@imbAF Well, for one, the question is: Why do you want to read it? But to answer the question as asked: Advanced familiarity with the Hamiltonian formalism and quantum field theory would be useful. This is not a textbook (though it contains lots of exercises).
@ACuriousMind apparently that kind of definition is used in Lang's diff geo book? but it's more like "a vector is an equivalence class of tuples" rather than "a vector is a tuple"?
@imbAF I'm not exaggerating: That may take you a lifetime :P QoGS is merely one viewpoint - the Hamiltonian, constrained one - and there are many others: The BV formalism, mathematical gauge theory with principal bundles (in the vein of Donaldson) and the various ad-hoc procedures of physicists (foremost Faddeev-Popov tricks) are others, for a start. It's a rich and subtle topic.
@qwerty you have to identify the tuple with the tuple "after coordinate transformation", yes. In the bundle view, it's the equivalence relation between different trivializations you quotient out.
But e.g. on $\mathbb{R}^n$ with a single chart, you don't really have to quotient everything out - the tangent bundle is $\mathbb{R}^n\times\mathbb{R}^n$, you can just treat the tangent vectors as tuples $(x_1,\dots,x_n)$ attached to a point and you're not doing anything technically wrong, as inelegant as it may seem
and lots of physics language really only makes sense if you primarily treat it as statements about spacetime = $\mathbb{R}^n$ and then you have to laboriously reconstruct the formal generalization to cases with more than one chart
i am reviewing electromagnetic waves in dielectrics. so if we write down the fields as plane waves, plug them into the macroscopic maxwells, we get that this situation supports transverse waves + a nontrivial dispersion relation. there is a comment in my notes that says [...]
[...] "note, $\frac{\omega}{\vert k \vert} = \frac{c}{\sqrt{\epsilon (\omega)\mu}}$ varies with $\omega$, so there is not a single phase velocity $\implies$ E is not in general a solution to the wave equation -- different frequencies travel with different speeds." but im confused about what this means when it says "E is not in general a solution to the wave equation"
Btw this problem came up when studying conformally invariant rel field theories when interpreting terms like $\partial^\mu$ as transforming contravariantly under a general coordinate transformation This leads to problems
It only transforms contravariantly under Lorentz transformations
@Relativisticcucumber It means that the E field alone, which in vacuum would satisfy the simple wave equation with fixed c, is no longer satisfying the simple wave equation when in a material. Instead, it has a non-trivial speed distribution.
@Relativisticcucumber yes; I hate EM in materials but instead of a constant $c$ you get a $c(\lambda)$ dependent on the wavelengths - the solutions to the wave equation in vacuum are not solutions to the equation in media and vice versa
i dont know why i always mix up jacobi and bianchi
but back to reality, i think i see @naturallyInconsistent @ACuriousMind. so it's like these are still plane wave solutions to the macroscopic maxwells, it's just that the macroscopic maxwells no longer lead to the wave equation due to the fact that the propagation speed depends on frequencies, am i getting this right? is there any important consequence to the fact that there is no longer a strict wave equation for em 'waves' in media?
These are still plane waves, and we usually still continue saying that they satisfy a generalised wave equation, and we would even omit the generalised.
There are many consequences; the speed of light in vacuum is the one that is the spacetime conversion factor, and none of these other speed of lights in materials is of any importance to physics. In a typical material, $c^2=v_pv_g$ and the information follows group velocity $v_g$ = envelope velocity, and this is slower than the spacetime conversion factor, which means that the unphysical phase velocity $v_p > c$
The photons in a material trivially gains mass, for example.
@naturallyInconsistent ah yes i have seen this. i did not recall that $c^2 = v_pv_g$ though so that is nice information
so if im getting this content right, there is a frequency that is the 'threshold frequency' for an EM wave to travel through a medium (i.e. this medium is transparent for this wavelength). i think this is the plasma frequency? so i see that if you change the frequency of the light that determines whether the light will make it through or not, but is there something that I can do to the material itself to make this threshold lower or higher? [...]
[...] i seem to be recalling that the plasma frequency comes from the phonon scattering within the material? or am i way off base
Also, the definition of $v_p=\omega/k\quad\bigwedge\quad v_g=\partial\omega/\partial k$ which really is just that we have a relation looking like $\omega^2=c^2k^2+$const
@Relativisticcucumber there is something like this, but it would only become sensible after you see band gaps and all.
@naturallyInconsistent i did see band gaps in a&m but alas i am not reading it today -- i am reviewing enm lecture notes to prepare for impending exam :P so u mean if i know band gaps i should know the answer to my above w? about adjusting the material itself to allow for more light penetration ?
Well, it is one of those things that are difficult to discuss in detail but exists a simplistic answer that is extremely easy to miss in a reading of an introductory text
In a metal, there is no band gap, and there is always a sea of electrons plasma. Then, any frequency lower than the plasma frequency will be mostly reflected and difficult to penetrate into the metal, exponential decay in penetration depth, etc
Above the plasma frequency, the electron sea waves are no longer able to keep up with the incident light, and so it will mostly pass through, but by then the energies are so large that you will start resolving the crystal structure
I mean, after UV rays, is X rays, and you know X ray crystallography is a thing
gamma rays could resolve nuclear structure; I'm defining the X ray v.s. gamma rays by energy and not the crazy definition that uses source (nucleus or not)
Whereas in a material with band gap, you need an insight that A&M mentions, but cannot make a big deal thereof: If you plot $\omega=ck$ in the usual band structure plots, you will realise that $c$ is so huge compared to the usual band structure curves, that they look like vertical lines. That is, when you try to conserve both energy and momentum in a one-photon interaction with electrons in the material, you will see that it is equivalent to saying that momentum of electron barely changed but
energy jumped vertically
The existence of the band gap, thus means that light at frequencies lower than the band gap cannot induce transitions, and that in turn means that it can pass through the material unimpeded, to first approximation.
@naturallyInconsistent so for the case of a metal can i do smth like adjust the temp or pressure of the metal to allow really low frequencies of light to pass through. i mean i havent fully internalized this scattering mechanism, but if i cool the metal down for instance, can this decrease the scattering enough to let smth like visible light through or is this absurd
That is how we understand crystalline see-through materials. Glass is amorphous and thus not treated in this way, but the idea is still borrowed from here
@Relativisticcucumber I dont think that will work, because it has no band gap. A normal metal should not ever decrease the scattering enough to become almost-transparent. However, if it is a semi-metal, then at very very cold, it might end up indistinguishable from having a tiny band gap, and thus you might have radio waves pass through. Nothing like visible light, though.
Low pressures should not change any picture here, since Fermionic matter is equivalent to having high pressures. However, if you press something so hard that its behaviour completely changes, that is possible. I mean, you can convert metals to insulator and vice versa just by cramming it so much that you shift the chemical potential (Fermi level) from middle of a band to the end of a band and vice versa
miao miao hath a meeting to go to in a while. But these should be way more than what you need.
@naturallyInconsistent wait i must be missing smth. if $v_g$ is physical, then $v_g \leq c$, and if $v_pv_g = c^2$, this would imply that $v_p \geq v_g$, but there is such a thing as anomalous dispersion, no? so is it that anomalous dispersion happens in 'atypical materials' and the relation you stated is, as you say, for typical? i am kind of confused about how this anomalous dispersion can happen
@Relativisticcucumber The message in which miao miao wrote $v_g<c$ and $v_pv_g=c^2$ is directly meant to imply $v_p>c$. Anormalous dispersion is just about whether you are on which side of the plasma frequency; if you have a typical metal whose plasma frequency is in UV, then you have normal dispersion, and if you have a atypical metal whose plasma frequency is in IR, then you have anormalous dispersion. See oceanopticsbook.info/view/theory-electromagnetism/level-2/…
@Relativisticcucumber Nothing particularly impressive; just from statistical thermodynamics, if you evaluate the equivalent pressures and temperatures of a simplistic ideal gas and compare that with Fermi level's, you will see that a degenerate Fermi material is equivalent to being very hot and very highly pressurised. You will also get to learn about how the degenerate electrons in metals only contribute to heat conduction but seems impermeable to heating; that's just because it is "already very hot"
Other textbooks, e.g. Kittel, would just cover the correct answers and all these phenomena. A&M is nice in that it would give you a hint that the pioneers were incredibly confused over all these paradoxical aspects as they slogged through the maths, yet never intending to make you suffer in the confusion.
@Relativisticcucumber Actually, that isn't true either; you can study surfaces with reflections, in which case you might just penetrate a few layers only. That is still sufficient to give you data on the crystalline underpinnings
@qwerty It's really more an accident and not that I like these books necessarily best, but the others are Quantum Physics by Glimm and Jaffe (the rigorous path integral book) and Weinberg's QFT I. I also have the BBS string theory book but that one I don't like at all :P
“The problem is that there are some ornery people who really like manifolds as they are, but sometimes have to work with things that are almost but not quite completely unlike manifolds. For these people, the further away from true manifolds they get, the more uncomfortable they feel. One of the biggest steps for such people is losing the underlying set. So diffeological spaces are a category in which those people can have most of the benefits of sheaves without having to discard their comfort blanket of something that still resembles manifolds in some way.
Does the perturbative expansion of the S-matrix depends on the number of particles in the final and initial state? We learned that if you have more field operator components $\phi^{+}(x)_N$ than particles at in a state $|i\rangle$ (initial state), you'd get zero when the field operators act on $|i\rangle$
Today we took an example of a 2electron to 2 electron collision process
We did the long calculations for first order perturbation theory
@imbAF Would what be a problem? It's not clear what you mean by the perturbative expansion "depending" on the number of particles.
@naturallyInconsistent the phenomena may be neat, but the math is boring (to me, I don't enjoy solving differential equations or arguing about the continuity of fields or whatever)
@imbAF I don't know what you mean by that - it is certainly not the case that fields acting on the vacuum would result in 0 because $n=0$ for the vacuum.
sure, but where's the problem? expectation values are linear, so that summand gives zero, but the other parts with more creation than annihilation operators (in the right order) are non-zero
For the s matrix we write: $S=T\exp[i\int_{t_0}^t H_i(t)dt]=T\exp[i\int_{t_0}^t L_i(t)dt]$ where H and L are the densities. I know the relationship between hamiltonian and lagrangian density via the canonical momentum, the expression, which I am not writing. So $\mathcal{H}= ...-\mathcal{L}$
But here we say that $\mathcal{H_I}=\mathcal{L_I}$
@imbAF I don't understand the question. Your expression is not the S-matrix, that's just a time-evolution operator for finite times $t_0$ to $t$ (if $H_i$ is some Hamiltonian), and what is $H_i,L_i$ in there? You can't just use random notation and expect people to know what you're talking about.
@imbAF Again, that's not the S-matrix, that's just a time evolution operator at finite times. There is at least a limit for the times to go to infinity missing.
And okay, it stands for "interaction" - but do you know the definition? If you start from some random Hamiltonian $H$, what is $H_I$? What is $L_I$? I believe if you unpack the definitions you should be able to see rather quickly how the two are related.
@imbAF No, you asked why $H_I = L_I$. Implying that you know what $H_I$ and $L_I$ are to begin with. If you don't know what $H_I$ is, why don't you ask "What is $H_I$?" (or, better read, re-read the text to figure it out)?
> Ces “nuages probabilistes”, remplaçant les rassurantes particules matérielles d'antan, me rappellent étrangement les élusifs “voisinages ouverts” qui peuplent les topos, tels des fantômes évanescents, pour entourer des “points” imaginaires
Looks like Mr. Grothendieck invented the QM topos :p
@Slereah do you expect me to be able to read that :P
@imbAF What I would have expected you to be able to present is the simple definition that for the full Lagrangian $L$ and Hamiltonian $H$, we have $L = L_0 + L_I$ and $H = H_0 + H_I$, where $L_0$ and $H_0$ are the free Lagrangian and Hamiltonian. If you plug that into the Legendre transform, you should be able to derive $H_I = -L_I$ (perhaps with one additional assumption if you notice a subtlety), i.e. your formulae are missing a minus sign.
@ACuriousMind you are correct it is missing a minus
additionally, in the beginning you said that what I wrote is the time evolution operator, and in order to have the S-matrix you need the limit to infinity
and you were correct to notice that as well
the boundaries are $\pm infty$
why is that you need limit to infinity, i.e asymptotic states
I think you're confusing the concrete space of functions/differential form on a given $M$ with the non-concrete (pointless) generalized universal space of forms
i.e. what they call the smooth moduli space of n-forms there
the moduli space is not the space of n-forms, it's the space that "modulates" the n-forms, i.e. you get a space of n-forms on some $M$ by considering the maps from $M$ into that moduli space
@ACuriousMind ACM I considered an arbitrary lagrangian density $\mathcal{L}=\mathcal{L}_0+\mathcal{L}_I$ and by using Legendre transformation I computed $\mathcal{H}$. The way I see it, the only way for me to have $\mathcal{L}_I=\mathcal{H}_I$ is if I can say that the interaction term doesn't contain any time differentiated term.
And so I checked the interaction terms for $\phi^4$ scalar field and for QED
And I noticed that in both cases the interaction term lacks time derivatives of field operators
Can someone help me understand one thing about the Moller scattering. In the lecture we considered until 2nd order PT. And without explicit calculation the lecturer said that the first order vanishes. I am currently trying to perform the following: $\lang 0|a_3a_4\bar{\psi}\psi A_\mu a^\dagger_1a^\dagger_2|0\rangle$. So I expand this expression by expressing the field operators via their annihilating and creating part. Then by that point I really don't know how to continue. But what I find odd is how the lecturer was able to immediately tell that this gives zero. Is it because the vector fi…
And is that because there are no incoming and outgoing photons
And as such when the anihilating part of the vector field operator acts on the vacuum gives zero?
@ACuriousMind they only generally do not depend on velocity because the "standard" curriculum chooses (consciously) to ignore more complicated physics, no?
the first time i actually treated ideal drag was in my ordinary differential equations course, not even a physics course :P
@SillyGoose not exactly - many dissipative systems, like those involving friction/drag, do not admit a Lagrangian description to begin with, and as Slereah said the standard curriculum contains at least one canonical example of a velocity-dependent potential: The Lagrangian for a particle moving in an EM field, since the Lorentz force depends on velocity.
I actually first encountered the hamiltonian for a particle in an EM field in condensed matter I when we introduced Landau levels lol. So i guess i never have encountered the lagrangian mentioned in a course yet.
@ACuriousMind maybe i misunderstand your point, but my point is that the standard curriculum is ignoring real phenomena (adapted to your argument) because they do not conform to the Lagrangian paradigm, in other words, are more “complicated”.
@SillyGoose I think "complicated" is the wrong word - there are also extremely simple equations which do not admit a Lagrangian description (e.g. $\ddot{x} = x+y, \ddot{y} = xy$) - the existence of the Lagrangian description is not a matter of complexity
i just realized that while the canonical example of a dipole (-q) --> (+q) is nice, it is somewhat misleading in that one has an (origin dependent) dipole for a distribution --> (+q) as well.
