The h Bar

11:27 AM

If I expand $(1+\lambda)^{-\Delta}$ for small $\lambda$ and get $(1-\lambda\Delta)$ to first order, what kind of expansion is this? Is it binomial?

oh it is, nvm

bolbteppa

More generally a first order Taylor expansion in $\lambda$

ACuriousMind

@Charlie it's only "binomial" if $\Delta$ is an integer, otherwise it's just Taylor expansion

Charlie

ok ty

ACuriousMind

technically I guess "binomial" would still fit since the "binom" is the $1+\lambda$, but we don't usually speak about "binomial" for non-integer exponents

Charlie

$\Delta$ here is the weight of a conformal primary so I'm pretty sure it's integer

Slereah

11:34 AM

in that case yeah you can do it via binomial expansion

no biggie

ACuriousMind

@Charlie if you mean the conformal dimension $\Delta = h + \bar h$, it is only a (half-)integer when one of the two weights $h$ or $\bar h$ is zero, since the spin $h-\bar h$ is constrained to be half-integer.

Charlie

Whatever the eigenvalue of the dilation operator on the state created by the primary on the vacuum is, that's what the $\Delta$ is, I don't really know anything else about it :C

ACuriousMind

Yes, that's $h + \bar h$, since the dilation generator is $L_0 + \bar L_0$

and it is not, in general, an integer.

Charlie

oh, I know even less than I thought :P

ACuriousMind

if you got the conformal operator from some classical field, then it will often be, but since there are CFTs with non-integral eigenvalues of $L_0 + \bar L_0$, there exist operators with non-integral conformal dimension

(if your text seems to be assuming it is (half-)integer, there's probably an assumption about the field being chiral (i.e. $\bar h = 0$) somewhere)

Charlie

11:49 AM

That was actually just me assuming it was an integer, probably because $P_\mu$ is supposed to raise it by an integer, but idk really

ACuriousMind

you can add integers to non-integers ;)

nothing forbidden about 2/3 + 1

Charlie

I think I assumed that if $K_\mu$ lowers it by an integer and the primary state has $K_\mu|\phi(0)\rangle=0$ that the primary weight had to be 1 otherwise $\Delta-1\neq0$

that being said I also know that there are lots of different primaries that all have different $\Delta$'s so in hindsight I wouldn't have known how to resolve that lol

ACuriousMind

@Charlie This is a common trap in thinking about lowering/raising operators. Consider the spin-1/2 representation of su(2): You have two states with -1/2 and 1/2 and there is a "raising" operator that maps the 1/2-state to 0 and a "lowering" operator that maps the -1/2-state to 0, but neither 1/2+1=0 nor -1/2-1=0!

Charlie

oh yeah

ACuriousMind

the property of a state being "end of the line" for the raising/lowering is entirely unrelated to the eigenvalue of that state

Charlie

11:56 AM

that is actually very good to know

satan 29

hello everyone

consider a classical harmonic oscillator

it can be easily shown that

$$ \sigma_{p} \sigma_{x} = \dfrac{E}{\omega} $$

where E= total energy of the system\

now, E is something I control

in theory, I can just let E be something like $\omega \dfrac{\hbar}{4}$

and that seems to me a contradiction of the HUP

infact, I can set E to be arbitrarily low: and the product of sp and sx would tend to zero..

Charlie

Isn't $E$ related to $\omega$ and $\hbar$ by $E=\hbar\omega(n+1/2)$

satan 29

its a classical HO

E can be whatever I want

Charlie

Oh right, well the HUP is a quantum phenomenon, why would you expect it to apply to a classical system?

satan 29

@Charlie right, this was were I was going.

John rennie sir also said the same

but my question is,

just what is the parameter for you to decide weather your system is "quantum-enough" , so that you can use the HUP ?

Charlie

12:09 PM

You're asking about crossing the bridge between QM and classical physics which is usually a bit awkward

usually it is said that $\hbar$ is related to the scale at which quantum corrections become sizeable, but that is not a precise statement

satan 29

maybe the answer lies in how the HUP is derived?

Charlie

Whether your system is "quantum enough" depends on how accurate you want your predictions ot be

ACuriousMind

@satan29 a good rule of thumb is that whenever anything involves $\hbar$ directly, you should probably do QM - as in this case when you let $E = \omega\frac{\hbar}{4}$ - since the heuristic classical limit of QM is when $\hbar \to 0$, which really means more something like "$\hbar$ is much smaller than any relevant quantity of action".

Charlie

you can use classical physics at the quantum scale you'll just get inaccurate predictions, so at what point you consider the divergence of classical predictions and quantum results "enough to swap to QM" depends on what you're doing

satan 29

@ACuriousMind hmm I see

ACuriousMind

12:13 PM

@satan29 The HUP is derived in the context of quantum mechanics, which is really a different physical theory than the classical one you derived your expression for $\sigma_p\sigma_x$ in. The HUP doesn't hold in classical physics and your expression doesn't hold in quantum physics.

satan 29

what I meant was

ACuriousMind

when you mix formulae from different incompatible theories, you'll get contradictions - this is expected

satan 29

s_p and s_x are just mathematical terms right? you dont expect them to change their meaning

weather you are doing Classical or QM

the formulation of the inequality by heisenberg may contain some arguments that reveal when exactly can we apply it..

thats what I meant/thought

ACuriousMind

@satan29 how did you define them in classical mechanics?

if you look at the definitions in CM and QM, they will be entirely different - they are not the same mathematical objects

satan 29

@ACuriousMind $$\sigma_{a}=\sqrt{ <a^2>-<a>}$$ ?

ACuriousMind

12:18 PM

in particular, classical mechanics doesn't really have "uncertainty" unless you're doing statistical mechanics - a classical solution of the equations of motion has all quantities like $x$ and $p$ well-defined with no uncertainty.

@satan29 yeah, but what does that expectation value there mean in the classical context?

satan 29

isnt that just the average?

ACuriousMind

A classical state is just a point in phase space, you can just plug it into any observable $f(q,p)$ to get a definite value - there is no "average" there

satan 29

is there not something called an "average position" in Classical mechanics?

ACuriousMind

you are probably averaging over time there, right?

satan 29

yes

ACuriousMind

12:21 PM

this is an entirely different average from the "expectation value" in QM - there you have a state at one fixed instant of time, and when you measure an observable on it, you can get different outcomes - the expectation value averages over these possible outcomes, not over time

satan 29

OHHHHH

yes, yes I remember reading the same notion of average in griffiths

I see, that explains a lot. What an eye opener

thanks a lot @ACuriousMind @Charlie

Physics Meta

12:44 PM

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Q: Posting and self answering basic derivations

It can be difficult at times to find quality derivations for 'some' of the basic formula used in HS physics, so I thought what if I post and self answer the basic derivations. Does this sound like a good idea/ Would the community be in support of me doing this? I think it is good because it will ...

discussion

Charlie

12:56 PM

I'm still slightly unsure about a point regarding "symmetry" in qft. Most of what ends up bugging me is the difference between evaluating the operator field at a point and then performing a spacetime transformation, $\phi(x)\to \phi(x')$ say... and acting with the infinite dimensional unitary Poincaré (or whatever group) rep on the fields, $\phi(x)\to \phi'(x)$. Because in the derivation of the "Ward identities" for CFT we take the path integral and act with some symmetry on the operators...

to get something like $\langle \phi(x_1)...\phi(x_n)\rangle\to \langle \phi'(x_1')\phi'(x_n')\rangle$, then we argue that the symmetry should leave the path integral measure and the action invariant to get:

$\langle \phi'(x_1')\phi(x_n')\rangle=\langle\phi(x_1')...\phi(x_n')\rangle$, and from this we get things like the fact that 2-point functions shouldn't depend on the points themselves, just their relative distance. And in CFT you find that the 2-point functions only evaluate in one of two ways based on their scaling dimension etc.

but I'm still confused by this notation that it's useful to act on both the operator and the spacetime point at which it is evaluated, $\phi'(x')$. This is still a strange thing to want to do to me

ACuriousMind

In most real-world cases you will have trivial transformations on the coordinates

Charlie

Trivial as in the identity?

ACuriousMind

but the prime example of where this happens is of course Lorentz transformation - they act on both

do you find it confusing why Lorentz transformations would act on both the fields and coordinates?

@Charlie yes

Charlie

@ACuriousMind Unfortunately yeah, this does confuse me for some reason

Slereah

Consider this simple example

Take the scalar field $f(x) = 1$

Rotate it, translate it, doesn't matter, it will still be the same after transformation

Now take the vector field $f(x) = (1,0,0)$

It is also "homogeneous" in some sense

But if you rotate it 90° around the $z$ axis, the field will change

To $(0,1,0)$

Because the new $x$ isn't the same as the old $x$

Charlie

1:06 PM

sure

maybe I can put it a slightly different way, if you calculate something useful in QFT like a scattering amplitude and you want to recalculate it in another frame, are you going to act with both the infinite dimensional Poincaré rep on the operator field $\phi$ and also act in the fundamental rep on the spacetime manifold on which the field is defined?

I guess if the answer to that is yes then I don't really have a problem, since it's effectively just a definition in qft

or whatever I guess the group action on the manifold can't be the fundamental rep but you know what I mean, the action of so(1,3) on Minkowski space

Physics Meta

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Q: Misuse of downvoting

I think this question is unsuitable for this site as opinion-based, and should be closed. Yet, at the time of writing it has only one closure vote (mine), but 6 downvotes. I think this is indicative of a larger problem: people use downvote as a dislike button of social networks, rather than as a ...

discussion down-votes

RewCie

@Charlie So, apparently, I visited a homeopath next day for psoriasis, he prescribed one Nat Mur 200 salts

the whole neck region cleared in three days but psoriasis again started

ACuriousMind

@Charlie No, it's either-or

RewCie

But, I was warned against overdose

ACuriousMind

If you transform with the unitary representation, then you don't have to act on the coordinates explicitly - when people write $\phi'(x')$ they mean the action on the classical target space and coordinates

this is (or should be, Wightman axiom) equivalent to doing $U\phi(x)U^\dagger$ on the operator in the unitary rep

@RewCie ...how is that a reply to Charlie's message?

Charlie

1:22 PM

So $\phi'(x')$ is not $U\phi(\Lambda x)U^\dagger$? If not I have definitely got something wrong in my head

ACuriousMind

@Charlie No, it is $U\phi(x)U^\dagger$

Charlie

And when you say either-or, you mean that acting with the Poincaré group on spacetime and re-evaluating the operator field should be the same as conjugating with the unitary rep and evaluating at the old point

ACuriousMind

The action of $U$ is supposed to be precisely so that the result of this is the same operator as doing the classical $\phi'(x')$ and then looking for the corresponding operator

Charlie

I feel like notes not being clear when we're talking about the classical $\phi$ and the operator $\phi$ is a significant source of my confusion

ACuriousMind

I phrased that a bit poorly

Charlie

1:26 PM

ok I need to think about this for a second

ACuriousMind

$\phi'(x')$ is also an operator

Charlie

oh

ACuriousMind

it is identical, as an operator, to $U\phi(x)U^\dagger$

the idea of why this must be is that $\phi'(x')$ is the "classical" way to transform the field and $U\phi(x)U^\dagger$ is the natural quantum mechanical way to transform the field

if the two were not the same, our theory would be inconsistent

Charlie

Ok so something like $U\phi(\Lambda x)U^\dagger$ is not a useful quantity

because you're basically doing two Lorentz transformations

ACuriousMind

well, it is a quantity, but it's not clear why you'd look at that - that's mixing the classical and the quantum mechanical notions of transformations

Slereah

1:33 PM

Technically as far as the Hilbert space is concerned, there isn't even necessarily an underlying space for fields

Charlie

If I can just write out the thing you posted a few days ago $\rho_C(g)\phi(\Lambda(g)x) = U_Q(g)\phi(x)U_Q^{-1}(g)$, the only thing that is now bothering me is that on the left it seems like we act in the active and passive picture on the classical field

I don't want to open up a whole thing about active/passive picture, but $\phi_C(g)\phi(\Lambda(g)x)$ seems like doubling up on transformations

Slereah

I recommend trying it out by hand?

ACuriousMind

@Charlie it's not - go back to Slereah's example. For a vector field you need to not only transform the coordinates but also do a transformation on the target space of the field if it's not a scalar

Slereah

Just pick the simplest non trivial operator you can

Charlie

oh in the scalar rep $\rho_C$ would be the identity

Slereah

1:36 PM

like idk pick a one dimensional QFT and perform some parity reversal on it

ACuriousMind

@Charlie yes

Slereah

that's about the simplest case you can get without getting too simple

Charlie

Do we not work in such a way that the underlying spacetime coordinates give a canonical basis for the tangent spaces?

ACuriousMind

coordinates never give a basis for the tangent space :P

Charlie

um

ok i find this troubling information

Slereah

1:38 PM

@ACuriousMind What about the coordinate basis

You FRAUD

Charlie

Is a canonical basis for the tangent and cotangent spaces not the derivatives and exterior derivatives of the coordinate functions?

Slereah

It's not canonical

It's a basis

ACuriousMind

@Charlie ah, that's what you meant - yes, sure. Why does the choice of basis matter?

there is nothing basis-dependent at all in the equation I wrote down

Charlie

I thought that was the whole thing behind the active/passive picture, if you're using a coordinate basis on the tangent space, then when you perform a coordinate transformation you get a change of basis on the tangent space for free and thus you don't need to act with a linear representation on the tangent space

or you can act with a linear rep on the tangent space and not change the underlying coordinates and this was the active/passive equivalence in a wy

ACuriousMind

Let's not get philosophical again :P $\rho_C(g)\phi(\Lambda(g)x) = U_Q(g)\phi(x)U_Q^{-1}(g)$ makes no reference to any basis whatsoever

Slereah

1:41 PM

Also for a start the fact that we're using $\lambda(g)x$ should make you suspiscious about talking about manifolds :p

That's for a vector space!

Minkowski space

Which you can define as a manifold if you want but you have to be careful about those statements

You don't need to bring up the tangent bundle at all with minkowski space, just field transforming under specific reps of the group

Charlie

man this topic makes me so sad, every time I put it to rest and think I've got it, it turns out I'm wrong :P

I need lunch, I'll have another think about it

Slereah

I mean uuuuuh

Imagine your space has two regions where the wavefunction has two values

And you consider $\langle \Psi, \Phi(x) \Psi \rangle$

ACuriousMind

@Slereah at some point you must not improve your knowledge of QM, but your definition of "understanding" :P

(joking but also not)

Slereah

@ACuriousMind One time I quoted "In mathematics, you never understand things; you just get used to them" to the math channel and they got mad :(

ACuriousMind

Heh. We've discussed that quote before around here

Slereah

1:55 PM

Actually let's pick the simplest example : in some region, $\hat{\Phi}(x) = \hat{I}$, and in another region, $\hat{\Phi}(x) = \hat{0}$

If you simply perform some kind of translation or whatever else on your field, the original will be $\langle \Psi, \Psi\rangle = 1$, and the transformed field will give you $\langle \Psi, \hat{0} \Psi\rangle = 0$

That is a Bad Symmetry

This means that in fact, you also need to transform your wavefunctions

Via some operators

Therefore you have that $\Phi(x) = A \Phi(f(x)) B$

which you can dispose however you wish

and later on show that they are unitary operators obeying such and such group properties

It's easier to show with non-relativistic QM tbh since you can use explicit wavefuntions

Although a bit trickier since the operators don't depend on space directly

Principle is essentially the same if you mess with $\hat{x}$ though

Not 100% sure my example makes sense bc I don't think you can transform away $0$ into not being $0$ via a linear operator, but you get the idea

Charlie

2:33 PM

I think I've realised what the problem was, I just assumed that we use a coordinate basis and so changing coordinates passively takes care of the basis on the tangent space. I suppose if you don't use a coordinate basis for the tangent space you have to manually act with some rep on the tangent vectors

So looking at expressions like $\rho(g)\phi(\Lambda(g) x)$ made me think that well we change coordinates with $\Lambda(g)$ and this induces a change of basis on the tangent space, and then we act with a linear rep of so(1,3) on the tangent vectors which seems like it doubles up the transformation on the vector space

maybe I was also confusing basis-using and basis-free equations

Slereah

Fundamentally the vector doesn't change, only its components

if the basis changes due to the transformation, then so do the components

Charlie

Yeah, I was assuming that the basis depends on the coordinate system, which I guess it doesn't actually have to. And more importantly it can't if we're working basis-free

the coordinate system of the underlying spacetime that is

ACuriousMind

2:52 PM

I think the underlying problem is that physicists often delude themselves into thinking symmetries like the Lorentz symmetry are about "changing coordinates" :P

Note that expressions like $\rho(g)\phi(\Lambda(g)x)$ don't have any coordinates - $x$ is a point (not "coordinates") in Minkowski spacei

this isn't a statement about what the field does when you "change coordinates" - this is a statement about doing the mathematical operation of applying a Lorentz transformation to $x$ and to the field, with no mention of any bases or coordinates

One reason I like index-free notation so much is precisely because it precludes any discussions about coordinates and makes manifest that we're talking about actual operations on well-defined mathematical objects, not a bunch of numbers in a coordinate system that may or may not be meaningful

Charlie

3:16 PM

I don't see where this leaves room for there to be "internal" and "spacetime" symmetries. Since in that expression we're transforming both the points of spacetime and then acting with some Lorentz rep on the field

ACuriousMind

that's because the Lorentz symmetry is a spacetime symmetry. For an internal symmetry, $x' = x$, i.e. you just have $\phi(x)\mapsto \rho(g)\phi(x)$

Charlie

oh

but if what you've written above, $\rho(g)\phi(\Lambda(g)x)$ is a Lorentz transformation, and Lorentz transformation is a spacetime symmetry, why are we additionally acting with $\rho(g)$? Doesn't this make it both a spacetime and internal symmery?

I'm having a hard time making consistent sense of this but I actually think it's finally getting there

If an internal symmetry is $\phi(x)\rightarrow \rho(g)\phi(x)$ I would expect a spacetime symmetry to just be $\phi(x)\rightarrow \phi(\Lambda(g)x)$

ACuriousMind

@Charlie no, because the $\rho(g)$ is determined by the $\Lambda(g)$ and the type of the field - for a spacetime symmetry, the thing in front of the field is directly related to the Jacobian of the action on spacetime

for a vector, $\rho(g)$ is just the $\Lambda(g)$. for a $(p,q)$-tensor, it's $p$ of the $\Lambda(g)$s and $q$ of the $\Lambda^{-1}(g)$s acting on the corresponding indices of the tensor, etc

you can't choose how a vector transforms - its transformation behavior under spacetime symmetries is already fixed by it being a vector

while internal symmetries are completely disjoint from what type of field you have

Charlie

Ok so the statement "spacetime symmetries act on the points of spacetime but not on the fields" is misleading because you still have to act with the $\rho(g)$ on the field

or maybe that's just absorbed into the definition

ACuriousMind

I think that's just an issue with natural language being not precise enough

there's an "action on the field" but it is one that is induced naturally by the action on the points - I can see why someone wouldn't want to call that "acting on the field"

Charlie

3:35 PM

I sort of don't see why the action of the points must naturally induce that action on the field

I think I now get the internal symmetry thing, for instance if you have a 3-vector field you might just want to rotate all the vectors at all the points, maybe by different amounts depending on their points, and that would be a local SO(3) transformation

ACuriousMind

@Charlie That's just the standard calculation every physicist text on relativity/tensors/anything pretending to be differential geometry does - when you do a transformation $x\mapsto y(x)$ (doesn't matter if you call it a "coordinate change", a "diffeomorphism" or whatever), the (co)vectors based in the derivatives/differential in $x$ change by the (inverse) Jacobian of $y$. Mathematically it's the pullback/pushforward along $y$.

Charlie

I guess I don't see why you would never just move the spacetime points around and then not act with the $\rho(g)$ on the field, is that just not a thing?

oh

wait, possible breakthrough

The mention of pushforward/pullback has maybe just solved all problems

ACuriousMind

math to the rescue :D

Ratman

3:54 PM

@ACuriousMind With "mathematically is the pullback along y". Do you mean that given the pullback $\phi^*$ that is a diffeomorphism between two manifolds M and N, then with this map we can construct a "natural map" between the tangent spaces (for example), that is the differential $d\phi$ ?

sorry if I disturbed your argument, but is a topic I am also interested in

Charlie

I wouldn't call it an argument, it's closer to therapy :p

Ratman

ahahahhahaha

well therapy is good for sure

Charlie

I'm going for a walk to think for a bit

ACuriousMind

@Ratman Yes. For a 1-form $\omega(x)$, the pullback along $f$ is defined by $(f^\ast \omega)(x)(v) = \omega(f(x))(\mathrm{d}f(v))$. If you write out in coordinates what this means for the pullback along $y^{-1}$ (where $y$ is my transformation from above), you'll find this turns out to be the same thing as what physicists call the transformation behaviour of a 1-form/covector field under the transformation $y$.

Ratman

5:04 PM

@ACuriousMind Thanks a lot, I'll go trough the argument later, for sure there are things I still don't get.

Charlie

So conformal symmetry is a spacetime symmetry, not an internal one

I think things are finally starting to add up a bit

ACuriousMind

now just 7 more and you have a byte

Charlie

>:(

ACuriousMind

:)

Jake Rose

6:41 PM

Can anybody help with this?

I've gotten as farm as minimal substitution but not sure how to do the rest

i.e. I've gotten $i\gamma^\mu \partial_\mu \phi - q \gamma^\mu A_\mu\phi - m \phi$

bolbteppa

6:53 PM

First turn it into an equation for $\overline{\psi}$ by taking the complex conjugate and transpose the way you set up the adjoint Dirac equation. Then take another transpose and you'll see you need to modify the equation further to get it back to a Dirac equation, which means multiplying it by some matrix $C$ which has to satisfy the properties in the problem statement

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