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Stream: theory: science

Topic: Quantum Interactions

Keith Elliott Peterson (Dec 08 2024 at 05:57):

I'm not sure where to ask this, but here seems appropriate.

What exactly do physicists mean when two fields interact (according to a probability amplitude)? Is it just a transfer of energy, a.k.a ability to do work, from one field to another? Is the universe fundamentally just playing hot potato?

John Baez (Dec 08 2024 at 16:48):

I think I count as a physicist for this purposes of this conversation. So:

We never say "interact according to a probability amplitude".

In either classical or quantum field theory, we say two fields interact if there's a term in the Lagrangian that involves a product of those two fields, or their derivatives. For example in QED the photon field $A$ interacts with the electron field $\psi$ because the Lagrangian is the sum of three terms:

• one for the photon
• one for the electron
• and an 'interaction Lagrangian' $\gamma^\mu \overline{\psi}\psi A_\mu$

The details don't matter much here, though I'll say $\gamma^\mu$ are not fields: they're a bunch of matrices called "gamma matrices".

What matters is that when you put a term in the Lagrangian that involves both $A$ and $\psi$, it means that changes in the photon field cause changes in the electron field, and vice versa! Thus, we say they interact.

Keith Elliott Peterson (Dec 08 2024 at 16:51):

John Baez said:

What matters is that when you put a term in the Lagrangian that involves both $A$ and $\psi$, it means that changes in the photon field cause changes in the electron field, and vice versa! Thus, we say they interact.

What is a 'change'? An observable/beable event in one of the fields?

John Baez (Dec 08 2024 at 16:52):

If you look at the Lagrangian for the Standard Model (which in this picture is written in much more expanded-out way than any sane physicist would normally write it) you'll see it's a big fat sum of products of fields and their derivatives. Most of these terms describe interactions between different fields (= particles) in the Standard Model.

John Baez (Dec 08 2024 at 16:54):

In physics, 'change' means that something has a nonzero time derivative. E.g. if a rock is moving we say its position is 'changing'.

In this case, we say the electron or photon field changes if it has a nonzero time derivative.

I have no idea what the heck is an 'observable/beable event'.

Keith Elliott Peterson (Dec 08 2024 at 17:04):

John Baez said:

In physics, 'change' means that something has a nonzero time derivative. E.g. if a rock is moving we say its position is 'changing'.

In this case, we say the electron or photon field changes if it has a nonzero time derivative.

How exactly does one measure the nonzero time derivatives of a field? How do we know the fields are interacting? Do we include self-interactions of a field?

The example of a rock with some acceleration, because something hit it, classically interacted with it, is intuitive, but a localized excitation of a field "moving" is a different ball game, no?

John Baez said:

I have no idea what the heck is an 'observable/beable event'.

Sometimes the hardest part is learning the language.
Well, after learning the maths.

John Baez (Dec 08 2024 at 17:08):

How exactly does one measure the nonzero time derivatives of a field? How do we know the fields are interacting? Do we include self-interactions of a field?

Too many questions for detailed answers. Suffice it to say if you can measure some number at two times $t_1 < t_2$ and it takes different values at those two times, we say it must have 'changed' at some time or times $t$ with $t_1 \le t \le t_2$. This lets us determine, through lots of work, whether changing one thing makes something else change.

Yes, the Lagrangian also includes 'self-interactions'.

John Baez (Dec 08 2024 at 17:11):

The example of a rock with some acceleration, because something hit it, classically interacted with it, is intuitive, but a localized excitation of a field "moving" is a different ball game, no?

A particle like a rock has a position $x(t)$ that's a function of time, while a field has a value $\phi(t,x,y,z)$ that's a function of time and space; in either case we say it's changing if its time derivative is nonzero.

John Baez (Dec 08 2024 at 17:13):

In quantum field theory the value $\phi(t,x,y,z)$ is not just a number: it's a whopping big matrix of numbers. But it's just numbers, so we can talk about the derivatives of $\phi(t,x,y,z)$, like $\frac{\partial}{\partial t} \phi(t,x,y,z)$.

John Baez (Dec 08 2024 at 17:20):

By the way, I'm using $\phi$ as a generic name for any field, like the photon field $A$ or the electron field $\psi$.

Keith Elliott Peterson (Dec 08 2024 at 17:22):

I take it then that the derivative here is a kind of matrix derivative.

Can such a derivative be seen as a kind of generalized metric (in the sense of a generalized metric space)? So the "metric" isn't taking values in $\mathbb{R}_{0\leq}$ but something else? That is to say. does QFT take place in an enriched category? Would the zero-valued derivatives then be a kind of identity morphism?

Keith Elliott Peterson (Dec 08 2024 at 17:31):

By the way, it's better to speak of "excitations of fields" rather than "wave-particles", yes?

Or as Steven Weinberg says,

A photon is an elementary excitation of the quantized electromagnetic field.

Photons illuminated: a gentle intro to quantum optics

John Baez (Dec 08 2024 at 17:31):

I take it then that the derivative here is a kind of matrix derivative.

A matrix is a bunch of numbers. If each number is a function of $x,y,z,t$ you can take derivatives like $\frac{\partial}{\partial t}$ of all these numbers. That's what I'm talking about.

Can such a derivative be seen as a kind of generalized metric (in the sense of a generalized metric space)? So the "metric" isn't taking values in $\mathbb{R}_{0\leq}$ but something else? That is to say. does QFT take place in an enriched category? Would the zero-valued derivatives then be a kind of identity morphism?

None of this makes sense to me, at all. And to be really blunt, I don't understand why you'd pivot so fast from trying to learn the basics of a subject (like what's an 'interaction' in QFT), to trying to slap some enriched category theory on it.

John Baez (Dec 08 2024 at 17:34):

I just don't think that ever works well. I've applied category theory to various subjects, but I try hard to get a really detailed understanding of those subjects first - or at least the math of those subjects, because category theory functions in the realm of math.

John Baez (Dec 08 2024 at 17:38):

You could learn the quantum field theory of a free scalar field without enormous amounts of work - this is where people usually start. It'd pay to learn Schrodinger's equation in quantum mechanics first. All this stuff is tons of fun.

Keith Elliott Peterson (Dec 08 2024 at 17:38):

John Baez said:

None of this makes sense to me, at all. And to be really blunt, I don't understand why you'd pivot so fast from trying to learn the basics of a subject (like what's an 'interaction' in QFT), to trying to slap some enriched category theory on it.

I guess I'm impatient to learn as much and as quickly as I can. :sweat_smile:
I feel I have a lot of ground to cover to catch up.

Also, this is the category theory Zulipchat, so I had to tie it in somehow.

John Baez (Dec 08 2024 at 17:42):

If you want to learn as much and as quickly as you can, figure out where you stand in the process of learning physics, grab some books at your level, and start reading them, working some of the problems. I don't think it pays to skip stuff here: if you don't know quantum mechanics and electromagnetism (which is a classical field theory), you'll have trouble understanding quantum field theory, since that's a blend of quantum mechanics and classical field theory.

John Baez (Dec 08 2024 at 17:45):

I think it really pays to remember that all this stuff is fun. Skipping ahead is like going to a fancy French restaurant and stuffing food down your mouth, or skipping some of the courses, because you're trying to get through the meal as fast as possible.

John Baez (Dec 08 2024 at 18:01):

By the way, I think all of us should pay a lot of attention to how @David Egolfand @John Onstead are using this Zulip to learn lots of category theory. It's really somewhat new. It shows that if someone here wants to teach themselves something, and they put enough work into it, demonstrating their willingness to suffer through the nitty-gritty details, not rushing too much, they will attract their own teachers. I should probably copy them and see if I can get someone to help me learn something I want to learn.

(What do I want to learn that's close to category theory? Maybe class field theory, or etale cohomology, or... yeah, Galois descent.)

John Baez (Dec 08 2024 at 21:41):

Anyway, returning to the original question

What exactly do physicists mean when two fields interact...?

Let me answer in a bit more detail with an example. To keep things simple say we have two fields that only depend on time, $f(t)$ and $g(t)$. Say they evolve according to differential equations, as everything in classical/quantum mechanics/field theory tends to do. We could have

$\displaystyle { \frac{d^2 } {dt^2 } f(t) = - f(t) }$

$\displaystyle { \frac{d^2 } {dt^2 } g(t) = - g(t) }$

and then we'd say $f$ and $g$ don't interact, because the solution of the first equation doesn't depend on the second one, and vice versa. Or we could have

$\displaystyle { \frac{d^2 } {dt^2 } f(t) = - f(t) - \lambda f(t) g(t) }$

$\displaystyle { \frac{d^2 } {dt^2 } g(t) = - g(t) - \lambda f(t) g(t)}$

where $\lambda$ is some number. Now we say the two fields interact, because the solutions affect each other: you can't solve the two equations separately. We call $\lambda$ a coupling constant and we also say the two fields are coupled.

John Baez (Dec 08 2024 at 21:44):

Another common feature you see here is that the original equations are linear and thus easy to solve, while the coupled equations are nonlinear and much harder to solve.

John Baez (Dec 08 2024 at 21:45):

In quantum field theory we don't usually write down the differential equations. Although we sometimes do, it's more common to write down the 'Lagrangian', which is a more compact expression that you can derive the equations from by turning a crank.

Notification Bot (Dec 12 2024 at 18:16):

This topic was moved here from #theory: philosophy > Quantum Interactions by Matteo Capucci (he/him).