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12:55 AM
does the course Electroweak Interaction and Extension of Standard Model need the knowledge of Quantum Field Theory to assimilate?
 
 
1 hour later…
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2:16 AM
Hi All...
 
2:34 AM
It is often stated that resolution is limited to about half the wavelength of the incident radiation. What is the motivation behind this statement?
 
2:59 AM
woah, it just occurred to me that i was in high school almost a decade ago
i wish GR gave a way to slow down time perception
 
@Bohemianrelativist Probably
 
3:11 AM
if I get banned for 1 hour go and buy lottery then you will probably win 1 million dollar so that you can leave this chat🤤🤪
 
 
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5:06 AM
0
Q: About using MathJax

Prof. MeowAs 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.

 
 
5 hours later…
9:54 AM
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 the inner product is just an operator, too
 
@ACuriousMind Okay, I need to think about what that means, what does the operator act on?
 
@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 $g_{ij}$
 
10:30 AM
@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?
 
 
1 hour later…
11:45 AM
@B.Brekke I don't really like the passive/active distinction so I'd prefer to remain silent on that point ;)
 
 
2 hours later…
1:39 PM
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 $\psi\sim e^{-ipx}$
 
@Charlie yes, that one of these is a creation and the other is an annihilation operator is relevant, you cannot choose differently
I'm not sure what you mean by the number of "basis solutions"
 
I'm assuming that's something along the lines of you want to get a positive energy solution from $\phi(x,t)|0\rangle$?
well just that the "most general solution" of the K-G equation is a linear combination of two functions, $e^{ipx}$ and $e^{-ipx}$, no more no less
 
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 $a_p\mathrm{e}^{\mathrm{i}px}$
 
Ok I had a feeling calling it the most general solution was a slightly odd way of putting it
 
2:16 PM
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
 
 
1 hour later…
3:20 PM
@ACuriousMind one of the most confusing things in physics
 
4:14 PM
@RyanUnger eh, I'm sure it doesn't even make the top 5, but it's pretty annoying, yeah
 
@ACuriousMind I bailed before things got worse
 
 
2 hours later…
6:26 PM
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
 
6:47 PM
What mean partial derivative with superscript notation? Example: d'Alambertian operator which equals to product of partial derivative with super and sub scripts.
 
7:43 PM
0
Q: How do the evanescent waves contribute/cause a diffraction limit?

schnIn 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...

 

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