve it you get the solution as x(t) = [(q_e / m_e) / (ω_0² - ω²Please Register or Log in to view the hidden image!] E cosωt The point is the restoring force is 180 degrees out of phase with the displacement. This is unlike a dielectric where they are in-phase. So we have our conducting electrons oscillating out of phase with the impinging light and they tend to reradiate little wavelets that tend to cancel the incoming electric disturbances causing them to oscillate in the first place. This lead to a rapidly decaying refracted wave. According to quantum mechanics, an electromagnetic wave transfers energy in quantized photons E = hv. Thus E = h-bar ω. The spin of a photon is either -h-bar or +h-bar where the signs indicate right or left handed spin. So there are two ways a photon can be 'spun' (hehe). A photon is a spin-1 particle and can either be +1 or -1. Whenever a charged particle (electron in a metal) emits or absorbs electromagnetic radiation, it will undergo a change of +- h-bar in its angular momentum (as well as change in energy and momentum of course). So, like you said, there are two orthogonal polarization directions attributed to photon spin. Take a pure left-circularly polarized plane wave. This will impart angular momentum to its target as if all the constituent photons in the wave had their spins aligned. Remember that each photon exists in either spin state with equal likelihood. LINEARLY POLARIZED LIGHT - No real majority of one type of photon spin. LEFT CIRCULARLY POLARIZED LIGHT - Photon spin in one direction RIGHT CIRCULARLY POLARIZED LIGHT - Photon spin in the other direction ELLIPTICALLY POLARIZED LIGHT - Unequal amounts of photon spin. Now you probably know this anyway, but as it is I do not know enough about quantum mechanics to relate it to quantum optics to answer the part about but I would say that it probably wouldn't be the Poynting vector but rather the wave propagation direction vector k. PS. Good questions! By the way what level of education are you at?