Looking Under the Hood of Feynman Diagrams

In summary: I don't know, correspondence between these quantum states and the wave-functions in QFT?In summary, I'm currently working my way through Griffith's Elementary Particles text, and I'm looking to understand what's going on with the underlying Hilbert space of a system described using a Feynman diagram. I'm fairly well acquainted with non relativistic QM, but not much with QFT. In particular, I'd like to know how we would represent the initial and final states within the space for different types of particles, as well as how we can explain certain types of constraints on the interactions based on the symmetry of the underlying Hamiltonian.- I'd first like to know how to represent states of free
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Jdeloz828
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TL;DR Summary
I'm looking to understand what's going on with the underlying Hilbert space of a system described using a Feynman diagram.
I'm currently working my way through Griffith's Elementary Particles text, and I'm looking to understand what's going on with the underlying Hilbert space of a system described using a Feynman diagram. I'm fairly well acquainted with non relativistic QM, but not much with QFT. In particular, I'd like to know how we would represent the initial and final states within the space for different types of particles, as well as how we can explain certain types of constraints on the interactions based on the symmetry of the underlying Hamiltonian.

- I'd first like to know how to represent states of free particles in the underlying Hilbert space. Take an electron for instance. I'd imagine the state vector would be represented as something of the form:

Ψ(r, t) ⊗ u(p),
where Ψ(r, t) represents the spatial part of the vector, encoding information about position and velocity, and u(p) is a bispinor encoding information about the spin state. Continuing in this way, a quark state would be represented as:

Ψ(r, t) ⊗ u(p) ⊗ I ⊗ c,
where Ψ(r, t) and u(p) are the same as the electron, I is the isospinor representing the quark flavor state, and c is the color state. A mediator, such as a photon would be represented as:

Ψ(r, t) ⊗ ε,
where ε represents the polarization vector encoding the photons spin state. My first (possibly dumb) question is this: How does a system evolve from an initial state vector, represented as a direct product of objects appropriate to the particles involved, into a final state with an different underlying representation. Take, for example, the process of pair annihilation:

e+ + e- → γ + γ
Our underlying state vector would have to evolve from a state representing an electron and positron to one representing two photons:

(Ψ(r, t)electron ⊗ u(p)) ⊗ (Ψ(r, t)positron ⊗ v(p)) → (Ψ(r, t)photon 1 ⊗ εphoton 1 ) ⊗ (Ψ(r, t)photon 2 ⊗ εphoton 2 ),
but how is this possible given that they're vectors from two fundamentally different vector spaces?

My second question involves how we explain certain constraints on interactions in terms of the Hilbert Space, using the following example. In his text, Griffiths makes the claim that if the color singlet gluon occurred in nature, it could be absorbed by baryons (also color singlets) resulting in a long-range component of the strong interaction. How do we explain this claim in terms of the symmetry of the underlying Hamiltonian and the evolution of the state vector?

I apologize if my questions seem somewhat naive. As I said, I'm not very familiar with relativistic QM. If anyone can help shed some light on these matters and/or point me to some good resources for learning more about it, it would be much appreciated.
 
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Maybe I'm misunderstanding you, but in the annihilation case, the space on each side is a product of photon and Fermion spaces. On the LHS, the state is a product of two Fermion states and the photon vacuum state.
 
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  • #3
Jdeloz828 said:
Take, for example, the process of pair annihilation:

e+ + e- → γ + γ
Our underlying state vector would have to evolve from a state representing an electron and positron to one representing two photons:

(Ψ(r, t)electron ⊗ u(p)) ⊗ (Ψ(r, t)positron ⊗ v(p)) → (Ψ(r, t)photon 1 ⊗ εphoton 1 ) ⊗ (Ψ(r, t)photon 2 ⊗ εphoton 2 ),
but how is this possible given that they're vectors from two fundamentally different vector spaces?

I'm at the same stage in the book at the moment. I'm not thinking like this at all. These are just my thoughts at this stage, as a fellow student, not a definitive answer!

QED is a wider theory than non-relativistic QM. It hasn't started with the axioms of QM. Why would the processes in QED be describable by standard QM? If they could, why would we need QED?

These are "new ideas".

What's happening in QED should be explicable from QFT, but not from QM. My assumption, which I made without thinking about it, is that we are dealing with processes not described by unitary evolution of a wave-function. We are dealing with the interaction of light and electrons.

I never assumed that we could derive QED from QM. Not in a simplistic evolution of a particle wave-function.

One problem with QM is that it fails relativistically without the antiparticles. Therefore, single-particle QM is a special case where you can ignore the antiparticle. One you start adding antiparticles, you can't go back to your single particle wave-function.
 
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  • #4
Well, there some things to say here, first of all, I will not discuss how the quantum states are written in QFT, but you need to know that they are not wave-functions. To write the quantum states you need to work with quantified fields (that's why it's called Quantum Field Theory). But surely there is some state associated with the electron that carries information of its position, its momentum, and it's "spin", and a similar state associated to the quark that carries all the information of isospin, colour, etc...
It's also true that a single electron lives in some Hilbert space and that a single quark lives in another Hilbert space, and you must consider their direct product.

But what do you mean by "fundamentally different"?
To put this in context let's return to the NR-QM, imagine two electrons, if we only care about spin, each lives in a 2D Hilbert space, and the total system is described by the direct product of the two HS. Imagine that the initial state is ##\left|01\right>##, I'm sure you won't be surprised if I tell you that I can put an interaction to the Hamiltonian to convert this state, after some time, to the state ##\left|10\right>##.
In QED it's the same, to describe the system of photons and electrons you need to work in the direct product of the electron space and photon space and, by adding the proper interaction in the Hamiltonian, you can change an "electron state" to a "photon state" exactly as in NR-QM. You don't even need to add photons, in QFT the state describing an electron with momentum ##\vec{p}## and the state describing an electron with momentum ##\vec{q}## belong to different Hilbert spaces. And what I called the "electron space" it's the direct product of infinite Hilbert spaces (what it's called the Fock space).

Your second question it's easier but also much more complicated. In a very naive way, all the constraints are given by the interactions, so if I give you the interaction for QCD you will find that there are only 8 fields describing 8 mediators and that any of them is a singlet of colour. Of course, then one can ask why the interaction are the way there are, etc... and all this can complicate until we arrive at the edges of physics.
If I understand the argument by Griffiths, goes like this;
You know that any baryon (or meson) can interact by exchanging gluons, but because gluons have colour, and there is some property called "confinement" that says that we cannot observe particles with colour, then this interaction is only possible to very small distances. But if there were a singlet gluon, confinement wouldn't forbid some strong interact with infinite range, so the interaction that mediates this exchange of singlet gluon would be identical to the interaction that mediates a photon, so we would observe that, for example, the Coulomb's constant of electrical force will be different for a proton-proton interaction and a proton-electron interaction (because in one interaction there would be a contribution of the gluons, and in the other no). That's one argument to forbid a singlet gluon.
 
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  • #5
Thanks for the great responses. I think it all clicked when Keith_McClary pointed out that that we can include vacuum states in the fields on both sides of the equation. Guassian97, what I meant by "fundamentally different" was that the Hilbert spaces vectors are the direct product of different types of objects on the LHS than on the RHS. But, as was pointed out, if we can include the photon vacuum states on the RHS and electron/positron vacuum states on the LHS when taking our direct product, the vectors in the Hilbert spaces are now constructed in the same way. Let me know if I'm thinking about this the right way.

As for the singlet gluon question, I understand the high-level argument made by Griffiths, but I think I'll need to wait until I'm actually exposed to the Hilbert Space/operator description of QCD to really understand what's going on under the hood here.

After researching a bit last night, I think it's time I start looking into second quantization as a starting point to understand these things a bit better. Anyone have any recommendations on a good resource for this topic? The wikipedia page was..okay..at best. Something with problems/exercises would be ideal.
 
  • #6
Jdeloz828 said:
as was pointed out, if we can include the photon vacuum states on the RHS and electron/positron vacuum states on the LHS when taking our direct product, the vectors in the Hilbert spaces are now constructed in the same way. Let me know if I'm thinking about this the right way.

This should work for a scattering process, which appears to be what you are trying to describe. The "in" and "out" Hilbert spaces should be the same, so the Hilbert space has to include all degrees of freedom that are occupied in either the "in" or the "out" state. The "vacuum states" you refer to are simply degrees of freedom that are unoccupied in either the "in" or the "out" state.
 
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  • #7
Jdeloz828 said:
In his text, Griffiths makes the claim that if the color singlet gluon occurred in nature, it could be absorbed by baryons (also color singlets) resulting in a long-range component of the strong interaction. How do we explain this claim in terms of the symmetry of the underlying Hamiltonian and the evolution of the state vector?

I don't think you need to consider the symmetry of the Hamiltonian here. The process being described, in terms of Feynman diagrams, would have the same lines going in and coming out (two baryon lines go in, the same two baryon lines come out), so the only evolution of the state vector is an exchange of momentum due to the interaction.

The simplest non-zero-order diagram would have a single color singlet gluon line going between the two baryon lines. This is basically the same as the corresponding Coulomb interaction diagram in QED (two electron lines go in, two electron lines come out, one photon line exchanged between them), and would produce the same result--a long range interaction (since the exchange particle--gluon or photon--is massless).

The actual color interaction in QCD, of course, does not have this property, because there are no color singlet gluons. In other words, the actual color interaction does not allow gluon lines that don't carry any color charge (note that the gluon line in the hypothetical diagram above doesn't carry any color charge, just as the photon lines in QED don't carry any electric charge). You might try drawing some actual QCD diagrams to compare: for example, try a single gluon exchange between an proton (three quark lines, uud) and a neutron (three quark lines, udd).
 
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  • #8
Jdeloz828 said:
After researching a bit last night, I think it's time I start looking into second quantization as a starting point to understand these things a bit better. Anyone have any recommendations on a good resource for this topic? The wikipedia page was..okay..at best. Something with problems/exercises would be ideal.

I would say that this Griffiths text is a lot easier than anything serious on QFT. I've got QFT for the Gifted Amateur, and I was hanging on to the material there by the skin of my teeth. I finished Part I. I'm much more comfortable with the material in Griffiths so far. I'm on the problems in Chapter 4 now.

My plan is to do Griffiths and something on Relativistic QM before going back to QFT. The other issue with QFT is that you can't really do much with it problems-wise until you've gone a long way. A lot of it is like pure maths without the rigour!

I don't want to put you off QFT, but it's on a different planet from the early chapters in Griffiths at any rate. There are Tong's notes online here, if you want to take a look:

http://www.damtp.cam.ac.uk/user/dt281/qft/qft.pdf
 
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Related to Looking Under the Hood of Feynman Diagrams

1. What are Feynman diagrams?

Feynman diagrams are visual representations of mathematical equations used to describe the behavior of subatomic particles in quantum field theory. They were developed by physicist Richard Feynman in the 1940s.

2. Why are Feynman diagrams important?

Feynman diagrams provide a way to visualize and calculate the interactions between particles in quantum field theory. They have been instrumental in the development of the Standard Model of particle physics and have led to numerous experimental predictions that have been confirmed by experiments.

3. How do Feynman diagrams work?

Feynman diagrams use lines and vertices to represent particles and their interactions. The lines represent the paths of particles and the vertices represent the points where interactions occur. By following the rules of Feynman diagrams, physicists can calculate the probabilities of different particle interactions.

4. Can Feynman diagrams be used to study all particles?

Feynman diagrams are primarily used to study subatomic particles, such as quarks and leptons, and their interactions. They are also used in the study of forces, such as electromagnetism and the strong and weak nuclear forces.

5. Are Feynman diagrams still relevant today?

Yes, Feynman diagrams are still widely used in modern particle physics research. They continue to play a crucial role in understanding the fundamental interactions of particles and have been applied to various areas of physics, including cosmology and condensed matter physics.

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