Why QFT still goes well while it lacks the notion of wave function?

[fxdung]

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)?

 

[PeterDonis]

in QFT it seems that there is a lacking of notion of wave function.

More precisely, the notion of "wave function" as it is understood in non-relativistic QM becomes frame-dependent in QFT.

I do not understand why QFT still goes well(it is a good theory to calculate any things)?

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.

 

[fxdung]

What do you mean when saying:"becomes frame dependent in QFT"?

 

[PeterDonis]

What do you mean when saying:"becomes frame dependent in QFT"?

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.

 

[A. Neumaier]

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)?

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]

What is state vector in QFT if there are interactings?Is it exp{Ht}/Fock state>?

 

[A. Neumaier]

In the interacting case the Hilbert space is a deformation of Fock space. In perturbation theory it is taken to be the Fock space of the corresponding free theory but this is an approximation only by Haag's theorem.

 

[fxdung]

Is the wave function that its variable is the field(Psi(Phi(x)) ) the same the vector states of Hilber space in QFT?

 

[vanhees71]

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.

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.

That's not the point, why wave functions have only a limited meaning in relativistic QT. A wave function describes a system with a fixed number of particles. Under relativistic conditions, i.e., if you consider particles at "relativistic energies" (i.e., if the energy-transfer in collisions becomes of the order of the rest energy of any particles which may be produced according to the conservation laws) the particle number is not conserved but you can create and annihilate all kinds of particles (only restricted by the conservation laws). That's why relativistic QT is most efficiently described as a quantum field theory, which allows an elegant description of such creation/annihilation processes.

Only for QED there has been an alternative formulation in terms of wave functions for the electrons and positrons, known as "Dirac's hole formulation". It is pretty complicated and confusing, but it is in fact mathematically equivalent to QED in the usual QFT formulation.

 

[A. Neumaier]

Is the wave function that its variable is the field(Psi(Phi(x)) ) the same the vector states of Hilber space in QFT?

The question as phrased makes no sense.

Read in Wikipedia about Fock space to inform yourself about how state vectors in free QFT look like.

 

[PeterDonis]

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).

Spinor/tensor fields can be frame-independent, but "wave function" as it is used in non-relativistic QM means the state of the system at some time; the "at some time" part must be frame-dependent for many-body systems (and even for single-particle systems if the particle is, say, a photon that has gone through a beam splitter).

 

[Keith_McClary]

You may have seen the "in-a-box" quantization of a classical field, where each mode of oscillation of the field is a harmonic oscillator. If you "cut off" the higher energy modes so that you only have a finite number of modes, then you have a multi-dimensional harmonic oscillator, so you do have a notion of wave function, but it is not a function of x,y,z. It is mathematically possible to do this in an infinite dimensional space (with out the energy cut off). I think @A. Neumaier has explained this somewhere better than I can.

 

[Demystifier]

But in QFT it seems that there is a lacking of notion of wave function.

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.

 

[HomogenousCow]

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.

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]

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.

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]

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.

I'm aware, what I'm saying is that the wave functional formalism is especially ill-defined even by QFT standards.

 

[samalkhaiat]

All good and bad results of the full interacting (even non-abelian) QFT have been derived (with the same level of “rigor”) from the formal solution of the Schrödinger equation in the “coordinate” |\psi \rangle basis (or the “momentum” |\pi \rangle basis): \Psi [\psi , t] = \int \mathcal{D}\psi_{0} \ \langle \psi |e^{-iHt}|\psi_{0}\rangle \Psi [\psi_{0} , 0] , where \langle \psi |e^{-iHt}|\psi_{0}\rangle \equiv G[\psi , t ; \psi_{0},0], is the amplitude for the field in the configuration \psi_{0} (\vec{x}) at t = 0 to evolve to \psi (\vec{x}) at t. In fact it is not hard to show that G[\psi , t ; \psi_{0},t_{0}] = \int \mathcal{D}\psi^{\prime} \ e^{iS[\psi^{\prime}]} , where S[\psi] is the classical action. And this is nothing but the Feynman path integral in QFT.

 

[fxdung]

Third, in relativistic QFT there is a wave function in the position space, but with a different interpretation than in QM.

What is the interpretation of wave funtion in position space in relativistic QFT?

 

[Demystifier]

What is the interpretation of wave funtion in position space in relativistic QFT?

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 obey
$$
\int d^3x |{\bf x}\rangle\langle {\bf x}|=1
$$
So if $\pi$ is a projector (for instance, it can be $\pi=|{\bf p}\rangle\langle {\bf p}|$ where $|{\bf p}\rangle$ is a momentum eigenstate), then the corresponding probability can be computed as
$$
p=\langle\psi|\pi|\psi\rangle=\int d^3x\int d^3x'\,
\langle\psi|{\bf x}\rangle\langle {\bf x}|\pi|{\bf x}'\rangle\langle {\bf x'}|\psi\rangle
=\int d^3x\int d^3x'\, \psi^*({\bf x}) \langle {\bf x}|\pi|{\bf x}'\rangle \psi({\bf x}')
$$
A similar generalization is also possible for the $n$-particle sector.

 

[vanhees71]

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.

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 is

B. Hatfield, Quantum Field Theory of Point Particles and
Strings, Addison-Wesley, Reading, Massachusetts, 10 edn.
(1992).

 

[A. Neumaier]

the wave functional formalism is especially ill-defined even by QFT standards.

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.

 

[fxdung]

Even though |x⟩ is no longer interpreted as a position eigenstate (because the corresponding position operator would not be Lorentz covariant)

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]

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?

Which part was not clear? Why is it not Lorentz covariant, or why the absence of Lorentz covariance is a problem?

 

[fxdung]

Please explain both of them.

 

[Demystifier]

Please explain both of them.

It's a complex subject, the proper treatment of which would require a separate thread. But here is a simple answer.

(i) The absence of Lorentz invariance:
Let $|{\bf x}\rangle$ be a position eigenstate. Then
$$\langle{\bf x}_1|{\bf x}_2\rangle=\delta^3({\bf x}_1-{\bf x}_2)$$
But $\delta^3({\bf x}_1-{\bf x}_2)$ is not a Lorentz invariant function. If you make a Lorentz transformation ${\bf x}\to{\bf x}'({\bf x},t)$, $t\to t'({\bf x},t)$, then
$$\delta^3({\bf x}_1-{\bf x}_2)\neq \delta^3({\bf x}'_1-{\bf x}'_2)$$

(ii) Why is that a problem?
In classical special relativistic physics, the particle position is a Lorentz 4-vector $x=(t,{\bf x})$. So naturally one would expect something similar in quantum physics, perhaps that the quantum position eigenstate is something like $|x\rangle$ satisfying
$$\langle x_1| x_2\rangle=\delta^4(x_1-x_2)=\delta^3({\bf x}_1-{\bf x}_2)\delta( t_1-t_2)$$
But this cannot be true is standard quantum theory, because time is not treated like that in standard quantum theory. Hence position of a particle cannot be Lorentz covariant in standard quantum theory in a way it is covariant in classical theory. That's a problem because one expects that a notion of a particle position should make sense in any Lorentz frame, not in just one. But in my opinion that's not such a serious problem if one thinks of the quantum position not as a property of the particle itself (which should be Lorentz covariant), but as a property of a response of the apparatus that measures the position (which defines a preferred Lorentz frame, namely the one with respect to which the apparatus is at rest).

 

[bhobba]

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.

Just to elaborate further see:
https://pdfs.semanticscholar.org/2fb0/4475228ff385a44a16e3ba42b432d3bf5b17.pdf

QFT is related more to formulation F with the added feature that a state can be in a superposition of particle numbers. It belongs to a rather strange space called a Fock space:
https://pdfs.semanticscholar.org/2fb0/4475228ff385a44a16e3ba42b432d3bf5b17.pdf

Thanks
Bill

 

[Tendex]

What is state vector in QFT if there are interactings?

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.

 

[king vitamin]

In the interacting case the Hilbert space is a deformation of Fock space.

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]

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)?

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]

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.

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.