3:54 AM
in my extremely naive opinion, it doesn't hurt to be reminded that all models are just that: models

2 hours later…
6:08 AM
@ACuriousMind I know electrons are weird but is there some group theory reason behind why their relative acceration is naively parity invariant?

1 hour later…
7:16 AM
@MoreAnonymous I'm not sure what you mean by "relative acceleration"

7:34 AM
@ACuriousMind Would it be okay for me to post the link to my latest question on PSE?
Also relative acccerlation is nothing more than the eigenvalue of acceleration of particle 1 - eigenvalue of acceleration of particle 2

@MoreAnonymous I can just look at it :P

@ACuriousMind Coolz :)

@MoreAnonymous As so often, you do a lot of complicated stuff in your question that you don't really explain any motivation for. The acceleration operator is (proportional to) $-V'(x_1 - x_2)$. This is invariant under parity if and only if $V'$ is invariant under parity. I'm not sure what all the other stuff in your question is doing, or what never-defined things like $a[\psi]$ are supposed to be

Yes , that potential is partity invariant
@ACuriousMind Sorry should have mentioned that
@ACuriousMind Thanks for pointing that out ... I've improved the question

@MoreAnonymous you're not really making things any clearer - "acceleration as a function of $\psi$" to me would just be the application of the operator $a$ to the state $\lvert \psi\rangle$. But you seem to think $a[\psi]$ is still an operator and it is entirely unclear what the $a_1,a_2$ are on the other side of the equation you write down for it, or why there would be the position basis coefficients $c$ involved in it

7:51 AM
@ACuriousMind So let me go into some detail .. We know relative acceleration of particles $1$ adn $2$ at postion $x_1$ and $x_2$ are classically given by $"x_{1} - "x_{2}$ you can see the operator I propose will give this answer

I'm not sure why your relative acceleration is not just $\hat{a}_1 - \hat{a}_2$

@ACuriousMind What happens if my wave function is in a superposition of postion?

then you're probably also in a superposition of eigenstates of rel. acc.
but this is QM, so that's nothing bad

@ACuriousMind Agreed

So what are you doing in this question? If you want a definitely value for rel. acc., you do what you do with any observable and just measure it, the result will be an eigenstate of rel. acc.

7:56 AM
@ACuriousMind Okay let my way function be $\psi\rangle = \frac{1}{2^{1/2}}(|x_1,x_2 \rangle + |x_1 + \delta,x_2 + \delta\rangle)$ the squareroot of $2$ will give you the wrong answer

you really will have to be more explicit about what you mean
also that's not really an admissible wave function, that's like the sum of two $\delta$-functions, which is not a square-integrable wavefunction
if you want to avoid running into problems with the ill-definedness of position eigenstates, you should demonstrate whatever you mean with a proper wavefunction, e.g. a Gaußian

@ACuriousMind Apply $a_1 - a_2$ on my $\psi$ you will get the relative acceleration wrong
@ACuriousMind You can take the momentum eigenket to make the same point

@MoreAnonymous the momentum eigenkets are similarly ill-defined
they're plane waves in position space, which aren't square-integrable either

@ACuriousMind Any ket which is in superposition of position will suffice

@MoreAnonymous I don't know what "wrong" means. Ignoring issues with the position kets, Your state is an eigenstate of $V(\hat{x}_1 - \hat{x}_2)$ with eigenvalue $V(x_1 - x_2)$

8:02 AM
@ACuriousMind Okay let $||\psi \rangle = \int_{-\infty}^\infty \int_{-\infty}^\infty c_{x_1,x_2} |x_{1} , x_{2}\rangle dx_1 dx_2$

nah, let's stay with your example. What's wrong here?

there is also a $c_{x_1,x_2}$ factor in your relative acceration here
@ACuriousMind which one?

@MoreAnonymous no there isn't - the value of an observable for a state is the eigenvalue of the state
@MoreAnonymous your $\lvert x_1, x_2\rangle + \lvert x_1 + \delta, x_2 + \delta\rangle$

@ACuriousMind This is not a normalised wavefuntioin

It's an eigenstate with eigenvalue $V(x_1 - x_2)$. The $\sqrt{2}$ does not play any role

8:05 AM
@ACuriousMind $\lvert \psi \rangle = \lvert x_1, x_2\rangle + \lvert x_1 + \delta, x_2 + \delta\rangle$$@MoreAnonymous Yes, I omitted the normalization. States are rays in Hilbert space, as long as you divide everything by the normalization at the end everything works. Normalization is just convenience You can normalize it and it is still an eigenstate with eigenvalue$V(x_1 - x_2)$if you compute$\langle V(x_1 - x_2)\rangle$for it, you get$V(x_1 - x_2)$, regardless of what normalization you chose One moment Wait is the eigenvalue of an operator is what you measure I think your problem isn't anything with the acceleration operator specifically, but that you're confused about how observables in QM work :P @MoreAnonymous yes, that's the Born rule that's literally the foundational principle of QM @ACuriousMind I might be forgetting something But do humour me? Ah your right ... Lemme edit some stuff @ACuriousMind Okay the question still remains (Actually a new version of it to be precise :P ) 8:25 AM Hi All.. Hi @JohnRennie Sir. @ACuriousMind Thanks for pointing that out sometimes you forget a lot outside academia ... How do you keep on top of your physics game? @MoreAnonymous you have to be precise what you mean by "swap" here and how this acts - as an operator,$a_{2,1} = -a_{1,2}$holds always and knows nothing about bosons or fermions you also have still some vestigial stuff in the question like expressing the state in the position basis that you don't need anymore @MoreAnonymous not sure I remain "on top" :P but physics.SE is the only physics I do these days @ACuriousMind I think it makes the like to acceleration eigenvalues clearer @ACuriousMind Let's say 2'nd experimentalist comes along And both of them disagree on the parity of coordinate systems choosen only Thats basically what I mean by swap I don't know what parity has to do with swapping the labels 1 and 2 @ACuriousMind So if I can express my wavefunction in the position basis I can think of it as swapping position 1 and 2 @ACuriousMind Seems to be enough for you ... I almost never answer questions ... Maybe I should give that a go sometime 8:38 AM any idea what this is. I know this not the place but will it be pizza? @MoreAnonymous sure, but parity means reversing one or three coordinate axes, that has nothing to do with swapping$x_1$and$x_2$in a function$f(x_1, x_2)$. Okay so maybe this is another error in thinking I have ... But I was under the impression for example:$P |x_1,x_2 \rangle \langle = |x_2,x_1\rangle $Swapping$1$and$2$:$P |x_2,x_1 \rangle \langle = |x_1,x_2\rangle $If$P$is supposed to be parity, then no. If it's "swapping particles", then yes (with a caveat for fermions). @ACuriousMind whats the difference between swapping particles and swapping their position? (and yes I should have written P |x_2,x_1 \rangle \langle = p |x_1,x_2\rangle$ where p is an eigenvlaue) ... Isn't this an active passive transformation thingy?

Note that the notation $\lvert x_1, x_2\rangle$ is ambiguous and often depends on context. It can mean $\lvert x_1\rangle\otimes \lvert x_2\rangle$ (in general), $\lvert x_1\rangle\otimes \lvert x_2\rangle + \lvert x_2\rangle\otimes \lvert x_1\rangle$ (for bosons) or $\lvert x_1\rangle\otimes \lvert x_2\rangle - \lvert x_2\rangle\otimes \lvert x_1\rangle$ (for fermions)
@MoreAnonymous my point is that "parity" does not mean "swapping position"
it means "reversing spatial dimensions", i.e. $x_1 \mapsto -x_1, x_2 \mapsto -x_2$
this is entirely different from swapping

8:46 AM
Ah your right ...

you have to be more careful about using the technical terms with their actual meaning, otherwise no one will understand you

@ACuriousMind Thanks ...
$\lvert x_1\rangle\otimes \lvert x_2\rangle - \lvert x_2\rangle\otimes \lvert x_1\rangle$ (for fermions).... But I also get the minus sign if I swap $1$ and $2$

@MoreAnonymous That you get the minus sign for fermions is precisely because that there is the expression for their $\lvert x_1, x_2\rangle$

Which is what I was alluding to

and you don't get the minus sign for bosons because their expression for $\lvert x_1, x_2\rangle$ is symmetric

8:48 AM
@ACuriousMind Yes but then the relative acceleration operator does not make sense
@ACuriousMind As we agreed

@MoreAnonymous If the fermions can be swapped like this, they are indistinguishable. It doesn't make sense to interpret the sign of "relative" quantities to begin with here since "1" and "2" are arbitrary labels and the states $\lvert x_1,x_2\rangle$ are not states where one particle is at position $x_1$ and the other at $x_2$ - it's one where they are equally mixed at both positions
the fermion state doesn't change the sign of its acceleration under swapping. That doesn't mean your acceleration operator doesn't make sense, it just means you unreasonably expect a quantum state of indistinguishable fermions to behave like a pair of classical distinguishable bosons :P

@ACuriousMind Are bosons distinguishable?

you probably think about $\lvert x_1, x_2\rangle$ as the state of "particle 1 is at $x_1$ and particle 2 is at $x_2$", but this is just wrong, for both bosons and fermions

@ACuriousMind Well if I apply $\hat x_1 \otimes 1 - 1 \otimes x_2 \lvert x_1, x_2\rangle$ why wont I get say the relative position
?

look at the states I wrote down - $\lvert x_1, x_2\rangle$ is always an equal superposition of $\lvert x_1\rangle\otimes \lvert x_2\rangle$ (particle 1 at $x_1$ and particle 2 at $x_2$) and $\lvert x_2\rangle\otimes \lvert x_1\rangle$ (particle 1 at $x_2$ and particle 2 at $x_1$), the only difference is the sign of superposition
@MoreAnonymous for bosons and fermions, only operators make sense if they preserve the symmetric/antisymmetric property (unless you break indistinguishability)

8:57 AM
@ACuriousMind but bosons are indistinguishable too

so?

why does it make sense there>
?

$x_1 - x_2$ does not have eigenstates that are symmetric in $x_1$ and $x_2$, so it is not a good operator for bosons
what you can measure is $\lvert x_1 - x_2\rvert$
because this actually has $\lvert x_1,x_2\rangle_B$ (where by the subscript I mean that this is the symmetric combination) as eigenstates
conversely, $\lvert x_1 - x_2\rvert$ is not a good operator for fermions, but $x_1 - x_2$ and $x_2 - x_1$ are, because $\lvert x_1,x_2\rangle_F$ are eigenstates for these operators
as soon as you claim to be able to measure an operator that's not an appropriate (anti-)symmetric combination, your two particles are not indistinguishable anymore, because you're claiming to have a device that can distinguish them

Okay what happens if I define an opertor $S$ which swaps $x_1$ and $x_2$ and define relative acceleration where $S$ is defined as:

$S\psi(x_1,x_2) = s \psi(x_2,x_1)$

where s is an eigenvalue

$a = S (x_1 - x_2)$
Does this evade the problem?

I don't understand the question
and what is $S(x_1,x_2)$?

9:05 AM
@ACuriousMind Sorry it was suppose to be $\psi$

$a = S\psi(x_1,x_2)$ doesn't make any sense either

Wait I trying to type to fast

no need to do that - chat is eternal
take your time to think about what you write actually makes sense

$a \psi(x_1,x_2)= S(a_1 -a_2)\psi(x_1,x_2)$
where

$S \psi(x_1,x_2)= \psi(x_2,x_1)$
Does this make more sense^ ?

Given two 1-particle operators $A_1$ and $A_2$, we can say in fully generality that $\lvert A_1 - A_2\rvert$ is a good operator for bosons and $A_1 - A_2$ and $A_2 - A_1$ is a good operator for fermions

9:07 AM
@ACuriousMind I see what your saying

The first question you need to answer if you're trying to do anything else is to answer why you're doing that

@ACuriousMind I was trying to describe relative acceleration for fermions

@MoreAnonymous sure, so take either $a_1 - a_2$ or $a_2 - a_1$

Okay lets say I choose $a_1 - a_2$

both are "equally good", neither will correspond to your classical idea of relative acceleration (or position!) because there is no antisymmetric state where one particle has a definite position and the other has the other - all antisymmetric states are entangled, you cannot think about the definite states of the individual particles

9:10 AM
@ACuriousMind You make sense ... I hate fermions

it's not specific to fermions - almost all symmetric (bosonic) states are entangled, too

@ACuriousMind But it works there?

what works there?

@ACuriousMind relative acceleration?

how does it work? Only $\lvert A_1 - A_2\rvert$ is a good operator for bosons, as I said

9:12 AM

and there, too, you have the same phenomenon - unless $x_1 = x_2$, the symmetric state is entangled and neither of the individual particles has a definite position
this is a general thing completely disjoint from anything specific about accelerations, fermions or bosons: If you try to think about quantum systems that are combined from two or more subsystems in terms of definite states of the subsystems, you're going to have a bad time

@ACuriousMind Sounds like me :P

that's because you're then ignoring entanglement, and generically almost all states of a combined system are entangled (the set of non-entangled states has measure zero)

Yea I think I get you now ... your right I forgot about entanglement and indistinguishability
Also one thing is still unclear to me why is $\lvert A_1 - A_2\rvert$ not an equally good operator for fermions?
Ah i think it has to be antisymmetric under permutations of $x_1$ and $x_2$
right

4 hours later…
1:54 PM
@DanielSank Wasn't me

2:06 PM
What the heck is the "instanton fluid" that Ron Maimon seems to love

I think that's what new coke was

@NiharKarve It's presumably a turn of phrase for the same fact that people sometimes describe as "the QCD vacuum is filled with instantons" - the proper vacua in QCD are the $\theta$-vacua, which are superpositions of vacua with pure instanton numbers
It's about as silly a phrase as calling a non-zero VEV of $\langle \bar \psi \psi\rangle$ a "fermion condensate"
I never really understood why people call that a "condensate" but it's a reasonably widespread notion

I remember being so disappointed to find out that a quark condensate wasn't a state of matter
I was imagining something with a bubble chocolate-like consistency

well, it's often an order parameter for a phase transition of the field theory, so I guess I can see why one might draw the analogue to a state of matter
@NiharKarve dark matter chocolate is my favourite

physics is a bit of a rollercoaster rly
thinking you know what something is until it isn't

3:12 PM
honestly by that point I'm not 100% how I would describe what a "proton" is in a satisfying way
and yet they're like most of all matter
I just drink them down

@Slereah you're thinking of protein drinks, not proton drinks ;)

Water's full of protons

By mass it's like 50% protons
and the other 50% is neutrons which I also wouldn't know how to describe
Imagine the blessed era when mecanism was a tenable idea and you could just imagine that everything was little marbles bumping into each other

1 hour later…
4:38 PM
In the electron-positron pair annihilation process, if I pass unpolarized incoming particles, what should I expect the spins of the outgoing particles to be in the non-relativistic limit?
I tried to work it out but I am getting an answer which looks weird to me. The outgoing particles retain the incoming spins if the both incoming particles had aligned spins. If the incoming spins are different, the spin along z vanishes for some and reappears in the x and y direction in the outgoing particles. This sounds very wrong.
Sorry, I meant the annihilation Feynman diagram in the Babha scattering process (pair annihilation followed by pair production), not pair annihilation process.

@Yashas why is it wrong?

It makes sense that outgoing particles retain the spins of the incoming spins when incoming spins are alinged since that's the only way spin can be conserved in the overall process. Spin is also conserved in when incoming spins are different since they are zero in the beginning and at the end.
@ACuriousMind I don't know. It feels wrong because spin vanished in z and reappeared in x and y. Why did it vanish in z?

overall spin is conserved - the spins of the outgoing particles still are oppositely aligned, right?

Intuitively, I was expecting the spins to be randomized. I can't find any reference for the non-relativistic case from google. So I am not sure if what I have is correct. I was hoping maybe someone would outright say it's wrong if it's wrong.
@ACuriousMind Yes, the spin vanishing in z is bothering me.

there's no conserved quantity that would make the outgoing particles "remember" that the incoming spins were oppositely aligned in the z-direction, so I'm not sure what the problem is
you can get vanishing total spin by opposite spins in arbitrary directions, so that's what you get - "everything that is not forbidden can happen"

4:53 PM
Maybe my interpretation of the terms is incorrect. I evaluated $\bar{u} \gamma^{\mu} v$ terms for each $\mu$ where $u$ and $v$ are spinors for particle & anti-particle. I got a zero for $\mu = {0, 3}$ but $2m$ for $\mu = 1$ and $-2mi$ (up-down) and $2mi$ (down-up) for $\mu = 2$. Now I'm not sure why I had to interpret each of these terms to indicate spin along an axis.

5:03 PM
ACM, could I just quickly ask about something you said here:
Mar 17 at 23:32, by ACuriousMind
@Charlie the CFT "primary fields" can be polynomial expressions in the actual "field" and its derivatives that appears in a Lagrangian. E.g. for the theory of a free 2d scalar boson field $\phi$, you have that $\phi$ is a primary, but $\partial\phi$ and $\bar\partial\phi$ are also primary.
You specifically say "appears in the Lagrangian", are you saying that the "scalar primaries" that create the primary states from the vacuum can only be terms that appear in the Lagrangian?
Because that would seem to very heavily restrict how many of them you have, like down to 2-3 possible primaries

No, it's supposed to be parsed "polynomial expressions in the actual (field ... that appears in a Lagrangian)" not "(polynomial expressions ... that appears in a Lagrangian)"
my habit of writing confusingly nested sentences strikes again :P
I'm saying if your CFT comes from a Lagrangian, the primary "fields" can be arbitrary functions of the stuff you'd usually call "field" in the context of the Lagrangian

Ah ok, do you in general have an infinite number of these? Like it seems you can make arbitrary polynomials of the field and its derivatives, provided you don't write stuff that's going to vanish on shell like $\partial^2\phi$ anywhere

yes of course there's an infinite numbers of these, $\phi^n$ for arbitrary integer $n$ for starters :P

@ACuriousMind and that's literally the foundational principle of QM (at least as foundational as the eigenvalue thing :P)

Ah

5:10 PM
another example of a family of primary fields in the free boson CFT are the vertex operators $:e^{i\alpha\phi}:$ which can actually be quite useful

$\phi^n$ was what I had in mind :P, maybe I'm doing a disservice to myself by constantly expecting "primaries" to be more mysterious than they actually are. Almost every source I've read so far (which admittedly is not a huge number) introduces these primary fields very, very offhandedly
I've not encountered anything like $e^{i\alpha\phi}$ yet, exponentials of field operators seems like a scary concept

it's not more scary than $\phi^n$ because if you try to think about what $\phi^n$ is supposed to be rigorously you run into the "you can't multiply distributions" issue rather quickly :P

oh yeah they're operator valued distributions huh

arguably a lot of renormalization is the price we pay for ignoring that as long as possible

well, fortunately that won't be in my exams so i'll pretend it isn't a problem for now >:)
rapidly running out of room under this carpet though

5:15 PM
that's what the extra dimensions are for

:0

5:35 PM
Dang it @ACuriousMind. Making answers that are more concise but saying just as much as mine ;)

@Charlie there's a lot of space between the $:\cdots:$

@BioPhysicist I probably wouldn't have written mine if yours had been there when I started, type faster next time ;)

6:00 PM
Or I need to get up earlier

3 hours later…
8:30 PM
I need some tips on developing a project I'm working on. I wanna formulate a 1-body orbital problem as a tangent vector field, considered as an associated bundle to the frame bundle. How should I choose my connection 1 form? I don't know much about symplectic geometry so I wanna see if it's possible to formulate it like this without symplectic structure

9:30 PM
0

I had this old question which was closed which I think I have now find a satisfactory answer to from Frank White's Fluid Mechanic book which is discussed exactly on page -6 (8th edition) We have already used technical terms such as fluid pressure and density without a rigorous discussion of thei...