# Tag Info

1

In the book "Plasmonics and Plasmonic Metamaterials: Analysis and Applications" edited by G. Shvets, Igor Tsukerman, we read in section 2.1: In other words - they clearly state that the enhanced reflectivity is a result of the presence of a inverted dye - that is, a dye with a population inversion, meaning that it can be subject to stimulated emission. ...

1

Google mathematical methods in the physical sciences pdf and you will be able to download an ebook by Mary Boas, which was written for people like yourself. As Jacob says above, calculus is a must learn, and lots of websites give you examples of different levels of calculus problems. Conceptually, a good textbook is Halliday and Resnicks Physics, which sets ...

1

The statement "On the cylindrical surface $\mathbf{J}\cdot\hat{n}=0$..." refers to just inside the wire so $\sigma\neq0$ and $\mathbf{E}$ cannot be whatever it wants.

1

I'm afraid that linear polarization is not as interesting an example as you may have hoped. First, the answer: in quantum optics, whether or not a quantum state exhibits linear polarization is independent of the photon count for that state. A single-photon state can be linearly polarized. Now, the explanation: in quantum electrodynamics (QED) it is ...

6

Correction, a single photon does not have a circular polarization. It has spin +1 or -1 to the direction of its motion. Qualitatively Left and right handed circular polarization, and their associate angular momenta. The way the classical wave emerges from the quantum mechanical level of photons is given in this blog entry, and it needs quantum ...

3

Let us look at coherent states $$|\alpha\rangle~=~e^{-|\alpha|^2/2}e^{\alpha a^\dagger}|0\rangle$$ $$e^{-|\alpha|^2/2}\sum_{n=0}^\infty \frac{(\alpha)^n (a^\dagger)^n}{n!}|0\rangle$$ If you have a classical system it means overlap between states is small. We then look at over lap $\langle\alpha'|\alpha\rangle$  \langle\alpha'|\alpha\rangle~=~e^{-(|\...

2

Light never completely behaves as a particle. Light never completely behaves as a wave. As pointed out by hsinghal, the Michelson interferometer showed that, even at the "single photon" level, we still see wave behaviors. These behaviors are well modeled by quantum mechanics, which treats light as neither a pure wave nor a pure particle. As you "add ...

18

Even though there is a single photon in a volume of your choice the light is still a wave. An experiment was performed which proved this. In this experiment a Michelson interferometer was set up and the incident light is so weak that only one photon was in the whole setup at a time. A photographic plate was used to detect the interference pattern. Now just ...

0

You can try Modern Problems in Classical Electrodynamics by Charles Brau. I think it is relatively clear in this book. At least I hope you can find the end-results in Zangwill or Brau. Then you should after trying be able to get there yourself. Start from the expression for the electric potential. In the integral there is a $\frac{1}{|\bar{r}-\bar{r}'|}$....

1

Momentum ($p$) is "really" $mv$, even for light and EM fields. This can be proven by the use of $E = mc^2$. The momentum for a photon (EM) is $p = mv$. Where the mass is given by $m = E/c^2$ and $v = c$. Substituting these into the equation, one obtains, $p = E/c$. Although this equation "looks" different from $p = mv$, because it was derived using the ...

2

You can find out $\dot{\alpha}=\sqrt{\sum_i(A^i+\partial^i\alpha)^2}-A^0$, thus, you can pose a Cauchy problem for $\alpha$. $\alpha_{,i}$ at $t=t_0$ is arbitrary ($i=1,2,3$). EDIT(6/29/16): A better wording would probably be that $\alpha$ at $t=t_0$ is arbitrary (but smooth enough).

Top 50 recent answers are included