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

[andresB]

https://arnold-neumaier.at/physfaq/topics/position.html

Quoting from the link: " Therefore, if we have a quantum system defined in an arbitrary Hilbert space in which a momentum operator is defined, the necessary and sufficient condition for the existence of a spatial probability interpretation of the system is the existence of a position operator with commuting components which satisfy standard commutation relations with the components of the momentum operator and the angular momentum operator. "

I can see the sufficient part, it's what we have in nonrelativistic QM (though, In the non-relativistic case, I think demanding the right commutation relations with the Galilean boost is also required).

But I fail to see why it is a necessary condition.

 

[pavsic]

What you appear to be trying to do here is pick out the hypersurface and hold that constant while making a Lorentz transformation. That's not what changing frames does. When you change frames, you change hypersurfaces: a different set of hypersurfaces become the "surfaces of constant time" that particle positions are supposed to be points in.

I wrote: "Recall what is "space". It is just a time slice, i.e., a hypersurface oriented so the that , . But it can be oriented differently (or observed from another reference frame)."
Here I meant the active (passive) Lorentz transformation.

Active LT: In a given reference frame one can have a surface $\Sigma$ and change it into another surface $\Sigma'$ related to $\Sigma$ by a Lorentz transformation. That is, the object of observation, namely, the hysperfusrface changes under the transformation.

Passive LT: The same object is observed from different reference frames. In one reference frame the surface components are $\Sigma_\mu$, in another reference frame the components of the same hypersurface are $\Sigma'_\mu = {L_\mu}^\nu \Sigma_\nu.$

To sum up, in the active case, the object changes, while the reference frame remains the same. In the passive case, the object remains the same, and the reference frame changes.

 

[pavsic]

No, they don't. You aren't using 3-D delta functions $\delta^3$. You are using 4-D delta functions $\delta^4$. That picks out points in spacetime, not space.

I am using $\delta^4 ({\bar x}' - {\bar x})$, where ${\bar x} \equiv {\bar x}^\mu$, and, as I explained in the previous reply, $\bar x^\mu= {P^\mu}_\nu x^\mu$ is the projection of the spacetime point onto the hypersurface $\Sigma$. This projection is also a spacetime point, it is given by four coordinates, therefore the 4-D delta function, but restricted to the hypersurface. Therefore, more precisely it should be written, e.g.,
$$\delta^4 ({\bar x}' - {\bar x})|_\Sigma.$$
Explicitly, it is
$$\delta^4 ({\bar x}' - {\bar x})|_\Sigma = n_\mu {\epsilon^\mu}_{\nu \rho \sigma} \delta (x'^\nu-x^\nu)\delta (x'^\rho-x^\rho)\delta (x'^\sigma-x^\sigma)$$
This is a covariant expression that replaces the non covariant 3-D delta function. In a particular reference frame in which the normal to $\Sigma$ is $n^\mu= (1,0,0,0)$ we have $\delta^4 ({\bar x}' - {\bar x})|_\Sigma =\delta^3 ({\bf x}'-{\bf x})$.

Thank you for pointing out that my notation was incomplete and that in fact it is a 3D delta-function, but, as I explained above, written covariantly, by using all four spacetime coordinates to designate position on the hypersurface $\Sigma .$

To your next comment I will reply in a separate post, because I do not know how to do with multiquotes.
I also intend to discuss why this "minithread" on position is necessary, and how it will lead us to the concept of wave function in QFT. For the time being, see my post number 36.

 

[PeterDonis]

Here I meant the active (passive) Lorentz transformation.

If we are talking about "position in space", which is what you claim to be trying to capture, then "space" means the appropriate hypersurface for a particular observer. Neither of your definitions captures this. If two observers are in relative motion, they each choose a different hypersurface as "space" (more precisely, "space" at an instant of their time). Typically they also each choose a different frame to describe what they observe, the frame in which their chosen "space" is a surface of constant coordinate time. So if we transform from one observer's description to the other, both the frame (coordinate chart) and the hypersurface change. Neither of your definitions captures this.

position on the hypersurface $\Sigma .$

But which hypersurface depends on the observer. Different observers will disagree about which hypersurface is "space" (at an instant of their time). That was my original objection, and you are not addressing it at all.

 

[Tendex]

depends on the observer. Different observers will disagree about which hypersurface

What definition of observer are you using and what change of observer are you referring to?
These are free quantum fields at a fixed time in the Schrodinger picture, the states don't change and don't pick specific particles positions.

 

[pavsic]

Well, but that's the problem. Wave functions do not make too much sense for relativistic QT, at least not the same sense as they make in non-relativistic QM, and that's the problem here.

Wave function makes sense in relativistic quantum field theory. Starting from the scalar quantum field $\varphi(x)$, where $x\equiv x^\mu = (t,{\bf x})$, we can construct the operators (see Jackiw: Diverse topics in theoretical and mathematical physics)
$$a(t, {\bf x}) = \frac{1}{\sqrt{2}} \left ( \sqrt{\omega_{\bf x}}\varphi + \frac{i}{\sqrt{\omega_{\bf x}}} \Pi \right) $$
$$a^\dagger(t, {\bf x}) = \frac{1}{\sqrt{2}}\left ( \sqrt{\omega_{\bf x}}\varphi - \frac{i}{\sqrt{\omega_{\bf x}}} \Pi \right) $$
which stisfy
$$[a(t,{\bf x}),a^\dagger (t,{\bf x}') = \delta^3 ({\bf x},\bf x'),$$
where $\Pi$ is the canonical momentum conjugated to $\varphi$ and $\omega_{\bf x}= \sqrt{m^2 -\nabla^2}.$
The above operators are just like the annihilation and creation operators of the harmonic oscillator. But here we have uncountabley infinite number of operators.
A state with with a definite ${\bf x}$ at time $t$, say $t=0$ is given by
$$a^\dagger({\bf x})|0\rangle,$$
where $a^\dagger({\bf x})\equiv a^\dagger(0,{\bf x}).$
A generic one particle state is
$$|\Psi \rangle = \int d^3 {\bf x}\, \psi (t,{\bf x}) a^\dagger ({\bf x})|0 \rangle.$$
Here $\psi (t,{\bf x})$ is a single particle wave function, satisfying the relativistic Schroedinger equation:
$$i \frac{\partial \psi}{\partial t}= \sqrt{m^2 -\nabla^2} \psi.$$
This is the Foldy equation. It comes from the Schroedinger equation for the state $|\Psi \rangle$:
$$i \frac{\partial |\Psi \rangle}{\partial t} = H |\Psi \rangle,$$
where $H$ is the usual Hamiltonian composed of the quantum field operators.

Position operator is
$$ \hat{\bf x}= \int d^3 {\bf x} \,a^\dagger ({\bf x}) {\bf x} \,a({\bf x}).$$.
Its eigenstates are $|{\bf x} \rangle \equiv a^\dagger ({\bf x}) |0\rangle.$

All those expressions can be cast into covariant forms. Namely,
${\bf x}$ are coordinates of a point on the hypersurface $t=constant.$ In a subsequent post I will discuss this in more detail and also address the point raised by PeterDonis.

 

[vanhees71]

You are dealing with field operaters, which of course makes sense. These are no wave functions though!

 

[Tendex]

You are dealing with field operaters, which of course makes sense. These are no wave functions though!

Sometimes Fock states in free quantum fields in the Schrodinger picture are referred to as "wave functions". These are not of course the same as the NRQM wavefunctions!

 

[pavsic]

You are dealing with field operaters, which of course makes sense. These are no wave functions though!

In my post 56, $a({\bf x})$ and $a^\dagger ({\bf x})$ are field operators, whilst $\psi (t,{\bf x})$ are complex valued wave functions ("wave packet profiles"), not operators.

 

[pavsic]

Sometimes Fock states in free quantum fields in the Schrodinger picture are referred to as "wave functions". These are not of course the same as the NRQM wavefunctions!

The Foldy equation of my post 56
$$i \frac{\partial \psi}{\partial t}= \sqrt{m^2 -\nabla^2} \psi$$,
in non relativistic approximation becomes the Schroedinger equation.

 

[PeterDonis]

What definition of observer are you using and what change of observer are you referring to?

You answered that yourself:

These are free quantum fields at a fixed time in the Schrodinger picture

"At a fixed time" requires picking a particular frame--a particular split of spacetime into "space" and "time", so that we have a set of spacelike hypersurfaces of "constant time" that foliate spacetime. There will be some observer who is at rest in that frame. That is the "observer" I speak of.

Changing observers means changing frames, i.e., changing which set of spacelike hypersurfaces we treat as being surfaces of "constant time". That must be the case by the principle of relativity: no set of such hypersurfaces is physically picked out over any other. So if there is a "Schrodinger picture" at all, there must be a different Schrodinger picture for each such choice of foliation. Which doesn't make sense, since the assumption behind the Schrodinger picture is that there is only one.

 

[Demystifier]

@pavsic In the past I have been working on similar approaches by myself (in fact you know me, I'm Hrvoje Nikolić), so I think I understand pretty well the advantages and disadvantages of such approaches. I don't need to talk to you about the advantages, so I would like to discuss with you the disadvantages.

Your notation $|\bar{x}\rangle$ is very elegant, but it hides the problem. So let me use a less elegant notation in which the problem is more transparent
$$|\bar{x}\rangle \equiv |x,\Sigma_x\rangle$$
where $x$ is the spacetime point not attached to any hypersurface, while $\Sigma_x$ is a hypersurface containing $x$. Similarly to @PeterDonis I am concerned with the meaning of the dependence on $\Sigma_x$. For a given $x$, the corresponding $\Sigma_x$ is not unique. There are at least 3 ways to interpret it:
1) There is a preferred foliation for the whole Universe, which makes $\Sigma_x$ unique.
2) $\Sigma_x$ is defined by the frame in which the measuring apparatus or observer is at rest.
3) $\Sigma_x$ is a free variable.

I'm not sure which of those interpretations do you prefer, but my impression is that you have in mind some variant of 3). But for 3) there is a problem equivalent to the problem found a long time ago by Newton and Wigner. Let $\Sigma_x$ and $\Sigma'_x$ be two different hypersurfaces associated with the same $x$. More specifically, let $\Sigma_x$ be the hypersurface of constant time $t$ in the Lorentz frame $S$ and let $\Sigma'_x$ be the hypersurface of constant time $t'$ in the Lorentz frame $S'$. Then the problem is that
$$\langle x,\Sigma_x|y,\Sigma_y \rangle \neq \langle x,\Sigma'_x|y,\Sigma'_y \rangle$$
In particular, if
$$\langle x,\Sigma_x|y,\Sigma_y \rangle = \delta^3({\bf x}-{\bf y})$$
then
$$\langle x,\Sigma'_x|y,\Sigma'_y \rangle = \delta^3({\bf x}'-{\bf y}')$$
which is a problem because $\delta^3({\bf x}-{\bf y})$ and $\delta^3({\bf x}'-{\bf y}')$ are not the same quantity expressed in two Lorentz frames. Instead, $\delta^3({\bf x}_1-{\bf x}_2)$ expressed in the $S'$ coordinates is
$$\delta^3({\bf x}_1-{\bf x}_2)=\delta^3({\bf x}_1({\bf x}'_1,t'_1)-{\bf x}_2({\bf x}'_2,t'_2))\neq \delta^3({\bf x}'_1-{\bf x}'_2)$$
In other words, the notion of localization depends on the choice of $\Sigma$, which is nothing but a covariant version of the Newton-Wigner result that the notion of localization depends on the Lorentz frame. Having a covariant version is definitely a progress, but it doesn't really solve the problem. How to interpret the dependence on $\Sigma$ in physical terms?

Anyway, my interpretation would be a combination of 1) and 2), while rejecting 3). At a more fundamental level there are some hidden variables obeying some kind of 1), while on practical instrumental level we can use a version of 2).

 

[Demystifier]

You are dealing with field operaters, which of course makes sense. These are no wave functions though!

You are missing the obvious. If there is the operator $\hat{a}({\bf x})$, then there is the state
$$|{\bf x}\rangle=\hat{a}^{\dagger}({\bf x})|0\rangle$$
and then there is the "wave function"
$$\psi({\bf x})=\langle {\bf x} |\psi\rangle$$
Refusing to call this object a "wave function" does not help much. At best you can argue (which is interpretation dependent) that this object does not have the same interpretation as the wave function in standard nonrelativistic QM.

 

[Tendex]

You are missing the obvious. If there is the operator $\hat{a}({\bf x})$, then there is the state
$$|{\bf x}\rangle=\hat{a}^{\dagger}({\bf x})|0\rangle$$
and then there is the "wave function"
$$\psi({\bf x})=\langle {\bf x} |\psi\rangle$$
Refusing to call this object a "wave function" does not help much. At best you can argue (which is interpretation dependent) that this object does not have the same interpretation as the wave function in standard nonrelativistic QM.

This is what I argued in #58 just with different words.
I believe the mistake here is treating free quantum fields as if they were physical. They aren't, and attributing them any of the 3 interpretations you gave in #62 is kind of moot. Just like trying to interpret physically what pavsic wrote in his first post as picking out physical events or not.

The fact is that free quantum fields are mathematical objects that serve to build physical interacting QFTs but it makes little sense to discuss them in physical terms, more so if, like pavsic seems to be doing, one tries to introduce some physical interpretations in the formalism that have more to do with a non-relativistic limit and an intent to somehow recover an old wave function from single particle relativistic quantum mechanics in the formal notation that simply is not there for properly quantized free fields.

 

[Tendex]

"At a fixed time" requires picking a particular frame--a particular split of spacetime into "space" and "time", so that we have a set of spacelike hypersurfaces of "constant time" that foliate spacetime. There will be some observer who is at rest in that frame. That is the "observer" I speak of.
Changing observers means changing frames, i.e., changing which set of spacelike hypersurfaces we treat as being surfaces of "constant time". That must be the case by the principle of relativity: no set of such hypersurfaces is physically picked out over any other. So if there is a "Schrodinger picture" at all, there must be a different Schrodinger picture for each such choice of foliation. Which doesn't make sense, since the assumption behind the Schrodinger picture is that there is only one.

This is correct in physical terms but for free quantized fields with charge there is trivially only one such foliation and no local observer or particular position changing of frames in this sense takes place with any physical consequences, quantization of the free field in flat Minkowski spacetime ensure constant global transformations and not having time or position observables do the rest. Of course things change and get much more complicated with interactions added, but for that there is not yet a clear mathematical non-perturbative description that completely "makes sense" of the Schrodinger picture(and therefore any other picture, however is physically much more "sensical" to use use the Heisenberg picture as almost all textbooks do) in the way we would like it to.

 

[pavsic]

@pavsic 2) $\Sigma_x$ is defined by the frame in which the measuring apparatus or observer is at rest.

I am adopting that interpretation. Namely, $\Sigma$ is defined by the frame in which the observer is at rest. But because there exist the observers who are in relative motion, this means that each of them is associated with a different $\Sigma$, the hypersurface on which the events observed by that observer are simultaneous.

The operators $a^\dagger({\bar x})$ for ${\bar x} \in \Sigma$ create the states with definite positions on a given $\Sigma$. In other words, $a^\dagger({\bar x})$ create the events that are simulatenoues on $\Sigma$. Relative to an observer $O'$ who is not at rest with respect to $\Sigma$, those events are not simultaneous. For the observer $O'$ simultaneous events are created by different operators, say $a'^\dagger({\bar x'}).$

The operators $a^\dagger (t=0,{\bf x})$ considered in the post 56 create the events at $t=0$, i.e., the events simulatenous for the observer $O$. The observer $O'$ whose moves relatively to $O$ can adopt analogous procedure for definting position states so that the associated events are all simultaneous relative to him. The $O'$'s simulateneity surface is different from the $O$'s simiulateneity surface, therefore he uses a different set of operators to create position states. Let they be denoted as $a'^\dagger (t'=0,{\bf x}')$.

This answers yout question
"Having a covariant version is definitely a progress, but it doesn't really solve the problem. How to interpret the dependence on $\Sigma$ in physical terms?"

The instruments used to determine positions by the observer in a Lorentz frame S are moving with respect to the instruments used by the observer in a Loreentz frame S'. Therefore, the position states depend on $\Sigma$ and the observer who is at rest with respect to $\Sigma$.

I will discuss later your other points. Now I have other things to do.

 

[Demystifier]

I am adopting that interpretation. Namely, $\Sigma$ is defined by the frame in which the observer is at rest.

That's perfectly OK for practical purposes. But I know that you are interested in deep questions in physics beyond pure pragmatism. So from a deeper point of view, do you think that quantum physics can be understood without referring to observers?

 

[pavsic]

Well, but that's the problem. Wave functions do not make too much sense for relativistic QT, at least not the same sense as they make in non-relativistic QM, and that's the problem here. All you can say from relativistic QFT are detection probabilities of a particle (i.e., an asymptotic free state, where a particle interpretation is possible) with a detector at a given position. This information is formally encoded in the field-operator autorcorrelation functions (aka $N$-point Green's functions).

In post 56 I wrote a single particle state

$$|\Psi \rangle = \int d^3 {\bf x}\, \psi (t,{\bf x}) a^\dagger ({\bf x})|0 \rangle, $$
where $\psi (t,{\bf x}) = \langle {\bf x}|\Psi (t,{\bf x}) \rangle$ is a complex valued wave packet profile, that is a wave function in the x-representation.

It satisfies the relativistic Schroedinger equation (known as the Foldy equation):
$$i {\dot \psi} = \omega_{\bf x} \psi (t,{\bf x}),$$
where $\omega_{\bf x} = \sqrt{m^2 - \nabla^2}$. Expanding
$$\sqrt{m^2 - \nabla^2} = m -\frac{\nabla^2}{2 m} + ... $$
and taking the ansatz
$$\psi (t,{\bf x}) = {\rm e}^{i m t} {\tilde \psi} ({\bf x})$$
we obtain the nonrelativistic Schroedinger equation as an approximation:
$$i {\dot {\tilde \psi}} = -\frac{\nabla^2}{2 m} {\tilde \psi}.$$

On the other hand, by inserting $\psi = \psi_R + i \psi_I$ into the equation $i {\dot \psi} = \omega_{\bf x} \psi (t,{\bf x})$, we obtain the following system of first order differential equation for the real and imaginary components of $\psi:$
$${\dot \psi} _R= \omega_{\bf x} \psi_I ,~~~~~~~~
{\dot \psi} _I= -\omega_{\bf x} \psi_R .$$
From the first equation we have
$$\psi_I = \omega_{\bf x}^{-1} {\dot \psi}_R .$$
Insertinf this into the second equation, we obtain
$$ \omega_{\bf x}^{-1}{\ddot \psi}_R = -\omega_{\bf x} \psi_R .$$
Multiplying the latter equation from the left by $\omega_{\bf x}$,
we obtain
$${\ddot \psi_R} + \omega_{\bf x}^2 \psi_R = 0,$$
i.e.,
$${\ddot \psi_R} + (m^2 - \nabla^2 )\psi_R = 0 ,$$
which is the Klein-Gordon equation for the real field $\phi \equiv \psi_R$. So we have found that a real Kleing-Gordon (c-number ) field takes place within the framework of the relativistic quantum field theory of a Hermitian scalar operator-valued field $\varphi$.
In other words, we have quantized a real scalar field by promoting it to the operator-valued field, constructed with them the creation operators, and position states, then taken a superposition of such position single particle state, where the superposition "coefficients"
were $\psi (t,{\bf x})$, satisfying the first order Foldy equation, which can be rewritten as the second order Klein-Gordon equation.

Thus, if we have a real Kleing-Gordon field (c-number, not operator) $\phi$, the role of the wave function has the superposition
$$\psi = \phi + i \omega_I^{-1} {\dot \phi},$$
of the field $\phi$ and its conjugated momentum ${\dot \phi}$.

 

[PeterDonis]

for free quantized fields with charge there is trivially only one such foliation

How so?

Note: In terms of the general objection I was making, @Demystifier has done a much better job of capturing it in his post #62.

 

[Tendex]

How so?

Note: In terms of the general objection I was making, @Demystifier has done a much better job of capturing it in his post #62.

Sure, your objection was physical, and he addressed it physically. I'm just pointing out the obvious, that free quantum fields are not physical, they are mathematical tools. I only opine about interpretations in the interpretations subforum. And I'm not very interested to be honest.

 

[vanhees71]

But all this is known for decades, and the standard solution is to use local realizations of the Poicare group in termx of field operators encoding both particles and antiparticles (Feynman-Stückelberg trick) leading to a Lorentz-covariant S-matrix.

 

[Tendex]

But all this is known for decades, and the standard solution is to use local realizations of the Poicare group in termx of field operators encoding both particles and antiparticles (Feynman-Stückelberg trick) leading to a Lorentz-covariant S-matrix.

Sure, that's good old perturbative RQFT, but I don't think pavsic and Peterdonis are discussing about that.