Difference between 'Quantum theories'

[StevieTNZ]

Glad I asked because I was definitely confused looking at Wikipedia pages on second quantization, QFT, etc.

 

[meopemuk]

But how does this transformation change the description of the Newton-Wigner annihilation field?

This is easy. Each operator transforms to the new reference frame by the same general formula. For the Newton-Wigner "annihilation field" this formula is

a(x) \to U_g a(x) U_g^{-1} = \int dp e^{ipx }U_g a(p) U_g^{-1}

Transformation formulas for the momentum-space annihilation operators a(p) can be found in Weinberg's textbook.

I was doing neither. I mainly tried teaching you that all these expressions fully deserve to be called quantum fields, as they are called by everyone except you.

So, basically you are saying that any operator expression that depends on arguments (x,t) can be called "quantum field". But I hope you wouldn't claim that all these various expressions can be used in the Weinberg's quote, from which we started our discussion:

''In its mature form, the idea of quantum field theory is that quantum fields are the basic ingredients of the universe, and particles are just bundles of energy and momentum of the fields."

Eugene.

 

[meopemuk]

Glad I asked because I was definitely confused looking at Wikipedia pages on second quantization, QFT, etc.

Wikipedia is terribly bad as a source of information about QFT. My best advise is to learn QFT from Weinberg's textbook. This is not an easy read for a beginner, but, at least, this book presents a clear and correct logic of what follows from what. If you complement the reading by articles from the literature lists in each chapter, then you will make it.

Eugene.

 

[A. Neumaier]

This is easy. Each operator transforms to the new reference frame by the same general formula. For the Newton-Wigner "annihilation field" this formula is

a(x) \to U_g a(x) U_g^{-1} = \int dp e{ipx }U_g a(p) U_g^{-1}

Transformation formulas for the momentum-space annihilation operators a(p) can be found in Weinberg's textbook.

Yes, messy formulas that don't tell what's going on. Compare my uniform, manifestly covariant formulas with his (5.1.11) and (5.1.12). I don't understand why you prefer the latter.

So, basically you are saying that any operator expression that depends on arguments (x,t) can be called "quantum field".

Yes. that s the standard usage. Just like any vector expression that depends on arguments (x,t) can be called ''vector field''.

But I hope you wouldn't claim that all these various expressions can be used in the Weinberg's quote, from which we started our discussion:

''In its mature form, the idea of quantum field theory is that quantum fields are the basic ingredients of the universe, and particles are just bundles of energy and momentum of the fields."

Insert ''certain'' before ''quantum'', and this problem is resolved. Which quantum fields qualify as ''basic ingredients'' is something that we'd need to discuss further.

I'd like to focus our further discussion by dissecting it into a number of more or less independent parts:

(i) Which fields represent the operational content of quantum field theory. This is directly related to your preceding comment. I'll open a new thread about this, in which I'll summarize the main issues from that part of our discussion here.

(ii) Indistinguishable particles and fields. I'll open a new thread about this, in which I'll summarize the main findings from that part of our discussion here. See
https://www.physicsforums.com/showthread.php?t=474321
https://www.physicsforums.com/showthread.php?t=474293

(iii) What deserves to be called a quantum field. Should this not yet be agreed upon by your remark above, I'd like to have this discussed in the present thread.

(iv) How to display and use Lorentz covariance. It seems to me that you only pay lip service to it by saying that given the transformation laws in Weinberg, everything is said about it and one doesn't need to consider it further. I'd like to have this discussed in the present thread. (By the way, I think that this attitude is part of the reason why you get certain unfounded, crackpottish results in parts of your book on QED - but I won't discuss the latter directly, unless you open a thread on your book in the Independent Research forum.)

 

[A. Neumaier]

One can also form the particle number operator for the point x as N(x) = a*(x)a(x). So, these operators are definitely associated with "points of space". However, their usefulness does not indicate that in reality there exists some continuous physical substance, called field. These operators are humble mathematical servants to the true masters - *particles*.

In which sense are your *particles* the true masters? Do you mean to imply with your statement that in reality there exists some discrete physical substance, called particles?

You agreed already that N(x) is not a number operator but a field density, and integrating N(x) over a region Omega of space will never result in an operator with integral spectrum, unless you take Omega=R^3.

Thus, how do you identify a particle, given a state of the whole system which is not just a single particle state?

What is discrete about the reality underlying QED? The only discrete thing I can see in QED is the spectrum of the object called (for historical reasons) the number operator. But this is an extremely nonlocal object that cannot be observed at all. So why should it be considered more real than the charge density, the current, or the electromagnetic field, which are observed every day?

Finally, why does this discrete reality (if it can be pinned down at all) deserve to carry the name particles?

 

[meopemuk]

In which sense are your *particles* the true masters? Do you mean to imply with your statement that in reality there exists some discrete physical substance, called particles?

Yes, this is exactly what I mean.

integrating N(x) over a region Omega of space will never result in an operator with integral spectrum, unless you take Omega=R^3.

Here I disagree. If we integrate N(x) over any region Omega of space we obtain an Hermitian operator with the spectrum 0,1,2,3,... whose interpretation is "the number of particles in the region Omega".

Thus, how do you identify a particle, given a state of the whole system which is not just a single particle state?

Any QFT state vector can be expanded as a linear superposition of 0-particle, 1-particle, 2-particle etc. states, according to the Fock space structure. So, I can always calculate the probability of finding N particles in any given state.

What is discrete about the reality underlying QED? The only discrete thing I can see in QED is the spectrum of the object called (for historical reasons) the number operator. But this is an extremely nonlocal object that cannot be observed at all.

Scattering amplitudes is the final result of most QFT calculations. When we calculate scattering amplitudes in QFT, we always specify the number of particles (and their observables, like momentum and spin) in both initial and final states. So, discrete particles are always present in their explicit form. I am not sure what you mean by the word "nonlocal", and why the particle number cannot be observed in your opinion?

So why should it be considered more real than the charge density, the current, or the electromagnetic field, which are observed every day?

As I wrote already, if you look at charge and current densities with high enough resolution, you'll see that they are not continuous quantities, but averaged descriptions for a large number of particles - electrons. The same is true for the free electromagnetic field. If the field has high frequency and low intensity, we can easily distinguish discrete particles there - photons. It is true that static electric and magnetic fields are really continuous, but they are not observable in the absence of charged particles. So, one can defend the position that there is nothing but charged particles directly interacting with each other via distance- and velocity-dependent forces.

 

[A. Neumaier]

Yes, this is exactly what I mean.

This is what I feared you mean. But there is no evidence at all for this point of view.

Here I disagree. If we integrate N(x) over any region Omega of space we obtain an Hermitian operator with the spectrum 0,1,2,3,... whose interpretation is "the number of particles in the region Omega".

That you make such claims without getting red in your face shows that your intuition about the behavior of field operators is very poorly developed. You'd make lots of calculations with quantum fields to get some practice...

Please check the action of N(x) (at time t=0) on the 1-particle sector. It is a simple exercise to check that when the support of the state is all of R^3 then the expectation of the integral of N(x) over a nonempty, open and bounded domain is always strictly between 0 and 1. This implies that the eigenvalues also must lie in this interval.

So your number interpretation is a figment of your imagination.

Any QFT state vector can be expanded as a linear superposition of 0-particle, 1-particle, 2-particle etc. states, according to the Fock space structure. So, I can always calculate the probability of finding N particles in any given state.

Please show me how you calculate, given an arbitrary Fock state, the probability of finding a single particle in a fixed bounded region Omega.

Scattering amplitudes is the final result of most QFT calculations. When we calculate scattering amplitudes in QFT, we always specify the number of particles (and their observables, like momentum and spin) in both initial and final states. So, discrete particles are always present in their explicit form.

Although you had often claimed the above, QFT is good for much more than calculating scattering amplitides.

I am not sure what you mean by the word "nonlocal", and why the particle number cannot be observed in your opinion?

The only integral over N(x) with an integral spectrum is the integral over all of R^3, a very nonlocal operator: One needs to gather information at every point x in space to be able to say something about this integral.

As I wrote already, if you look at charge and current densities with high enough resolution, you'll see that they are not continuous quantities,

I cannot see this. Please demonstrate it for me.

 

[meopemuk]

Please check the action of N(x) (at time t=0) on the 1-particle sector. It is a simple exercise to check that when the support of the state is all of R^3 then the expectation of the integral of N(x) over a nonempty, open and bounded domain is always strictly between 0 and 1. This implies that the eigenvalues also must lie in this interval.

So your number interpretation is a figment of your imagination.



Please show me how you calculate, given an arbitrary Fock state, the probability of finding a single particle in a fixed bounded region Omega.

Operator for the number of particles in the volume \Omega is

N(\Omega) = \int_{\Omega} dx a^{\dag}(x)a(x)

Consider an N-particle state vector in the Fock space, with each particle having a definite position

\Psi = a^{\dag}(y_1) a^{\dag}(y_2) \ldots a^{\dag}(y_N) |0 \rangle...(1)

Then one can prove that

N(\Omega) \Psi = M \Psi

where M is an integer number, which tells us how many particles lie inside the volume \Omega. States of the type (1) form a basis in the Fock space. If \Psi is a one-particle state

\Psi = a^{\dag}(y_1) |0 \rangle

then

N(\Omega) \Psi = \Psi if y_1 \in \Omega
N(\Omega) \Psi = 0 if y_1 \not{\in} \Omega

For a state that is not well-localized, the expectation value of N(\Omega) gives the probability of finding the particle in the volume \Omega.

Although you had often claimed the above, QFT is good for much more than calculating scattering amplitides.

QFT can also calculate energies of bound states (e.g., Lamb shifts). What else?

I cannot see this. Please demonstrate it for me.

I hope you wouldn't argue that the current in metals is produced by discrete electrons. Continuous charge and current fields are just very rough approximations. The discrete nature of the radiation field was demonstrated by Einstein in his 1905 paper about the photo-electric effect.

 

[A. Neumaier]

N(\Omega) = \int_{\Omega} dx a^{\dag}(x)a(x)
Consider an N-particle state vector in the Fock space, with each particle having a definite position
\Psi = a^{\dag}(y_1) a^{\dag}(y_2) \ldots a^{\dag}(y_N) |0 \rangle...(1)

Unfortunately, your proposed state is not a Fock state, not even when you symmetrize it. For it is not normalizable.

Then one can prove that
N(\Omega) \Psi = M \Psi
where M is an integer number

However, I see that the situation N=1 that I had considered was too simple to decide things since there is nothing yet to symmetrize. Indeed, on the 1-particle sector, N(Omega) is an orthogonal projector with eigenvalues 0 and 1 only. This shows that my previous argument was flawed, and that in an infinite-dimensional Hilbert space one can conclude only that the spectrum is in [0,1], not that the endpoints are excluded.

Upon reexamining the situation, I can see that a precise formulation of your idea is the following:

If Omega is open, the kernel of N(Omega)=integral_Omega dx a(x)^*a(x) is the Fock space F_0(Omega) whose 1-particle space is the Hilbert space of wave functions whose support is disjoint with Omega. And for M>0, we have N(Omega) Psi = M Psi for every Psi in the (non Fock) space F_M(Omega) obtained from F_0(Omega) by applying to its elements M smeared creation operators whose wave function has support in the closure of Omega. This defines a direct decomposition of the full Fock space. Therefore, yes, you are right, N(Omega) has spectrum {0,1,2,...}.

So it is me who is getting red in the face. I apologize, and thank you for having learned something new. (I should have known from Haag's local field theory, but I had never seen discussed that certain spectra are preserved under localization. Maybe DarMM can point to a place in the literature?)

Thus there is _something_ discrete about QFT, namely the existence of number operators localized in some region. Thus we can consistently talk about the expected number of massive particles in a region Omega of space at a given time t (which was zero in the above discussion).

But of course, this is a nonrelativistic situation, and it is the relativistic treatment that is relevant to the foundations of quantum field theory. That something nontrivial happens is that the above can be replicated in the relativistic situation only for massive particles since one needs a Newton-Wigner position operator to bring the Fock space in a representation where the annihilator fields satisfy the nonrelativistic spatial CCR. In particular, nothing of the above discussion applies to massless particles; so QED is not covered since photons are massless.

In any case, the problem of which operators are associated to observable reality gets nontrivial in relativistic field theory and must be discussed carefully. This is a situation that I haven't yet fully analyzed, so I need more time to prepare for this discussion, which may take a few days. When I am ready, I'll start a new thread.

QFT can also calculate energies of bound states (e.g., Lamb shifts). What else?

I had answered this already:

You don't realize how much is done with quantum field theory. People calculate a lot more: Thermodynamic properties of equilibrium states, hydrodynamic equations for flowing fields, kinetic transport equations governing the behavior of semiconductors, etc.. The fact that this is not in Weinberg's book doesn't mean that it is not done. If scattering of particles were the only application of QFT, the latter wouldn't play the fundamental role it plays.

Indeed, _all_ of the many applications of quantum field theory to macroscopic matters are done with little or no reference to scattering processes. Look at any book on nonequilibrium statistical mechanics. Note that there is little difference in principle between relativistic and nonrelativistic treatments - in both cases one uses the same field theory. The nonrelativistic case differs only in that
- antiparticles and particle-number changing processes are neglected, and
- interactions are nonlocal.
Nothing else.

I hope you wouldn't argue that the current in metals is produced by discrete electrons. Continuous charge and current fields are just very rough approximations. The discrete nature of the radiation field was demonstrated by Einstein in his 1905 paper about the photo-electric effect.

I answered this one earlier today ago in a separate thread,
https://www.physicsforums.com/showthread.php?p=3147920
and ask you to reply there concerning this aspect.

See also my previous post #37 in the present thread.

 

[meopemuk]

Unfortunately, your proposed state is not a Fock state, not even when you symmetrize it. For it is not normalizable.

Sorry, I don't get it. I thought that this is the standard QFT method for building symmetrized and normalized N-particle states. See, for example, (4.2.2) in Weinberg's book. The only difference in my formula is that I am using position representation instead of Weinberg's momentum representation.

That something nontrivial happens is that the above can be replicated in the relativistic situation only for massive particles since one needs a Newton-Wigner position operator to bring the Fock space in a representation where the annihilator fields satisfy the nonrelativistic spatial CCR.

This is true. A satisfactory position operator for photons has not been constructed yet. Perhaps, it cannot be defined at all. Perhaps, photons cannot be exactly localized in the position space. I don't have a certain opinion about that. However, they can be perfectly localized and counted in the momentum space. So, they still look like discrete particles.

Indeed, _all_ of the many applications of quantum field theory to macroscopic matters are done with little or no reference to scattering processes. Look at any book on nonequilibrium statistical mechanics. Note that there is little difference in principle between relativistic and nonrelativistic treatments - in both cases one uses the same field theory. The nonrelativistic case differs only in that
- antiparticles and particle-number changing processes are neglected, and
- interactions are nonlocal.
Nothing else.

Yes, QFT applications to condensed matter processes use many of the same techniques as the fundamental relativistic QFT of elementary particles. However, at the fundamental level these two frameworks are quite different. Condensed matter QFT is inherently approximate as it ignores the discrete atomistic structure of matter. For example, the continuous phonon field is not a good approximation at distances comparable with interatomic separations in the crystal. On the other hand, relativistic QFT is supposed to be exact at all distances.

Of course, there is also an "effective field theory" school of thought in relativistic QFT, which claims that relativistic fields are not exact concepts, and that there is some yet unknown space-time granularity, which plays the role similar to the "crystal lattice" for fundamental particle fields. However, in my opinion, this is a wild hypothesis without any experimental support.

The field approximation in condensed matter is justified, because we deal with systems containing a huge number of particles. So, using the particle-based picture would be mathematically hopeless. In the fundamental relativistic QFT, we usually focus on processes involving small number of particles, which can be treated separately.

Another difference between the condensed matter and fundamental field theories is that the former does not use the concept of (Galilean) relativity, because there is always one distinguished frame of reference - the one connected with the underlying crystal lattice.

Eugene.

 

[A. Neumaier]

Sorry, I don't get it. I thought that this is the standard QFT method for building symmetrized and normalized N-particle states. See, for example, (4.2.2) in Weinberg's book. The only difference in my formula is that I am using position representation instead of Weinberg's momentum representation.

This was a minor remark only (and only partially correct, since, indeed, your states are symmetrized. Sometimes I am a bit too fas and then make small mistakes.) Fock space is a Hilbert space and does not contain unnormalizable states. Your states live in a corresponding rigged Hilbert space.

This is true. A satisfactory position operator for photons has not been constructed yet. Perhaps, it cannot be defined at all.

There are theorems that it cannot be defined in a unique way. (If one assumes all the properties that Newton and Wigner assume, there is none; if one relaxes the requirement on rotations, there are infinitely many, so none of them could be distinguished as ''the physical one'' - moreover, these wouldn't lead to a good position representation.) See the entry ''Particle positions and the position operator'' of Chapter B2 of my theoretical physics FAQ at http://arnold-neumaier.at/physfaq/physics-faq.html#position)

Perhaps, photons cannot be exactly localized in the position space. I don't have a certain opinion about that. However, they can be perfectly localized and counted in the momentum space. So, they still look like discrete particles.

The momentum representation is completely nonlocal, in any physical sense of localization.

Yes, QFT applications to condensed matter processes use many of the same techniques as the fundamental relativistic QFT of elementary particles. However, at the fundamental level these two frameworks are quite different. Condensed matter QFT is inherently approximate as it ignores the discrete atomistic structure of matter. For example, the continuous phonon field is not a good approximation at distances comparable with interatomic separations in the crystal. On the other hand, relativistic QFT is supposed to be exact at all distances.

In the nonequilibrium statistical mechanics of a weakly interacting) boson and fermion gas, nothing is approximated, except for the usual approximations of perturbation theory.
The electron field in condensed matter theory and the photon field in quantum optics are also the same as in (relativistic or nonrelativistic) QED; only the nuclear structure is approximated.

Both approximation sare also used in the application of QED to the hydrogen atom; so there is no difference in principle.

Of course, there is also an "effective field theory" school of thought in relativistic QFT

I was not referring to that.

Another difference between the condensed matter and fundamental field theories is that the former does not use the concept of (Galilean) relativity, because there is always one distinguished frame of reference - the one connected with the underlying crystal lattice.

This is a minor difference only. The QED treatment of the hydrogen atom also does not use the concept of Poincare invariance.