What Are the States in Quantum Field Theory?

[hamster143]

Do you remember that, in QM, particles are indistinguishable?

So, a state with two particles in point A and point B could be described by a field that's nonzero in A and B and zero everywhere else.

To allow correlations between particle positions, for example, to describe a system with two particles separated by distance X, you can make a functional that gives you nonzero for field states that contain two particles at distance X and zero for all other field states.

Could someone please show the steps connecting the state in the "field eigenbasis" to the state in the "Fock basis" a bit more explicitly? Not just the math of "here is an equation", but the procedure ... ie. how we derive the connection between the two.

First you need to compute position representation of the vacuum wavefunctional. samalkhaiat did that in post #28.

Then, as you know, you can create all multiparticle states by acting with particle creation operator on the vacuum wavefunctional. That operator can be expressed through $\hat{\phi}$ and $\hat{\pi}$. We have position representations of them all (posts #30 and #31).

The process is analogous to construction of energy eigenstates of 1-D harmonic quantum oscillator in QM. First you prove that the lowest-energy state is

$$\psi = C \exp{(-\frac{m\omega x^2}{2})}$$

Then you act on $\psi$ with

$$a^{\dag} = \sqrt{\frac{mw}{2}} (x - \frac{1}{m\omega} \frac{\partial}{\partial x})$$

Because as soon as we try to add interactions, I have a feeling this is going to get even more confusing.

That's not even half of it. Things get _really_ confusing when you realize that, for fermions, "the field" is Grassmann-valued.

 

[meopemuk]

JustinLevy,

I think the only way to understand QFT is to accept that quantum fields are NOT some "physical objects" that can have "states" and "observables". Quantum fields are just purely mathematical constructs (abstract operators in the Fock space) which appear to be useful for building relativistic Hamiltonians. I suggest you to re-read Weinberg's vol. 1. This book is excellent in everything except its title. QFT is not about dynamics of fields. QFT is a theory about systems with varying numbers of particles. Quantum fields play only a technical role there.

Eugene.

 

[TriTertButoxy]

JustinLevy,

I think the only way to understand QFT is to accept that quantum fields are NOT some "physical objects" that can have "states" and "observables". Quantum fields are just purely mathematical constructs (abstract operators in the Fock space) which appear to be useful for building relativistic Hamiltonians. I suggest you to re-read Weinberg's vol. 1. This book is excellent in everything except its title. QFT is not about dynamics of fields. QFT is a theory about systems with varying numbers of particles. Quantum fields play only a technical role there.

Eugene.

Oh! I wholeheartedly disagree here. The quantum fields are physical objects. The quantum electric and magnetic fields are most certainly observables. The relationship of photons and the electromagnetic fields lies at the heart of quantum optics.

In fact, I view quantum fields in precisely the manner that samalkhaiat has described in his lengthy expositions. And furthermore, particles, are an emergent phenomenon of quantum field theory.

 

[Physics Monkey]

What is the operational (experimental) definition of this "probability"? Experimentalists know pretty well how to measure particle observables (position, momentum, spin, etc.) and how to define associated probabilities. What is your advise for experimentalists about how to measure "the probability density for the field to be in the configuration $\phi(\vec{x})$ at time t"?

Eugene.

We don't have to restrict our attention to particle scattering experiments only. For example, we can ask about the physical properties of a quantum magnet consisting of many spins. Such a material is well described by quantum field theory (potentially without Lorentz invariance) in certain regimes. If we imagine the quantum spins are Ising like variables then the wavefunctional of the corresponding scalar field actually gives something perfectly physical: the probability to find a particular local configuration of spins (in a course grained approximation). Of course, the entire magnetization as a function of space may be difficult to measure in practice, but it certainly makes physical sense.

 

[Physics Monkey]

That makes intuitive sense, until this:

I get lost there.
In the Fock basis, there would be a function of N positions for the N particle basis, right? And then there'd be a sum over all N.

This seems to allow correlations between particle positions that don't seem possible to me with a function of only one position.

Something like that, although I don't understand your second comment.

Could someone please show the steps connecting the state in the "field eigenbasis" to the state in the "Fock basis" a bit more explicitly? Not just the math of "here is an equation", but the procedure ... ie. how we derive the connection between the two. Because as soon as we try to add interactions, I have a feeling this is going to get even more confusing. And how are antiparticles represented?

Let's consider a single real scalar field so there are no anti-particles to worry about. You agree that an arbitrary state may be written in the form $$| \psi \rangle = | \text{no particles} \rangle + | \text{one particle} \rangle + | \text{two particles} \rangle + ...$$ where $$| \text{two particles} \rangle$$ means any state consisting of two particles (and similarly for the other terms). For example, the most general form of $$| \text{two particles} \rangle$$ is something like $$\sum_{k_1 k_2} c_{k_1 k_2} | k_1 k_2 \rangle$$. To convert this Fock basis representation into the field eigenbasis you need to know the overlap between states like $$|k_1 k_2 \rangle$$ and $$| A \rangle$$ where $$| A \rangle$$ is a field eigenstate.

This overlap is something you can calculate using the definition of $$|k_1 k_2 \rangle$$ in terms of creation operators and the relation between creation operators of definite momentum and the field operator. I'll show you how its done for the one particle state schematically. $$| k \rangle = a_k^+ | \text{vac} \rangle$$ and $$\langle A | k \rangle = \langle A | a_k^+ | \text{vac} \rangle \sim \langle A | \int d^3\,x\, e^{i k x} \phi(x) | \text{vac} \rangle = \int d^3\, x \,e^{i k x} A(x) \langle A | \text{vac} \rangle$$. This last equality follows from the fact that $$| A \rangle$$ is a field eigenstate. All you need to know is the vacuum overlap and you can calculate the overlap with any single particle state. You should be able to work out the pattern for multiparticle states and fill in the other details. One important detail is that you really need to use both the canonical momentum and field operators, but the canonical momentum operator also has a relatively simply action on field eigenstates (it acts as a functional derivative).

Is this what you had in mind?

 

[meopemuk]

We don't have to restrict our attention to particle scattering experiments only. For example, we can ask about the physical properties of a quantum magnet consisting of many spins. Such a material is well described by quantum field theory (potentially without Lorentz invariance) in certain regimes. If we imagine the quantum spins are Ising like variables then the wavefunctional of the corresponding scalar field actually gives something perfectly physical: the probability to find a particular local configuration of spins (in a course grained approximation). Of course, the entire magnetization as a function of space may be difficult to measure in practice, but it certainly makes physical sense.

Non-relativistic QFT as used in condensed matter applications is quite different from the relativistic QFT in particle physics. The fields used in condensed matter physics usually have well-defined physical meanings (magnetization, elastic deformation, etc.). The same cannot be said about particle fields in, e.g., QED.


To the best of my knowledge, relativistic quantum field theories are good only for calculating the S-matrix = scattering cross-sections of *particles* and energies of bound states of *particles*. I don't know any distinctive field-related quantity that was predicted by a relativistic QFT and successfully measured in experiment.

Eugene.

 

[Physics Monkey]

Non-relativistic QFT as used in condensed matter applications is quite different from the relativistic QFT in particle physics. The fields used in condensed matter physics usually have well-defined physical meanings (magnetization, elastic deformation, etc.). The same cannot be said about particle fields in, e.g., QED. To the best of my knowledge, relativistic quantum field theories are good only for calculating the S-matrix = scattering cross-sections of *particles* and energies of bound states of *particles*. I don't know any distinctive field-related quantity that was predicted by a relativistic QFT and successfully measured in experiment.

Eugene.

I have to disagree with you here. For example, the electromagnetic field features in both high energy physics and condensed matter physics, and I would argue that it has a perfectly well defined meaning. One can do all kinds of things with this theory that don't involve calculating S-matrix elements.

Or think about early universe cosmology. At finite temperature there can be no S-matrix since particles cannot get far away. Even the concept of a reasonably well defined particle like excitation may disappear. Similar considerations apply at finite density like in neutron stars, etc. In all these cases one is interested in hydrodynamic response or collective modes which are not directly related to particle concepts. Of course, you may also be interested in some particle like observables as well. A more down to Earth example is the RHIC fireball where one has a strongly interacting soup of quarks and gluons, deconfined but not really free.

 

[meopemuk]

One can do all kinds of things with this theory that don't involve calculating S-matrix elements.

My point is that in QED one cannot calculate anything beyond the S-matrix, i.e., the correlation between asymptotic states in the remote past and future. Only when you consider this infinite time interval there is a cancelation between divergent terms in the S-matrix expansion and divergent counterterms in the QED Hamiltonian. So, you can obtain accurate renormalized scattering amplitudes.

If you decide to calculate the time evolution at finite times you'll see that the Hamiltonian of QED is basically divergent, that the time evolution operator does not exist, and that results are meaningless.

I don't know any first-principle calculation of the time evolution in electron-positron-photon systems based on the QED Hamiltonian. If you know one, I would appreciate a reference.

Eugene.

 

[schieghoven]

Hello all, I missed a whole lot of this conversation, but I'm delighted to see the group's thoughts on this topic.

Regarding this point which seemed to be left open...

That means that states do not transform in a trivial way.
To compute the outcome after a boost, we have to compute the wavefunction on the "new" hypersurface from the wavefunction on the "old" hypersurface. To do that, we need dynamics of the field and the exact Hamiltonian.

To correct my earlier remark, I think that the transformation law is still a unitary representation, but it's not like any usual representations we normally see.

I agree with this point, and even better, I'll provide the representation for the free Dirac field.
$$\begin{align} \mathcal{P}_j &= \partial_j \\ \mathcal{J}_j &= -\epsilon_{jkl}x_k \partial_l + J_j \\ \mathcal{P}_0 &= -iH = -\alpha_k \partial_k -i\beta m \\ \mathcal{K}_j &= x_j \mathcal{P}_0 + K_j = -x_j \alpha_k \partial_k -ix_j \beta m + K_j \end{align}$$
These should be regarded as operators on the single-particle state space $L^2(\mathbb{R}^3,\mathbb{C}^4)$, i.e. square-integrable Dirac amplitudes as a function of space. Here J_j and K_j are the local part transforming the spin-amplitudes,
$$J_j = \frac{-1}{2}\begin{pmatrix} i\sigma_j & 0 \\ 0 & i\sigma_j \end{pmatrix}, \qquad K_j = \frac{1}{2}\begin{pmatrix} \sigma_j & 0 \\ 0 & -\sigma_j \end{pmatrix}$$

The basic idea is that the generator of time translations is the (free) Dirac hamiltonian, and this must be complemented with three boost generators $\mathcal{K}_j$ so that the set of ten provides a rep of the Poincare algebra... for example,
$$\begin{align*} [\mathcal{P}_i,\mathcal{K}_j] &= [\partial_i, x_j\mathcal{P}_0 + K_j] \\ &= [\partial_i, x_j\mathcal{P}_0] \\ &= [\partial_i, x_j] \mathcal{P}_0 \\ &= \delta_{ij} \mathcal{P}_0 \end{align*}$$
as required; all the other commutators can be computed with a little effort and agree with the standard results of e.g. Foldy 1956 or Weinberg's text. Integrating the $\mathcal{K}_j$ (i.e. solving the corresponding PDE) would provide the correspondence between amplitudes in different inertial frames, just as integrating $\mathcal{P}_0$ provides the correspondence between amplitudes at different times. As @hamster143 noted, the boost operators are intimately related to the dynamics.

Moreover, this is a unitary representation: all of the expressions (1-4) are anti-Hermitian with respect to the standard norm on L^2, so they exponentiate to unitaries. As far as I know this is unique to the Dirac system: spin-one and scalar KG admit Poincare reps, but fail to be unitary.

These expressions show that the probabilistic interpretation of a one-particle *fermion* configuration space is relativistically covariant, with the function space L^2 providing a concrete realisation of state space. Not sure how things work in the photon sector though... obviously a big sticking point.

Any thoughts?

Dave

 

[schieghoven]

My point is that in QED one cannot calculate anything beyond the S-matrix,
I don't know any first-principle calculation of the time evolution in electron-positron-photon systems based on the QED Hamiltonian. If you know one, I would appreciate a reference.

Gallavotti has indicated for some years that QFT perturbative series seem to have many features in common with the so-called Lindstedt series describing perturbations of classical Hamiltonian integrable systems. I think this is a terrific line of inquiry speculating whether integrable systems theory might have something to say about quantum theory. In order to apply these ideas he would need a QED Hamiltonian in precisely the sense you have suggested, but I don't think he has one. And if such a formalism existed, he would probably find out pretty quick... he's an editor of Commun. Math. Phys. (!)

This is quite an interesting review by him
(Gallavotti07) Gallavotti, G., class-phys/0711.2544, 2007

Best

Dave

 

[Physics Monkey]

My point is that in QED one cannot calculate anything beyond the S-matrix, i.e., the correlation between asymptotic states in the remote past and future. Only when you consider this infinite time interval there is a cancelation between divergent terms in the S-matrix expansion and divergent counterterms in the QED Hamiltonian. So, you can obtain accurate renormalized scattering amplitudes.

If you decide to calculate the time evolution at finite times you'll see that the Hamiltonian of QED is basically divergent, that the time evolution operator does not exist, and that results are meaningless.

I don't know any first-principle calculation of the time evolution in electron-positron-photon systems based on the QED Hamiltonian. If you know one, I would appreciate a reference.

Eugene.

Once again I must disgree. First of all, QED has a cutoff and is a perfectly reasonable low energy effective theory from which all kinds of things can be computed. Second, as a matter of principle no one has ever measured an S-matrix element because no one has ever waited for infinite time. It seems a silly point, but it's important to realize that we are not really measuring S-matrix elements. We measure something very close, but it can't and doesn't require the limit of infinite time. Quantum field theory with a cutoff is a perfectly reasonable system to consider finite time evolution and this is what QED is.

 

[meopemuk]

First of all, QED has a cutoff

What is the value of this momentum cutoff and why?

Quantum field theory with a cutoff is a perfectly reasonable system to consider finite time evolution and this is what QED is.

Creation operators in QED (even with a cutoff) create "bare" electrons, whose states are not eigenstates of the full Hamiltonian, so they do not resemble the real physical electrons at all. The time evolution of "bare" states is completely unphysical. Traditional QED cannot calculate the time evolution of real "dressed" states. I've advertised the alternative "dressed particle" theory on this Forum many times. In my opinion, it is the only consistent approach to the time evolution in QFT.

Eugene.

 

[Physics Monkey]

What is the value of this momentum cutoff and why?

QED is certainly cutoff at the electroweak scale where it becomes part of a larger non-abelian gauge theory. You'll find this is true of almost everything. Regardless of what you may think the ultimate high energy theory is, the low energy physics tends to look like a quantum field theory as long as you use the concept of effective field theory.

Chiral perturbation theory, heavy quark effective theory, soft collinear effective theory, Fermi theory of beta decay, etc are all effective theories. We even know higher energy completions of these theories, but it is still often the effective theories that we use to make predictions. QED as applied to the real world is such a theory. We know it isn't good to arbitrary high energy, but we use it quite successfully anyway.

Creation operators in QED (even with a cutoff) create "bare" electrons, whose states are not eigenstates of the full Hamiltonian, so they do not resemble the real physical electrons at all. The time evolution of "bare" states is completely unphysical. Traditional QED cannot calculate the time evolution of real "dressed" states. I've advertised the alternative "dressed particle" theory on this Forum many times. In my opinion, it is the only consistent approach to the time evolution in QFT.

Eugene.

Sorry, Eugene, but I don't agree with you here. The question of whether bare particles and dressed particles are similar is a question for the theory. Sometimes they are. In QED the bare and dressed particles are actually pretty similar. I also don't agree that the time evolution of bare particles is unphysical. For example, suppose I produce a high energy nearly free quark in some collision. This quark is effectively a bare particle. Now a very interesting question is how this quark dresses itself into physical hadrons and mesons as it loses energy. This dressing happens quite fast (a time scale set by lambda qcd), and the whole finite time process is relevant for jet physics, RHIC physics, and much more.

On a more positive note, I think the dressed particle notion is more widespread than you realize. I certainly learned about it. Perhaps it isn't presented as "purely" as you might like, but it's out there. I regard it as part of the physics of quantum field theory (or really quantum physics in general) where the physical states get dressed up by interactions. One can even observe things like this in real time in condensed matter systems where the interaction strength is controllable.

 

[meopemuk]

I also don't agree that the time evolution of bare particles is unphysical. For example, suppose I produce a high energy nearly free quark in some collision. This quark is effectively a bare particle. Now a very interesting question is how this quark dresses itself into physical hadrons and mesons as it loses energy. This dressing happens quite fast (a time scale set by lambda qcd), and the whole finite time process is relevant for jet physics, RHIC physics, and much more.

I don't know much about quarks, so let us stick to (bare) electrons in QED. From your logic it follows that once a single electron is prepared, the next time instant we are dealing with the electron "dressed" with infinite number of photons and electron-positron pairs. So, your claim is that this "dressing" is a real physical process that can be measured in experiments eventually. There is no such thing as single isolated electron moving peacefully through space. Is it so?

Eugene.

 

[Physics Monkey]

I don't know much about quarks, so let us stick to (bare) electrons in QED. From your logic it follows that once a single electron is prepared, the next time instant we are dealing with the electron "dressed" with infinite number of photons and electron-positron pairs. So, your claim is that this "dressing" is a real physical process that can be measured in experiments eventually. There is no such thing as single isolated electron moving peacefully through space. Is it so?

Eugene.

I didn't say anything about instantaneous dressing or infinite numbers of photons at the "next time instant", I said the real time dynamics of the dressing process can be quite interesting. I think it's fair to say that the process is less interesting in QED than QCD where something really dramatic happens.

I also didnt say that there is no such thing as a "single isolated electron moving peacefully through space". Clearly we can prepare such states, but we can prepare other kinds of states as well. Let's not forget that the real electrons we deal with in the lab can be annihilated, localized, etc and therefore do not exactly correspond to the mathematical idealization of a momentum eigenstate representation of the poincare group.

 

[meopemuk]

I didn't say anything about instantaneous dressing or infinite numbers of photons at the "next time instant", I said the real time dynamics of the dressing process can be quite interesting. I think it's fair to say that the process is less interesting in QED than QCD where something really dramatic happens.

The dynamics of a single "bare" electron in QED is very curious too. If you formally make the unitary time evolution operator out of QED Hamiltonian (even ignoring the fact that the true Hamiltonian of QED contains divergent counterterms) and apply this operator to one-bare-electron state you'll see that the time evolution is a complicated process of creation and annihilation of photons and electron-positron pairs.

As far as I know nobody has seen these kinds of processes in experiments. My conclusion is that "bare" electrons and the associated Hamiltonian do not provide adequate description of particles seen in nature. In order to study the time evolution one should switch to the "dressed particle" description.

Eugene.

 

[Physics Monkey]

The dynamics of a single "bare" electron in QED is very curious too. If you formally make the unitary time evolution operator out of QED Hamiltonian (even ignoring the fact that the true Hamiltonian of QED contains divergent counterterms) and apply this operator to one-bare-electron state you'll see that the time evolution is a complicated process of creation and annihilation of photons and electron-positron pairs.

I feel like you're not actually reading what I write. I have argued that QED has a cutoff, but your posts continue to ignore that fact and speak about divergences. I have also said that the dressing process occurs with all kinds photons and particle-antiparticle pairs appearing, I only took exception with your statement that it was somehow instanteous.

As far as I know nobody has seen these kinds of processes in experiments. My conclusion is that "bare" electrons and the associated Hamiltonian do not provide adequate description of particles seen in nature. In order to study the time evolution one should switch to the "dressed particle" description.

Eugene.

In my opinion this statement just doesn't make sense. I am not saying that bare electrons are the same as dressed electrons. Sure they have the same quantum numbers, but they are different states. On the other hand, for any finite regulated theory one can prepare states identical to the state of the bare particle and study their time evolution. Just because bare electron states are potentially hard to prepare doesn't mean their evolution is unphysical. Conceptually identical experiments are carried out routinely in atomic physics and condensed matter physics labs. I have also argued that this kind of thinking is experimentally relevant for QCD.

 

[meopemuk]

... one can prepare states identical to the state of the bare particle and study their time evolution. Just because bare electron states are potentially hard to prepare doesn't mean their evolution is unphysical.

There is a contradiction in your position. When scattering amplitudes are calculated in QED (and compared with real observable cross-sections) they are calculated exactly for "bare" particle states. So, QED assumes that electrons available in routine experiments are "bare". This disagrees with your claim that "bare" states are difficult to prepare.

Eugene.

 

[Physics Monkey]

There is a contradiction in your position. When scattering amplitudes are calculated in QED (and compared with real observable cross-sections) they are calculated exactly for "bare" particle states. So, QED assumes that electrons available in routine experiments are "bare". This disagrees with your claim that "bare" states are difficult to prepare.

Eugene.

This is also not true. Even your beloved Weinberg makes this point (p 110), that we must include some interactions in the description of asymptotic states so that these states are the physical particles. At the level of perturbation theory this is related to the amputation of external lines in Feynman diagrams. The issue is particularly evident in qcd where the bare quarks are not at all like good asymptotic states. The question of proper asymptotic states is a dynamical one that qft solves just fine. Note also that these comments have little bearing on how hard or easy it is to prepare a state describing a bare particle.

 

[meopemuk]

The issue is particularly evident in qcd where the bare quarks are not at all like good asymptotic states. The question of proper asymptotic states is a dynamical one that qft solves just fine. Note also that these comments have little bearing on how hard or easy it is to prepare a state describing a bare particle.

I agree that free quarks do not exist. That's why I was talking about QED not about QCD. Let me summarize what we've discussed so far and see if we can agree on that.

1. In QED the state of one electron is described as

$$a^{\dag}_p |0 \rangle$$......(1)

Perhaps being naive, I think that this state corresponds to a single free electron that can be easily prepared in the laboratory. This view is supported by the fact that when calculating scattering amplitudes in QED we take matrix elements on states exactly as (1). The results of these calculations are very accurate.

2. It is also true that "bare" states like (1) are not eigenstates of the total QED Hamiltonian. In particular, the time evolution of such states (generated by the total QED Hamiltonian) is rather non-trivial. It involves creation and annihilation of multiple (virtual) photons and electron-positron pairs.

3. My point is that this non-trivial dynamics has never been observed. A single free electron always looks like a single free electron and nothing else. In my opinion, this is a serious contradiction in the QED formalism.

4. In my understanding, you think that the non-trivial dynamics of a single "bare" electron can be observed if we learn how to prepare such states. This implies that states participating in (well-described) scattering experiments are not "bare" states like (1), but some other states. What are they?

Eugene.

 

[DrDu]

Maybe as usual in PF when somebody asks a question related to QFT it is immediately highjacked. I think, most of the problems which have been addressed are highly peculiar to relativistic field theories and not to qft in general. E.g. the basic states of a free non-relativistic electron gas are simply Sslater determinants of Bloch waves. Or, to bring a bosonic example, for a set of Einsteinian phonons, just products of harmonic oscillator functions.

 

[genneth]

I think we should return to the original poster's question. I believe he has not realized that it is possible to talk about probability distributions (or in this case, amplitudes) over field configurations. That is, the object $$\Psi[\phi]$$ is the corresponding object to $$\psi(x)$$ in usual QM. Just like x labels a possible configuration of a single particle, $$\phi$$ labels the possible configuration of a whole field (in some space-like surface). Thus, the states could be written $$\left|\phi\right\rangle$$, analogous to $$\left|x\right\rangle$$. These would then form a complete basis. In practise, it's hard to construct these things, and instead we use the Fock representation, where the particle nature of things are more apparent.

 

[DrDu]

Exactly, genneth. Maybe an example would be helpfull, e.g. the linear chain of particles sitting in equilibrium at equally spaced positions labelled by index i. Then the deviations from these positions is u_i. The set of all displacements is the field phi and the basis states would be $$|\phi>=\prod_i |u_i>$$ the |u_i> corresponding to the position eigenstates |x> of genneth, phi itself would be the vector of all the displacements u_i or the function u(r) in the continuum limit. This basis is seldomly used, especially in the continuum limit, as we are usually interested in the solutions where the u_i vary continuously. Hence one uses often an alternative basis of Fourier transformed states $$|k>=\sum_l |u_l>\exp ilk$$ the field is then characterized by $$|\tilde{\phi}>=\prod_k |u_k>$$. The latter basis has the advantage that it remains discrete in the continuum limit. Furthermore, we can get rid easily of the non-smooth configurations by cutting of the product at high momenta k. That's called regularization.

 

[meopemuk]

Maybe an example would be helpfull, e.g. the linear chain of particles sitting in equilibrium at equally spaced positions labelled by index i. Then the deviations from these positions is u_i. The set of all displacements is the field phi and the basis states would be $$|\phi>=\prod_i |u_i>$$ the |u_i> corresponding to the position eigenstates |x> of genneth, phi itself would be the vector of all the displacements u_i or the function u(r) in the continuum limit.

All this is definitely true in condensed matter physics. However, I don't think that the same ideas are valid for fundamental relativistic quantum fields. There is no physical "lattice" or "medium" whose "displacements" can be interpreted as particle quantum fields and whose "excitations" can be interpreted as particles.

Eugene.

 

[Count Iblis]

All this is definitely true in condensed matter physics. However, I don't think that the same ideas are valid for fundamental relativistic quantum fields. There is no physical "lattice" or "medium" whose "displacements" can be interpreted as particle quantum fields and whose "excitations" can be interpreted as particles.

Eugene.

But you can pretend that there is such a lattice and then work in the scaling limit where the effects of any such lattice are invisible. It then doesn't matter whether or not in reality such a lattice exists.

QFT is only an effective theory, like thermodynamics was before we knew that it is based on statistical physics (you can do thermodynamics without knowing that a gas consists of molecules).

 

[genneth]

All this is definitely true in condensed matter physics. However, I don't think that the same ideas are valid for fundamental relativistic quantum fields. There is no physical "lattice" or "medium" whose "displacements" can be interpreted as particle quantum fields and whose "excitations" can be interpreted as particles.

Eugene.

I don't think the Count was trying to say that things are exactly the same, more just giving a concrete example of how the formalism works. After all, it's Hilbert spaces all the way down...

 

[meopemuk]

But you can pretend that there is such a lattice and then work in the scaling limit where the effects of any such lattice are invisible. It then doesn't matter whether or not in reality such a lattice exists.

If it doesn't matter, then I would prefer to think that this lattice doesn't exist. It is always good to reduce the number of unprovable assumptions to a minimum.

QFT is only an effective theory, like thermodynamics was before we knew that it is based on statistical physics (you can do thermodynamics without knowing that a gas consists of molecules).

But today we *do know* that gas consists of molecules and we *do not know* if there is "the lattice". So, it is well established that thermodynamics is an effective theory. The claim that relativistic QFT is also an effective theory does not have any experimental support. My opinion is that this claim is just another attempt to "sweep problems under the rug".

Eugene.

 

[Count Iblis]

Don't you think the renormalizability of the Standard Model is indirect evidence for it to be an effective theory? Similarly to why you have a renormalizable phi^4 theory for the Ising model in the scaling limit: Because all the irrelevant operators flow to zero in the scaling limit.

 

[meopemuk]

Don't you think the renormalizability of the Standard Model is indirect evidence for it to be an effective theory?

"indirect evidence" is not the same as "proof".

Eugene.

 

[genneth]

If it doesn't matter, then I would prefer to think that this lattice doesn't exist. It is always good to reduce the number of unprovable assumptions to a minimum.

But today we *do know* that gas consists of molecules and we *do not know* if there is "the lattice". So, it is well established that thermodynamics is an effective theory. The claim that relativistic QFT is also an effective theory does not have any experimental support. My opinion is that this claim is just another attempt to "sweep problems under the rug".

Eugene.

I'd say reason works the other way. To assume that you have a "true" theory, without any upper cut off, is wildly optimistic. The question of where that cut off is, and what is beyond, are good questions.

 

[meopemuk]

ITo assume that you have a "true" theory, without any upper cut off, is wildly optimistic.

I am optimistic because I know a theory which

1. does not have ultraviolet cutoff
2. does not have divergences in S-matrix calculations
3. produces the same scattering amplitudes as the traditional renormalized QFT

E. V. Stefanovich, "Quantum field theory without infinities", Ann. Phys. (NY), 292 (2001), 139.

E. V. Stefanovich, "Relativistic quantum dynamics", http://www.arxiv.org/abs/physics/0504062

Eugene.

 

[xepma]

So you're basically slowly but surely hijacking this thread to promote your own book?

 

[DrDu]

meopemuk, in some sense for an infinite system, the wavefunction never exists, but one has to define states as functionals on the space of operators. So maybe the difference between relativistic and non-relativistic qft is not so fundamental after all.

 

[meopemuk]

meopemuk, in some sense for an infinite system, the wavefunction never exists, but one has to define states as functionals on the space of operators. So maybe the difference between relativistic and non-relativistic qft is not so fundamental after all.

I do not agree with the usual characterization of QFT as a theory studying systems with infinite number of degrees of freedom. I think it is more appropriate to call them "systems with uncertain number of degrees of freedom". The number of degrees of freedom is proportional to the number of particles in the system, and this number can change in the process of time evolution. For example, within QFT one can describe a state with only two particles. There is no problem in describing such a state by a 2-particle wave function. The only "problem" is that in the course of time evolution the number of particles in the system can change, so the form of the wave function becomes more complicated and the number of degrees of freedom increases.

Eugene.

 

[A. Neumaier]

In the Fock basis, there would be a function of N positions for the N particle basis, right? And then there'd be a sum over all N.
Could someone please show the steps connecting the state in the "field eigenbasis" to the state in the "Fock basis" a bit more explicitly? Not just the math of "here is an equation", but the procedure ... ie. how we derive the connection between the two.
Can someone show what the "wavefunctional" equation would be for QED? I still don't understand how you are deriving these things since there are fields in the Lagrangian, but no wavefunctional.

First consider the single-particle case, i.e., ordinary quantum mechanics.
Working in the position representation just means working with the wave function whose values are the coefficients in the eigenbasis |x> of the commuting position operators x_1, x_2, x_3, short x:

$$|\psi> = \int dx \psi(x) |x>,$$ where $$\psi(x)=<x|psi>$$

Working in the momentum representation just means working with the wave function whose values are the coefficients in the eigenbasis |p> of the commuting momentum operators p_1, p_2, p_3, short p:

$$|\psi> = \int dp \psi(p) |p>,$$ where $$\psi(p)=<p|psi>$$

To translate between the two, one needs to know how to find the eigenstates
of p in the x-representation, and the eigenstates of x in the p-representation.
This is given by the Fourier transform.

Now consider a field theory. To get a representation one needs to pick a maximal commuting family of operators and their eigenstates. Depending on the choice, one gets different but isomorphic representations. By diagonalizing momenta, one gets
the traditional Fock representation in terms of eigenstates $$|p_1,...,p_N>$$, where N=0,1,2,... and each p_k is in R^3, and the wave functions are the coefficients $$\psi_N(p_1,...,p_N)$$ in

$$|\psi> = \sum_N \int dp^N \psi_N(p_1,...,p_N) |p_1,...,p_N>,$$

Note that any 1-particle operator A lifts to the resulting Fock space by means of
$$\hat A = \int dp a^*(p) A a(p)$$ with the corresponding c/a operators,
giving in particular for the total momentum

$$\hat p |p_1,...,p_N>=(p_1+...+p_N) |p_1,...,p_N>,$$

showing that we have indeed eigenstates. This is the appropriate representation for
scattering experiments, where the input configurations are prepared in momentum eigenstates.

By diagonalizing instead Hermitian field operators at time t=0 (which commute because of the canonical commutation relations), one gets the functional Schroedinger representation in terms of eigenstates $$|\phi>$$ (one state for each possible classical field configuration \phi at time t=0), and the wave functions $$\psi(\phi)$$ are coefficients in a corresponding functional integral over all fields \phi.
(To get references, go to scholar.google.com and enter the key words functional schroedinger.) This is the appropriate representation when the field was prepared at time t=0.

Again conversion from one to the other representation requires the solution of the corresponding eigenproblems, but I haven't seen anyone do this explicitly.