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fxdung
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In QM by virtue of wave function we calculate any things. But in QFT it seems that there is a lacking of notion of wave function.I do not understand why QFT still goes well(it is a good theory to calculate any things)?
fxdung said:in QFT it seems that there is a lacking of notion of wave function.
fxdung said:I do not understand why QFT still goes well(it is a good theory to calculate any things)?
fxdung said:What do you mean when saying:"becomes frame dependent in QFT"?
In QFT one has state vectors belonging to a Hilbert space, in the free case a Fock space. And one has linear operators acting on it, e.g., the smeared fields. Thus everything that makes QM work is still present in QFT.fxdung said:In QM by virtue of wave function we calculate any things. But in QFT it seems that there is a lacking of notion of wave function.I do not understand why QFT still goes well(it is a good theory to calculate any things)?
If you define relativistic single-particle or many-body wave functions, which is possible to some extent at least for free massive particles, these are all frame-independent mathematical objects (spinor/tensor fields). Of course their components change as spinor/tensor-field components change under Poincare transformations.PeterDonis said:QFT is a relativistic theory; "frame-dependent" in QFT means the same thing it means in any relativistic theory: that the quantity in question depends on your choice of reference frame.
The question as phrased makes no sense.fxdung said:Is the wave function that its variable is the field(Psi(Phi(x)) ) the same the vector states of Hilber space in QFT?
vanhees71 said:If you define relativistic single-particle or many-body wave functions, which is possible to some extent at least for free massive particles, these are all frame-independent mathematical objects (spinor/tensor fields).
That's wrong at several levels. First, in non-relativistic QFT there is a wave function in the position space, with the same interpretation as in QM. Second, in relativistic QFT there is a wave function in the momentum space, with the same interpretation as in QM. Third, in relativistic QFT there is a wave function in the position space, but with a different interpretation than in QM. Fourth, in QFT a natural space is neither the position nor the momentum space, but the field space. In the field space, the quantum state is represented by a wave functional, which is a generalization of a function.fxdung said:But in QFT it seems that there is a lacking of notion of wave function.
Demystifier said:Fourth, in QFT a natural space is neither the position nor the momentum space, but the field space. In the field space, the quantum state is represented by a wave functional, which is a generalization of a function.
Well, many things in QFT aren't rigorous. For instance, the famous calculation of ##g-2## is also not rigorous, because it is based on renormalization and regularization of ill-defined divergent loop integrals. Yet the final result agrees with experiments up to many digits. My point is that the absence of rigor in QFT is not such a big problem from a practical point of view because one can always make things well defined by some (more or less ad hoc) regularization.HomogenousCow said:This is often repeated, but from what I've seen there isn't a rigorous way to do any quantum mechanics with a space of wave functionals. Even in the case of the free klein gordon fields, all the states of interest are horribly ill-defined distributional integrals where the normalization is simply hand-waved away.
Demystifier said:Well, many things in QFT aren't rigorous. For instance, the famous calculation of ##g-2## is also not rigorous, because it is based on renormalization and regularization of ill-defined divergent loop integrals. Yet the final result agrees with experiments up to many digits. My point is that the absence of rigor in QFT is not such a big problem from a practical point of view because one can always make things well defined by some (more or less ad hoc) regularization.
What is the interpretation of wave funtion in position space in relativistic QFT?Demystifier said:Third, in relativistic QFT there is a wave function in the position space, but with a different interpretation than in QM.
Even though ##|{\bf x}\rangle## is no longer interpreted as a position eigenstate (because the corresponding position operator would not be Lorentz covariant), in 1-particle sector of QFT one can still define the states ##|{\bf x}\rangle## which obeyfxdung said:What is the interpretation of wave funtion in position space in relativistic QFT?
Functional methods are among the most important tools to study QFTs. Of course you have to renormalize the divergences due to the difficult handling of distributions and products of distributions as with any other method. The most elegant way in the path-integral approach is the heat-kernel approach, which is closely related to Schwinger's world-line formalism. Of course, you can also use the operator formalism to formulate the functional approach. A nice book isHomogenousCow said:This is often repeated, but from what I've seen there isn't a rigorous way to do any quantum mechanics with a space of wave functionals. Even in the case of the free klein gordon fields, all the states of interest are horribly ill-defined distributional integrals where the normalization is simply hand-waved away.
No. See, e.g., the work by Jackiw cited here. There is a lot one can do with wave functionals that cannot be done easily with perturbation theory. The quantization of solitons and instantons are well-known examples.HomogenousCow said:the wave functional formalism is especially ill-defined even by QFT standards.
Please explain more detail why the /x> is no longer interpreted as a position eigenstate in relativistic QFT.Why can we say that because corresponding position operator is not Lorentz covariant?Demystifier said:Even though |x⟩ is no longer interpreted as a position eigenstate (because the corresponding position operator would not be Lorentz covariant)
Which part was not clear? Why is it not Lorentz covariant, or why the absence of Lorentz covariance is a problem?fxdung said:Please explain more detail why the /x> is no longer interpreted as a position eigenstate in relativistic QFT.Why can we say that because corresponding position operator is not Lorentz covariant?
It's a complex subject, the proper treatment of which would require a separate thread. But here is a simple answer.fxdung said:Please explain both of them.
PeterDonis said:Because there are other ways of mathematically modeling quantum systems than wave functions. The fact that standard non-relativistic QM uses them does not mean any quantum theory has to use them.
If you are more ok with the wave-function for free fields, you can maybe make the transit to interacting fields more easily looking at the non-perturbative result of the Källén-Lehmann decomposition that obtains interacting QFT two-point correlation functions from the sum of free propagators that can always be Fourier transformed between momentum and position.fxdung said:What is state vector in QFT if there are interactings?
A. Neumaier said:In the interacting case the Hilbert space is a deformation of Fock space.
The massless case is less well understood because a nontrivial superselection structure appears. But I think my characterisation is still valid since the term deformation is not very specific. For example, asymptotic particles turn into infraparticles with nonintegral anomalous dimensions. But the deviations from integrality go to zero with the interaction strength, justifying the term deformation.king vitamin said:Apologies if this is a question which merits its own thread (and if it is, I'm happy to start one), but this isn't always true, correct? One needs some further assumptions (say, a mass gap)?
A. Neumaier said:The massless case is less well understood because a nontrivial superselection structure appears. But I think my characterisation is still valid since the term deformation is not very specific. For example, asymptotic particles turn into infraparticles with nonintegral anomalous dimensions. But the deviations from integrality go to zero with the interaction strength, justifying the term deformation.
The 2D case is quite well understood, as there are many exactly solvable models in 2 spacetime dimensions, and many others. Many exactly solvable models do not have a Fock space like structure, and cannot be deformed into Fock spaces.king vitamin said:Interesting, thanks. I would guess things get hairier in cases where one doesn't have any sense of a weak-coupling, like certain CFTs. I would also guess that 2d CFTs are at least better understood since mathematicians have spent a lot of time with them, but as far as I'm aware their Hilbert space is largely studied on the cylinder where the spectrum is gapped anyways.
Demystifier said:The absence of Lorentz invariance:
Let be a position eigenstate. Then
pavsic said:The quantum position states are ##|{\bar x} \rangle##, and they satisfy
$$\langle {\bar x}_1|{\bar x}_2 \rangle = \delta^4 ({\bar x}_1 - {\bar x}_2) ,$$
which is a covariant relation.
Why notPeterDonis said:But this doesn't work as soon as you have more than one particle. There is no way to covariantly write down a state with two particles at different positions.