@Mr.Feynman i am doing a research project that involves some qft, and one of the grad students involved was recommending me schwartz and said it was the standard book so i was just curious ! i have also heard of P&S, zee, and tong's lecture notes, but i personally was not a fan of zee when i tried it out.

@fqq I came up with that because apparently what is called QFT II in my uni is basically QFT in other places and my QFT I is some kind of transition course!

@Relativisticcucumber I see. Schwartz is definitely a renowned book. It's recommended in all of my courses but I already got the other books :P

Regarding Tong's notes, like Schwartz I know they're out there and I may occasionally consult them. There's still plenty of books, I don't know if they are to be considered introductory though.

Oh, one more thing. There are some Sidney Coleman's lecture notes on arXiv

bloop

@SillyGoose honk

Come on chaps, reopen this. How it got closed escapes me since the meaning was perfectly obvious.

@Mr.Feynman regarding our discussion about negative energy solutions to dirac equation a few days back, one of my profs said that the negative energy solutions to the dirac equation for hydrogen atom correspond to positron solutions and that, though this is a valid solution mathematically, we dont see it because of a lack of antimatter. i think someone who answered ur q on stack was saying smth like this, though i didnt parse through their entire response? do you oppose this explanation xD

@Mr.Feynman plenty of fish in the sea as they say -- yeah i actually found before i started my sakurai endeavor the qft books were a bit too advanced, but im feeling a bit better about it after going through sakurai. going to try again after i finish a few more sak chapters

@SillyGoose MERP

@Relativisticcucumber MERP?

To ask the question more formally, can't the dressed propagator be obtained by a free theory with a modified dispersion relation dependent on the self energy

i.e. a free theory with dispersion relation $\tilde{E}(k) = E(k)- \Sigma(k)$

where $\tilde{E}(k)$ is the modified dispersion relation and $E(k)$ is the standard dispersion relation

I mean it is a standard exercise to compute QED corrections to the electron self-energy via the difference between the free and fully summed interacting propagator

it's how you get to mass renormalization

Sorry, I meant if I have $E(k) = k^2$ as my dispersion relation and I have a free theory, if I have another free theory with dispesion $E(k) = k^2 - \Sigma(k)$, I will get a free propagator that is the dressed propagator of the 1st theory

what do you mean "with dispersion relation"

quantum (field) theories are defined by their Hamiltonians/Lagrangians, not dispersion relations

The dispesion relation enters in the Hamiltonian through $H=\int E(k) A^\dagger_k A_k$

for a free theory

ah, sure

that's just the concept of the "quantum effective action", usually denoted $\Gamma[\varphi]$ - it's an action whose tree-level amplitudes are the same as the fully summed amplitudes of the original theory

Ah okay, so this is a correct way of interpretating everything?

it's an interesting object in theory but it's not a very practical one because in typical scenarios the only practical way to compute the terms in the effective action, like the self-energy $\Sigma(k)$, is order-by-order in perturbation theory via Feynman diagrams

but yes, I think in principle you are interpreting this correctly

Ah okay thanks

What if I have a dressed propagator $$\frac{Z}{\omega - \tilde{E}(k) + i \epsilon}$$

Accounting for the $Z$ may be different

Unless all the states in the effective theory are normalized differently

Also, I am doing non-rel QFT here

Actually, this can just scale the creation/anihilation operators

by $\sqrt{Z}$

So the effective Hamiltonian will be $H= \int \mathrm{d} k Z \tilde{E}(k) A^\dagger_k A_k$

I'm not sure what the question is

note that the difference between free and interacting theories is more than just the self-energy

your effective action isn't that, it's that + a bunch of interaction terms

Even for non-rel physics?

why would relativity matter?

if you have an interacting QFT, you have at least one vertex

Am I making the mistake of assuming the 2pt function describes the whole theory i.e. I need to consider all higher order Greens functions?

this doesn't just modify the propagator, it also allows for scattering

@DIRAC1930 yes, apparently :P

the propagator does not encode the whole physics

at some order in perturbation theory, the interaction terms will allow for scattering, i.e. getting different particles out of an interaction than are going into it

this is not described merely by the fully summed propagators

it is all n-point functions together that determine all of the physics - one way of axiomatizing QFTs is calling them Wightman functions and saying a QFT is defined by a list of n-point functions or an algorithm to compute them and there's a bunch of reconstruction theorems that say QFTs whose n-point functions agree everywhere are the same QFT

So for a non interacting theory, the 4 point correlator will be 1

I'm guessing it only deviates slightly for small interactions

depends on your notion of slightly :P

but sure, the limit of taking the coupling constants to 0 is supposed to be smooth, so you can find values for which the interaction is smaller than whatever $\epsilon$ you like

but generally the scattering terms are at least of the same order as the corrections to the propagator

Will the 6 pt function be of smaller than the 4pt function?

So can I just worry about the 4 pt function in the limit of weak coupling?

not in general, no

I mean, if you have a theory that only has an interaction term like $\phi^8$, then it's the 8-point function that is non-trivial at tree-level and the others will only follow at higher orders

but if you have $\phi^3$, it's the 3-point function that's non-trivial at tree-level

Okay so I essentially have the equivalent of a $\phi^4$ interaction

sure, in that case the 4-point function is the "most important" non-trivial function since it's non-trivial at tree-level

but it's pretty rare that you want to stop at tree-level, so I'm not sure why you're trying so hard not to have to do QFT :P

Will the first excitation just be described by the dressed 2 pt function?

i.e. if I just have one excited mode

it's dynamics will be described by the dressed 2 pt function

So in the regime of just above absolute 0, will I just essentially have 1 excitation?

what is a "first excitation" in this context?

I'm trying to understand Landau's Fermi liquid theory and I am using the reference Statistical Physics II from A Course of Theoretical Physics

I'm afraid I don't know anything about that

until now it sounded as if you were trying to understand some general features of QFT :P

The language is the same i.e in chapter 14 he has the vertex functions you describe

and there is wavefunction renormalization given by Z etc.

sure, condensed matter often uses QFT but things like "first excitation" are not phrases that make sense in a general context

I'm not sure they even use it lol I'm just trying to figure out what is going on lol

also the meaning of several aspects like renormalization differs greatly between the hep-th and the cond-mat applications

e.g. in cond-mat there is often some sort of literal physical meaning to renormalization (some definite energy scale of the system that's going up, or a length/resolution scale that's changing) while in hep-th renormalization is a bit more elusive in its concrete meaning

it really sounds to me like you want to talk to someone who knows about Fermi liquids specifically, not someone who knows generic QFT :P

So the self energy fully determine the dispersion relation of a single particle (in general QFT)? And (this is a guess) will the vertex function do something to give a new coupling constant?

Given that a representation maps a group to $\mathrm{GL}(V)$ instead of $\mathrm{GL}(n,\mathbb{C})$ (although they're isomorphic chosen a basis), should I always think of the linear actions of groups as active transormations? Is there any instance where the distinction actually gets relevant?

@DIRAC1930 yes, but from my point that's tautological (I would define the self-energy in QFT to be the higher-order corrections to the propagator)

@Mr.Feynman I personally think that whenever you think the distinction between "active" and "passive" might matter you have already committed some conceptual error and should start over :P

there is nothing about $\mathrm{GL}(V)$ to me that would make it "more active" than $\mathrm{GL}(n)$ you can let both act "on the basis vectors" or "on the other vectors"

Okay so from the QFT point of view, the 4pt function describes scattering and will be of the form $\langle \dots\rangle=1- A$ where $A$ is the correction from the free theory. What does this $A$ represent if I were to write an effective Hamiltonian?

in the quantum effective action it becomes a coupling constant

but note that this "constant", like the self-energy, depends on the momentum/energy scale

this is what people mean when they talk about a running coupling

But then wont this further effect the dressed propagator i.e. change the self energy even more?

@DIRAC1930 no, this isn't an infinite feedback loop - the quantum effective action, by definition, reproduces the fully summed amplitudes of the original theory at tree-level

Ah okay

the summed propagator, after all, already contains all the effects of the vertices of the original theory

trying to use the quantum effective action as a "normal" action and do another sum to all orders gives you nothing meaningful, as far as I know

Ah yes of course

The self energy has diagrams that have a vertex

but you need to be careful: the quantum effective action is far weirder than you might think

for the free part we could just talk about the dispersion relation but the interacting term produce in general non-local terms

in fact I've never seen anyone write down a quantum effective action explicitly

if you want to think about a "normal" action that reproduces the QFT amplitudes at a specific energy scale at tree-level, we need to switch to the Wilsonian effective action

that one depends on an energy scale $\Lambda$ and only works at tree-level for processes at scale $\Lambda$, but it's much more tractable

and in fact the Wilsonian effective action is the thing where the running coupling occurs directly

Something you may or may not find interesting, for a Fermi liquid, the 2nd order correction to the self energy is imaginary therefore the elementary excitations decay like unstable particles

complex energies mean decaying states, yes, in hep-th contexts that usually occurs for resonances

@ACuriousMind The wiki article disagrees with this iirc but I see your point. If anything, it feels at least more natural to think of the former as active and the latter as passive because I have the tendency to only consider the action on "the other vectors" :P

What physically does counterterm renormalization mean for something like a Fermi gas?

I'm a bit confused because IIRC, there are no infinities but I may be wrong

Is it correct to say that the mass doesn't get renormalised and that we only have wavefunction renormalization and the coupling constant renormalization

So I have a question about the unitary time evolution operator (UTEO)

In Sakurai, we merely state that we should have the UTEO should have the "composition property". Reading this now, it seems like the composition property is some expression of "path independence" or only one path existing

is this an intelligible interpretation :P

@SillyGoose by "composition property" you mean $U(t_1,t_2)U(t_2,t_3) = U(t_1,t_3)$?

yes

that's just the statement that time translation should be a group

hm but why should it be a group

okay, so all this property is saying is that it shouldn't matter if I evolve in small steps or in one giant step

so this is unrelated at all to an interpretation of time as being an "arrow" :P because it doesn't seem like it'd allow for another interpretation if you're saying it only matters what your end points of time are

it's a statement about the continuous divisibility of time: I can arrive at the physical state at time $t_1$ either by considering the full time evolution from some start time $t_0$ to some end time $t_1$, or I can think about $N$ small steps from $t_0$ to $t_0 + (t_1 - t_0)/N$, then to $t_0 + 2 (t_1 - t_0)/N$ etc.

these should give the same result

without this the concept of "the final state at $t_1$" wouldn't even make sense since the resulting state would differ depending on how I choose to think about the time interveral

@SillyGoose I would say this is unrelated to any "arrow" of time, yes, because the unitary time evolution is, by virtue of being unitary, reversible!

evolve from $t_0$ to $t_1$ with $U(t_0,t_1)$, evolve in the reverse direction with $U(t_0,t_1)^\dagger$

oh also so why is there a reverse time operator (anti-unitary, i think?) and also the hermitian conjugate of the UTEO

what is the conceptual difference :P

@SillyGoose I mean, this is just like classical mechanics

if you run the classical equations of motion backwards you can evolve "final condition" back to initial conditions at earlier times

there's nothing special to quantum mechanics here

time evolution is a reversible map in both cases

and it's a map that obeys such a "composition property" in both cases, too

heh I have no exposure to classical mechs other than intro level (kinematics, force, etc.)

so it's just like a "duality"?

that's why I keep saying the more classical mechanics you know the easier QM gets :P

to have both operators be distinct but have the same function?

there's nothing quantum about this idea of time evolution being reversible and infinitely divisible/composable

well to be fair i don't think my confusion is stemming from treating it in a quantum context

surely any concept if you've seen it once before will be easier to digest on latter seeings. but i just have not seen the idea of time evolution in any context before :P

I mean you've solved equations of motions, right?

yes

in classical Hamiltonian physics, your solutions are functions $x(t),p(t)$ for some initial conditions $(x_0,p_0)$

you can get from this a notion of classical time evolution by considering, for fixed $t_1$, the map that sends any point $(x,p)$ in phase space to the point $x(t_1),p(t_1)$ that you get by using $(x,p)$ as initial conditions at $t=0$

(mathematically this is called the Hamiltonian flow of the Hamiltonian)

that seems reasonable indeed

quite the name :P

this gives you a map $\phi_t$ on phase space for each $t$

and now, just because the solutions to the equations of motions are unique, you have that $\phi_{t}\circ\phi_{t'} = \phi_{t + t'}$

because when you "look along" a solution of the equations of motion, when you first look ahead by $t$ and then by $t'$ from there that must be the same point as looking ahead by $t + t'$

the quantum property $U(t_1,t_2)U(t_2,t_3) = U(t_1,t_3)$ is just this but we're being more cautious in that we don't assume that time evolution is "always the same", i.e. depends only on the difference between the start and end times

(you can make this necessary in classical mechanics, too, if you make your Lagrangian/Hamiltonian time-dependent)

@SillyGoose the name is silly but you can consider a "Hamiltonian flow" for any function on phase space not only the Hamiltonian, so it's not quite as redundant as it first seems

Hm I am confused about your point that we are being more cautious.

@SillyGoose Yes, the exact analogy to my classical mechanics example would be a time evolution operator that just depends on a single time parameter $t$ and then the composition is $U(t)U(t') = U(t+t')$

people will often write this like $U(t_1,t_2) = U(t_2 - t_1)$, time evolution depends only on the duration passed and not on when we started

this is true (both in classical and quantum mechanics) whenever your Lagrangian/Hamiltonian is time-independent

which makes sense: If the thing that determines the equations of motion never changes, it shouldn't matter when we start evolving things with it

oh I see. so encoded in the quantum notation is that we are caring about the actual start and end times, not just the difference

Ah okay I see this makes sense

Because if your hamiltonian (generator of time evolution) is time dependent then of course the way in which "time evolves" will depend on where in time you are, right?

but when it is time-dependent (e.g. the world explodes at $t_0$ and physics ceases to function but before that everything is normal) it matters a lot whether our attempt to time-evolve things crosses $t_0$ or not

XD

well so this is just a general property of time evolution, that we should really care about the actual times. but in the special case of time-independent time evolution generators, this detail can be disregarded?

yes

So then really is it not even similarities between classical mechanics and quantum mechanics? it is just we are applying the same broad function-based definitions (or trying to) to whatever theory of mechanics we'd come up with. e.g. start with definitions of a system and state. then you can talk about a map that sends states to the state at a later time (+ perhaps other properties); slap a name on it "time evolution map"

I do find it pretty strange that many QM texts feel the need to discuss this explicitly when almost no classical mechanics text discusses this

@SillyGoose yes, I would say all these properties of time evolution are so essential to what we mean by "time evolution" that they need to hold in every physical theory

it seems to be a popular UG course question as well :P as in use the time evolution operator to find the state at a later time for whatever purpose; i am thinking of when you have spin precession in a magnetic field

what does mechanics mean anyways? i always vaguely associated it with a theory on how things move or perhaps how things interact with other things

the "issue" is probably that the time evolution operator features a bit more prominently in QM; you don't get quite as far with "solve the Schrödinger equation and don't think about it" in QM as you do with "solve the equations of motion and don't think about it" in CM

@SillyGoose well, originally it was just the science of how tools (like levers and pumps) work

think Archimedes and his levers and fluids

but then it turned out levers just follow the same rules as everything else in the universe :P

lol

quite a different notion of levers we have in sakurai eh

in unrelated news i am getting a temperature control water boiler >:D i am slowly accruing everything i need to consistently make the perfect cup of tea

so what I'm hearing is you're a tea mechanic

maybe one day i'll rise to that title

I'm sure you will (and then relentlessly criticize the terrible tea all the other people drink :P)

if a tea is good then a tea is good :)

i think i shall work through some of halmos's linear algebra book this break. it is about time to properly learn direct sum, tensor product, spectral theorem, and so on :P

Did someone mention tea?

I can sense it XD