The h Bar

Obliv

02:24

Is there a way to understand Huygen's principle without running into overflow errors in the brain?

I get the gist, but if i extend the principle to literally treating every point as a source then I try to imagine infinite spheres emanating and infinite spheres emanating from them etc

just total chaos

bolbteppa

03:15

Yes, first of all if you're trying to use it in a classical context things are more subtle, see Feynman's comments here:

Huygens–Fresnel principle

The Huygens–Fresnel principle (named after Dutch physicist Christiaan Huygens and French physicist Augustin-Jean Fresnel) states that every point on a wavefront is itself the source of spherical wavelets, and the secondary wavelets emanating from different points mutually interfere. The sum of these spherical wavelets forms a new wavefront. As such, the Huygens-Fresnel principle is a method of analysis applied to problems of luminous wave propagation both in the far-field limit and in near-field diffraction as well as reflection. == History == In 1678, Huygens proposed that every point reached...

In a QM context, you can think of a wave function at $x' = (t',\mathbf{x}')$ (i.e. a 'wave' at $x'$), $\psi(x')$, as a linear combination of waves that began at some earlier time $t$, $t'>t$, emanating from all points $\mathbf{x}$, where the strength/amplitude of the wave $\psi(x) = \psi(t,\mathbf{x})$ that arrives at $x'$ will be proportional to the original amplitude $\psi(x)$, where $K(x';x)$ is the proportionality.

Thus we are saying $\psi(x')$ is a sum of waves $K(x',x)\psi(x)$ i.e. $\psi(x') = \int d^3 \mathbf{x} K(x';x) \psi(x) = \int d^3 \mathbf{x} K(t',\mathbf{x}';t,\mathbf{x}) \psi(t,\mathbf{x})$, where $t' > t$ is assumed, which can be made explicit by writing this as $\theta(t' - t) \psi(x') = \int d^3 \mathbf{x} K(x';x) \psi(x)$

A non-trivial illustration of this arises when a wave function has stationary states, in which case it can be expanded as $\psi(x) = \sum_n c_n \psi_n^*(\mathbf{x}) e^{-iE_nt}$, where $c_n = \int d^3 \mathbf{x} e^{i E_n t} \psi(x) = c_n(t)$.

We can now insert $c_n(t_1)$ into $\psi(x_2)$ and get $\psi(x_2) = \int d^3 \mathbf{x}_1 \sum_n \psi_n(\mathbf{x}_2) \psi_n^*(\mathbf{x}_1) e^{iE_n (t_2 - t_1)} \psi(t_1,\mathbf{x}_1)$. Thus
$$\theta(t_2 - t_1) \psi(x_2) = \int d^3 \mathbf{x} [\theta(t_2 - t_1) \sum_n \psi_n(x_2) \psi_n^*(x_1)] \psi(x_1) = \int K(x_2;x_1) \psi(x_1)$$

Usually it's written as $K(x_2;x_1) = i G(x_2;x_1)$ because when you apply the Schrodinger equation to it, the $i$ in $i \partial_t$ cancels the $i$ in front of the $G$ when you apply the operator to $\theta(t_2 - t_1) \psi(x_2)$

Things get even more subtle when you apply this simple picture to relativity, where you now have 'negative energy' stationary states to deal with, which I wont go into

bolbteppa

03:42

(In Ket notation this is $\theta(t_2 - t_1) \psi(x_2) = \theta(t_2-t_1) <x_2|\psi> = \int d^3 \mathbf{x} \theta(t_2-t_1)<x_2|x_1><x_1|\psi> = \int d^3\mathbf{x} K(x_2;x_1) \psi(x_1)$, where $|x> = e^{i \hat{H} t}|\mathbf{x}>$)

Shashaank

@naturallyInconsistent Oh

Silly Goose

@Slereah dirac cites a similar reason for pursuing what is now the dirac equation

Shashaank

03:58

What is the physics sign convention for Thermo?

Silly Goose

05:36

is there a formal way to think about a given a differential equation, e.g. $\frac{d}{dt}\psi(x,y,z,t) = \nabla^2\psi(x,y,z,t)$, and various "complexifications" of it? For instance, we might want to solve $i\frac{d}{dt}\psi(x,y,z,t) = \nabla^2\psi(x,y,z,t)$ instead using knowledge that we have of the original DE. Or vice versa.

user 85795

06:54

hi

Ryder Rude

hi

user 85795

honk

Mr. Feynman

07:15

@SillyGoose are you trying to relate the heat equation and SE?

Silly Goose

@Mr.Feynman this is the motivating example for the question

naturallyInconsistent

H O N K

@Mr.Feynman what sound is it that you make?

meow

Silly Goose

the first 5 minutes of this talk talk about a quantum state as being a section youtube.com/watch?v=sshJyD0aWXg; i think this is the type of formulation of textbook quantum that i was trying to find earlier.

the related paper is here: arxiv.org/pdf/2302.10778.pdf

Slereah

08:29

Trying to figure out a bit what Kerr's singular curves are like and as an example I thought I'd look at the AdS null curves at infinity, but then I can't seem to find anything about whether or not these curves are singular or anything about them???

It's like people never did anything with finite affine length singular curves

Closest I can find is some Penrose paper on ideal points where he seems to discuss curves like that a bit, calling them null-finite $\infty$-TIP, but he doesn't really give much in the way of examples

Slereah

08:54

Time dilation experiment

John Rennie

Hafele–Keating experiment?

Slereah

that's the one

Hafele on theleft, and that intelligent computer is Keating I assume

Mr. Feynman

09:13

@naturallyInconsistent the sound produced by waving hands

@SillyGoose well, if you consider an analytically extend functions of time $t$ to a complex variable $s$ and you restrict to imaginary $s=i\tau$ with $\tau\in\mathbb{R}$ (known as imaginary time) you have a correspondence between the two cases

Basically this turns imaginary exponentials into decreasing exponential and comparing the propagators of the respective equations you can see you're passing from one to the other

naturallyInconsistent

09:28

@Mr.Feynman voom voom?

Ryder Rude

10:31

@SillyGoose i dont like this approach personally. it's too close to Bohmian mechanics

they've replaced a deterministic beable with a stochaistic beable

they also seem to be using decoherence to solve the measurement problem

but beables r not needed if u r using decoherence anyway

beables have problems with QFT, as there is no position basis. there is no preferred choice of what basis to use as a beable

Slereah

13:53

> A magnitude if divisible one way is a line, if two ways a surface, and if three a body. Beyond these there is no other magnitude, because the three dimensions are all that there are, and that which is divisible in three directions is divisible in all. For, as the Pythagoreans say, the world and all that is in it is determined by the number three, since beginning and middle and end give the number of an 'all', and the number they give is the triad.
And so, having taken these three from nature as (so to speak) laws of it, we make further use of the number three in the worship of the Gods.

It's all connected

Slereah

14:05

Artistotle's On the heavens is a much nicer read than his physics book rly

not much metaphysical faffing about, he starts off like it is

All bodies are made of fire, air, etc

as we all know

Slereah

14:19

> Further, this circular motion is necessarily primary. For the perfect is naturally prior to the imperfect, and the circle is a perfect thing. This cannot be said of any straight line:-not of an infinite line; for, if it were perfect, it would have a limit and an end: nor of any finite line; for in every case there is something beyond it, since any finite line can be extended.

I'm not sold on his arguments

Proof : circles are great?

Daniel Underwood

14:59

lines are clearly gross

Slereah

15:19

I don't know why people think Aristotle is so great really

Bar was pretty low apparently

naturallyInconsistent

Aristotle is great for the other stuff, not for this part

Slereah

I'd say something that stupid casts doubts on the rest

Ryder Rude

16:03

it is not that stupid from the past's perspective

u can see urself writing this if u were living in the past

most children arent into empiricism. they just create a philosophy of the world in their heads

Slereah

I'd argue it is v. stupid

Ryder Rude

when did the idea that we r sticking to the earth like it's a magnet come along?

was it the same time when we found that the earth is a sphere

Aristotle said that the natural motion of Earthly things is to go down

and i think he believed in geocentrism with a spherical earth

so he must have had this magnet idea

so he knew that down wasnt absolute

Slereah

He very much did believe in geocentrism with a spherical earth

Ryder Rude

sorry the "didnt" was added by my subconscious

Slereah

16:19

He defines down in this very book

As towards the center

Ryder Rude

so had a correct idea

with a spherical earth, it's impossible to get this idea wrong

in flat earth models, there can be somehing like "everything falls down in the universe and the earth is falling down slower than the objects on earth"

did anyone belive in such notions of down in ancient history

spherical earth is extremely old knowledge, so unlikely

Slereah

There was some notion of the earth falling down, although the more usual flat earth model was just that it was just floating on a big sea

Flat earth models were usually not amazingly complex

Ryder Rude

oh

Slereah

It was usually something like

The rarer half flat half round model

Ryder Rude

lol

i used to believe something similar to the first model in middle school

i read that it was a sphere but all i saw around me was flat

@Slereah so this picture is what i believed in. a dome with a flat surface

the dome being the sky

anyone have similar experiences?

Slereah

16:35

No I wasn't raised in Arkansas

Ryder Rude

lol

i also thought that rockets have to pierce through the sky

Silly Goose

22:18

@ACuriousMind @naturallyInconsistent I think I have a concrete example that is physically relevant concerning the discussion earlier. In particular, consider a physical system with hamiltonian that is not equal to the total energy of the system $H$. What would it mean to treat this system quantum mechanically.

Consider a charged particle in an electromagnetic field. It seems like the classical Hamiltonian of a charged particle in an electromagnetic field is not identified with the total energy of the system.

ACuriousMind

@SillyGoose You just quantize it according to the rules of (constrained) quantization

why does it matter whether the Hamiltonian is "the energy"?

Silly Goose

I guess when I say "treat this system quantum mechanically" I mean to include the usual identification of Hamiltonian eigenstates with states of definite energy.

And so how to interpret Hamiltonian eigenstates in this situation is what I am interested in knowing

ACuriousMind

depends on the system!

but saying that Hamiltonian eigenstates have to be identified with states of definite energy is definitely not what "treating a system quantum mechanically" means :P

Silly Goose

let me revise to treating it at as systems are treated in an "introduction to quantum mechanics level textbook" :)

Claudio Menchinelli

Sorry guys I wrote down in my notes that the ground state eigenfunction is always an even function. The problem is I know my professor said this so it must be true, nevertheless I can't think of a reason why this should be true and how I should show this. Anybody got any clues?

ACuriousMind

22:24

@SillyGoose you will not meet a system whose Hamiltonian is not energy in an intro to QM textbook

Silly Goose

well so then per the discussion we had before, "energy" ceases to be "useful" when we want to consider a charged particle in an electromagnetic field?

ACuriousMind

because typically these show up either in relativity (reparametrization-invariance forces zero Hamiltonian) or gauge theories (which require either ad-hoc methods or developing a theory of constrained quantization), neither of which you want to do in an intro context

Silly Goose

what happens to statistical mechanics in these contexts :P...goodness

@ClaudioMenchinelli were any assumptions made about the potential?

ACuriousMind

and the Hamiltonian for the EM field is energy, it's just not $T+V$ where the Lagrangian is $T-V$ (because the magnetic potential isn't a "real" potential associated with energy since the magnetic force does no work)

Claudio Menchinelli

Let me see

symmetric potential

wrt to the x=0 axis

Oh I see maybe why you asked this

ACuriousMind

22:30

this is kind of the inverse example you're looking for - the Hamiltonian is energy, but it doesn't fit into the T+V/T-V framework, see e.g. Qmechanic here

Claudio Menchinelli

$\hat{p}^2$ is even since $\hat{p}$ is an odd operator

no wait the position operator is odd

Relativisticcucumber

@ClaudioMenchinelli i think theres an exercise in griffiths that proves this

Claudio Menchinelli

Cohen said something about even hamiltonians

Oh I found it

Relativisticcucumber

:)

ACuriousMind

@ClaudioMenchinelli why does it matter that the position operator is odd? There's no odd powers of the position operator in your Hamiltonian here (since by assumption it has a symmetric potential)

Claudio Menchinelli

22:31

it's the last paragraph

Silly Goose

@ClaudioMenchinelli you can start with your definition of Schrödinger equation and consider the two cases: 1) suppose $\psi(x)$ is a solution, 2) then what is $\psi(-x)$?; the hypothesis that $V(x) = V(-x)$ must also be used

Claudio Menchinelli

I thought the simmetry of $V(x)$ depended upon the position operator

ACuriousMind

@ClaudioMenchinelli So? No matter how it depends on it exactly, your assumption here is explicitly that $V(x)$ is symmetric/even

Silly Goose

@ACuriousMind hm sorry what do you mean by is energy?

Claudio Menchinelli

yep, I see now thanks, I was a bit distracted

ACuriousMind

22:34

@SillyGoose it's the thing that will show up in the work-energy theorem

what do you mean by energy? :P

Silly Goose

i'm not sure what energy is anymore...

ACuriousMind

now we're getting somewhere - if you can't even say what the definition is, why does it matter whether the Hamiltonian "is energy" or "is not energy" :P

Silly Goose

I think it matters because I (~wrongly) took as definition that the Hamiltonian is the generator of time translations and that we call the quantity that is conserved under time translations energy. But this way of framing things seems inappropriate outside of an introductory textbook setting

Relativisticcucumber

what does it mean to say that a state is relativistically normalized? tong goes through this derivation to find the normalization term (i think) based on the measure being lorentz invariant, but i dont get the point of this? why cant we just normalize our states according to $\langle p \vert q \rangle = (2\pi)^3\delta(p - q)$ and be done with it?

ACuriousMind

@Relativisticcucumber you could but then you'd just get annoying factors of $2m$ all over the place

it's just the convention in relativistic contexts to not do that :P

Relativisticcucumber

22:43

okay i see. so this is just how i should normalize the states in qft in general or is this a specific context?

Silly Goose

I think my broad goal here is to tease out to what extent this Hamiltonian is useful. Previously, it seemed like quite a fundamental and ubiquitous tool in physics, but now it seems more like just another tool. I can adjust the way I see things in light of this to something along the lines of: we just want to find a language to write the physics in that is most convenient, i.e. most natural. it is a dream to imagine that one could find a single language in which all of physics is natural.

Relativisticcucumber

@SillyGoose a dream of Jaffe

Silly Goose

and to clarify, I do not wish or mean to consider pathological cases when looking for these sorts of generalizations of concepts like energy and what not.

ACuriousMind

@SillyGoose What do you mean? The Hamiltonian still predicts the correct physics even when it's not energy!

if you hadn't been through years of intro physics where we drill this reified idea of energy into your head you would never have found this weird in the first place :P

Silly Goose

@ACuriousMind I guess then I would like to know what this statement means. Moreover, if the Hamiltonian, whether it is energy or not, lets us make these accurate predictions, why even define such a thing as energy at all! Just call it the Hamiltonian

Another way of saying: If "being energy" is not what fundamentally makes it a valuable concept, then I would not like to call it energy as if to emphasize that energy is what is important :P

ACuriousMind

22:47

@SillyGoose you want to explain Hamiltonian mechanics to students who first learn about collisions and how they obey energy and momentum conservation???

don't throw the baby out with the bath water - just because at some point you have to say goodbye to our naive idea of energy that doesn't retroactively invalidate all the cases where it was useful

Silly Goose

I guess this suggests to me that Hamiltonian as energy is distinct from Hamiltonian in general. That is, the Hamiltonian is useful when it is identified with energy and it is useful when not but for utterly different reasons

but I would expect that the Hamiltonian is useful because of the same set of reasons for both cases. so the above is not intuitive to understand

well I guess I see it like this now and I am not sure if this is an accurate way to see things. Just like in say Tong's QFT lecture notes where the field and its conjugate field happen to be related by fourier transform, but this fourier relation is not generally true between a field and its conjugate; so is the relation between Hamiltonian and total energy of the system. It is convenient and it is useful because of its convenience, but strictly speaking the Hamiltonian has nothing

to do with total energy

ACuriousMind

I mean the problem is really that you have no abstract definition of "total energy"

it is definitely true that the notion of the Hamiltonian grew out of us realizing this conserved quantity called "energy" was kinda useful

but then the Hamiltonian formalism generalized to settings where suddenly the Hamiltonian wasn't anymore what we'd like "energy" to be (like identically zero), and the link between the Newtonian notion of "energy" and the Hamiltonian breaks

Silly Goose

@Relativisticcucumber :-)

ACuriousMind

I don't think there's anything deep going on here - just one concept (the Hamiltonian) generalizing another (the energy)

Silly Goose

well okay I can accept it being that way

Relativisticcucumber

23:06

here, a and b are creation and annihilation operators, but i dont see why the solution to the KGE should include such operators?

i mean how do they appear in the general solution to that equation?

ACuriousMind

@Relativisticcucumber they're not "creation and annihilation operators" when you look at the solution to the classical KG equation, they're just coefficients for the two basic solutions $\mathrm{e}^{\mathrm{i}p_\mu x^\mu}$ and $\mathrm{e}^{-\mathrm{i}p_\mu x^\mu}$

that these turn out to be creation and annihilation operators after quantization isn't really relevant here

Relativisticcucumber

@ACuriousMind hm, so how is it proper to view these coefficients as the creation and annihilation operators?

ACuriousMind

@Relativisticcucumber what do you mean "proper"?

quantization turns classical observables into operators

Relativisticcucumber

i just mean purely in terms of looking at this as solving an ode

ACuriousMind

the classical observables of field theory are the fields, so your $\phi(x)$ becomes an operator, and so must the $a(k)$ and $b(k)$

Relativisticcucumber

23:16

i dont see how the coefficients could be operators?

ACuriousMind

You can invert this Fourier transform to get an explicit formula for the $a(k)$ that looks like $a(k) = \phi(k) + \mathrm{i}\pi(k)$. Quantization turns the $\phi(k)$ and $\pi(k)$ into operators by definition, so the $a(k)$ become operators, too

what does it matter that this idea of decomposing the field into $a(k)$ and $b(k)$ was originally inspired by solving the KG equation?

the claim that $a(k)$ so defined is an operator is true no matter whether you even know about the KG equation or not

Relativisticcucumber

23:29

@ACuriousMind so fields can act on states also, right?

ACuriousMind

what does being an operator mean if not that it's an operator on the space of states?

Relativisticcucumber

@ACuriousMind i thought a field is an operator valued function, so i more thought of it as a function but bleh so the fields are a kind of creation operator? conceptually i mean?

ACuriousMind

@Relativisticcucumber oh, that's just a language issue, we really don't keep saying "operator-valued function" all through QFT :P

both the $\phi(x)$ and $a(k)$ are "operator-valued functions", if you want to be pedantic

(if you want to be even more pedantic the $\phi(x)$ isn't even a function but a distribution but let's not get into that)

Relativisticcucumber

okay i see. so i am looking at two examples. it would seem that for free fields we have these two operators $b(k)$ and $a(k)$ but then when we do a complex scalar field, we have four of these operators. it's my understanding that this corresponds to having particles and antiparticles and the ability to create and annihilate them. im wondering how we show that these operators are unique? [...]

[...] like how we show that indeed the 4 operators in the complex scalar field case are four ind. operators that create and annihilate matter and antimatter?

ACuriousMind

...shouldn't the texts you're reading explain exactly that? :P

who's teaching you that these are creation and annihilation operators without showing why

Relativisticcucumber

23:37

its possible that im just missing it. im referring to the complex scalar field example on page 13 here damtp.cam.ac.uk/user/tong/qft/two.pdf

13-14 i mean

ACuriousMind

and what exactly is the question here? You did the real scalar field before this, now you have a complex scalar field that's made out of two real scalar fields and hence has double the number of independent variables

Relativisticcucumber

i think i see how it workd for scalar field by looking at CRs but the antimatter and matter is confusing me as well as how we show these are two distinct types of matter

like how do i know the operators create a type of particle thats truly different than the matter counterpart?

because since the fields are just conjugates they arent even rigorously independent

so im skeptical

ACuriousMind

@Relativisticcucumber you're trying to run before you can walk - we're still just doing free field theory here

in a free scalar field theory there isn't really any meaning to saying something is "matter" or "antimatter" or whatever at all

Relativisticcucumber

i think my confusion is the statement that we have 4 independent operators. im only convinced that two are independent

because $\psi$ and $\psi^*$ dont introduce new info?

ACuriousMind

@Relativisticcucumber see physics.stackexchange.com/q/89002/50583

Relativisticcucumber

23:44

reading now thanks

ACuriousMind

note also that we do this in complex analysis all the time: $\frac{\partial z^\ast}{\partial z} = 0$

it's not really a physics trick, just how complex derivatives work

Relativisticcucumber

@ACuriousMind i think im still confused. so the answer by qmechanic says that we can solve this in 3 different ways and get the same physical result, right? and so of course we can rewrite it however we want, but in order to get 4 truly independent operators, we would need 4 degrees of freedom. what im seeing here is that regardless of how we write it, we still only have two degrees of freedom, as the first two methods state? maybe im missing the point?

well the difference between scalar and complex is the introduction of antimatter, right? so there must be a new DOF if we are now able to make and get rid of antimatter ? @ACuriousMind

ACuriousMind

@Relativisticcucumber where do you think the 2 operators for the real scalar field come from?

Relativisticcucumber

specifically a new DOF that comes from introducing "complexness"

ew why is it backwards

ACuriousMind

I'm confused why you accept the $a$ and $a^\dagger$ as two operators for the one real scalar field $\phi$ but then complain that we get $a,a^\dagger,b,b^\dagger$ for a complex scalar field, which is two real scalar fields

one set of c/a operators for a real scalar field, two sets of c/a operators for the complex one, which is just the complex sum of two real scalar fields

what's the problem?

Relativisticcucumber

23:59

@ACuriousMind because its not two real scalar fields its the same field?

or i mean

by taking the complex conj of smth we get no new information, right?

ACuriousMind

@Relativisticcucumber $\phi = \phi_1 + \mathrm{i}\phi_2$ looks like two real scalar fields $\phi_1,\phi_2$ to me

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