I personally find the style quite nice. i only read several pages, but halmos tells you when he is making an abuse of notation or inaccuracy for the sake of pedagogical convenience and also tells you when there are alternate paths to develop a concept in linear algebra and tells you why he develops it the way he does if that makes sense
Hmm, interesting. Will definitely give it a try. My LA course starts next semister (approx 30 days left) and the primary recommendation is Hoffman kunze. Many of my seniors said halmos is for a 2nd reading
@nickbros123 oh interesting. where are you based? i read through the first three chapters of hoffman and kunze before stopping :P. you may have a different experience, but I thought that the notation gets in the way of the content at times.
i took a course in linear algebra based entirely on prof lectures. then i read some of H&K. Now, I'm patching up some holes with Halmos :P.
@SillyGoose I'm based in India. This year (in my institute) one cannot exactly trust courses based on prof lectures, since semesters are cut short, and the instructors' main goal becomes completion of the list of course objectives rather than covering nuances in every subtopic.
@Relativisticcucumber Sorry, I've been quite busy. They way I'd put is that when we're studying hydrogen as the relativistic limit of the system we know non relativistically we're interested in the positive energy solutions. The negative energy solutions are a problem in the one-particle relativistic theory, which what we're concerned with here. I don't think we should even think to attribute any meaning to negative energy solutions in the one-particle theory because they're just nonsense
Arising from the fact we've decided to treat a relativistic theory as a one particle theory
Think of this easier example. The solutions of the free Dirac equation might have either positive or negative solution. When you perform the non relativistic reduction of this, you consider energies of the form $E\approx mc^2$ which is what you have in the nR limit you know for a free particle, instead of $E\approx -mc^2$
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According to [this](https://en.wikipedia.org/wiki/Dirac_bracket#Generalized_Hamiltonian_procedure) wiki article, the weak equality in the context of constrained Hamiltonian mechanics is defined as follows:
> Two functions on phase space, $f$ and $g$, are weakly equal if they are equal when the constraints are satisfied, but not throughout the phase space, denoted $f\approx g$. If $f$ and $g$ are equal independently of the constraints being satisfied, they are called strongly equal, written $f=g$. It is important to note that, in order to get the right answer, no weak equations may be used before evaluating derivatives or Poisson brackets.
On the other hand, a book I'm reading defines it as:
@nickbros123 oh yes I would not trust the lectures I had :). I certainly left with many holes in my knowledge. i hope you enjoy the course and whichever book you end up using
@Mr.Feynman They're not exactly definitions of the same thing
your first quote defines "weak equality" generally while the second just says that "the constraints are weakly zero" means the constraints are relations that may not be invariant under general Poisson brackets
@ACuriousMind So, is it just the same symbol for two different things? Does it mean that "the contraints are weakly zero" is not the weak equality $f\approx g$ between $f:=\varphi_m$ and $g:=0$?
Because if that were the case, then I wouldn't see how that $f\approx g$ in this specialized case would lead to the statement about the PB
@Mr.Feynman A function that has zero PB with every canonical coordinate is constant on the entire phase space. The constraints are not constants on the entire phase space, so they have non-zero bracket with at least one canonical coordinate
i.e. generally, something that is weakly zero but not zero will have at least one non-zero bracket
@geocalc33 If your flat space is a lattice and has a finite volume, then the set of trajectories will be discrete, yes. In tht case, the analogue of the path integral would be to integrate some function from the set of trajectories to scalers
So we are not "adding up" the trajectories themselves. We are only adding the scaler function values
The trajectories work as the index of the summation. Like in usual integration, it's "dx"
Actually, there is something called the "functional Fourier transform". So, integral transforms have generalisations to functional integration
"Functional" is the term used for functions of functions @geocalc33
The usual fourier transform is defined for functions. It employs usual integration. There is a generalisation to functionals. It employs functional integration
If I have an electron whose spin is in some superposition, can I rotate the direction of the spin, via magnetic field or something like that, without destroying the superposition?
@RyderRude I understand that more or less, but at the instant when the interaction starts, is it correct that the magnetic field immediately collapses the spin to a certain direction?
@RyderRude I think I get it now. You basically take a functional that assigns a unique scalar to a unique trajectory - and you sum up these outputs with respect to the space of trajectories
@ACuriousMind Okay, and if I further suppose that this spin is a part of an entangled pair, does it follow then that the possibly distant spin of the other electron is rotating in correlation with the first spin?
@ACuriousMind Well, since if you say that superposition is not destroyed I am basically asking whether the superposition of the pair (entanglement) is destroyed
the two can stay entangled without the spin of the second partner changing
let's say you start with an entangled state $\lvert \uparrow_z\rangle \otimes \lvert \downarrow_z\rangle + \lvert \downarrow_z\rangle \otimes \lvert \uparrow_z\rangle $
now you rotate the spin of the first particle, and you get e.g. $\lvert \uparrow_y\rangle \otimes \lvert \downarrow_z\rangle + \lvert \downarrow_y\rangle \otimes \lvert \uparrow_z\rangle $
this is still entangled, and nothing happened to the second particle
Vaidman of the Elitzur Vaidman bomb experiment is a proponent of MWI. I found it amusing in relation to that experiment... that bombs will go off but not always in our world lol
If I have a Hamiltonian $\hat{H}=\sum_k \frac{k^2}{2m} \hat{c}^\dagger_k \hat{c}_k + g H_{int}$ what exactly gets renormalized? Is it just the energy dispersion relation and g without any mass renormalization?
Also, the Hamiltonian I wrote above has $E(k)=k^2/2m$ but this is the dispersion of a free particle (I think). Shouldn't this change once we add interactions?
As in we are going to observe a dispersion relation when doing experiments in real life I imagine
which is going to be different from the free one in the Hamiltonian
@DIRAC1930 so there are two different aspects of "renormalization" - one is to get rid of divergent quantities and the other is to take an interacting theory and express it so that you can talk about the particles in it like you're used to
you can/have to do the latter even when there's nothing necessarily diverging; I just didn't get that was what's meant here
so what's the question? You'll "renormalize" the fields/wavefunctions by multiplying $Z_k^{1/2}$ just so that their propagator has the right numerator again
The other renormalization is a choice becuz u can work with the bare quantities too. The infinity renormalization is necessary becuz the bare quantities are infinite and un-workable
@DIRAC1930 okay, so what I think is happening here is that you need to re-read the assumptions of your model here:
"The main idea of the Fermi liquid theory is that, in a Fermi gas, even in the presence of interactions, the low-lying structure of excitations is the same as in the non-interacting one."
from the notes you linked
that means you're not interested in anything the interaction term does except for what it does to the propagator (which contains the "non-interacting physics", after all)
so in your version of "renormalization" you're just by assumption mapping the interacting theory back to the non-interacting one
so what happens to your coupling constant is that it becomes zero by the very specific assumptions of your model
Hmm but IIRC scattering is a big part of Fermi Liquid Theory i.e. the vertex function is mentioned as being important and is calculated in LL for example
that you're neglecting the interactions in the theory you're "renormalizing towards" doesn't mean that you can just forget about the interactions completely
In equation (3.4.18), why does he write +regular part
Sorry the +regular part is from inserting a complete set of states into the Greens function
But that part doesn't appear in the free Greens function so how can you just assume that the dressed GF acts like the free one when they have a different form
Hmm actually, maybe it's because the Fermi liquid theory is only valid near the Fermi surface and that dressed GF has a pole near there so that term is the one that dominates
it's just how functions work? The fully summed propagator is $\frac{\mathrm{i}}{p - m_0 - \Sigma(p)}$. This has a lowest pole at some $p = m$, so its $\frac{i Z}{p - m} + \text{regular terms}$, where "regular terms" means they don't have poles at $m$
this is a fully general thing in QFT, for once :P
and the lowest pole of the propagator is more or less the definition of what the (renormalized) mass is
So I have to somehow extract the part that is not dependent on momentum
Or I just find the pole
I think the interpretation is different since the propagator is $\frac{1}{\omega - E_k}$ where $E_k$ seems to be a particular energy so maybe the energy level gets renormalized instead of the mass
Let $V$ be a vector space and $V'$ be its dual space. Given a subset of a vector space $S \subset V$, the annihilator of $S$, denoted $S^0$ is the set of all $y \in V'$ such that $[x, y] = 0$ for all $x \in V$. @ACuriousMind
basically the annihilator of a subset of vectors is a subset of covectors that send the subset of vectors identically to $0$
