From observable to operators in QFT

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They are different things - the fields and the states its operates on.

The mathematical connection is it can be shown the field is the same as creation and annihilation operators.

Thanks
Bill
speaking of creation and annihilation operators. I want to ask this since days ago. In every point of the quantum fields, are there creation and annihilation operators that is in addition to the say spin or momentum operators or are creation, annihilations, spin, momentum operators can only be used one point of the field at a time? For example, one point of the field lower and rise particle by creation and annihilation operators.. while in another point of the field, it's spin operator acting on the state, while in another, it's momentum operator acting on the state.. or does the same momentum operator also have creation annihilation in *same* point of the field?
 

atyy

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speaking of creation and annihilation operators. I want to ask this since days ago. In every point of the quantum fields, are there creation and annihilation operators that is in addition to the say spin or momentum operators or are creation, annihilations, spin, momentum operators can only be used one point of the field at a time? For example, one point of the field lower and rise particle by creation and annihilation operators.. while in another point of the field, it's spin operator acting on the state, while in another, it's momentum operator acting on the state.. or does the same momentum operator also have creation annihilation in *same* point of the field?
The quantum field operator is a creation operator. psi(x) creates a particle at x.
 
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The quantum field operator is a creation operator. psi(x) creates a particle at x.
But if you don't need to create a particle at x or psi(x), then the quantum field operator can be a momentum operator at x (or psi(x))?
 
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But if you don't need to create a particle at x or psi(x), then the quantum field operator can be a momentum operator at x (or psi(x))?
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
 

atyy

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But if you don't need to create a particle at x or psi(x), then the quantum field operator can be a momentum operator at x (or psi(x))?
I don't have an intuitive explanation for this, but the usual momentum observable is an integral over some weird function of the field operator psi(x). You can see the expression in http://www-physique.u-strasbg.fr/~polonyi/mbt.pdf [Broken] (Eq 44).

Edit: Oops, wrong! Apparently the momentum of one particle cannot be expressed in the second quantization formalism. https://books.google.com/books?id=UM3mUcXV830C&dq=single+particle+observable+second+quantization&source=gbs_navlinks_s (bottom of p23)
 
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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.
 
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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

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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

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

 
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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.
 
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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?
 
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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
 
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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)?
 
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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.
 
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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"?
 
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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
 
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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?
 
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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 cant 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
 
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Lucas - please, please can you put your thinking cap on. Because the field consists of operators it does not follow it cant 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.
 
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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
 
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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."
 
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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 cant 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
 
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Come again. What do you think is measured when you measure it? The difference is you cant 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.
 

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