7

Your problem -- as far as I can see -- is a simple calculation error: while you are correct that $\sin(k_0 x) = \mathcal{I}(e^{ik_0 x})$, it does not follow that $$\mathcal{I}(e^{i(k_0-k) x}) = \sin(k_0 x)e^{ikx} \quad \text{(Wrong!)},$$ as you should quickly be able to see since the left-hand side should be real but the right hand side has an imaginary part....


6

Well, the wave equation is $$ \frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2} $$ where $c$ is just a constant, and has the interpretation of wave velocity (by simple dimensional analysis). Putting in $u(t, x) = e^{i(\omega t−kx)}$: $$ \frac{\partial^2}{\partial t^2} e^{i(\omega t−kx)} = c^2 \frac{\partial^2}{\partial x^2}e^{i(\...


3

While you can't use a spatial Fourier transform because of $l(x)$ and $c(x)$ not being constant, you can still use a Fourier transform in time to turn the PDE into an ODE. So using the Fourier transform $$\tilde{V}(x,\omega) \equiv \int_{-\infty}^\infty dt~V(x,t) e^{-i\omega t},$$ your equation transforms to $$\frac{\partial}{\partial x}\Big(\frac{1}{l(x)}\...


2

Note that $\int dp'\ p' e^{-ip'(x'-x'')} = 2\pi i\delta'(x'-x'')$, where $\delta'$ is the distributional derivative of the delta function defined by $$\int dx \ f(x) \delta'(x-a) := -f'(a)$$ Note also that for a complex number $a$, we have that $a-a^* = 2\color{red}{i}\mathrm{Im}(a)$, not just $2\mathrm{Im}(a)$. Those two pieces of information, plus the ...


2

This is a basic property of Fourier transforms. Quoting from Cohen-Tannoudji (Photon and atoms, introduction to QED, beginning of section I.B.1): Since $\mathbf{E}(\mathbf{r}, t)$ is real, it follows that $$\mathscr{E}^*(\mathbf{k}, t) = \mathscr{E}(-\mathbf{k}, t).$$ Note that $\mathbf{E}(\mathbf{r}, t) \leftrightarrow \mathscr{E}(\mathbf{k}, t)$ are ...


1

Fourier-transform spectroscopy is not just any spectroscopy. Fourier transform involves conversion from the time domain to the frequency domain and back, so it is necessarily present in all spectroscopic applications - whether the "conventional" spectroscopy or the Fourier-transform spectroscopy. However, in the "conventional" ...


1

Since $T_{\mu \nu}(x)$ is symmetric in $\mu$ and $\nu$, the RHS is also symmetric by default. If it is not manifestly symmetric (like in your example), you can always symmetrize it over the corresponding indices. So in your integral, change $k_\mu q_\nu \to \frac{k_\mu q_\nu+k_\nu q_\mu}{2}$. In order to find $T_{\mu \nu} (p)$ (where $p=$ momentum) from $T_{...


1

I think this paper may answer your question, although I couldn't find the whole paper available anywhere, unless you have privileges to download from the American Journal of Physics: Monochromators as light stretchers. It showed that indeed the prism does hold on to light for a time period that is commensurate with the uncertainty relation that you wrote. ...


1

You're integrating over all possible values of the momenta, so in this case the minus sign is not needed.


1

As Emilio Pisanty already said, you don't need to. It is actually possible to determine the reciprocal lattice for any lattice in an arbitrary number of dimensions: Let $V$ be a $n$-dim. real vector space and let $g\colon V\times V\to\mathbf{R}$ be a non-degenerate bilinear map (we don't need to assume that $g$ is symmetric). If $(a_1,\ldots,a_n)$ is a basis ...


Only top voted, non community-wiki answers of a minimum length are eligible