Anonymous at Mon, 28 Oct 2024 18:59:45 UTC No. 16453873

>>16453803
Its all explained in physics books but i guess you want cleverer insight.
The gist is just to say you are going to do a quantization of an extended object, not just something with a few variables like position or speed, but of something extended, so a field. You need an infinite amount of numbers to describe an extended object, think of something as simple as a tense string (like a common guitar string), its a one dimensional object made of an infinite amount of points, and each point can take a different value, as far as displacement and speed goes.
So to quantize this, you just have to create an infinite amount of operators. Every observable gets an operator, and theres infinite amount of them.
The extended object doesnt have to be a wavefunction but in practice it is. Scientists pretend that the wavefunction is a classical field, that something like the Dirac equation describes a common field like a pressure or voltage field, and then quantize it. Its a game, as if there were ever classical electron wave equations akin to those for photons in classical electromagnetism
You just say "lets say the electron is like a photon, a classical wave and not a point, lets quantize this extended object".

Anonymous at Mon, 28 Oct 2024 19:39:27 UTC No. 16453911

>>16453803
>>wave functions become operators
It doesn't make much sense so don't worry about it.
It's mostly useful as a mnemonic, since the single-particle part of a QFT Hamiltonian looks like the QM expectation value with the wave functions replaced by field operators.

Anonymous at Mon, 28 Oct 2024 19:43:08 UTC No. 16453915

>>16453803
>>wave functions become operators
They don't. Fields become operators. Wavefunctions are projections of states onto real space, ie [math]\langle x | \psi\rangle[/math]. Quantized fields are operators, whose eigenstates are momentum eigenstates [math]|p\rangle[/math] + spin stuff. The wavefunction of a momentum eigenstate is [math]\langle x | p \rangle = e^{ixp}[/math], ie a traveling plane wave.
>I just don't get what this means in terms of physical effects
Non-commuting fields result in addition interactions during propagation. In the path integral language, since the action S is composed of non-commuting operators, its exponential doesn't follow the usual composition law [math]e^xe^y = e^{x+y}[/math]. Instead, one has to use the Baker–Campbell–Hausdorff formula [math]e^Xe^Y = e^{X+Y+[X,Y]/2 + \ldots}[/math]. These additional terms manifest themselves as loops in Feynman diagrams, which we interpret as virtual pair creation and annihilation.

Anonymous at Mon, 28 Oct 2024 20:04:33 UTC No. 16453935

>>16453873
>>16453911
>>16453915
wrong

Anonymous at Mon, 28 Oct 2024 22:18:44 UTC No. 16454080

>>16453873
>simple as a tense string (like a common guitar string), its a one dimensional object made of an infinite amount of points, and each point can take a different value,
Isn't it limited simply by virtue of it being taut, sets limits as to the ways in which any point can possibly move, as well as variables like the material of the string

Anonymous at Mon, 28 Oct 2024 22:21:22 UTC No. 16454083

>>16453873
>>16454080
I want to strangle every fucking retard who describes harmonic oscillators as string. Pointless bullshit. Free quantum fields indeed behave like quantum harmonic oscillators locally. But all this overphilosophizing about le extended objects without the supporting math outs you as pseuds.

Anonymous at Mon, 28 Oct 2024 22:23:19 UTC No. 16454087

>>16454083
>Free quantum fields
Such as? How much might an expensive one be?

Anonymous at Mon, 28 Oct 2024 22:34:21 UTC No. 16454096

>>16454087
tree fiddy

Anonymous at Tue, 29 Oct 2024 00:13:19 UTC No. 16454203

>>16453803
>lust provoking image

Anonymous at Tue, 29 Oct 2024 01:37:48 UTC No. 16454297

>>16453803
>I just don't get what this means in terms of physical effects
Position and momentum bases are only non-relativistic approximations If you try to localize a particle below the Compton wavelength then pair production has to be accounted for. If you try to localize the entire interaction region to a point you get a black hole.

Anonymous at Tue, 29 Oct 2024 04:46:29 UTC No. 16454452

>>16453803
>wave functions become operators
No, that's not how to think of it. Here's the difference in paradigm.

Ordinary single particle quantum mechanics ("first quantization")
>Spatial position is represented by an operator X
>Time doesn't correspond to an operator, rather all operators in the Heisenberg picture depend on time t as a parameter
>The ground state of the Hamiltonian is represented by some ket |0>.
> If we act on the ground state by X we get a new state X|0>. In the harmonic oscillator system this is the first excited level.
> There is a "propagator" G(t_2,t_1)=<0|X(t_2) X(t_1)|0>, which in the harmonic oscillator case is the amplitude for the first excited level to "propagate" from t_1 to t_2.
>If we consider a more general system besides the harmonic oscillator, information about the higher excited energy levels is mixed into the propagator too.

Now consider the "second quantized" paradigm
>The operator X is considered to be an abstract field that has nothing to do with space
>If as before X only depends on t, this is considered to be a quantum field theory on a spacetime with a single time dimension and zero space dimensions
>But we can add spatial dimensions as arguments X(x,t) to get a genuine quantum field theory
> The ground state |0> is often called the vacuum, and the first excited level X|0> (in the free theory) is considered to be a single particle state in this paradigm
> In the interacting theory X|0> and the propagator again includes information about the higher excited states, which are thought of as multiparticle states

Anonymous at Tue, 29 Oct 2024 06:02:04 UTC No. 16454498

>>16454080
>Isn't it limited simply by virtue of it being taut
The amount of points in the string isnt limited

Anonymous at Tue, 29 Oct 2024 06:07:41 UTC No. 16454503

>>16453803
Second quantization is exactly like first quantization except you have an infinite amount of things to quantize
Classical mechanics has the same, theres a mechanic for point particles, a mechanics for solid rigid bodies and then theres the classical field mechanics where you use concepts like a Lagrangian density. Fundamentally describing the mechanics of some vibrating membrane (Brane?) is just describing the single point mechanics of every point in the membrane.

Anonymous at Tue, 29 Oct 2024 09:02:41 UTC No. 16454593

>>16454452
>>The operator X is considered to be an abstract field that has nothing to do with space
>>But we can add spatial dimensions as arguments X(x,t) to get a genuine quantum field theory
What the fuck do you think a field is, you fucking retard? A field is a map between vector spaces. If it’s a scalar field, the map is phi : R^(1,3) -> C. Your first example isn’t an “abstract field” because it’s not a field at all.

And we don’t just “add” spatial dimensions willy-nilly. Any operator must be a unitary, projective representation of the Poincare group. The translation subgroup demands that these operators are fields in the same way the Lorentz subgroup demands they have a definite spin and orientation (C and P numbers). Spacetime coordinates function as indices of an infinite-dimensional group.

Anonymous at Tue, 29 Oct 2024 12:48:16 UTC No. 16454696

>>16454593
>you fucking retard?
Not surprised the reply I get here begins with this.

>If it’s a scalar field, the map is phi : R^(1,3) -> C. Your first example isn’t an “abstract field” because it’s not a field at all.
You say this, when I said
>If as before X only depends on t, this is considered to be a quantum field theory on a spacetime with a single time dimension and zero space dimensions
>single time dimension and zero space dimensions
>not R^(1,3)
Try reading and comprehending first before you type your reply

Anonymous at Wed, 30 Oct 2024 12:25:24 UTC No. 16455792

>>16453803
Quantum fields have a wave function just as particles do. Pure states of a quantum field are superpositions of classical field configurations.
The reason QFT usually focuses on operators instead of wave functions is really about fermions, imo. Fermions demand antisymmetrized wave functions like

|(1,2,3) > = |123> +|312> + |231> - |213> - |321> - |132>

The approach is to represent this with creation operators acting on the vacuum:

a_1 a_2 a_3 | 0>

Then to enforce the antisymmetry in these operators by making them have fermionic commutation relations.
First and Second quantization are old fashioned terms, I think they are unhelpful. The state of QFT teaching and textbooks is abysmal, there is a lot of confusion in the subject even amongst mid-level experts.

Anonymous at Wed, 30 Oct 2024 22:56:53 UTC No. 16456372

>>16455792
>antisymmetry
How many symmetries are there?
And how many antisymmetries?