As I was just writing my answer, I felt the need for the in-built MathJax toolbar for ease in writing equations and formulas without adding commands like \pi R^2 inside double $s. It will just create smooth writing experience for all, particularly the new ones.
If I want to evaluate how a vector $\vec{v}$ transforms under an operator $\mathcal{O}$ I can evaluate $\mathcal{O}\vec{v}$. If I want to see how an operator $A$ transforms, I could evaluate $\mathcal{O}A\mathcal{O}^{-1}$. If I want to see how an inner product transforms, what should I evaluate then?
@B.Brekke It eats two vectors to produce a scalar, depends how you want to "curry" that. So you can think of it as something that maps a vector to a dual vector, or you might just think of it as the operator in $\langle v,w\rangle = w^T M v$, or...
or you think of it with indices - a vector is $v^i$, a matrix is ${A^i}_j$, a metric/inner product is $g_{ij}$
@ACuriousMind Thanks, I think I understand now. So if my question is, how does the inner product transform, I should evaluate the matrix $M$. Or I could evaluate the vectors $w^T$ and $v$ instead, and this would be this passive/active transformation correspondence?
What determines the number of "basis solutions" when writing what we call the most general solution in field theories. e.g. $\phi(x,t)=\int \tilde d^3p \left(a_p e^{-ipx}+a^\dagger_p e^{ipx}\right)$ in qft, or the "left/right moving modes" in string theory? It can't just be a general result that only two functions are needed right
And actually as a side note, does it matter which $a_p$ or $a^\dagger_p$ gets matched with $e^{ipx}$ or $e^{-ipx}$ in the above? I've always wondered, the only difference I can see it making is in the position space representations of the state $\phi(x,t)|0\rangle$, you're either going to get something like $\psi\sim e^{ipx}$ or $\psi\sim e^{-ipx}$
You can write every field like this, it's just a Fourier transform. The special thing about the free theory is that the Hamiltonian is diagonal in Fourier space and so the $a_p, a^\dagger_p$ have special meaning
it's really not the "general form of the solution", and it's also not a linear combination of two functions, it's a Fourier transform which is the integral over uncountably many functions $a_p\mathrm{e}^{\mathrm{i}px}$
It's defined in a way so that in the non-relativistic limit things sync up with the usual definitions which are natural in-and-of-themselves
e.g. $e^{-i(Et - \mathbf{p} \cdot \mathbf{r})}$ is the plane wave that, in the non-relativistic limit, corresponds to a free particle, and in the non-relativistic case one finds it natural to define $\hat{\psi} = \sum_{\mathbf{p}} \hat{a}_{\mathbf{p}} e^{-i(Et - \mathbf{p} \cdot \mathbf{r})}$ and $\hat{\psi}^{\dagger} = \sum_{\mathbf{p}} \hat{a}_{\mathbf{p}}^{\dagger} e^{+i(Et - \mathbf{p} \cdot \mathbf{r})}$ so the relativistic case is set up to reduce to these conventions
From a relativistic perspective one finds $E = \pm |E|$ so including both cases in the plane wave expansion $\hat{\psi} = \sum_{\mathbf{p}} (\hat{a}_{\mathbf{p}}^{(+)} e^{-i(|E|t - \mathbf{p} \cdot \mathbf{r})} + \hat{a}_{\mathbf{p}}^{(-)} e^{-i(-|E|t - \mathbf{p} \cdot \mathbf{r})})$ it's natural to leave the first term as an annihilation operator, and then do something to the second term to make sense of it
Anyone mind sharing some insight as to why resolution is limited to about half the wavelength of the incident light? See for instance: physics.stackexchange.com/q/389599/236145
What mean partial derivative with superscript notation? Example: d'Alambertian operator which equals to product of partial derivative with super and sub scripts.
In the following article, under the section Physics of the Superlens, it is stated:
The light emitted or scattered from an object includes not only
propagating waves but also evanescent waves, which carry the
subwavelength detail of the object. The evanescent waves decay
exponentially in any med...
As I was just writing my answer, I felt the need for the in-built MathJax toolbar for ease in writing equations and formulas without adding commands like \pi R^2 inside double $s. It will just create smooth writing experience for all, particularly the new ones.
