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The state space in quantum field theory

by nicf
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nicf
#1
Jan1-12, 03:33 PM
P: 4
I'm a grad student in math, and I've been trying to learn some physics on the side by taking some classes and reading books. I took a class on quantum field theory last semester that was taught out of Srednicki; the class was very good, but I found myself at the end with a conceptual question that I haven't been able to find an answer to. I've had a little trouble explaining what my question is in the past, but I'm going to do my best.

In quantum mechanics, there are states, which are elements of some pre-specified Hilbert space, and observables, which are operators on that Hilbert space. Most treatments I've seen have used this framework, with some adjustments, to set up QFT. The example everyone likes to start with is the (free) Klein-Gordon field. You find that an operator that satisfies the Klein-Gordon equation has to have a Fourier expansion that looks just so, and you call the operators that appear in it creation and annihilation operators for the particles represented by your field. You can get expressions for these operators in terms of your field just by pushing symbols around. You then take your state space to be the smallest Hilbert space on which these operators can act nontrivially and which contains something that's killed by the annihilation operators. This is, as I understand it, called a "Fock space." The states are all given physical interpretations that I feel pretty good about.

This all seems to go off without a hitch if you have more than one free field in your theory. Just tensor the Fock spaces together and there you go. But the impression I've been given is that once you introduce interactions, this all goes to hell. Certainly the analysis that allowed us to come up with the forms of the creation and annihilation operators doesn't work anymore. So my questions are:

(1) Did we change the state space when we changed the Lagrangian? This isn't how it worked in QM, is it? What's the new state space?

(2) Do we still have creation and annihilation operators? Do they act the same as in the free picture? Should I assign them the same physical interpretation?

(3) Somewhat relatedly, what is the physical interpretation of the operator that has the same name as the field? It's not an observable, is it? It's not even Hermitian in general.

I suspect the problem is that I'm just thinking of this whole thing wrong somehow. Anything anyone could say to set me straight would be greatly appreciated.
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sheaf
#2
Jan1-12, 03:44 PM
P: 203
Have you had the chance to look at this thread yet ?

http://www.physicsforums.com/showthread.php?t=388556

There is some interesting and relevant discussion there. There is a lot of discussion of the states in the Schroedinger picture, which isn't often discussed in the textbooks, which are mostly concerned with perturbation theory in the interaction picture in order to do scattering problems.
Chopin
#3
Jan1-12, 09:11 PM
P: 351
I'm still learning this stuff as well, so take what I say with a grain of salt, but I think I can answer at least some of your questions.

1) I'm not exactly sure what you mean when you say "state space". But fundamentally, what you're always doing when you do QFT is trying to find eigenstates of a given Hamiltonian. If you change the Hamiltonian, these states will of course change as well, just like they would in normal QM. So if that's what you mean, then yes, the state space will change.

Think of it like perturbation theory in normal QM--in that case, you're trying to describe the effects of a change to the Hamiltonian by expressing the new states as linear combinations of the old states. QFT is the same way--the perturbation is the interaction term that you add to the Hamiltonian, and all of the machinery of Dyson series, Wick's theorem, and Feynman diagrams is basically a way of expressing the effects of this interaction in terms of the original free-field eigenstates.

2) No. The fact that we're able to find creation and annihilation operators in free theory essentially means we've solved the theory--by acting with combinations of these operators on the vacuum, we're able to inductively construct every eigenstate of the Hamiltonian. As you probably know, it's impossible to exactly solve any interacting theory (with the exception of a couple of obscure toy examples), so that right away should illustrate that it's not possible to come up with creation and annihilation operators in those cases.

There are still analogues of one- and multi-particle states, but they work differently than free theory. In particular, there are still one-particle states [itex]|k\rangle[/itex], which obey the relation [itex]\langle 0|\phi(x)|k\rangle = e^{ikx}[/itex] just like they do in free theory. However, since there's no creation operator, there's no way to easily extend this to two-particle states. This is to be expected, because while an interacting theory can still have a concept of one particle travelling through space, the concept of two particles is different, because they won't just travel through space along their own trajectories forever--they may interact with each other. So what looks like two particles with momenta [itex]k_1[/itex] and [itex]k_2[/itex] at one point in time, may at some later point look like a different two-particle state, say with momenta [itex]k_3[/itex] and [itex]k_4[/itex] instead.

So clearly, things are more complicated than in free field theory. When you do scattering theory, there are a couple techniques for creating states that look like two-particle states in the far far past, or in the far far future, which you use build your S-matrix. These are called "in" states and "out" states, and are basically as close to two-particle states as you can get.

3) I'm still hazy on this. But as far as I can tell, [itex]\phi(x)[/itex] doesn't really have a simple physical interpretation. In the case of one-particle states, as I mentioned above, you have [itex]\langle 0|\phi(x)|k\rangle = e^{ikx}[/itex], so in that sense it sort of acts like it creates a particle at position [itex]x[/itex]. But when you get into multi-particle states, things get weird. Because in QFT the number of particles can change, the state [itex]\phi(x)|0\rangle[/itex] doesn't just create a single particle at position [itex]x[/itex], it also creates a whole host of multi-particle states as well. So as far as I can tell, the notion of having multiple particles at definite positions doesn't really make sense anymore when you're dealing with an interacting field theory, and the operator [itex]\phi(x)[/itex] has no real physical interpretation--it's just a building block that you can use to build other operators.

nicf
#4
Jan1-12, 09:43 PM
P: 4
The state space in quantum field theory

Might as well keep using the same numbers.

(1) By the state space I mean the Hilbert space where the states live. In, say, spin-0 single-particle quantum mechanics this is (once you pick a basis) just the space of all the wavefunctions. This doesn't depend on the Hamiltonian. The eigenstates of the Hamiltonian certainly depend on the Hamiltonian, but they live in the same big Hilbert space. Surely there is some analogous description in QFT; the Fock space seems to be it in the free-field case, but none of the authors I'm reading seem to want to talk about it anymore once interactions enter the picture, which makes me think I'm not understanding something.

(2) This is more or less where I landed too, but there's something that feels off about it. We say, for example, that a hydrogen atom consists of an electron and a proton. There's certainly interaction going on in this description --- that's the whole point --- but I'm somehow still allowed to say that that electron and that proton are there, and that's what the state looks like. Maybe the electron shoots off a photon every once in a while and reabsorbs it later, or maybe there's an electron-positron pair off in the corner somewhere. I imagine that the right description of the state of the hydrogen atom is some sort of superposition of all these things. What I'm complaining about isn't that I don't know exactly what that superposition is; it's that I don't know how I would even start to write it down.

The in and out state business is something that makes me feel a little queasy for exactly this reason: the state "looks like" two free particles for |t| large; what does it "look like" otherwise? What vector space does it even live in? What objects are we even talking about?

(3) This feels right to me too, but I share your feeling of haziness. How exactly did we go from "function that tells us what value the field has at each point" to "operator that's only good for bookkeeping"?
Chopin
#5
Jan1-12, 10:06 PM
P: 351
Edit: After re-reading your response, I think I misinterpreted it when writing this. I'll leave this post around, but it probably won't answer your question. I'll write another one with a couple more thoughts.

1) Ok, I think I see the source of some of your confusion. You need to make a distinction between the Hilbert space that the states live in, and the set of eigenstates of the Hamiltonian. You're making the assertion that once we find the eigenstates, we can just treat that set as the only states in existence, but that's not true, because that set isn't a complete vector space. For instance, if we have [itex]\phi_1[/itex] and [itex]\phi_2[/itex] as eigenstates of some Hamiltonian, then [itex]0.75\phi_1 + 0.25\phi_2[/itex] is also a permissible state of the system (although not one with definite energy). If we just restricted ourselves to the set of eigenstates, this state would not be one of them, meaning that our set of states wouldn't be a complete vector space.

The eigenstates are basis states of the Hilbert space, but they don't form the entire Hilbert space. So changing the Hamiltonian creates a new set of eigenstates, but they all still live in the same big Hilbert space.

2) I think the reason you can get away with talking about a hydrogen atom that way is that the interaction is weak enough that neither particle loses its identity. The same thing happens if you scatter two particles with very low energy--they'll bounce off of each other, but there will always be two particles going in and two coming out. However, if you scatter them with a higher energy, there starts to be enough energy for pair production, so you might have different particles come out than went in. In that case, it doesn't really make sense to talk about states with a fixed number of particles, because the interaction causes that number to change.

To the second part of this question, the state lives in the same Hilbert space as all the other states (see #1), but it just doesn't look like a specific free-field multi-particle state during the middle of the interaction--it looks like a superposition of a lot of different multi-particle states all put together. This is what Feynman diagrams are illustrations of.

3) I think the answer is "because we never really had any business asking where each individual particle was in the first place." In the specific case of a single particle, you happen to be able to, but in general I think the lesson of QFT is that the idea of multiple individual particles that each have their own identity is actually an illusion.
Chopin
#6
Jan1-12, 10:24 PM
P: 351
Ok, I'll try this again now that I read your question again. The central point of your question is, what sort of object do the states live in. The answer is: just one big Hilbert space.

When you construct a free-field theory, you pull states out of one Hilbert space, construct the eigenstates, and observe that they form a fairly regular subset of that space which we call Fock space. As you said, you can construct a multi-species system just by taking the tensor product of multiple Hilbert spaces.

The trick is, this tensor product is itself a Hilbert space, too. It just happens to be one with a special structure--the fact that it can be broken down into sub-spaces (I think those are called quotient groups, right? It's been a while since my abstract algebra course...) However, other than that special structure, there's nothing different about this Hilbert space than the other ones--it's still just a big vector space.

Adding interactions means that the eigenstates can't be described as the tensor products of smaller states anymore, because they interact with each other. But there are still elements of the big Hilbert space, they just can't be classified quite as cleanly as the eigenstates of the original Hilbert space could.

Does that clarify things at all? I know exactly what you're going through right now--I spent a lot of time trying to get my head around this same concept, and still don't think I completely get it. The more I understand it, though, the more I can see why everybody has such a hard time explaining it--it's just that things are a lot different in reality than one would naively think when coming from a classical perspective, where we think about the universe as having a fixed number of individual particles. The reality is that it only seems that way in certain cases--in others, the notion just doesn't make sense at all.
nicf
#7
Jan1-12, 11:03 PM
P: 4
There's an analogous affront to our naiveté when we go from (nonrelativistic) classical to quantum mechanics, though, and it's one we can completely resolve: the question "where is the electron?" in, say, the double-slit experiment doesn't have a well-defined answer, but the electron does have a state, and using that state you can read off the probability of finding the electron in some region of space at some specified time.

Similarly, although we may not be able to answer the question "how many particles are here right now?", is there some way to represent the state so that it contains some information about something I might conceivably care about, as we could do in QM? It seems strange to me that the end of the story would be "It's complicated, don't worry about it," especially if we're going to be making statements about what the state "looks like" if the interaction is "weak enough"; without knowing what's doing the looking-like I have trouble even understanding what such a statement means.
Chopin
#8
Jan1-12, 11:23 PM
P: 351
Well, first of all, you can ask questions about conserved quantities, like "how much energy is there?", or "how much charge is there?". In fact, since QFT is a local theory, you can even do one better and say "how much energy is there in this specific area?" You may not know how many particles are carrying the energy or what kinds, but you know how much there is, and you can watch it flowing around different areas of the universe.

So I guess the way to define a two-particle state when one particle is here and the other particle is in Alpha Centauri is to say that it's a state which has a packet of energy here, and another over there. When the particles are well-localized wave packets that are widely separated, then this notion makes sense. However, once they get close together, talking about the situation this way no longer has any meaning.
meopemuk
#9
Jan2-12, 01:54 AM
P: 1,746
nicf,

Let me tell you how I understand this. Though, I'm sure not everybody would agree with me.

Quote Quote by nicf View Post
(1) Did we change the state space when we changed the Lagrangian? This isn't how it worked in QM, is it? What's the new state space?
The space of states does not depend on the presence/absence of interaction. The same Fock space can be used to describe both interacting and non-interacting QFT systems.

Quote Quote by nicf View Post
(2) Do we still have creation and annihilation operators? Do they act the same as in the free picture? Should I assign them the same physical interpretation?
The same creation/annihilation operators can be used in both interacting and non-interacting case. The interaction manifests itself only by the presence of an interaction operator in the full Hamiltonian. In this sense QFT is not different from ordinary quantum mechanics.


Quote Quote by nicf View Post
(3) Somewhat relatedly, what is the physical interpretation of the operator that has the same name as the field? It's not an observable, is it? It's not even Hermitian in general.
There is no physical interpretation for the quantum field operator. This operator is simply a mathematical object, which is convenient for building relativistically invariant interaction Hamiltonians. See S. Weinberg, "The quantum theory of fields" vol. 1.

Eugene.
tom.stoer
#10
Jan2-12, 02:29 AM
Sci Advisor
P: 5,369
Introducing different fields with interactions simply means that you take their individual Fock spaces (electrons, photons, quarks, gluons, ...) and merge them together (tensor product).

The only big caveat is the quantization of gauge fields which means that you start with unphysical degrees of freedom, i.e. a large Hilbert space which has to be reduced to the physical Hilbert space; this is technically rather involved, especially in non-abelian gauge theories like QCD, but the physical picture is clear: you start with a 4-component photon field Aμ but you know that there are only 2 physical polarizations, so 2 components are unphysical and have tobe eliminated.
ytuab
#11
Jan2-12, 07:28 AM
P: 189
The quantum field theory is originally based on Klein-Gordon, and Dirac equation.
And as we know, the electron moves unfer Lorentz force, which can be expressed as interaction term of A (= covariant derivetive).

We can not ignore this covariant derivative, which expresses Lorentz force, so developed this form into no-abelian gauge theories such as weak force and strong force.

All these theories need to use Feynman diagrams and interaction term (+ creation and annihilation operators).
(I think there are no other forms in QFT, because the starting point is very limited.)

For example, in the beta minus decay (of the weak force), the neutron emits electron, and changes into proton.
(To be precisely, the down quark (-1/3e) of the neutron emits the electron (-e), changes into the up quark (+2/3) of the proton.)

[tex]d \to u + e^- (+ \bar{\nu})[/tex]

But considering the interaction term of QFT, the emitted electron must become very heavy W- boson, first, before becoming the very light electron. Because both of quarks and electrons are fermions, which needs interacting boson. So,

[tex]d \to u + W^- \quad and \quad W^- \to e^- (+ \bar{\nu})[/tex]

This means the minus charge of the down quark changes into the very heavy W- boson temporarily, and then changes into very light electron, according to the rules of the interaction term in QFT.
(The electron can not go out of down quark directly.)
And both of down quark and up quark are elementary particles, so the equation of down quark = up quark + electron is not satisfied. Because in this case, the elementary particle down quark is made of two particles, which means down quark is "composite" particle.

This is the reason why W boson and quarks are expected in QFT ?

Of course, the photon (= vector potential A) has no charge, we can not use this vector potential A as the interacting boson in the upper beta decay.
(The charge of W boson originates from "mathematical" Pauli matrices.)

But I have one question here. the electron and proton attract each other by Coulomb force. (Especially when they are close to each other, the attractive Coulomb force is very strong.)
In the beta dacay, this strong attractive Coulomb force is blocked by some "strong ?" repulsive force for them to decay ?
(Can the "weak" force block this Coulomb attractive force, and cause decay, though it is weak ?)
Chopin
#12
Jan2-12, 10:17 AM
P: 351
Ok, so after reading some of these other responses, I think maybe I should revise my answer to #2 a bit. The creation/annihilation operators are still there, they just don't do exactly the same thing in an interacting theory.

Certainly you can still operate on the vacuum with them, and that will still produce states in the Hilbert space just like before, although now that the Hamiltonian has changed, these states will no longer be eigenstates of it, which is what was useful about the creation/annihilation operators to begin with. However, in some situations, the states they give are still close approximations.

For instance, assume that we parametrize the interaction term in the Lagrangian with a coupling constant: [itex]\mathcal{L} = \mathcal{L}_0 + g\mathcal{L}_{int}[/itex]. Now consider the case where [itex]g[/itex] is vanishingly small. The Lagrangian is now very close to a free theory, where the C/A ops can construct the eigenstates, so it's reasonable to assume that the new eigenstates are nearly the same as the old, and that the overall structure of those eigenstates (i.e. lining up into regular patterns that we call Fock Space), is still basically right. As we turn up [itex]g[/itex] higher and higher, who knows what will happen, but at least in the low limit, that structure should still hold up.

Now we get to the part where I've seen texts start doing some handwaving. Even though we don't exactly know what position means in this theory yet, if it is to be any sort of approximation to the real world, there must be some sense that when two particles are very far apart, they don't interact very strongly, meaning that if we want to build a two-particle state where one particle is on Earth, and the other is in the Crab Nebula, we can basically just assume that [itex]g=0[/itex]. Under that assumption, we ought to be able to build localized wavepackets using the C/A operators, and the resulting state should be very close to an eigenstate of the full Hamiltonian as well, as long as the packets we build are very far apart spatially.

Now say we do that at some specific time, and then press play and let things move forward. Intuitively, those two particles will get closer and closer to each other, and it will start to become less and less correct to ignore the interaction. So describing the system at later times by using the product of two creation operators will be less and less of a good approximation. However, if we continue running things forward, past the collision, then eventually things will get far apart again (in some intuitive sense), and so we would once again be justified in describing the situation with just a pair of creation operators.

What I've just described are "in" states and "out" states. For an "in" state [itex]\Psi[/itex], if you take any operator [itex]A(x)[/itex], then the expectation value of the state [itex]\langle\Psi|A(x)|\Psi\rangle[/itex] gets closer and closer to [itex]\langle a_{k1}a_{k2}|A(x)|a^{\dagger}_{k1}a^{\dagger}_{k2} \rangle[/itex] as [itex]x_0\rightarrow -\infty[/itex], but looks crazier and crazier as you move forward in time. Similarly, an "out" state looks more and more like a state made of creation operators as [itex]x_0\rightarrow \infty[/itex], and looks weird in the far past. Finding the S matrix amounts to figuring out what set of "out" states you have to add up in order to build a certain "in" state, which basically means finding the superposition of free-field states that the system will asymptotically approach in the future, if it is set up such that it asymptotically approaches one specific free-field state in the past.
tom.stoer
#13
Jan2-12, 10:21 AM
Sci Advisor
P: 5,369
ytab, your post has not much to say about the original question and is partially wrong

Quote Quote by ytuab View Post
The quantum field theory is originally based on Klein-Gordon, and Dirac equation ...
... and Maxwell's equations

Quote Quote by ytuab View Post
And as we know, the electron moves unfer Lorentz force, ...
classical description w/o relevance here

Quote Quote by ytuab View Post
We can not ignore this covariant derivative, ...
we can't; it's at the core of QED

Quote Quote by ytuab View Post
All these theories need to use Feynman diagrams and interaction term
interaction terms - yes (constructed e.g. via gauge-covariant couplings); but Feynman diagrams - no! they are no construction principle but a specific approach towards a solution via a an special approximation scheme, perturbation theory

Quote Quote by ytuab View Post
Because in this case, the elementary particle down quark is made of two particles, which means down quark is "composite" particle.
no, the down quark is one elementary excitation in Fock space
nicf
#14
Jan2-12, 11:12 AM
P: 4
Quote Quote by meopemuk View Post
The space of states does not depend on the presence/absence of interaction. The same Fock space can be used to describe both interacting and non-interacting QFT systems.
This would be lovely. I have books that seem to imply that we can no longer assign a particle interpretation to the states when we're working with an interacting theory. Is this just, as Chopin said, because the states are no longer eigenstates of the Hamiltonian? Or is there something more complicated going on?

Quote Quote by meopemuk View Post
The same creation/annihilation operators can be used in both interacting and non-interacting case. The interaction manifests itself only by the presence of an interaction operator in the full Hamiltonian. In this sense QFT is not different from ordinary quantum mechanics.
How do you define the creation and annihilation operators then, say for a real scalar? It's certainly no longer true that (these equations are from Srednicki) [itex]\phi(x)=\int\frac{d^3k}{(2\pi)^32\omega}(a(k)e^{ikx}+a^\dagger(k)e^{-ikx})[/itex]; that depended on [itex]\phi[/itex] being a solution of the Klein-Gordon equation. (Incidentally, I'm still a little unclear on what role the operator version of [itex]\phi[/itex] is playing here and why it has to satisfy the classical field equation, but that's a different question.) Do we just define [itex]a(k)=\int d^3xe^{-ikx}(i\partial_0\phi(x)+\omega\phi(x))[/itex]? There seems to be something funny going on here with the time dependence. I'm a little lost.

Quote Quote by tom.stoer View Post
The only big caveat is the quantization of gauge fields which means that you start with unphysical degrees of freedom, i.e. a large Hilbert space which has to be reduced to the physical Hilbert space; this is technically rather involved, especially in non-abelian gauge theories like QCD, but the physical picture is clear: you start with a 4-component photon field Aμ but you know that there are only 2 physical polarizations, so 2 components are unphysical and have tobe eliminated.
This I think I'm totally fine with; it doesn't seem to affect the answer to my question all that much, in that there shouldn't be anything going on here that isn't already going on with the free fields. Right?
meopemuk
#15
Jan2-12, 02:59 PM
P: 1,746
Quote Quote by nicf View Post
This would be lovely. I have books that seem to imply that we can no longer assign a particle interpretation to the states when we're working with an interacting theory. Is this just, as Chopin said, because the states are no longer eigenstates of the Hamiltonian? Or is there something more complicated going on?
Yes, once you add interaction (e.g., the [itex] \phi^4 [/itex] term) to the free Hamiltonian, you get a funny situation that the vacuum state [itex]|0 \rangle [/itex] and one-particle states
[itex]a^{\dagger}(k)|0 \rangle[/itex] are no longer eigenstates of the Hamiltonian. In a sense, this means that single particle "interacts with itself", which is totally unacceptable. This also means that scattering matrix (or operator) calculated with this interaction is non-trivial in the 0-particle and 1-particle sectors of the Fock space. The latter problem is fixed by renormalization theory, which simply adds certain "counterterms" to the interaction Hamiltonian. The role of these counterterms is to suppress unwanted terms in the S-operator and make sure that there is no scattering in 0-particle and 1-particle sectors. In realistic theories, like QED, this renormalization is sufficient to obtain the S-matrix in a very good agreement with experiments. So, most textbooks do not bother to investigate the question of states in QFT any further. However, there are still a few important unanswered questions.

The renormalization trick cancels 1-particle and vacuum self-scattering. But it does not address the self-interaction in these states. Even in a renormalized theory the vacuum and 1-particle states are *not* eigenstates of the total Hamiltonian. There are two ways one can try to fix this problem. One way is to say that our original states, like [itex]|0 \rangle [/itex] and [itex]a^{\dagger}(k)|0 \rangle[/itex] are not true vacuum and 1-particle states and that [itex]a(k), a^{\dagger}(k) [/itex] are not true particle annihilation and creation operators. Then the task is to find so-called "dressed particle" states, which *are* eigenvectors of the renormallized Hamiltonian.

The other way to deal with this problem (and I like this way more than the other) is to say that our Hamiltonian is not good and that we need to modify the Hamiltonian so that [itex]|0 \rangle [/itex] and [itex]a^{\dagger}(k)|0 \rangle[/itex] become proper eigenstates of the new energy operator. This goal can be achieved in perturbation theory. Moreover, this modification (dressing) of the Hamiltonian can be done in a way which does not affect the S-matrix, so all good renormalized scattering results obtained with the traditional Feynman diagram approach remain valid.

If you want to read more about these ideas, you can start with

O. W. Greenberg, S. S. Schweber, "Clothed particle operators in simple models of quantum field theory", Nuovo Cim., 8 (1958), 378

and use Google Scholar's citation service to find more recent articles on this theme.

Quote Quote by nicf View Post
How do you define the creation and annihilation operators then, say for a real scalar? It's certainly no longer true that (these equations are from Srednicki) [itex]\phi(x)=\int\frac{d^3k}{(2\pi)^32\omega}(a(k)e^{ikx}+a^\dagger(k)e^{-ikx})[/itex]; that depended on [itex]\phi[/itex] being a solution of the Klein-Gordon equation. (Incidentally, I'm still a little unclear on what role the operator version of [itex]\phi[/itex] is playing here and why it has to satisfy the classical field equation, but that's a different question.) Do we just define [itex]a(k)=\int d^3xe^{-ikx}(i\partial_0\phi(x)+\omega\phi(x))[/itex]? There seems to be something funny going on here with the time dependence. I'm a little lost.
As I've said above, in a theory with a renormalized dressed Hamiltonian, creation and annihilation operators remain exactly the same as in the free theory. All interaction effects are present in the Hamiltonian. Once you have the Hamiltonian you can answer any physically relevant question. The *interacting field* and its connection to c/a operators is totally irrelevant, so you shouldn't bother thinking about it.

Eugene.
ytuab
#16
Jan2-12, 06:11 PM
P: 189
Quote Quote by tom.stoer View Post
ytab, your post has not much to say about the original question and is partially wrong

interaction terms - yes (constructed e.g. via gauge-covariant couplings); but Feynman diagrams - no! they are no construction principle but a specific approach towards a solution via a an special approximation scheme, perturbation theory

no, the down quark is one elementary excitation in Fock space
Hm, I think I should have added more explanation to #11.

According to the interaction term and creation (+ annihilation) operator methods, (of course, using Feynman diagram, it is very easy to imagine), the next interactions are allowed. For example,

[tex]e^- + e^+ \to \mu^- + \mu^+ [/tex]
This case means the annihilation of electron and positron, photon, and the creation of muon and antimuon.
Or

[tex]e^- + \mu^- \to e^- + \mu^-[/tex]
This case means the electron and muon interact with each other by Coulomb force (= photon ).
Or

[tex]d^- \to u^+ + e^- (+ \bar{\nu})[/tex]
This beta minus dacay shows that -1/3e down quark changes into +2/3e up quark and -e electron.
OK, the total charge is conserved.

But the third case of the beta decay is completely differernt from the upper two cases.
Of course, the down quark and up quark are elementary particles, which can not be divided.
But in the third case, clearly the down quark (of neutron) emits the one minus charge (-e) from its body, because its charge changes from -1/3e to +2/3e.
(This minus charge moves into W- boson, and then electron.)

I wonder this process means the down quark (or neutron) is a kind of the composite particle ?
(Because the thing such as "one minus charge" is clearly separated from the down quark.)

And as I said, the plus charge and minus charge attract each other stronger as they are closer to each other.
Especially, in the very narrow area of weak force, the Coulomb attractive force is very strong.
So, for them to decay, the strong "repulsive" force is needed to cancel this strong attractive force ?
But the rules of the interaction term method in QFT do not allow this complex and concrete process.
So simply, they define the up and down quarks are elementary fermions, and the weak force is mediated only by W- boson to match the interaction term ?
George Jones
#17
Jan2-12, 06:40 PM
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Quote Quote by nicf View Post
I'm a grad student in math, and I've been trying to learn some physics on the side by taking some classes and reading books. I took a class on quantum field theory last semester that was taught out of Srednicki;
A little off-topic, but
Quote Quote by George Jones View Post
Given your background, you might be interested in Quantum Field Theory: A Tourist Guide for Mathematicians by Gerald Folland

http://www.amazon.com/Quantum-Theory.../dp/0821847058

Although Folland doesn't cover as much as Peskin and Schroeder, Folland does cover a lot more than most rigourous math books on quantum field theory. Folland uses rigour where possible, and where physicists quantum field theory calculations have yet to be made mathematically rigourous, Folland states the mathematical difficulties, and then pushes through the physicists calculations.

I've been waiting for fifteen years for someone to write this book, and now I don't have time to read it.
tom.stoer
#18
Jan3-12, 01:43 AM
Sci Advisor
P: 5,369
@ytuab: again, Feynman diagrams are a mathematically tool for book-keeping, nothing else, and they have nothing to do with the definiton of the state space


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