The h Bar

imbAF

20:21

When we considered the canonical quantization of the real scalar field, one of the requirement was that the following commutation relations should be valid:

$[\phi(x),\Pi(y)]_{x_0=y_0}=i\delta^{(3)}(\vec x- \vec y)$
$[\phi(x),\phi(y)]_{x_0=y_0}=0$
$[\Pi(x),\Pi(y)]_{x_0=y_0}=0$

These were just given, no explanation how we arrive here, or why this is necessary or even what it means to consider the commutation of two field operators. At least in QM, the commutator was a way to tell if the physical quantities could be measured at the same time, or if the operators had a joint basis. But here …

ACuriousMind

if you understand why we do $[x,p] = \mathrm{i}$, you understand this. It's just the field version of the Poisson brackets turned into commutators.

imbAF

I understand the consequence of it in QM (not how we come about it), but in QFT is there an interpretation

ACuriousMind

I don't know how often I can say that the operators are still operators like in QM

so yes, operators that commute still have joint eigenbases and still can be measured simultaneously (if they're observables) etc.

imbAF

Ok but $\phi$ is an operator and it assigns an operator value at any spacetime point. So that is not an observable

At least that is what I believe

or suspect

ACuriousMind

I don't know what you're trying to say. An observable is a self-adjoint operator, my caveat is merely because e.g. a complex field has $\phi \neq \phi^\dagger$ and so is not self-adjoint.

imbAF

20:30

Ok, I'll be blunt. I don't know what evaluating $\phi(x)$ at a spacetime point should mean? Or should be? Is it a scalar, a complex value

I don't know

ACuriousMind

(At the physical level of rigor), the quantum field is simply an operator-valued function. That is, for each $x$, $\phi(x)$ is an operator.

imbAF

Ah, ok

And what would $\langle 0| \phi(x)|0 \rangle$ be?

Like what is meant with the expectation value of an operator-valued function?

Is there some physical significance to it ?

ACuriousMind

I'm not sure where the problem is. $\phi(x)$ is an operator for fixed $x$. So that's the expectation value of the operator $\phi(x)$ in the state $\lvert 0\rangle$.

imbAF

But isn't it different evaluating $\phi(x)$ at some point x^mu and calculating the VEV at some point x^\mu ?

ACuriousMind

The physical significance depends on what the physics significance of the field is to begin with. If it's the electric field $\vec E(x)$, the meaning should be rather obvious. If it's the Higgs field, it's less so :P

imbAF

20:34

They are the same thing?

ACuriousMind

@imbAF of course it's different. $\phi(x)$ is an operator, the VEV is a number.

expectation values of operators are numbers, again, it's just QM

imbAF

@ACuriousMind Yes but you say:

by "that"

you mean what?

ACuriousMind

I have no idea what you mean. Which 'that' are you referring to? Which message?

imbAF

@ACuriousMind This one

ACuriousMind

"that" refers to your $\langle 0\vert \phi(x)\rvert 0\rangle$

imbAF

20:38

Ok, but you did say that the physical significance of VEV depends on the field

i.e $\vec E(x)$ it would be the electric field strength at that point

if I am not wrong

ACuriousMind

I mean, it depends just like the significance of the expectation value of an operator in ordinary QM depends on the significance of the operator

there's literally no difference

imbAF

Yes, but of course I am assuming a case, where such a thing is possible

In QM i am not considering an operator, that doesn't represent a physical quantity

that ofc wouldn't make sense to talk about

ACuriousMind

you could

Tobias Fünke

oO

ACuriousMind

no one forbids you to e.g. talk about the expectation value of the self-adjoint operator $x+p$

imbAF

20:40

But because QFT is more abstract, and QM is like a reference point, I tried to give a meaning to expectation value of the field operator.

ACuriousMind

it's rarely useful, but the formalism allows it perfectly fine

imbAF

x+p ?

I have never encountered that

Tobias Fünke

@imbAF Roughly speaking, QFT is just QM with some additional postulates/an additional structure, namely to make QM compatible with SR

imbAF

what physical quantity would that be? sum of position and momentum O.o

Tobias Fünke

@imbAF it was meant as an example

ACuriousMind

20:42

@imbAF my entire point is that that is an example of a self-adjoint operator that usually has no physical meaning

yet it is formally an observable, and the formalism of QM allows you to talk about its expectation values, eigenvectors, etc.

imbAF

but for whom you can find an expectation value. I see

Ok I didn't know that

ACuriousMind

there's not really anything to "know" here. If you understand how QM works, you can understand that all those notions formally make sense for any operator

imbAF

but yeah, when I talk about significance of expectation values, I, ofc am cosidering something measurable, standard to say so,

@ACuriousMind True, but I make a distinction in QFT when it comes to operators. Namely, you can have all sorts of operators, who might represent a physical quantity i.e Hamilton operator in QFT or not, and field operators. This distinction doesn't exist in QM

I don't know if it is right to do

Or if it makes sense to do so

ACuriousMind

I have no idea what this "distinction" is supposed to be

I just gave you an example from plain QM of an operator without obvious physical significance

the difference is only that QFT forces you to perhaps consider such operators a little more often

imbAF

It is because, in my view, field operators, generate excitations of the field

And i consider them unique in that regard

ACuriousMind

20:46

we're just back at the point where I tell you to stop looking for direct physical meaning for every intermediate expression that may appear somewhere

the VEV of a scalar field is a really pointless example anyway because it will always be zero or a constant that you then set to zero by some tricks

imbAF

Ok, if I were to ask you what is the difference between the hamilton operator and field operator in QFT, for some theory, how would you word the distinction or difference between the two?

ACuriousMind

I don't understand the question. What's the difference between the Hamiltonian and position or momentum?

Allie

Hi all

I made some progress in understanding my resesrch :)

imbAF

If I were to classify operators, as caring meaning, i.e representing a physical quantity, then they do not differ in that regard.

If you were to list an operator, who represents no physical quantity, I would point that as a distinction between the two

ACuriousMind

I would suggest you stop trying to classify operators that way, at least at the beginning. The meaning of the operators becomes clear as you see what they are used for in the formalism

Tobias Fünke

20:49

@Allie nice :) you want/are allowed to share a bit?

imbAF

@ACuriousMind Then should I for the moment disregard their "meaning"?

ACuriousMind

at least in the sense you seem to use the term, yes :P

Allie

@TobiasFünke sure!

You know what OFDFT is?

imbAF

Is it because it's faulty ?

What is the issue with that?

Tobias Fünke

imbAF: The Hamiltonian is the Hamiltonian. The field is the field. The latter are used, for multiple reasons, to build observables, as the Hamiltonian. Roughly: The fields are the "building blocks" in QFT

imbAF

20:50

I am curious to know

Allie

DFT but not kohn sham basicallyb

Tobias Fünke

@Allie only superficially

yeah

ACuriousMind

@imbAF it's just not useful

Allie

So you would need a functional that is solely of density

KS ofc uses orbitals to get the non-interacting KE

imbAF

@ACuriousMind Can you elaborate ?

ACuriousMind

20:51

you are so far away from seeing how the entire actual machinery of QFT works that it's impossible to tell you the meaning of anything in it; you have to learn QFT to understand QFT

imbAF

@ACuriousMind And that would mean, a pure mathematical approach?

Allie

So one way to do OFDFT is to try to get approximations for the non-interacting Kohn-Sham KE

T_s[n] + Hartree[n] + XC[n] is ir total energy functional

That T_s would usually be the sum of the KEs of the KS orbitals right

DIRAC1930

@imbAF I think one of the only ways to learn QFT at least conceptually is to learn 2nd quantisation from L&L 3, and then follow L&L 4 to learn about the quantisation of the electromagnetic fields, Klein-Gordon fields, and then the Dirac fields. Ultimately, one has to make the leap that Pauli made in 1934 (which is eluded to in L&L 4) regarding the iterpretation of $j_\mu$.

ACuriousMind

@imbAF not exactly; but you are stuck at "what does the VEV of a field mean" when it will become perfectly evident what it means once you learn that a) only non-scalar field have non-zero VEVs anyway and b) non-zero scalar VEVs lead to symmetry breaking

but to understand the significance of that you have to learn all the steps in-between

Tobias Fünke

@Allie yes, I am listening :) I understand so far.

DIRAC1930

20:54

There is not much more conceptually to rel QFT compared to non-rel QFT. One has a field operator, that obeys certain field equations. In the non-rel case, that field equation is the Schrodinger equation, in the rel case, it is the Klein-Gordon, Dirac etc field equations

Allie

Okay, and so obviously getting T_s[n] is difficult just from n

DIRAC1930

The only difference is that one has both these positive and negative modes in the field expansion

Tobias Fünke

@DIRAC1930 Well, this is only partially true. Relativistic QFT is more complicated; for example, due to the lack of a well-defined position basis

@Allie yes

Allie

So one way you might think about it

DIRAC1930

However this is resolved by interpreting the coefficients correctly (as eluded to in L&L 4)

Allie

20:56

Is take the von Weiszaker functional

You aware of that is?

imbAF

@ACuriousMind The two things that you listed, I believe I don't understand them. But I want to. So idk, just reading different books and notations, will make me reach the point of understanding this? It feels like you need 1.0 across every BA physics, to be able to understand.

Tobias Fünke

@Allie yes

DIRAC1930

@TobiasFünke Okay then just use momentum basis in non-rel QFT

ACuriousMind

@Allie I assume you mean Weizsäcker ;P

Allie

It basicallt is the KE of a system of bosons with, of course, wave function sqrt(n(r)/N)

imbAF

20:57

@DIRAC1930 His way of treating 2nd Qunatization is messy or difficult to understand. But I get your point

Allie

He was a nazi ill spell his god damn name wrong if i want

DIRAC1930

@imbAF Follow a different source so you can at least follow the latter pages

You probably only need the "one-particle" operators

Allie

Anyways, so you can actually break up T_s[n] into T_vw[n] + V_P(r), where that v_P is the so called Pauli potential. It basically is the potential that will recreate the fermion system

So that you dont get KE of bosons you get KE of fermions

imbAF

@DIRAC1930 It's just the time constrain I have.

DIRAC1930

@imbAF I understand, I didn't have any time to do any of this when I was first learning QFT

Allie

21:00

So my research is all about trying to find (approximations of) v_P

Tobias Fünke

@imbAF my two cents: get one or two sources, study them step by step. You should be fine

@Allie interesting

imbAF

@DIRAC1930 I need to create time for 2nd quantization

Allie

The way Im looking at it is very different though

Im looking at it in terms of the so called exact elevtron factorization

Analogous to the nuclear electron factorization

Tobias Fünke

ah

yes

interesting! :)

ACuriousMind

@imbAF I have no idea what grades have to do with it, but yes, my position is that a lot of QFT becomes only clear in retrospect, once you have seen it at least once in its entirety until you arrive at the practical computation of scattering amplitudes for QED at least and then can appreciate different setups and approaches for it

Tobias Fünke

21:01

I am sure I will read some day a paper of you!

Allie

But instead you factor the N electron wf into a one electron wf and an N-1 electron wf that depends on what the one electron coord os

Maybe lol

Tobias Fünke

interesting

Allie

And it gives a really interesting new way of viewing the Pauli potential

Tobias Fünke

and regarding OFDFT: is it somewhat related with RDMFT?

because you could get the kinetic energy from the 1RDM

ACuriousMind

@Allie this is a horrible argument; do you really want to potentially confuse some guy named Weiszaker with a Nazi? :P

Allie

21:02

In the view of the EEF the pauli potential is not the correction from a boson system to a fermion system, but its a potential that actually captures the effects of the N-1 electrons around

imbAF

@ACuriousMind Well at least is good to know that If I keep blasting my head against the wall, one day I'll see the other side..hopefully. ACM I am banking a lot of trust at Weingand, when i start in Mars.

Tobias Fünke

@Allie hmmh interesting

Allie

Im really not sure, it honestly could be &tobias

Tobias Fünke

ok

Allie

So for example whaat i learned today is

In the view of the EEF, one of those terms in the Pauli potential is actually a correction to the one electron KE that accounts for the fact that if the environment (N-1 electron wf) needs to move in order for you to move the one electron, then there’s a “resistance” to movement

So the more the N-1 wf changes wrt changes in the 1 electron coordinate, the more reisstance you see, and that part of the potential is directly proportional to that resistanxe

You define the change in the N-1 electron wf by the fubini-study metric

Tobias Fünke

21:07

do you have a nice reference, by chance?

I think I want to learn more about the basics of OFDFT

Allie

Yes!

Tobias Fünke

hehe

Allie

OFDFT or EEF?

pubs.acs.org/doi/abs/10.1021/acs.chemrev.2c00758 Heres a good OFDFT paper

Tobias Fünke

both ;)

Allie

EEF: journals.aps.org/prresearch/abstract/10.1103/…

One day i will be as smart as u tobias

Tobias Fünke

21:10

thank you, much appreciated! I'll have a look tomorrow

@Allie haha, I bet you are much smarter already

Allie

Uhhh idk about that. Im still going through griffiths electrodynamics lolol

Tobias Fünke

I hate ED lol

Allie

Really

I dont mind it so far

Tobias Fünke

I am not good at it, no.

Allie

Not my favorite thing but

Tobias Fünke

21:12

Yes, it is for sure interesting!

Allie

I need to just catch up on my physics knowledge

I basicallt started learning physics like a year ago at this point

I have so much i need to cover and want to cover before i start my phd

Tobias Fünke

Today I dreamed that I met a former colleague, who recently got his PhD, and now he is offered, a few months after graduation, a professorship in NYC lol

@Allie yeah, you will do. You will also learn on the fly :) no worries

you will be fine

Allie

I wass about to say ong thats awesome

And then i read “I dreamed”

Tobias Fünke

xd

weird dream

imbAF

22:09

@TobiasFünke One question. When we consider this:
$[\phi(x),\Pi(y)]_{x_0=y_0}=i\delta^{(3)}(\vec x- \vec y)$.
What is the meaning of the commutator being zero?

Tobias Fünke

what is the meaning of $[X,P_y]=0$?

I don't know what to say other than: "it means both operators commute"

imbAF

which in QM means that measuring one does not influence the other

the same is the case in QFT?

measuring a field???? doesn't influence the other???

Operators commute is a mathematical description of what happens anyway

ACuriousMind

I don't know why you keep asking the same question, acting as if QFT was so completely outlandish. It's just operators, the same as in QM. If the operators $\phi(x)$ and $\Pi(y)$ commute, it means exactly the same as when two operators in QM commute.

imbAF

I could elaborate, but we'd end up in the same discussions

DIRAC1930

22:25

@imbAF Are you aware of Tong's lecture notes in QFT?

They might be of help

imbAF

@DIRAC1930 I am not

DIRAC1930

damtp.cam.ac.uk/user/tong/qft/qft.pdf

imbAF

canonical quantization ?

DIRAC1930

Thats on page 21

imbAF

Yes I am reading it

From what it says, my problem, stems from the fact that we never did, as far as I remember, canonical quantization from classical mechanics to QM. We started with the Heisenberg inequality

So I never got to see poison brackets changing to commutators and all that

But thanks

Allie

23:15

Poisonous brackets

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