Concept of wavefunction and particle within Quantum Field Theory

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I am struggling to properly understand this two concepts.
-1st: Could someone give me some insight on what a ket-state refers to when dealing with a field? To my understand it tells us the probability amplitude of having each excitation at any spacetime point, but I don't know if this is accurate. Also, we solve the free field equation not for this wavefunction but for the field itself. The latter sounds rather strange to me, since the field is indeed an operator. To me it looks as if we solved the Schrodinger equation not for the wavefunction but for the operator X, which is just an observable.

Once you solve the equation for a free field you see that applying the creation operator (p) to a ket-state creates a particle with momentum p.
-2nd: As far as I know the only thing we know is that applying the creation operation increases the momentum of the system by a quantity p. It is fair then to think that this operation is analogous to creating a particle, but what do we know about this particle? For instance, where do we have created it?
 

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Demystifier
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A ket ##|\psi\rangle## is not a probability amplitude of anything. To get a probability amplitude you must consider a quantity of the form ##\langle a|\psi\rangle##, which is the probability amplitude that the measurement of observable ##A## will give the value ##a##. The observable ##A## may be almost anything you like, e.g. energy, momentum, electric field at a point, or particle position (in relativistic QFT the particle position is somewhat tricky and requires additional clarifications).

The fact that we solve the equation for the field operator ##\phi({\bf x},t)## just means that we work in the Heisenberg picture. In nonrelativistic QM it is analogous to solving the harmonic oscillator for the position operator as ##x(t)=ae^{-i\omega t} +a^{\dagger}e^{i\omega t}##. Alternatively, in QFT one can also work in the Schrodinger picture in which the field operator ##\phi({\bf x})## does not depend on ##t##, but it is rarely used in practice.

If you create particle with a definite momentum, then you cannot know its position. There is an operator that creates a particle at definite position (as I said it's tricky in relativistic QFT, but it's not problem at all in nonrelativistic QFT used e.g. in condensed matter physics), but in this case you know nothing about its momentum.

For more details see also my https://arxiv.org/abs/quant-ph/0609163 Secs. 8 and 9.
 
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  • #3
Jufa
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Many thanks for your answer, it helped me a lot! Also I checked your paper and found it really interesting.
 
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  • #4
CuriousLearner8
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A ket ##|\psi\rangle## is not a probability amplitude of anything. To get a probability amplitude you must consider a quantity of the form ##\langle a|\psi\rangle##, which is the probability amplitude that the measurement of observable ##A## will give the value ##a##. The observable ##A## may be almost anything you like, e.g. energy, momentum, electric field at a point, or particle position (in relativistic QFT the particle position is somewhat tricky and requires additional clarifications).

The fact that we solve the equation for the field operator ##\phi({\bf x},t)## just means that we work in the Heisenberg picture. In nonrelativistic QM it is analogous to solving the harmonic oscillator for the position operator as ##x(t)=ae^{-i\omega t} +a^{\dagger}e^{i\omega t}##. Alternatively, in QFT one can also work in the Schrodinger picture in which the field operator ##\phi({\bf x})## does not depend on ##t##, but it is rarely used in practice.

If you create particle with a definite momentum, then you cannot know its position. There is an operator that creates a particle at definite position (as I said it's tricky in relativistic QFT, but it's not problem at all in nonrelativistic QFT used e.g. in condensed matter physics), but in this case you know nothing about its momentum.

For more details see also my https://arxiv.org/abs/quant-ph/0609163 Secs. 8 and 9.
Nice paper. Thank you for sharing.
 
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  • #5
gentzen
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Nice paper. Thank you for sharing.
It is indeed a nice paper. I really like it. That's why he is the Demystifier!
 
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It is indeed a nice paper. I really like it. That's why he is the Demystifier!
Usually, although he's currently mystifying us with his free quark at rest!
 
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  • #7
Demystifier
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Usually, although he's currently mystifying us with his free quark at rest!
There is no demystification without prior mystification. I just want to understand this stuff better and I feel that the currently existing literature does not explain this stuff clearly. I am not (yet) in a phase to demystify this stuff, I'm just trying to think of it from different angles. Eventually I hope that I will be able to write something about it in a demystifying way, but first I have to look at it from different angles, including the wrong ones.
 
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vanhees71
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Usually, although he's currently mystifying us with his free quark at rest!
But here he makes us think about very subtle issues with local gauge symmetries, and I think there are many myths in the standard literature (and not only textbooks!) which have to be resolved!
 
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