From observable to operators in QFT

[lucas_]

That makes no sense. Its Its like saying the EM field can be its momentum. From the EM field you can get the momentum via Noethers theorem - but its not the field. The same with QM. From the field you can derive the momentum operator - but its not the field.

The creation and annihilation operators derived from the quantum field creates particles with a certain momentum when you chug through the math.

Thanks
Bill

You mean in QFT, there are only creation and annihilation operators and no spin or momentum operator. Note in QM we have spin and momentum operator that gives us the value when we get the eigenvalues of the observable.
BTW.. all the 4 books you suggested is available in the net free. How can the authors earn money when all available free. Poor them. I will read them anyway. With some basic concepts clear up initially its easier to read heavy loaded books. Thanks.

 

[bhobba]

You mean in QFT, there are only creation and annihilation operators and no spin or momentum operator.

I said:

From the field you can derive the momentum operator - but its not the field.

Thanks
Bill

 

[lucas_]

Regular QM of many identical particles can be rewritten as a QFT, so the observables are the same in QM. This sort of QFT is non-relativistic, but there are no conceptual differences for relativistic QFT, just many more technical difficulties because of the relativistic symmetry.

http://hitoshi.berkeley.edu/221b/QFT.pdf
These notes quantize the Schroedinger equation as a classical field theory to get a QFT, which is then shown to be QM of many identical particles.

http://www.colorado.edu/physics/phys7450/phys7450_sp10/notes/2nd_quantization.pdf
These notes take QM of many identical particles and rewrite it as a QFT.

I can only understand the verbal part above in your refereces above. But just a bird view idea. You said regular QM of many identical particles can be rewritten as a QFT. But there are no creation and annihilation operators in QM. So you need to create these.. then it's not the same as rewriting it.. or you need to put the creation and annihilation operators artificially.. it's not there in the original QM.

 

[atyy]

I can only understand the verbal part above in your refereces above. But just a bird view idea. You said regular QM of many identical particles can be rewritten as a QFT. But there are no creation and annihilation operators in QM. So you need to create these.. then it's not the same as rewriting it.. or you need to put the creation and annihilation operators artificially.. it's not there in the original QM.

It is the same theory in a new language.

 

[atyy]

As a simple example, the harmonic oscillator in QM can rewritten using ladder operators.

 

[lucas_]

I said:Thanks
Bill

Ok. As I see it clearer now. QM and QFT are just mathematical operations to perform on the quantum state. But the quantum state of QFT is more complex than that in QM. This is the reason Bohmian mechanics can easily be done in standard Quantum Mechanics but hard in Quantum field theory because of the creation and annihilations of particles and because the quantum state in QFT are more complicated (are you saying it's the same?). That said, can you still say, Bill, that the ignorant essemble interpretation can work in the quantum state of QFT?? You don't have just superpositions of positions or paths in QFT, you have quantum fluctuations and other stuff that can't just be modeled as essembles, can it.

 

[bhobba]

(are you saying it's the same?).

No.

That said, can you still say, Bill, that the ignorant essemble interpretation can work in the quantum state of QFT??

Why would you think it wouldn't?

Thanks
Bill

 

[lucas_]

No.
Why would you think it wouldn't?

Thanks
Bill

In QFT, forces result from exchange of virtual particles.. like virtual photons.. so using the ensemble interpretations, each position of the virtual photons are part of the ensemble? also during particle creation and annihilations like photons giving rise to electron-positron pair, why do you model this using the ignorant ensemble?

 

[bhobba]

In QFT, forces result from exchange of virtual particles.. like virtual photons.. so using the ensemble interpretations, each position of the virtual photons are part of the ensemble? also during particle creation and annihilations like photons giving rise to electron-positron pair, why do you model this using the ignorant ensemble?

Your reasoning escapes me.

All the Fock space is the union of a state with no particles, one particle, two particles etc etc. A general state is a superposition. When you observe it the superposition becomes a mixed state of zero particles etc.

It's not hard.

Thanks
Bill

 

[lucas_]

Your reasoning escapes me.

All the Fock space is the union of a state with no particles, one particle, two particles etc etc. A general state is a superposition. When you observe it the superposition becomes a mixed state of zero particles etc.

It's not hard.

Thanks
Bill

Maybe that's where the problem lies. Our computational methods are perturbation like.. even the interactions are done indirectly (Fock like).. this is the reason why ensemble can be used... but when using another analytical scheme like non-perturbation where the forces are dealt with directly, then they may no longer be seen as part of ensemble. Can you share any papers along this line (alternatives to QFT)?

 

[bhobba]

Can you share any papers along this line (alternatives to QFT)?

Sorry - don't know any.

The perturbation methods used (look into lattice gauge theory) - yes - but not to the principles.

Thanks
Bill.

 

[lucas_]

Sorry - don't know any.

The perturbation methods used (look into lattice gauge theory) - yes - but not to the principles.

Thanks
Bill.

What? You mean ensembles can be used because of the perturbation methods used? and what do you mean "not to the principles"?

 

[bhobba]

What? You mean ensembles can be used because of the perturbation methods used?

No.

and what do you mean "not to the principles"?

Perturbation theory is used heavily in QFT. It introduces things that don't actually exist like virtual particles. You can find many threads in this forum about it.

Thanks
Bill

 

[lucas_]

No.
Perturbation theory is used heavily in QFT. It introduces things that don't actually exist like virtual particles. You can find many threads in this forum about it.

Thanks
Bill

Ensemble may make sense in quantum mechanics.. but in QFT it's more unintuitive. so you are saying you can also interpret the quantum state as saying when you perform the experiment a million times, the probability distribution of the quantum state will make manifest?

I'm beginning to read your recommended book "QFT for the gifted amateur" after getting your assurance it's not really for the gifted as I'm not gifted. But reading the first sense in chapter 1 where the book highlights "Every particle and every wave in the Universe is simply an excitation of a quantum field that is defined over all space and time" it's not entirely accurate. I mean I know the book is not for layman but for more advanced laymen almost a textbook. In formal QFT, the quantum field is operator field, you can't say it is defined over all space and time.. with the exception of the EM field where you can say it has existence.. but the fermionic field, this can't be detected with an antenna. So quantum field can't an object or beable in spacetime, it is purely mathematical unless you are saying since the universe is mathematical anyway.. you agree to say the quantum field is really there over space and time?

 

[bhobba]

In formal QFT, the quantum field is operator field, you can't say it is defined over all space and time

Lucas - please, please can you put your thinking cap on. Because the field consists of operators it does not follow it can't be defined over all space and time any more than the EM field being defined by real numbers says anything about such.

Its a proper textbook but can be studied with just a first course in QM. But you will need to think and not jump to conclusions like that.

Thanks
Bill

 

[lucas_]

Lucas - please, please can you put your thinking cap on. Because the field consists of operators it does not follow it can't be defined over all space and time any more than the EM field being defined by real numbers says anything about such.

Its a proper textbook but can be studied with just a first course in QM. But you will need to think and not jump to conclusions like that.

Thanks
Bill

Ok. I'll put my thinking cap on while reading the book. The operator field may be defined over all space and time but it is not real (or objectively there in space at all). For us laymen. It means a lot whether to visualize it is there or not (for example, the wave function is not there in space but can be defined over all space and time).. so maybe we should treat operator field akin to wave function where it is only in the maths in paper to compute and don't really exist. Thanks for all the small bird view, it can spell the difference when we laymen read rigorous books.

 

[bhobba]

The operator field may be defined over all space and time but it is not real (or objectively there in space at all).

It's as real or not real as things like EM fields. But that type of categorisation is not physics - its philosophy. Physics considers it real because it has energy and momentum.

And if you don't think it real, since everything is composed of Quantum Fields - all matter - everything - then you would consider that not real - a rather silly position.

Thanks
Bill

 

[lucas_]

It as real or not real as things like EM fields. But that type of categorisation is not physics - its philosophy. Physics considers it real because it has energy and momentum.

And if you don't think it real, since everything is composed of Quantum Fields - all matter - everything - then you would consider that not real - a rather silly position.

Thanks
Bill

EM field is real because it has energy and momentum. But quantum fields or operator fields don't have energy and momentum. What may have energy and momentum may be the quantum state it is acting on. But you can't say the operator fields have energy and momentum. Anyway. Reminds me of the book by M.Y. Han called "A story of light".. the following he mentioned about the emulation of light and whether this holds true in the fermionic fields. Interestingly, he removes this passage in his second edition after the discovery of the higgs boson (maybe he is convinced second quantization works):

"The first leap of faith is the introduction of the concept of matter fields, as discussed in Chapter 7. The quantization of the electromagentic field successfully incorporated photons as the quanta of that field and - this is critical - the electromagnetic field (the four-vector potential) satisfied a classical wave equation identical to the Klein-Gordon equation for zero-mass case. A classical wave equation of the 19th century turned out to be the same as the defining wave equation of relativistic quantum mechanics of the 20th century! This then led to the first leap of faith - the grandest emulation of radiation by matter - that all matter particles, electrons and positrons initially and now extended to all matter particles, quarks and leptons, should be considered as quanta of their own quantized fields, each to its own. The wavefunctions of the relativistic quantum mechanics morphed into classical fields. This conceptual transition from relativistic quantum mechanical wavefunctions to classical fields was the first necesary step toward quantized matter fields. Whether such emulation of radiation by matter is totally justifiable remains an open question. It will remain an open question until we successfully achive completely satisfactory quantum field theory of matter, a goal not yet fully achieved."

 

[bhobba]

EM field is real because it has energy and momentum. But quantum fields or operator fields don't have energy and momentum.

Come again. What do you think is measured when you measure it? The difference is you can't say it has that before measurement - not that it doesn't have that property.

This is a philosophical issue that has been thrashed out ad nausium and the above will be my last comment on it. Think about it as you like - and its got nothing to do with understanding QFT.

BTW your comment about a quantum state having energy or momentum etc is incorrect. A quantum state is like probabilities on a dice. If each side has a probability of 1/6th how does it have the property of a side is up? Of course there are interpretations where its real - but there are also interpretations where its anything but.

Thanks
Bill

 

[lucas_]

Come again. What do you think is measured when you measure it? The difference is you can't say it has that before measurement - not that it doesn't have that property.

This is a philosophical issue that has been thrashed out ad nausium and the above will be my last comment on it. Think about it as you like - and its got nothing to do with understanding QFT.

BTW your comment about a quantum state having energy or momentum etc is incorrect. A quantum state is like probabilities on a dice. If each side has a probability of 1/6th how does it have the property of a side is up? Of course there are interpretations where its real - but there are also interpretations where its anything but.

Thanks
Bill

In plain quantum mechanics. The energy and momentum are in the hamiltonians of the wave function. It is not in the operators. Therefore in QFT, the energy and momentum are not in the operators (or operator fields), but in whatever the operators are acting on.. which is the quantum state or state vector. I thought this was clear. Operators are like square root. How can square root have energy or momentum.. common sense.

 

[bhobba]

In plain quantum mechanics. The energy and momentum are in the hamiltonians of the wave function.

That's incorrect.

QM is a theory about observations. It only has properties like energy or momentum when its observed to have those properties. To determine the statistical outcome of those observations you need both the observable and the state then you use the Born rule. A QFT field is as real or not real as any observable in QM. It's not real in the sense of having that value - it is real in the sense it allows us to determine things that are very real - observations.

Thanks
Bill

 

[lucas_]

That's incorrect.

QM is a theory about observations. It only has properties like energy or momentum when its observed to have those properties. To determine the statistical outcome of those observations you need both the observable and the state then you use the Born rule. A QFT field is as real or not real as any observable in QM. It's not real in the sense of having that value - it is real in the sense it allows us to determine things that are very real - observations.

Thanks
Bill

Oh. I read in the thread yesterday written by Eugene https://www.physicsforums.com/threads/what-are-the-states-in-qft.388556/page-2 that:

"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."

In case you know Eugene. How was his QFT different from the mainstream? I am using his context as I learned it yesterday from him. Do you also believe the above? But he said opposite to you.

 

[bhobba]

"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.

That's his view - not mine. I managed to find the following that gives mine pretty well:
http://www.quora.com/Are-quantum-fields-real-or-merely-a-mathematical-tool-used-to-describe-elementary-particles

However it makes zero difference to the theory itself.

Thanks
Bill

 

[lucas_]

In quantum mechanics, you need to get the operators from your pocket in terms of a calculator to compute the eigenstate (eigenvalues) of the observable.
But in quantum field theory, the operators are labels in points of spacetime, hence the operators are just floating in the vacuum (before measurements) instead of being located inside your calculator.

This is an accurate way to imagine it right guys (others beside Bhobba)? What you think?

 

[strangerep]

In quantum mechanics, you need to get the operators from your pocket in terms of a calculator to compute the eigenstate (eigenvalues) of the observable.
But in quantum field theory, the operators are labels in points of spacetime, hence the operators are just floating in the vacuum (before measurements) instead of being located inside your calculator.

This is an accurate way to imagine it right guys (others beside Bhobba)? What you think?

What do I think? Well,... for most of the statements you make, I feel a desire to say: "rubbish" -- so I bite my tongue and (mostly) stay silent.

But I'll try to be more constructive: I think you should put QFT aside for a while and go back and study ordinary QM properly. Basic stuff: like how a Hilbert space is constructed to be a representation of a (physically relevant) dynamical symmetry group; like how the "unitary irreducible representations" of the rotation group give the observed half-integral angular momentum spectrum; and like how a tensor product of Hilbert spaces is used to model multiparticle systems.

Can you access a copy of Ballentine's QM text? If you can cope with the level of math therein, I guarantee that such study would be much more profitable than persisting with all the ill-informed heuristic flubbing around that's been happening in this thread.

------------------------------
BTW, I know Eugene reasonably well, and I was here when those older threads were in progress. The underlying points of difference between Eugene's view and the view of some others are on a rather advanced level -- concerned with difficult convergence issues in advanced QFT, and how useful QFTs may be constructed. I don't recommend that any QFT beginner get embroiled in that stuff, but rather master a basic textbook level of QFT first before confronting the more archane aspects.

 

[atyy]

In quantum mechanics, you need to get the operators from your pocket in terms of a calculator to compute the eigenstate (eigenvalues) of the observable.
But in quantum field theory, the operators are labels in points of spacetime, hence the operators are just floating in the vacuum (before measurements) instead of being located inside your calculator.

This is an accurate way to imagine it right guys (others beside Bhobba)? What you think?

The construction of the non-relativistic quantum mechanical momentum and position operators from the quantum field operators is given in David Tong's QFT notes http://www.damtp.cam.ac.uk/user/tong/qft.html, section 2.8.1.

 

[QuasiParticle]

Apparently the momentum of one particle cannot be expressed in the second quantization formalism.

I'm a little confused as to what this means. Surely in second quantization one can use the momentum representation, where the momenta are analogous to the positions in the position representation.

 

[atyy]

I'm a little confused as to what this means. Surely in second quantization one can use the momentum representation, where the momenta are analogous to the positions in the position representation.

Me too. It looks like the total momentum operator is the usual momentum operator of one particle - see Tong's notes linked two posts above.

 

[vanhees71]

I do not understand what you mean. The momentum operator is given by the corresponding momentum density which is given by the (Belinfante) energy-momentum tensor. For the free Klein-Gordon field it's given by
$$\vec{\pi}=-:\dot{\phi} \vec{\nabla} \phi^{\dagger}+\dot{\phi}^{\dagger} \vec{\nabla} \phi:.$$
The colons mean normal ordering.

 

[atyy]

I do not understand what you mean.

Neither do I, I was quoting a reference given in post #30.

The question I was trying to answer is: if we use the second-quantized language and write non-relativistic quantum mechanics of many identical particles as a quantum field theory, how are the usual momentum and position operators of single-particle quantum mechanics written in the second quantized form?

As I understand the answer is given in section 2.8.1 of Tong's notes linked in post #56.