The h Bar

LucashWindowWasher

04:35

If we only integrate over a finite region of space, then the derivation of the EL equations changes. Certain boundary terms that are usually ignored cannot be anymore, and they will end up deciding what the appropriate boundary conditions the fields can have.

Captain Bohemian

05:04

I feel my dreamland is a better place than real world.

Captain Bohemian

05:15

@LucashWindowWasher even integrating over infinite space may not dispense with trouble.

It may diverge so that we need to add some boundary integral to renormalze it.

Captain Bohemian

07:11

I feel my dreamland is like another universe. When I fall asleep, I experience a wormhole which takes me to that universe which is far more vivid than the universe I encounter when I wake.

Slereah

11:46

Morning

Charlie

Afternoon

Charlie

12:05

In P&S, when looking at "the K-G field as harmonic oscillators", we say the operator $a^\dagger_p$ creates momentum $p$ and energy $E=\sqrt{|p|^2+m^2}$ and we end up with the state $a^\dagger_p|0\rangle$, is $|0\rangle$ an element of some Hilbert space? So there exists a hilbert space at every point?

Or should I not worry where exactly $|0\rangle$ lives at this point

Oh wait it literally is just effectively a hilbert space at each point

ACuriousMind

@Charlie not "at every point" - the state with definite momentum $p$ has completely indefinite position

(if you even manage to say what "position" is - this isn't straightforward in relativistic theories)

The Fock space spanned by the states created by the $a_p^\dagger$ operators is the spaces of states for the theory, it has no association to any particular position

Charlie

12:26

hmm

and there is just one fock space?

Slereah

The Hilbert space of Klein Gordon isn't $L^2(\mathbb{R}^2)$

You can't calque it on "a function at points of the spacetime"

Charlie

we have an operator at every point in spacetime, but they all act on the same state space?

hmm

Slereah

Yes

Basically you select a point, you apply it to your operator, and that operator is applied to the Hilbert space

$\langle 0 | \phi(x) |0\rangle$ is a number

and that's your measurements

The Hilbert space you're using is the Fock space, ie it's a sum of every power of the one particle state

Charlie

If all operators act on the same state space, how do I distinguish a two particles at two different points?

or two different regions

Slereah

$$H = \bigoplus_{n = 0}^\infty \bigotimes_{i = 0}^n H$$

Charlie

12:33

I think I'm happy with what fock space is

at least basically

Slereah

@Charlie Talking about individual particles is stricky in QFT, but if you want a rough idea, you can consider them as little wavepackets

They're all just bumps in the same scalar field

Charlie

smooth bumps right

Slereah

ideally, yes

Charlie

hmm ok

Slereah

Coherent states, as they are called

They're the closest equivalent in quantum theory to "classical" particles

Charlie

13:04

so if the state of our quantum field is described a some vector in the fock space, can we (like in regular QM) choose a basis for the fock space like the position basis, and this is where we get the pop-science depiction of a particle in a quantum field as a "little wavepacket", ie. a bump in the scalar field as just mentioned

or rather maybe the analogue of the "position basis" is the basis generated by the field operator $\phi(x)$ hmm

ACuriousMind

13:15

it's not "the state of the field"

the field is an operator, not the state

Charlie

is state of the system more accurate?

ACuriousMind

what "system"? :P

Charlie

I don't know :c

ACuriousMind

Ontology in QFT is hard. In condensed matter contexts you can think about the state of the chunk of matter under consideration, but in the high-energy context there isn't really any system except "the universe"

Charlie

It's actually very relieving to see that the state space construction in qft is basically an extension of that of qm

now it feels a bit more approachable

Slereah

13:19

Well it is, but beware that it's also not unique

Charlie

oh

as in there are several different formalisms of qft?

Slereah

The Fock space feels like it's just point particles, but on the other hand, $L^2(D(\mathbb{R}^3))$ also is a Hilbert space for QFT

and the two are isomorphic

Charlie

is this somewhat analogous to the $L^2(\Bbb R)$ space from qm?

Slereah

Well yes, in the sense that both are the space of square-integrable functions on the classical configuration space

Charlie

I'm wondering just how much carries over, like can we construct a basis on the fock space from the eigenvectors of the field operators?

ok that's good

Slereah

13:21

The configuration space of a classical point particle is $\mathbb{R}^3$, the configuration space of a classical scalar field is $D(\mathbb{R}^3)$

Charlie

I'm getting a bit ahead of myself here but still

I'm not familiar with the notation $D(\Bbb R^3)$

Slereah

That's the space of Schwartz functions

ie functions where the field and its derivatives go to zero at infinity

Charlie

oh I hadn't thought about that but I guess it makes sense

Slereah

It's not mandatory to use, but it's more practical

Charlie

is any assumption made about the field operators $\hat \phi(x)$? are they hermitian and stuff?

Slereah

13:24

If it's a real scalar field, it should be yes

Charlie

that seems to suggest that when we're not dealing with the simplest possible field theories the field operators aren't necessarily hermitian and so the "finding a basis for the hilbert space from the eigenvectors of some operator" isn't used in qft

Slereah

13:39

Well I mean, even if it's a complex field, you can always split it

Just make it into a real part and imaginary part

Boom, it's two real observables

Charlie

:o

I'm reading atm that interacting theories mess up the fock space formalism

but that seems like something I shouldn't worry about atm

ACuriousMind

@Charlie you're right that that isn't much used, but that's because for interacting theories we usually don't really know the Hilbert space (Haag's theorem)!

Slereah

We just pretend it's the Fock space :p

And hope nobody notices

Charlie

is qed an "interacting theory"? I suspect it is, am I going to have the hilbert space formalisms forcibly taken from me soon?

Slereah

It is, yes

Charlie

13:45

damn

well, I'll enjoy it while it lasts

Slereah

although it's a bad idea to pick EM as your first interacting theory

It has a lot of complications

Usually the first one is the interacting scalar field

Charlie

I'm just following p&s atm so whatever comes first in there is what I plan to cover

Slereah

PS uses scalar fields yeah

Charlie

14:36

I haven't seen this explicitely said in p&s yet, if $\hat\phi(\mathbf x)$ creates particles, which operator annihilates them?

ACuriousMind

$a^\dagger$ creates particles, not $\phi$.

but when you apply $\phi \sim a + a^\dagger$ to the vacuum, then the $a$ part does nothing, so $\phi(x)\lvert 0\rangle$ is just the action of the creation operators "in" $\phi(x)$

Charlie

ahhh

what if we act a second time with $\phi(x)$?

ACuriousMind

then you get some state that doesn't have a definite number of particles

Charlie

is it common to act twice with $\phi$?

ACuriousMind

I don't know what that means :P

We don't usually "act" with operators all that much because (see above) we don't really understand the Hilbert space we're in

Charlie

14:40

oh

ACuriousMind

QFT is largely computing expectation values in the vacuum or some ensemble, not mucking about with specific states

Charlie

oh ok that is good to know then

Slereah

Well

It depends what field you study in QFT

High energy physics, quantum optics, statistical mechanics and relativistic chemistry all have different goals

ZeroTheHero

15:12

@BioPhysicist are you there?

BioPhysicist

15:34

@ZeroTheHero Kind of

rob

18:00

0

Q: If 2 metal spheres are in contact and a charged rob is brought closer, what would the charge on the spheres be when the are separated?

Two charged metal spheres L and M are placed in contact. Before contact L had a net charge of +10e and M a net charge of -10e. A positively charged rod is brought close to L, but does not touch it. The two spheres are then separated and the rod is withdrawn. What is then the charge on each sphere...

When I read "charged rob" in that question title, I was a little ... shocked.

8

Nike Dattani

21:24

This question may be of interest to anyone in experimental physics (or even some theorists too), especially since it seems that Raymond Birge was the one that popularized the technique while trying to fix inconsistencies with measurements of "fundamental physical constants"

2

Q: When was the "Law of Propagation of Error" first stated?

This is my first time posting on HSM, so please bear with me if it's off-topic. I can move it to Stats.SE or Mathematics.SE if necessary. A widely cited 1966 paper (with currently 1030 citations) mentions the "law of propagation of error" but does not actually state it or give any citation direct...

mathematics physics reference-request discoveries calculus

Charlie

22:04

Is $SO(3)\subset O(1,3)$?

ACuriousMind

22:37

@Charlie yes, all spatial rotations are also Lorentz transformations

Charlie

ah of course thanks

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