4:42 AM
Why is renormalized self energy taken to be zero at $p^2=m_{\text{physical}}^2$?
This is what I mean by renormalized self energy (in one loop QED)
The book (Collins) says that the procedure to renormalize is to "add" a counterterm so as to cancel the divergent piece, and then the finite piece is fixed by the condition that the renormalized self energy vanishes at $m^2=m_{\text{physical}}^2$. I don't understand why the latter is required
5:32 AM
I feel like this is a choice, but I don't understand the motivation behind it

2 hours later…
7:56 AM
Hello Everyone...

2 hours later…
9:27 AM
@DannyuNDos What does "overly concerned" mean?
@NairitSahoo it's just your renormalization condition; renormalization conditions are solely chosen so that certain properties or formulae come out nice, they are not physical conditions in any sense
for the self-energy, that it vanishes at $m_\text{phys}$ is the condition that at $p^2 = m_\text{phys}^2$ your bare and dressed propagators are equal
9:44 AM
@SirCumference yes, it is particularly confusing for German speakers because Impuls means momentum and the word for impulse is Kraftstoß (lit. "force push", sadly not the Jedi one) but German courses often not even teach this; frankly you never see the "impulse" again because it's a pretty irrelevant quantity
this is part of a larger phenomenon where the anglophone world has a strangely...formulaic approach to basic mechanics and gives a bunch of stuff names (other examples include "SUVAT equations", "work-energy theorem") that I've never seen anyone in German dignify with separate names (the two I mentioned are just equations of motion and energy conservation for us)
I think it's a result of a higher prevalence of standardized testing in English - if you want to do standardized testing on these topics, having separate names for these things makes it easier (and also provides an additional potential test question :P)
10:03 AM
for a given finite dimensional representation of $\mathfrak{su}(2)$, is there a unique (up to isomorphism) decomposition into irreps?
how could it not be unique?
10:15 AM
well I am obtaining the irrep decomposition in question by just simultaneously diagonalizing $S^2$ and $S_z$.
The eigenstates I get $\lvert j, m\rangle$ are literally different (as in the values in the concrete vectors are different) from another resource. I am using these eigenstates to write up some jump operators to put into a lindblad master equation. I wrote up the jump operators in the basis I got and the one in the resource. the two sets give different dynamics for a fixed initial state and hamiltonian.
I think this is to be expected because the state and hamiltonian are the same but the jump operators are literally different
but I also think that there shouldn't really be a physical difference due to using a different basis of simultaneous eigenstates of $S^2$ and $S_z$, so I am a bit confused about the difference in dynamics
that's a completely different question as the one you first asked about decompositions :P
for one, notice that even for 1-dimensional eigenspaces you always have a free choice of phase $\mathrm{e}^{\mathrm{i}\phi}$ in front of "the" eigenvector
so two different people can get different results for "the" eigenvector without disagreeing about the eigenspace itself (which is what matters in the abstract decomposition)
or, if you really just mean "the concrete values are different" because you chose a different basis to express them in...then that's just how linear algebra works and I don't know what the question is :P
as usual, you'll have to be much more specific and much less prematurely abstract for anyone to tell what's going on here

1 hour later…
11:38 AM
@ACuriousMind I mean, she seems to be fearing the batteries.
@DannyuNDos you've explained one vague term by another vague term :P
I mean, it's healthy to have some amount of respect for something that can explode and burn down your house, especially when handled improperly.
when you say overly concerned, you imply she is more worried than is appropriate. How does that inappropriate worry show?
11:51 AM
She said she won't buy wireless stuffs anymore.
that doesn't seem to have anything directly to do with Li accus :P
Yeah, she doesn't know what's Li-ion and what's not.

2 hours later…
1:36 PM
@ACuriousMind Are parastatistics due to the spin-statistics theorem having extra projective reps of the symmetric group in 2D, or is it due to something else entirely
nlab seems to imply that it is unrelated to spin statistics
@Slereah what are parastatistics :P
Anyon stuff
you sure?
Quasistatistics I mean
Or one of the other one idk
because I've never heard the term outside of its Wiki article and that article seems to imply parastatistics are not anyonic statistics
1:41 PM
The quantum objects of many layers
I know nothing about either "parastatistics" or "quasistatistics" :P
Well then just anyons
What is this, an Abbott and Costello routine
anyons are special to 2d/2+1d because the universal cover of the rotation/Lorentz group there is an infinite cover instead of a double cover
but you can also exhibit this difference classically by looking at the braid groups, the original paper on anyons is pretty good on that topic
The nlab is seemingly implying that it is unrelated to the spin statistics theorem :
> On the other hand, anyonic braiding is conceptually different from boson/fermion statistics – if it were on the same footing then the spin-statistics theorem would rule out anyonic braiding.
@ACuriousMind yes they are different things and parastatistics are much more niche
1:44 PM
@Slereah I think that depends very much on what you think the spin-statistics theorem says :P
they're not being super precise in what they say about that
even stating the theorem for arbitrary dimensions isn't that straightforward (you can't talk about "half-integer" and "integer" spin in general dimensions, at least not without defining "spin" first)
Which seemed weird to me because I vaguely remember reading that it does indeed derive from it, although I need to find those papers again
https://en.wikipedia.org/wiki/Fine_structure#:~:text=The%20Darwin%20term%20changes%20potential,demonstrated%20by%20a%20short%20calculation.

Why in this article, we say that the relativistic energy expression is only equal to the kinetic energy? In most notations that I've seen it's $E=\sqrt{m^2c^4 + (pc)^2}$
2:50 PM
@imbAF hi. the rest energy would only contribute a constant term to the hamiltonian in this conext, maybe thats y they only care about kinetic energy
also, theyre doing relativistic corrections to the non rel Hamiltonian, and the non rel Hamiltonian only has a kinetic energy term. so they're correcting the kinetic energy
3:53 PM
hello
4:07 PM
yo
4:22 PM
@RyderRude and why it has only kinetic terms the non-relativistic Hamiltonian? The particle, is exposed to the potential of the nucleus
and It also given as $H=\frac{p^2}{2\mu} + V(\vec r)$
So the non-relativistic Hamiltonian has the potential present
And the Hamiltonian is a representation of the energy of the system, which has kinetic and potential terms
But for some reason when we consider the relativistic dispersion relation, we only consider the kinetic term
5:09 PM
5

Given a Dirac spinor $\psi$, its "charge conjugate" spinor is given by $\psi^c = C\psi^\ast$, where $C$ is a charge conjugation matrix defined by a certain convention (e.g. $C^\dagger \gamma^\mu C = -(\gamma^\mu)^\ast$). Meanwhile, the anti-particles are associated with the Dirac adjoint (or Dri...

And I have the same doubt as the user Bcpicao's last comment.
What is the answer to that? $\psi^c$ or $\bar{\psi}$: which represents the positron field?
@NairitSahoo the question is ill-formed; the spinor fields in QFT are operators after all, not states, they do not "represent particles".
if you can write down a technical definition of what it means for the field to "represent the positron", then the answer should be obvious
@ACuriousMind Is there any sense in which physicists associate one particular particle and not its antiparticle to $\psi$ $\bar{\psi}$ or $\psi^C$? Because from the mode expansion I can see that both kind of particles are being produced with any of these field operators, right?
yes (to the second question), which is why I find the question of "which of these" is the particle and which the antiparticle field strange and meaningless
@ACuriousMind Good point, work-energy theorem is another term I've hardly ever used since freshman year
You're probably right regarding the standardized testing explanation. Kind of a shame though that we're overcomplicating introductory physics classes with an overload of rather useless terms
the other thing I found weird about physics education is how noninertial reference frames are only taught in second year (at a lot of universities, at least)
@ACuriousMind I found this question when I had the following doubt: EW theory the interaction term is $i \bar{\psi} \gamma^\mu D_\mu \psi$ where $\psi$ is a column matrix of a lepton and its corresponding neutrino. How exactly do I restrict the presence of phenomenologically unallowed left handed antineutrinos?
5:24 PM
like that is a very useful thing to know. No reason it shouldn't be in intro physics classes
it's not like fictitious forces are that crazy of a concept to explain
I mean... in my $\psi$ there is a left handed neutrino field. But it can produce (left handed) antineutrinos too, right?
@NairitSahoo I don't understand what you mean - when $\nu$ is purely left-handed, then $\bar{\nu}$ is purely right-handed
in that sense the anti-particles are associated with the Dirac conjugate
@ACuriousMind No you just said that whenever I see a $\nu_L$, since they are field operators so they must produce/annihilate both particles and antiparticles
In that sense, there would be states which represent left handed antineutrinos, right?
no
I didn't say anything about a $\nu_L$, the context above was a full Dirac field
also if you look at the mode expansion for the spinor field, it has the creation operator for one type of particle and the annihilation operator for the other inside of it
the annihilation operator for the anti-particle of course has the opposite handedness to the particle it annihilates (since the combination with the particle needs to result in the "non-handed" vacuum)
there are no left handed antineutrinos in your expression if you only start from a left-handed neutrino field $\nu_L$
just work out the expressions
@NairitSahoo now that I think of it, I revise my answer to this to "no": It is indeed the case that $\psi(x)$ creates particles and $\bar{\psi}(x)$ creates anti-particles from the vacuum
5:44 PM
@ACuriousMind What about the $\bar{\nu}_L$ in the Lagrangian?
the $\nu_L$ has the creation operator for the l.h. neutrino and the annihilation operator for the r.h. antineutrino. The $\bar{\nu}_L$ has the creation operator for the r.h. antineutrino and the annihilation operator for the l.h. neutrino
i have a stupid question:
so, hydrogen's ionization energy is 13.59844 eV, meaning that if you give the electron that much energy, then the electron can escape to infinity right
but that's not actually the "amount of energy" that the electron has right
(that's related to how you define the "zero energy", i suppose)
so how much energy does the electron really have?
i guess i first need to clarify what kind of energy i'm talking about
yes :P
what does the energy of the wavefunction mean?
is it a physical quantity?
what is "the energy of the wavefunction"?
is any energy a "physical quantity"?
energy is just a number, see e.g. physics.stackexchange.com/q/138972/50583 and its answers
5:59 PM
it's the eigenvalue of the S. equation right
sure, that's just a number
oh really
btw is that related to the de Broglie equation?
also does the Schrödinger equation have momentum?
and then on top of all that there's Einstein's equation that relates E, p, and m0 right
Einstein is relativistic, this is all non-relativistic
don't throw together incommensurable things
I also don't know what it means for an equation to "have momentum"
as in, is there a quantity that is usually associated with the wavefuction that we call "momentum"
sure, both position and momentum are observables in QM
6:10 PM
and they're related by some formula right
are you trying to learn QM without actually reading any intro to QM book? Any QM book should discuss this
as in, the usual 1s orbital is the position wavefunction right
hmm, lemme have a look
like, the Heisenberg uncertainty principle is extremely famous and is a relation between (the uncertainties of) position and momentum
do you have a recommendation?
6:14 PM
from wiki it looks like they are two operators that satisfy a commutator equation
$[\hat{x},\hat{p}] = i\hbar$
ok thanks
yeah if that's the first thing you hear about it, it is not useful to answer specific questions about the hydrogen atom, you need to learn the basics of QM and operators first
well, from what little i know, basically that means you can apply the momentum operator to the 1s wavefunction, and then if you take the absolute value squared then that gives you the probability density function of the "measured" momentum right
almost :P
but there's enough wrong with it that you wouldn't get the right result with this, that's why one needs to learn the basics firmly first
ok, thanks
it first started out with electromagnetism right
and you can apply it to individual particles right
and so does that mean, when the electron sits in the 1s orbital then you can't ask any relativistic question about it, but once you isolate it from the atom then you can start to ask those questions?
no, not really
of course the whole world is "really" relativistic and you can do relativistic quantum mechanics (quantum field theory, really)
just the normal QM we learn first and use all the time is to that like Newtonian mechanics is to special relativity classically
6:27 PM
ok
right
so,
special relativity is ok, it's the gravity part (of general relativity) that is the verboten area right
no, not really :P
the way in which QFT and GR are incompatible is a rather technical affair; at low everyday energies there is not really any problem
ok
@ACuriousMind if the newtonian mechanics is the low-speed approximation (v -> 0?) to special relativity, then what is QM in relation to QFT?
...the same?
why worry about this if you don't even yet have a proper grasp of how QM works?
you couldn't understand any of the actual technical answers anyway
motivation :P
6:43 PM
surely not knowing would be even more motivation
@ACuriousMind ok, but the difference between these energies (i.e. the eigenvalues) of the electron orbitals do correlate experimentally with the observed absorption spectrum of the hydrogen right (i.e. by E=hf)
@LeakyNun I never said anything to the contrary, I just refused to commit to the nebulous philosophical notion of a "physical quantity"
i wasn't implying you did, i was just making sure
and i wasn't being philosophical, i was asking if this (absolute) quantity is something you can observe
How does one
observe
@LeakyNun the question of what it means to "observe" a quantity is again philosophy :P
6:49 PM
well the kinetic energy is E=0.5mv^2 and you can measure the v
Can you
well you can record the location at t1 and location at t2
but in the same vein as the energy differences showing up on the absorption spectrum, the total binding energy of course also shows up during absorption - above that energy you knock the electron off the atom, i.e. ionize it
that's why that energy is also called the ionization energy of hydrogen
7:05 PM
When we talk about the FS and HFS, how exactly we force the splittings? Or make the finer splittings apparent ?
Normally we have the unperturbed Hamiltonian, and via the absorption or emission spectrum we can detect the different energy values, is that correct?
So how can we "force" the finer splitting present in FS consideration?
@imbAF the split into "unperturbed" and "perturbation" here is entirely artificial; obviously reality has the full Hamiltonian all the time, you can't "turn off" relativistic corrections in the real world
Exactly
so how does one become more detailed
I mean, how do you make the finer splitting visible
no, you didn't understand what I said: The splitting is there all the time
So all the time you get the full splitting
you don't need to do anything special to "make it visible" except to do some experiment that's sensitive enough to detect the small differences
7:11 PM
I mean I lack the vocabulary to say what I want but you get me
@ACuriousMind Ahaaa
So depending on the experiment we can show these small details
in fact we measured the fine structure splitting long before we had any notion of quantum Hamiltonians or perturbation theory of the hydrogen atom
Ok
One additional question on this topic
Say we perform an experiment in which you can only get the spectral lines which correspond to the unperturbed Hamiltonian
(funnily enough, that was also Michelson and Morley, of the fame of trying to detect the aether)
Is my statement accurate, before I go on?
@imbAF I don't know what that means
7:13 PM
Meaning an experiment where you cannot detect the small differences corresponding to the FS or HFS
again, the split between "unperturbed" and "perturbed" is entirely artificial in this case, you cannot turn off the perturbation in reality here, since you can't just shut off relativity :P
I know but you said
depending on the experiment we perform we can detect these small changes
shifts
can i rephrase it as "you just haven't zoomed in enough"?
@imbAF sure, you can have experiments that are not sensitive enough to see the difference reliably
Yes, and we consider initially the spectral lines which correspond to the degenerate energy eigenvalues
7:15 PM
but I want to be clear that this is conceptually different from an experiment that just works with the unperturbed systems, it is just the case the error you incur by neglecting the perturbation is irrelevant for the case at all
Yes I fully understand
That the full hamiltonian is present all the time
but depending on the experiment and precision
one can either consider or ignore this extra small/ finer changes
Well I had a question which doesn't align with what we discussed really
Would it be wrong for me to say
When performing such an experiment where the small changes corresponding to the FS are present, one can say that the degeneracy of the energy level is lifted. Of course it's not lifted because we did something different, we just performed a more accurate experiment.
Normally nothing is lifted. But could one make this simplistic perhaps inaccurate argument
I don't really know what the question is; if we're being accurate there is nothing to lift because the degeneracy was always there, we just couldn't measure it; what exactly do you want to know about that?
I don't know if it's correct to ask the following:
When ignoring nucleus spin:
I consider that the degeneracy of an energy eigenvalue = nr. of electrons on that energy level. i.e n=1 2 electrons so max degeneracy is 2, n=2 8 so max. degeneracy is 8.
For n=1 if there's such an experiment to split this level into to sub-levels, where each electron occupies one, then the nr. of splittings = nr. of electrons.

Now when you consider proton spin for H-atom, the ground state is 4 times degenerate. Let's say that there's such an experiment that can fully measure the degeneracy, you'd have 4 spectra
7:33 PM
no, all that sounds rather confused
you're not measuring "the degeneracy"
the whole point of (hyper)fine splitting is that the levels are not actually degenerate
I understand that
@imbAF sorry, but if you do, you did a really bad job of expressing that :P
you keep claiming that you understand things when you just said stuff that sounded as if you don't
Well there's nothing to lift anyway
the splittings are there
it's just that depending on the experiment we can witness those or not
also I have no idea where the 4 for the ground state comes from, even with hyperfine structure the n=1 has only 2 states
@ACuriousMind That is why I said "I don't know if it's correct to ask the following:" after you properly articulated the split into "unperturbed" and "perturbation" here is entirely artificial; obviously reality has the full Hamiltonian all the time, you can't "turn off" relativistic corrections in the real world
@ACuriousMind Hold on, with nuclear spin you'd have $|n,l,m_l,m_s,m_I$
so you have 4 whatever that is called
7:36 PM
no, that would mean you could have 4 electrons in the innermost shell, which would be news to most chemists
n=1, l=0,m_l=0, m_s=\pm 1/2 and m_I=\pm 1/2
@ACuriousMind Exactly, it's not possible
?????
sorry, I have no idea what you mean then
It's better to show you the text, so you wont think Idk what I am talking about
there is no fine splitting of the n=1 level, and the hyperfine splitting is into two distinct levels
Yes
and still it was stated a 4 fold degeneracy
7:41 PM
ah, one has to be careful here: this 4-fold degeneracy is for the electron + proton
this is a different kind of state than when we just talk about the electron
Since I am considering proton spin, then the state/eigenstate is of the atom as a system
not of the electron, which is what we mostly consider, but call it eigenstate of the Hamiltonian of the H-atom, which, we already discussed
does a proton change energy by absorbing/emitting a photon?
@LeakyNun protons usually don't absorb or emit photons; what process do you mean here?
well since you guys are discussing the proton as well
i was just wondering how we can measure that
@ACuriousMind In this new scenario, what I asked does it make sense or no?
And just to reiterate I have fully understood FS and HFS that we discussed a bit earlier
7:46 PM
@imbAF I don't really know what you're after, sorry
you didn't say anything wrong in that case but I have no idea what you want to know from me
Do you agree that the ground state is 4 fold degenerate?
@imbAF of the electron + proton system? Yes, of course
Is there an experiment that can showcase this ?
not really
Would it make sense to, in theory consider such an experiment?
7:48 PM
I don't know how to consider an experiment "in theory" that doesn't exist :P
3
Ok
Since it's not possible then that ends my question
@LeakyNun measure what? The proton? just do a Rutherford and shoot it at stuff
@ACuriousMind 😂
I feel modern students are perhaps insufficiently taught that we experimentally found all the building blocks for QM and atomic theory before we knew how they fit together
is that to me or to the other person
7:52 PM
no one predicted the proton or the fine structure splitting and then went looking for it, people found these things rather unexpectedly in their labs and then someone had to figure out what the heck was going on
@LeakyNun both and neither, I'm just ranting :P
@ACuriousMind well yeah we're taught that the scientific method is to make predictions first and then do experiments to validate the predictions; in some sense i guess this is exactly backwards of the historical process you have described
well, that's not wrong (Popper wasn't an idiot), it's just missing the initial step where the theory by which we generate the prediction is motivated by some very specific observation, and then vindicated by its predictions explaining a bunch of other observations, too
I believe the best way would be to consider a specific case, try to solve it and then generalize the solution, so that you can adapt it to any other possible particular scenario . But this is somehow to difficult for people to do /s
8:11 PM
The scientific method is a lie
It's all chaos
@ACuriousMind there's also the realistical element of that - where negative results are not published 🙃
8:45 PM
i have a comparison comparing string theory to the Huygen's theory of light. Huygen's theory models specific aspects of light, and there is an underlying theory (Maxwell's theory) which models general things about light. Similarly, string theory models a specific aspect of the universe (the S matrix), but the underlying theory to that is completely missing rn.
:66104364"the $\nu_L$ has the creation operator for the l.h. neutrino and the annihilation operator for the r.h. antineutrino". I didn't understand the "annihilation operator for the r.h. antineutrino" part
Why r.h.?
it's how it works if you do the math :P
I mean mode expansion is just a one liner, right?
on a high level, I already gave the plausibility argument: $\nu_L$ is left-handed and:
3 hours ago, by ACuriousMind
the annihilation operator for the anti-particle of course has the opposite handedness to the particle it annihilates (since the combination with the particle needs to result in the "non-handed" vacuum)
Well I just can't hope to right that one line.
@ACuriousMind How do you send texts like this? What you have already sent and with a big | in white in this way
8:57 PM
you just post a link to the chat message and nothing else, the full text of the message I sent there is just https://chat.stackexchange.com/transcript/71?m=66104320#66104320
oh thanks
@ACuriousMind How do u know that the annihilation operator is left handed?
...did I not explain this precisely in the message you responded to? :P
> since the combination with the particle needs to result in the "non-handed" vacuum
@ACuriousMind But then I am assuming that the antiparticle is right handed
I thought that using the fact that annihilation operator is left handed you deduced that the antiparticle must be right handed
again, you can do all the tedious math here and it works out the right way
I said explicitly this is just a plausibility argument
@ACuriousMind I mean... what's the math. I am usually used to write down the mode expansion in one line
What math should I do? Any hints?
Should I start with a Dirac field mode expansion and act with projection operators or something?
9:02 PM
@NairitSahoo what do you think "left-handed" or "right-handed" means?
Also, I'm not really sure what there is to derive: We started from the assumption that $\nu_L$ is a left-handed spinor, no?
@ACuriousMind Exactly what I was saying. I was asking what math should I do. I usually write this in one line. But I never had to face the issue of handedness before this
then I have no idea what the question is
@ACuriousMind When we write down an equation, there is some logic to it, right? I don't understand that logic
I don't know what that means, either
what equation
@ACuriousMind mode expansion of $\nu_L$
9:06 PM
Why is there a _right_ handed antineutrino in it? Don't you see we are going in circles?
You say that we have to have a non-handed vacuum. Then?
there is no "right-handed anti-neutrino" in it
there is the annihilation operator for the r.h. anti-neutrino in it
@ACuriousMind Yeah, sorry meant that. Why for that and not for l.h. anti-neutrino?
because the associated anti-particle field $\bar{\nu}_L$ is a right-handed Weyl spinor?
there's just no field here that could create the left-handed anti-neutrino
@ACuriousMind How do I see this?
9:09 PM
if you have a massless Weyl spinor, you have that spinor with its handedness and its anti-particle with the opposite handedness
@ACuriousMind Yeah this this... How do I see this?
@NairitSahoo if you unpack the definition of the Dirac conjugate $\bar{\dots}$, you can see it inverts handedness
it's just algebra
@ACuriousMind Do you mean that if $\psi$ is L.H. then $\bar{\psi}$ is R.H.?
yes
Then I will try to prove it\
Ok... yeah I see it
then?
oh
ok
Now I see it